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Second Thoughts about Shakespeare's As You Like It
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Audio recording of a lecture given by Grant Franks on June 14, 2023, as part of the Graduate Institute Summer Lecture Series. The Graduate Institute provided this description of the event: "As You Like It is a muchloved fixture in the Shakespearean canon, generally staged as a joyous woodland romcom. However, if one dares to look a little more closely, – as the sophomore seminar that I colead did recently – doubts begin to surface. The seeming frivolity of action is dogged by questions. Why does a light comedy include patches of morose darkness like those in Jaques’s muchacclaimed “Seven Ages of Man” speech? Is Touchstone really a “touchstone” in any genuine sense, or is his suggestive name just meaningless fluff? In the romantic Forest of Arden, populated by shepherds speaking blank verse, why do Rosalind and Orlando conduct their offbeat courtship entirely in prose? Most importantly, why is the major plot tension of Act I – namely, the murderous sibling rivalries of Orlando and Olivier and of the two dukes – virtually ignored during the middle of the play and then casually, almost carelessly, disappeared in Act V?
This lecture will not explain everything, but it aims to outline some interesting ways to puzzle over As You Like It.
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Franks, Grant H.
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Shakespeare, William, 15641616
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Text
The Heptadecagon
Grant Franks
October 8, 2019
Review
We’ve looked at the first encounters with the 1 , the development of complex arithmetic, the roots
of unity, constructibility of points on the complex plane (considered algebraically) and followed the
construction of the pentagon. These points will be reviewed a final time next Wednesday in the final
lecture. (Remember: Wednesday, October 16, Junior Common Room, 3:15 pm.)
Let’s move on to the Heptadecagon.
Remember the Pentagon?
We’ve seen how to approach the algebraic construction of the pentagon. The process begins with the
equation:
x5  1 = 0
One factors out the one real solution that all “roots of unity” equations share, namely (x  1):
1 + x + x2 + x3 + x4 = 0
This equation is irreducible so long as one allows only rational numbers, Q, as solutions. (In mathjargon, it is “irreducible over the rationals.”) But if one constructs a finite quadratic field extension
Q( 5 ), it can be factored into two quadratics. A second finite quadratic field extension allows it to be
factored fully into four linear factors from which you can read oﬀ the solutions readily.
�2 ���
5 The Heptadecagon.nb
0.31 + 0.95 ⅈ
0.81 + 0.59 ⅈ
0.81  0.59 ⅈ
0.31  0.95 ⅈ
x4 + x3 + x2 + x + 1 = 0
Irreducible over Q
Then adjoin
5
x2  η2 x +1
x2  η1 x +1
(x  ζ1 )
Irreducible over F1
Irreducible over F1
Then adjoin ζ1
Then adjoin ζ2
(x  ζ4 )
Do the Same Thing, But More O�en
(x  ζ2 )
(x  ζ3 )
�5 The Heptadecagon.nb
���
3
We’ll follow the same basic plan to construct the heptadecagon. However, the procedure has a few
additional complications due to the greater number of steps.
The Equation
For the heptadecagon, we start with the equation:
x 17  1 = 0.
Again we factor out the one real solution (x  1) to obtain:
1 + x + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 + x13 + x14 + x15 + x16 = 0
This 16th degree equation has sixteen roots, all complex, that we designate:
ζ1 , ζ2 , ζ3 , ζ4 , ζ5 , ζ6 , ζ7 , ζ8 , ζ9 , ζ10 , ζ11 , ζ12 , ζ13 , ζ14 , ζ15 , ζ16
Graphically, these roots appear on the complex plane as vertices of a regular 17gon, making equal
angles at the center:
ζ5
ζ4
0+1i
ζ3
ζ6
ζ2
ζ7
ζ
ζ8
1 + 0 i
1+0i
ζ9
ζ 16
ζ 10
ζ 15
ζ 11
ζ 12
0  13
1i
ζ
ζ 14
Unsurprisingly, these occur in eight pairs of complex conjugates. (In the diagram above, complex
conjugates are joined by orange dotted lines).
Arrangement of the Sixteen Roots: the EightPeriods
�4 ���
5 The Heptadecagon.nb
Following the general procedure seen with the pentagon, we are going to split these roots up into two groups of eight, then
four groups of four, then
eight groups of two, then
sixteen individuals.
At each stage, we will make numbers by taking the sums of the members in each group. With the
pentagon, we found that even though we didn’t know the values of any of the roots (the ζ’s), we could
figure out a quadratic formula for the values of the two intermediate sums:
η1 = ζ 1 + ζ 4
η2 = ζ 2 + ζ 3
because we could figure out their sum and the product, η1 + η2 and η1 ×η2 .
When working on the pentagon, the way in which to subdivide the four roots presented little trouble.
We had reason to believe that the complex conjugates had to stay together, so there was only one
possible subdivision of the four roots into two pairs. The second division separated the two pairs roots
from their conjugate mates.
Now, however, we have sixteen roots and eight pairs of conjugates. For the first subdivision, there are
8 x 7 x 6 x 5 = 1,680 possible ways to separate the eight pairs into two groups of four. We don’t know a
priori whether some or all, or not all or possibly only one will work. On the surface, it seems that trial
and error might not work.
Sorting out the roots properly is more than half the battle in doing this construction. It will require a
small detour.
Half the Problem is Pretty Easy
First, some good news. We’re looking for a sorting of the roots that will allow us to find the sum and the
product of the two groups. In that quest, the sum of the two groups will pose no problem. No matter
how we divide the sixteen roots into two bunches, we will be able to get their total sum. Say we just
sort out the first eight and the last eight:
A = ζ1 + ζ2 + ζ3 + ζ4 + ζ5 + ζ6 + ζ7 + ζ8
B = ζ9 + ζ10 + ζ11 + ζ12 + ζ13 + ζ14 + ζ15 + ζ16
Now, when we add A + B, we get the sum of all sixteen roots. And we know that to be equal to negative
one. In fact, no matter what subdivision we make, the sum of the two divisions will be negative one.
Remember:
�5 The Heptadecagon.nb
���
5
1 + x + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 + x13 + x14 + x15 + x16 = 0
which is to say:
x + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 + x13 + x14 + x15 + x16 = 1
Just put in ζ for x:
ζ + ζ2 + ζ3 + ζ4 + ζ5 + ζ6 + ζ7 + ζ8 + ζ9 + ζ10 + ζ11 + ζ12 + ζ13 + ζ14 + ζ15 + ζ16 = 1
So the issue that has to be addressed is η1 × η2 , the product of two eightterm sums.
Modular Arithmetic
Sorting out the other half of the problem will involve us in modular arithmetic. It’s not at all diﬀicult for
anyone who has read clock.
�
ζ6
ζ
ζ7
ζ2
ζ5
ζ3
ζ8
ζ4
Remember that multiplying roots of unity by themselves  that is, raising them to powers  will move
the solution around the unit circle in the complex plane like a clock hand. (In this case the clock hand
goes counterclockwise; all analogies have problems!). For example, there are five fi�h roots of unity. If
I square the first one, then cube it and so forth, the result moves around the unit circle. Also, when the
�6 ���
5 The Heptadecagon.nb
hand has gone completely around, all later solutions are equivalent to one or another of the first five
solutions. Thus, ζ 6 is equivalent to ζ 1 .
The technical term for this sort of equivalence is “congruence”; we write ζ 6 ≡ (ζ 1 )Mod 5 , “zeta to the
sixth is congruent with zeta one, modulo 5.”
Congruence of this sort will be very useful for us, for Gauss’s solution to the sorting problem involves
some very high powers of the 17th roots of unity.
Primitive Roots
One observation about modular arithmetic before we go on. Suppose we are working in modulo 17 as we will be doing. Take some number, a, and raise it to successive powers. In ordinary arithmetic, it
will grow continually. In modular arithmetic, it will go around the cycle of available numbers. Some
numbers in doing to touch all the values available; some do not.
Take 2, for instance:
TableFormTablen, 2n , Mod2n , 17, {n, 1, 16},
TableHeadings → None, "n", "2n ", "(2n )mod 17 "
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
2n
2
4
8
16
32
64
128
256
512
1024
2048
4096
8192
16 384
32 768
65 536
(2n )mod 17
2
4
8
16
15
13
9
1
2
4
8
16
15
13
9
1
If I use “n” to designate the power to which one raises the root, notice that 2 cycles through seven
values before coming to n=1 and repeating itself.
On the other hand, 3 cycles through all the possible values before repeating itself
�5 The Heptadecagon.nb
���
7
TableFormTablen, 3n , Mod3n , 17, {n, 1, 16},
TableHeadings → None, "n", "3n ", "(3n )mod 17 "
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
3n
3
9
27
81
243
729
2187
6561
19 683
59 049
177 147
531 441
1 594 323
4 782 969
14 348 907
43 046 721
(3n )mod 17
3
9
10
13
5
15
11
16
14
8
7
4
12
2
6
1
It can be shown that, for prime numbers, there is always at least one such value. For our purposes, with
the 17gon, we only need one. The number three will work for us. (There are others; 2, 4, 8, 9, 13 and 15
don’t work; 3, 5, 6, 7, 10, 11, 12, and 14 do.)
Ordering of the Sixteen Roots
Gauss ordered the sixteen roots in accordance with the expression:
n
ζ (3 )
Since 3n modulo 17 cycles through all values from 1 to 16 before repeating, this ordering will encompass all sixteen complex roots. The ordering looks like this:
n
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
3n
1
3
9
27
81
243
729
2187
6561
19 683
59 049
177 147
531 441
1 594 323
4 782 969
14 348 907
43 046 721
(3n )mod 17
1
3
9
10
13
5
15
11
16
14
8
7
4
12
2
6
1
ζ3
ζ
ζ3
ζ9
ζ10
ζ13
ζ5
ζ15
ζ11
ζ16
ζ14
ζ8
ζ7
ζ4
ζ12
ζ2
ζ6
ζ
n
mod 17
�8 ���
5 The Heptadecagon.nb
Notice that each term is obtained from the previous one by successive powers of three. Notice also,
that if one takes every other term, one has a succession by powers of nine:
1, 9, 81, 729 …
or
3, 27 = 3 x9, 243 = 3 x 81, 2187 = 3 x 729 …
This will be useful in what follows.
Two Eight Periods
The 16period is divided into two 8periods by taking alternate members of the series and summing the.
η1 = ζ + ζ9 + ζ13 + ζ15 + ζ16 + ζ8 + ζ4 + ζ2
η2 = ζ3 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6
Notice that each period contains four pairs of complex conjugates.
ζ5
ζ
ζ4
ζ3
6
ζ2
ζ7
ζ
ζ8
ζ9
ζ 16
ζ 10
ζ 15
ζ 11
ζ 12
ζ 14
ζ
13
Complex 17th roots of unity  Two EightPeriods
Sum of the 8periods
η1 + η2 = 1, as explained above.
Product of the 8periods
The product η1 η2 requires some calculation. We have the multiplication of two eightterm sums, which
will yield sixty four terms:
�5 The Heptadecagon.nb
ζ + ζ9 + ζ13 + ζ15 + ζ16 + ζ8 + ζ4 + ζ2
ζ3 + ζ10 + ζ5 + ζ11 + ζ14 + ζ7 + ζ12 + ζ6 = …
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
exponent1
1
1
1
1
1
1
1
1
9
9
9
9
9
9
9
9
13
13
13
13
13
13
13
13
15
15
15
15
15
15
15
15
16
16
16
16
16
16
16
16
8
8
8
8
8
8
8
8
4
4
4
4
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
exponent 2
3
10
5
11
14
7
12
6
3
10
5
11
14
7
12
6
3
10
5
11
14
7
12
6
3
10
5
11
14
7
12
6
3
10
5
11
14
7
12
6
3
10
5
11
14
7
12
6
3
10
5
11
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
sum
4
11
6
12
15
8
13
7
12
19
14
20
23
16
21
15
16
23
18
24
27
20
25
19
18
25
20
26
29
22
27
21
19
26
21
27
30
23
28
22
11
18
13
19
22
15
20
14
7
14
9
15
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
summod 17
4
11
6
12
15
8
13
7
12
2
14
3
6
16
4
15
16
6
1
7
10
3
8
2
1
8
3
9
12
5
10
4
2
9
4
10
13
6
11
5
11
1
13
2
5
15
3
14
7
14
9
15
���
9
�10 ���
5 The Heptadecagon.nb
53
54
55
56
57
58
59
60
61
62
63
64
4
4
4
4
2
2
2
2
2
2
2
2
+
+
+
+
+
+
+
+
+
+
+
+
14
7
12
6
3
10
5
11
14
7
12
6
=
=
=
=
=
=
=
=
=
=
=
=
18
11
16
10
5
12
7
13
16
9
14
8
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
≡
1
11
16
10
5
12
7
13
16
9
14
8
This is a little hard to digest. However, there are some patterns and repetitions. Looking just at the
exponents of the terms:
1 + 3 = 4 ;
9 + 10 = 19 ; 13 + 5 = 18 ;
1 + 10 = 11 ; 9 + 5 = 14 ;
1 + 5 = 6 ;
15 + 11 = 26 ; 16 + 14 = 30 ; 8 + 7 = 15 ;
13 + 11 = 24 ; 15 + 14 = 29 ; 16 + 7 = 23 ;
9 + 11 = 20 ; 13 + 14 = 27 ; 15 + 7 = 22 ;
1 + 11 = 12 ; 9 + 14 = 23 ; 13 + 7 = 20 ;
1 + 14 = 15 ; 9 + 7 = 16 ;
1 + 7 = 8 ;
13 + 12 = 25 ; 15 + 6 = 21 ;
9 + 12 = 21 ; 13 + 6 = 19 ;
15 + 3 = 18 ;
1 + 12 = 13 ; 9 + 6 = 15 ;
13 + 3 = 16 ;
1 + 6 = 7 ;
13 + 10 = 23 ; 15 + 5 = 20 ;
9 + 3 = 12 ;
8 + 12 = 20 ; 4 + 6 = 10 ;
16 + 12 = 28 ; 8 + 6 = 14 ;
15 + 12 = 27 ; 16 + 6 = 22 ;
16 + 3 = 19 ;
8 + 3 = 11 ;
4 + 3 = 7 ;
2 + 3 = 5 ;
2 + 10 = 12 ;
4 + 10 = 14 ; 2 + 5 = 7 ;
8 + 10 = 18 ; 4 + 5 = 9 ;
16 + 10 = 26 ; 8 + 5 = 13 ;
15 + 10 = 25 ; 16 + 5 = 21 ;
4 + 12 = 16 ; 2 + 6 = 8 ;
2 + 11 = 13 ;
4 + 11 = 15 ; 2 + 14 = 16 ;
8 + 11 = 19 ; 4 + 14 = 18 ; 2 + 7 = 9 ;
16 + 11 = 27 ; 8 + 14 = 22 ; 4 + 7 = 11 ;
2 + 12 = 14 ;
Or, with the sums reduced modulo 17:
1 + 3 = 4 ;
9 + 10 = 2 ; 13 + 5 = 1 ;
1 + 10 = 11 ; 9 + 5 = 14 ; 13 + 11 = 7 ;
1 + 5 = 6 ;
15 + 11 = 9 ;
16 + 14 = 13 ; 8 + 7 = 15 ; 4 + 12 = 16 ; 2 + 6 = 8 ;
15 + 14 = 12 ; 16 + 7 = 6 ;
9 + 11 = 3 ; 13 + 14 = 10 ; 15 + 7 = 5 ;
8 + 12 = 3 ; 4 + 6 = 10 ;
16 + 12 = 11 ; 8 + 6 = 14 ; 4 + 3 = 7 ;
2 + 3 = 5 ;
2 + 10 = 12 ;
1 + 11 = 12 ; 9 + 14 = 6 ; 13 + 7 = 3 ;
15 + 12 = 10 ; 16 + 6 = 5 ;
8 + 3 = 11 ; 4 + 10 = 14 ; 2 + 5 = 7 ;
1 + 14 = 15 ; 9 + 7 = 16 ; 13 + 12 = 8 ;
15 + 6 = 4 ;
16 + 3 = 2 ;
8 + 10 = 1 ; 4 + 5 = 9 ;
1 + 7 = 8 ;
15 + 3 = 1 ;
16 + 10 = 9 ;
8 + 5 = 13 ; 4 + 11 = 15 ; 2 + 14 = 16 ;
1 + 12 = 13 ; 9 + 6 = 15 ; 13 + 3 = 16 ;
15 + 10 = 8 ;
16 + 5 = 4 ;
8 + 11 = 2 ; 4 + 14 = 1 ;
1 + 6 = 7 ;
15 + 5 = 3 ;
16 + 11 = 10 ; 8 + 14 = 5 ; 4 + 7 = 11 ;
9 + 12 = 4 ; 13 + 6 = 2 ;
9 + 3 = 12 ; 13 + 10 = 6 ;
Or, again, more graphically:
2 + 11 = 13 ;
2 + 7 = 9 ;
2 + 12 = 14 ;
�5 The Heptadecagon.nb
ζ4
ζ 11
ζ6
ζ 12
ζ 15
ζ8
ζ 13
ζ7
ζ2
ζ 14
ζ3
ζ6
ζ 16
ζ4
ζ 15
ζ 12
ζ
ζ7
ζ 10
ζ3
ζ8
ζ2
ζ 16
ζ6
ζ9
ζ 12
ζ5
ζ 10
ζ4
ζ
ζ8
ζ3
ζ 13
ζ6
ζ 11
ζ5
ζ2
ζ9
ζ4
ζ 10
ζ 15
ζ3
ζ 14
ζ 11
ζ
ζ 13
ζ2
ζ5
ζ 16
ζ 10
ζ7
ζ 14
ζ9
ζ 15
ζ
ζ 11
���
11
ζ8
ζ5
ζ 12
ζ7
ζ 13
ζ 16
ζ9
ζ 14
Notice that the yellow rows have all the same members; so do the green rows. Also, that the yellow
rows and the green rows have all diﬀerent members, so that a yellow row and a green row together have
all sixteen elements. And all sixteen elements together equal negative one, so that the total of all sixtyfour terms is … (drum roll, please) … negative four.
This is not an accident. It was carefully orchestrated by Gauss’s arrangement of the two eightgroups.
In fact, the necessity that leads to this arrangement can be seen through examination and analysis of
the patterns of the terms entering into the multiplication together with about a week of practice with
modular multiplication. I can’t undertake that dissection in more detail here; any one interested can
pursue a more complete presentation in the texts referred to in the handout.
For our purposes, what is essential is that we know the sum and the product of the eightperiods,
η1 and η2 .
Constructing and Solving an Appropriate Quadratic
We now know that the sum η1 + η2 = 1 , and the product η1 η2 = 4. We can therefore make a
quadratic equation with these two numbers as its solutions. Using y as a variable, we have:
y2 
(1) y +
( 4) = 0
η1 + η 2
η1 η2
whose solutions will be η1 and η2 . This equation can be solved with the quadratic formula.
y=
1 ±
1 + 16
2
=
1 ±
2
17
Beautiful. So I know values of the η’s, which are the sums of the 8periods. The two values are:
�12 ���
5 The Heptadecagon.nb
η1, 2 =
1 ±
2
17
The approximate values for η1 and η2 are 1.56155 and 2.56155.
The Four Periods
Next we make four periods of four by taking every fourth root from the original series, starting with the
first, second, third and fourth, respectively:
ζ, ζ3 , ζ9 , ζ10 , ζ13 , ζ5 , ζ15 , ζ11 , ζ16 , ζ14 , ζ8 , ζ7 , ζ4 , ζ12 , ζ2 , ζ6
ζ13 ,
Period 1 = ζ,
Period 2 =
Period 3 =
ζ3 ,
ζ16 ,
ζ5 ,
ζ9 ,
Period 4 =
ζ4 ,
ζ14 ,
ζ15 ,
ζ10 ,
ζ12
ζ8 ,
ζ11 ,
ζ2
ζ7 ,
ζ6
Sums of the Four Periods
We make the sums of each of the fourperiods:
μ1 = ζ + ζ4 + ζ13 + ζ16
μ2 = ζ3 + ζ5 + ζ12 + ζ14
μ3 = ζ2 + ζ8 + ζ9 + ζ15
μ4 = ζ6 + ζ7 + ζ10 + ζ11
Again, the sums present no diﬀiculy: μ1 + μ3 = η1 and μ2 + μ4 = η2 , since the the fourperiods
are gotten by segregating elements of the two eightperiods.
Graphically, the four periods are pictured below. Notice that the 8periods have been subdivided: the
red 8period into red and green 4periods; the blue 8period into blue and orange 4periods. Once
again, notice that each fourperiod includes two pairs of complex conjugates:
�5 The Heptadecagon.nb
ζ5
ζ
���
ζ4
ζ3
6
ζ2
ζ7
ζ
ζ8
ζ9
ζ 16
ζ 10
ζ 15
ζ 11
ζ 12
ζ 14
ζ 13
Inner dots = 8 periods; Outer dots = 4 periods
Products of the 4Periods
As for the products of the μ’s, we can work out the terms directly. First, take μ1 times μ3 :
Expand[μ1 μ3]
ζ3 + ζ6 + ζ9 + ζ10 + ζ12 + ζ13 + ζ15 + ζ16 + ζ18 + ζ19 + ζ21 + ζ22 + ζ24 + ζ25 + ζ28 + ζ31
Which, when simplified by reexpressing the exponents modulo 17:
ζ3 + ζ6 + ζ9 + ζ10 + ζ12 + ζ13 + ζ15 + ζ16 + ζ1 + ζ2 + ζ4 + ζ5 + ζ7 + ζ8 + ζ11 + ζ14
Put in numerical order of the exponents:
ζ1 + ζ2 + ζ3 + ζ4 + ζ5 + ζ6 + ζ7 + ζ8 + ζ9 + ζ10 + ζ11 + ζ12 + ζ13 + ζ14 + ζ15 + ζ16 = 1
And for μ2 times μ4 :
Expand[μ2 μ4]
ζ9 + ζ10 + ζ11 + ζ12 + ζ13 + ζ14 + ζ15 + ζ16 + ζ18 + ζ19 + ζ20 + ζ21 + ζ22 + ζ23 + ζ24 + ζ25
Again, reduced by reexpressing the exponents modulo 17:
ζ9 + ζ10 + ζ11 + ζ12 + ζ13 + ζ14 + ζ15 + ζ16 + ζ1 + ζ2 + ζ3 + ζ4 + ζ5 + ζ6 + ζ7 + ζ8
Put in numerical order of the exponents
ζ1 + ζ2 + ζ3 + ζ4 + ζ5 + ζ6 + ζ7 + ζ8 + ζ9 + ζ10 + ζ11 + ζ12 + ζ13 + ζ14 + ζ15 + ζ16 = 1
13
�14 ���
5 The Heptadecagon.nb
Cool.
Solving for the μ’s
Once again, we know the sums and products of pairs of variables, in this case μ1 and μ3 and also
μ2 and μ4 :
μ1 + μ3 = η1
μ1 × μ3 = 1
μ2 + μ 4 = η 2
μ2 × μ4 = 1
With these sumsandproducts, we can make two quadratic equations; we use v and w as variables:
v 2  η1 v  1 = 0
μ1 , μ 3 =
η1 ±
whose solutions are μ1 and μ3
η1 2 + 4
2
w2  η 2 w  1 = 0
And
μ2 , μ 4 =
η2 ±
whose solutions are μ2 and μ4
η2 2 + 4
2
Just to show where we are at this point, we can identify the values of the μ’s:
μ1 =
η1 +
η1 2 + 4
2
=
1
2
1
2
1 +
17 +
4 + 14 1 +
17
μ2 =
η2 +
η2 2 + 4
2
=
1
2
1
2
1 
17 +
4 + 14 1 
17
μ3 =
η1 
η1 2 + 4
2
=
1
2
1
2
1 +
17 
4 + 14 1 +
μ1 =
η2 
η2 2 + 4
2
=
1
2
1
2
1 
17 
4 + 14 1 
2
≈ 2.04948
2
≈ 0.344151
17
2
≈ 0.487928
17
2
≈ 2.9057
The Two Periods
With this in hand, we look at the twoperiods, obtained as before but this time taking every eighth root
from the original list:
β1 = ζ + ζ 16
β2 = ζ 3 + ζ 14
β3 = ζ 8 + ζ 9
�5 The Heptadecagon.nb
β4
β5
β6
β7
β8
=
=
=
=
=
���
ζ 7 + ζ 10
ζ 4 + ζ 13
ζ 5 + ζ 12
ζ 2 + ζ 15
ζ 6 + ζ 11
It may be worth noting that each pair of roots that make up a β is a complex conjugate pair. This is
importanT, although its special importance won’t appear until the next stage.
As in previous steps, these twoperiods come about by separating elements of the fourperiods. Their
sums thus lead us back to the variables of the previous step:
β1
β2
β3
β4
+
+
+
+
β5 = ζ + ζ 16 + ζ 4 + ζ 13 = μ1
β6 = ζ 3 + ζ 14 + ζ 5 + ζ 12 = μ2
β7 = ζ 8 + ζ 9 + ζ 2 + ζ 15 = μ3
β8 = ζ 7 + ζ 10 + ζ 6 + ζ 11 = μ4
We thus have sums of pairs of the β’s. It remains to figure out the products of the same pairs.
Products of the 2Periods
With a little labor, we can figure out the products of the 2periods paired in way given above. The
products are given below, including the reduction of the exponents modulo 17:
β1 β5 = ζ + ζ16 ζ4 + ζ13 = ζ5 + ζ14 + ζ20 + ζ29 = ζ5 + ζ14 + ζ3 + ζ12 = μ2
β2 β6 = ζ5 + ζ12 ζ3 + ζ14 = ζ8 + ζ15 + ζ19 + ζ26 = ζ8 + ζ15 + ζ2 + ζ9 = μ3
β3 β7 = ζ2 + ζ15 ζ8 + ζ9 = ζ10 + ζ11 + ζ23 + ζ24 = ζ10 + ζ11 + ζ6 + ζ7 = μ4
β4 β8 = ζ6 + ζ11 ζ7 + ζ10 = ζ13 + ζ16 + ζ18 + ζ21 = ζ13 + ζ16 + ζ1 + ζ4 = μ1
Now we have defined the products as well as the sums of the four pairs of 2periods. Again, the way in
which this multiplication works out is not an accident; it follows from Gauss’s original ordering of the
roots that combinations taken by twos, by fours, and so forth will always multiply so as to produce
these intermediate periods.
Constructing Four Quadratic Equations
With that in mind, we can construct four quadratic equations with the β’s as roots:
q2  μ1 q + μ2 = 0 whose roots are β1 and β5
15
�16 ���
5 The Heptadecagon.nb
r2  μ2 r + μ3 = 0 whose roots are β2 and β6
s2  μ3 s + μ4 = 0 whose roots are β3 and β7
t2  μ4 t + μ1 = 0
whose roots are β4 and β8
Their solutions can be obtained with the quadratic formula. Since we know the values of the μ’s, we
can calculate the values of the β’s:
β1 =
μ1 +
μ1 2  4 μ2
2
= 1.86494
β5 =
μ1 
μ1 2  4 μ2
2
= 0.184537
β2 =
μ2 +
μ2 2  4 μ3
2
= 0.891477
β6 =
μ2 
μ2 2  4 μ3
2
= 0.547326
β3 =
μ3 +
μ3 2  4 μ4
2
= 1.47802
β7 =
μ3 
μ3 2  4 μ4
2
= 1.96595
β4 =
μ4 +
μ4 2  4 μ1
2
= 1.20527
β8 =
μ4 
μ4 2  4 μ1
2
= 1.70043
The Singletons
One more step remains: dividing the 2periods into individual roots. This is in some ways the easiest
step of all.
Their Sums
There are sixteen individual roots:
ζ, ζ2 , ζ3 , ζ4 , ζ5 , ζ6 , ζ7 , ζ8 , ζ9 , ζ10 , ζ11 , ζ12 , ζ13 , ζ14 , ζ15 , ζ16
These, take pairwise in a particular order, constitute the β’s:
β1
β2
β3
β4
β5
β6
β7
β8
=
=
=
=
=
=
=
=
ζ + ζ 16
ζ 3 + ζ 14
ζ8 + ζ9
ζ 7 + ζ 10
ζ 4 + ζ 13
ζ 5 + ζ 12
ζ 2 + ζ 15
ζ 6 + ζ 11
Here we see that we already have the sums of the sixteen ζ’s, taken pairwise.
Their Products
�5 The Heptadecagon.nb
���
This time, the product of the roots just as simple as the sums. Since, as already noted above, each β
pair constitutes a pair of complex conjugates, their product  obtained by adding their exponents  is
always seventeen or, on the unit circle in the complex plane, +1.
ζ ζ16 = ζ17 =
ζ3 ζ14 = ζ17 =
ζ8 ζ9 = ζ17 =
ζ7 ζ10 = ζ17 =
ζ4 ζ13 = ζ17 =
ζ5 ζ12 = ζ17 =
ζ2 ζ15 = ζ17 =
ζ6 ζ11 = ζ17 =
1
1
1
1
1
1
1
1
WIth this information, we have the sum and the products of the roots (taken in this special order) we
can construct eight quadratic equations whose solutions are the ζ’s.
r 2  β1 r + 1 = 0
s2  β 2 s + 1 = 0
t2  β3 t + 1 = 0
v 2  β4 v + 1 = 0
w2  β 5 w + 1 = 0
x 2  β6 x + 1 = 0
y 2  β7 y + 1 = 0
z2  β8 z + 1 = 0
whose solutions are ζ and ζ16
whose solutions are ζ3 and ζ14
whose solutions are ζ8 and ζ9
whose solutions are ζ7 and ζ10
whose solutions are ζ4 and ζ13
whose solutions are ζ5 and ζ12
whose solutions are ζ2 and ζ15
whose solutions are ζ6 and ζ11
We can apply the quadratic formula to find the solutions. Since we have the values for the β’s, we can
obtain values for the ζ’s:
ζ and ζ16 =
β1 ±
β1 2  4
2
= 0.932472 ± 0.361242 ⅈ
ζ3 and ζ14 =
β2 ±
β2 2  4
2
= 0.445738 ± 0.895163 ⅈ
ζ8 and ζ9 =
β3 ±
β3 2  4
2
= 0.982973 ± 0.18375 ⅈ
ζ7 and ζ10 =
β4 ±
β4 2  4
2
= 0.850217 ± 0.526432 ⅈ
ζ4 and ζ13 =
β5 ±
β5 2  4
2
= 0.0922684 ± 0.995734 ⅈ
ζ5 and ζ12 =
β6 ±
β6 2  4
2
= 0.273663 ± 0.961826 ⅈ
ζ2 and ζ15 =
β7 ±
β7 2  4
2
= 0.739009 ± 0.673696 ⅈ
ζ6 and ζ11 =
β8 ±
β8 2  4
2
= 0.602635 ± 0.798017 ⅈ
Shown graphically:
17
�18 ���
5 The Heptadecagon.nb
ζ4
ζ5
ζ
ζ3
6
ζ2
ζ7
ζ
ζ8
ζ9
ζ 16
ζ 10
ζ 15
ζ 11
ζ 14
ζ 12
ζ 13
As advertised.
Full Algebraic Presentation of the Sixteen Complex Roots of Unity
The stack of quadratic equations involved in calculating the heptadecagon vertices is diﬀicult to grasp
when they are all assembled into a single formula. One of the roots is represented as:
1
2
1
4
1
2
17 
17 +
1
8
1 +
1
17 +
34+6
4 +
1
64
1 +
17 +
34  2
+
2
4
17 +
57834
17 + 2 34 + 6
17

342
17
8
2 17+
17
17 +
2
578  34
17 
34  2
17  8
2 17 +
17
It’s a challenge to grasp such a thing, but even a casual inspection shows that it consists exclusively of
stacks of rational numbers and square roots combined with rational functions (addition, subtraction,
multiplication and division). That alone is enough to guarantee its constructibility.
�5 The Heptadecagon.nb
���
19
An abbreviated graphic representation of the process of constructing the heptadecagon might look like
this:
ζ1 … 16 =
β1, 2, 3, 4, 5, 6, 7, 8 =
βn 2  4
βn ±
2
μ1, 2, 3, 4 2  4 μ2, 3, 4, 1
μ1, 2, 3, 4 ±
2
μ1, 3, 2, 4 =
η1,2 =
η1,2 ±
η1,2 2 + 4
2
17
1 ±
2
The sixteen roots were subdivided successively as follows.
ζ , ζ 3 , ζ 9 , ζ 10 , ζ 13 , ζ 5 , ζ 15 , ζ 11 , ζ 16 , ζ 14 , ζ 8 , ζ 7 , ζ 4 , ζ 12 , ζ 2 , ζ 6
The whole
t
The μ's
The ζ's
ζ
ζ 16
ζ 13 , ζ 4
ζ 13
ζ 3 , ζ 5 , ζ 14 , ζ 12
ζ 9 , ζ 15 , ζ 8 , ζ 2
ζ , ζ 13 , ζ 16 , ζ 4
ζ , ζ 16
The β's
ζ 3 , ζ 10 , ζ 5 , ζ 11 , ζ 14 , ζ 7 , ζ 12 , ζ 6
ζ , ζ 9 , ζ 13 , ζ 15 , ζ 16 , ζ 8 , ζ 4 , ζ 2
The η's
ζ4
ζ 9, ζ 8
ζ9
ζ8
ζ 15 , ζ 2
ζ 15
ζ 3 , ζ 14
ζ2
ζ3
ζ 14
ζ 10 , ζ 11 , ζ 7 , ζ 6
ζ 5 , ζ 12
ζ5
ζ 12
ζ 10 , ζ 7
ζ 10
ζ7
ζ 11 , ζ 6
ζ 11
ζ6
The sum of the whole set was 1. The sum of the η’s required a finite quadratic field extension to
include
1 ± 17
2
. Each successive subdivision required another finite quadratic field extension of what
went before. But (and?) that is the sine qua non of constructibility: a point is constructible if (and only
�20 ���
5 The Heptadecagon.nb
if) it is defined by numbers that are either rational or the result of a succession of finite, quadratic field
extensions from the rationals.
Extension
This is a pretty remarkable result: the 17gon is constructible using the techniques available in Euclid’s
Elements. In fact, Gauss’s result is even more remarkable than that. It has both a positive and a negative side, which I can report although we haven’t done quite enough work to demonstrate both sides
fully.
The Positive Side
The positive side is that any figure is constructible if it the number of its sides is either,
(a) a prime number equal to 2n + 1, or
(b) some multiple of 2p times such a number, or
(c) the sum of two of the primes in (a), or 2p times that sum.
So, 20 = 1, 1 + 1 is 2, which is prime. You can’t make a 2gon, really, but you can make polygons that
are multiples of 2 times it: a 4gon (square), and 8gon (octagon), etc.
Next, 21 = 2. 2 + 1 is 3, which is prime. You can make a 3gon (triangle), or a 6gon (hexagon), etc.
A�er that, 22 = 4. 4 + 1 is 5, which is prime. You can make a 5gon (pentagon). Euclid could do this.
These are all the constructible prime ngons that Euclid knew. He doesn’t say so much, but if he had
known the construction of another it is hard to believe he would not have given it.
A�er that, 24 = 16. 16 + 1 is 17, which is prime. You can make a 17gon (heptadecagon). This was
Gauss’s great discovery. I cannot believe that Euclid knew that the 17gon was constructible.
But there are more!
Consider: 28 = 256. 256 + 1 = 257, which is prime. The 257gon is constructible.
Consider: 216 = 64 536. 64,536 + 1 = 64,537, which is prime. The 64,537gon is constructible.
In fact, if you can find another number of the form 2n + 1 which is prime, it too will be constructible.
These are the socalled “Fermat numbers,” named for Pierre Fermat who conjectured that all numbers
of the form 2n + 1 were prime, provided that n itself is a power of 2. Such numbers are:
�5 The Heptadecagon.nb
���
21
3, 5, 17, 257, 64537, …
The next one would be 4,294,967,297 … but this on turns out not to be prime. (It is 641 times
6,700,417). As of last year, only the first eleven such numbers have been fully tested. The last one,
11
22 + 1, has 617 digits; it has two prime factors. The next one has 1,234 digits; whether it is prime or
not is still undetermined. Las Vegas is not giving odds. In fact, now it is conjectured that apart from the
first five, no other Fermat numbers are prime, although as far as I know that guess hasn’t been proven
or disproven.
The Negative Side
The negative claim is that only these polygons are constructible. Gauss did not demonstrate that his
construction method was the only one possible. The negative claim was demonstrated later by Pierre
Wantzel in 1837. The sevengon, which Euclid just skips, cannot be constructed. Neither can the 9gon,
the 11gon or the 13gon. Euclid constructs the (nonprime) 15gon by the combination of the triangle
and the pentagon. His leap from the hexagon to the 15gon is completely unexplained in the Elements,
and to my knowledge few students remark on it. (They should.)
So What?
What follows from this exercise? That is the subject of next week’s talk.
Further Reading
For further explication of Gauss’s sorting of the roots and modular congruences, you may wish to
consult
Hadlock, Field Theory and Its Classical Problems (Mathematical Association of America, 1978)
�
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Text
The Pentagon and the Heptagon
Recapitulation
Two weeks ago, we saw how Rafael Bombelli began to suspect that imaginary numbers might be
meaningful as he worked on the cubic equation
x 3  15 x  4 = 0
Using the formula Cardano stole from Tartaglia, got
x=
3
2 + 11
1
+
3
2  11
1
which he was then able to solve by intuiting that
2 + 11
1 = 2 +
3
1 .
The second lecture described Caspar Wessel’s graphic presentation of the arithmetic of complex
numbers. On the complex number plane (i) complex numbers can be expressed in polar coordinates by giving a distance (modulus) and an
angle (argument);
(ii) multiplication of complex numbers amounts to
(a) multiplication of their distances (moduli) and
(b) adding their angles (arguments); and
(iii) the solutions of equations of the form x n  1 = 0, known as the “Roots of Unity,” appear
graphically as the vertices of an equilateral ngon in the unit circle on the complex plane.
Last week, we encountered the idea of “Constructible Numbers.” We showed that Euclid’s postulates
allowed construction of lengths that correspond to the field of rational numbers, a collection of numbers that is closed under the operations of addition, subtraction, multiplication and division. In addition, Euclid’s postulates allow the construction of incommensurable magnitudes (which correspond to
irrational numbers). However, Euclid’s postulates do not permit construction of all incommensurable
magnitudes. We can only construct those that correspond to numbers that can be found in towers of
finite quadratic field extensions, that is, field extensions that have a degree of 2n over the rational
numbers. Plenty of numbers are not included. For example, 2 is not constructible; neither are the
nonalgebraic (transcendental) numbers, which form an uncountable infinity far greater than the
countable infinity of the algebraic numbers.
3
�2
bers that is closed under the operations of addition, subtraction, multiplication and division. In addition, Euclid’s postulates allow the construction of incommensurable magnitudes (which correspond to
irrational
numbers).
However, Euclid’s postulates do not permit construction of all incommensurable
4 The
Pentagon and
the Heptagon.nb
magnitudes. We can only construct those that correspond to numbers that can be found in towers of
finite quadratic field extensions, that is, field extensions that have a degree of 2n over the rational
numbers. Plenty of numbers are not included. For example, 2 is not constructible; neither are the
nonalgebraic (transcendental) numbers, which form an uncountable infinity far greater than the
countable infinity of the algebraic numbers.
3
We now turn to an application of what we have seen so far: construction of a regular pentagon in a
given circle.
The LesserKnow Impossibility Problem
Ancient geometry knew several classical problems that seemed impossible; the three most famous
were trisecting the angle, doubling the cube and squaring the circle. These were daunting challenges.
No one had found how to accomplish any of them, but the ancients did not know whether they were
really impossible or only difficulties awaiting clever solutions
Only in the late 18th and 19th centuries did we learn that these three problems really are impossible, at
least with the tools of Euclidean geometry.
In addition to these three, lesserknown but equally interesting problem is that of the heptagon, the
regular sevensided polygon. In book IV of the Elements, Euclid shows how to construct in a given
circle an equilateral triangle (IV.2), a square (IV. 6), a pentagon (IV. 11) and a hexagon (IV. 15), regular
figures with three, four, five and six sides. He then shows how to construct a regular 15gon (IV. 16).
Then he stops.
The reader might be expected to wonder, why? Why jump from 6 to 15? Euclid, in his customary
laconic way, says nothing. Some of the figures he skips over were easily constructible. The octagon is
easily made by bisecting the angles of the square. The 10gon can be gotten similarly from the pentagon and the 12gon, from the hexagon. But the orderliness of Euclid’s sequence really falls apart with
the 7gon. With what we learned last week, it is easy to see that the 7gon is impossible to construct. It
is obtained from the polynomial:
x 7  1 = ( x  1) x 6 + x 5 + x 4 + x 3 + x 2 + x 1 + 1 = 0
That sixthdegree polynomial is irreducible and, since its degree over the rationals is not a power of
two, we can see right away that these complex roots are not constructible.
Did Euclid know that the 7gon was impossible? He probably suspected it. He surely knew that he
couldn’t do it, which is not quite the same thing.
Beyond the Impossible: the Unsuspected Possible
In 1796, at the age of 19, Carl Friedrich Gauss realized the impossibility of constructing the 7gon; what
is mor, he realized at the same time that there are other polygons that can be constructed. Looking
only at those with a prime number of sides, in his book Disquisitiones Arithmeticae, he not only showed
that the 17gon is constructible, he showed how to do it. This is remarkable advance beyond what
Euclid knew.
To help us get to Gauss’s result, it will be helpful to begin with a slightly simpler project: the algebraic
�4 The Pentagon and the Heptagon.nb
3
In 1796, at the age of 19, Carl Friedrich Gauss realized the impossibility of constructing the 7gon; what
is mor, he realized at the same time that there are other polygons that can be constructed. Looking
only at those with a prime number of sides, in his book Disquisitiones Arithmeticae, he not only showed
that the 17gon is constructible, he showed how to do it. This is remarkable advance beyond what
Euclid knew.
To help us get to Gauss’s result, it will be helpful to begin with a slightly simpler project: the algebraic
construction of the pentagon.
Euclid’s Construction of the Pentagon
Of course, Euclid knew how to construct a regular pentagon in a given circle. To begin, let’s review how
Euclid’s construction works.
First, a Special Triangle
He begins with construction of a very special triangle, one that is isosceles and whose base angles are
both twice as big as its vertex angle.
θ
2θ
2θ
A little reflection shows why this triangle might be important to the construction of a regular pentagon:
the three angles of the triangle total up to 180°, of course, but they also add up to five times the vertex
angle. This triangle creates one angle that is onefifth of 180°, and two that are onefifth of 360°. If this
triangle can be made, it will be the key to constructing the pentagon.
However, constructing this triangle is no simple matter.
To make it, Euclid recalls that back in book II, proposition 11, he had shown how to cut a line at a point
so that the square on one portion of the line is equal to the rectangle contained by the whole line and
the remaining portion of the line.
�4
4 The Pentagon and the Heptagon.nb
Digression
This kind of division is known as one into "mean and extreme ratio," sometimes also referred to as the
"Golden Ratio." It has many cool features, including connection to Fibonacci numbers and logarithmic
spirals, but we haven' t time to get into all these things right now.
If we take the whole AB to be “1” and the distance AC to be “x”, then finding this ratio can be understood as analogous to solving the equation:
x 2 = (1  x )
x2 + x  1 = 0
or
whose solutions are:
1±
1  4 (1)
2
=
1±
5
2
You may note that these values are not rational, since they contains the square root of five. They are, of
course, constructible, which we know because (a) we are dealing with a quadratic extension of the
rationals and (b) because Euclid in fact constructs one of them. (No surprise there.)
�1±
1  4 (1)
2
=
1±
5
4 The Pentagon and the Heptagon.nb
2
5
You may note that these values are not rational, since they contains the square root of five. They are, of
course, constructible, which we know because (a) we are dealing with a quadratic extension of the
rationals and (b) because Euclid in fact constructs one of them. (No surprise there.)
Returning to the Construction
Euclid takes a line divided in this way and, using one end as a center, draws a circle with the whole line
as a radius:
A
C
B
He then makes a chord in the circle equal to the larger segment of the divided line:
A
C
B
D
He completes the triangle ABD, and joins CD:
�6
4 The Pentagon and the Heptagon.nb
A
C
B
D
Finally, he draws a circle that goes through points A, C and D:
A
C
B
D
Thanks to a proposition from earlier in Book III, he knows that when from a point outside a circle (like
point B) a line cuts a circle (as line BCA), and another line is draw to the circumference of the circle (as
line BD), and when the rectangle on AB, AC is equal to the square on BD, then the line (BD) is tangent to
the circle (ACD).
With that established, another proposition of Book III allows him to say that the angle CDB (angle 1) is
equal to the angle CAD (angle 2):
�4 The Pentagon and the Heptagon.nb
A
C
2
B
4 5
3 1
D
Add angle CDA to both. Thus angles 2 + 3 are equal to angles 1 + 3. But because AB = AD (in the circle
around A), angles 1 + 3 are equal to angle 5 . So:
angle 5 = angles 1 + angle 3 = angle 2 +angle 3
and because of exterior angles in triangle CBD
angle 4 = angle 2 + angle 3
Therefore, triangle ABD is isosceles and line DB = line DC. And line DB = line AC.
Therefore, angle 3 = angle 2 = angle 1.
This, then, is the isosceles triangle with its base angles equal to twice the vertex angle.
The Pentagon
With the isosceles triangle having the base angles equal to the vertex angle now available, the rest is
easy.
7
�8
4 The Pentagon and the Heptagon.nb
Simply bisect the arcs standing on the longer sides, which are each twice the arc on the shorter side.
Now you have five equal sides and your pentagon is complete.
Join the vertices and you have not only a pentagon, but a pentangle (a regular fivepointed star).
�4 The Pentagon and the Heptagon.nb
9
This construction is completely rigorous and very clever. However, it offers no clues at all about how to
pursue construction of other such primesided polygons, such as the 7gon, the 11gon, the 13gon, etc.
Preliminary: the Pentagon
The algebraic construction of the pentagon amounts to finding the roots of the fifth degree cyclotomic
polynomial. That is, we begin with the equation:
x5 = 1
or
x 5  1 = 0.
The number 1 is evidently a solution to this equation. It is, in fact, the only rational solution. Therefore,
the equation can be factored by removing the factor (x  1):
x 5  1 = ( x  1) x 4 + x 3 + x 2 + x + 1 = 0
The Fifth Order Cyclotomic Polynomial
The second expression, x 4 + x 3 + x 2 + x + 1, is irreducible “over the rationals”; that is, it can’t be
simplified by showing it to be the product of factors of lower degree among the rationals. We can be
completely sure that this expression is irreducible because we know that the four roots of the polynomial x 4 + x 3 + x 2 + x + 1 = 0 are complex with imaginary components. They are the four nonreal fifth
roots of unity.
�10
4 The Pentagon and the Heptagon.nb
ζ1
ζ2
ζ3
ζ4
But being irreducible over the rationals doesn’t mean that this thing can’t be factored in an extended
field. In fact, it has been shown that every polynomial of nth degree can be factored into n linear factors
in the full complex number field. Our challenge is to find which factors need to be appended to the
rationals in order to factor or “split” this fourth degree polynomial.
Complex Conjugates
We haven’t discussed complex conjugates, but this diagram presents the idea nicely. Notice that the
complex roots of this polynomial appear as two pairs of complex numbers, symmetrically arranged
above and below the real number axis. Root ζ1 is paired this way with root ζ4 and root ζ2 is paired
with root ζ3 . Being so arranged, these roots are written in this form:
a+bi
and
a  bi
The expressions are the same except for the positive and negative signs attached to the imaginary
portions.
Complex conjugates have this handy feature: when a pair of complex conjugates are added, their sum
is a real number. Also, when a pair of complex conjugates are multiplied together, their product is a
real number.
This feature is handy because we are often looking for roots of polynomials whose coefficients are
rational (or, in any case, do not involve imaginaries). Of course, you can construct an arbitrary polynomial with a random selection of complex roots:
(x  (2 + 7 i)) (x  (9  3 i)) (x  (15 + 4 i)) = …
But if you multiply this trio out, you will have some imaginary coefficients.
�is a real number. Also, when a pair of complex conjugates are multiplied together, their product is a
real number.
4 The Pentagon and the Heptagon.nb
11
This feature is handy because we are often looking for roots of polynomials whose coefficients are
rational (or, in any case, do not involve imaginaries). Of course, you can construct an arbitrary polynomial with a random selection of complex roots:
(x  (2 + 7 i)) (x  (9  3 i)) (x  (15 + 4 i)) = …
But if you multiply this trio out, you will have some imaginary coefficients.
(x  (2 + 7 i)) (x  (9  3 i)) (x  (15 + 4 i)) =
(813 + 699 ⅈ)  (142  41 ⅈ) x + (4  8 ⅈ) x2 + x3
In fact, the only way to eliminate the imaginary components from the expanded polynomial is if the
coefficients occur in pairs of complex conjugates. That way, when the conjugates are multiplied, the
imaginary components disappear.
Return to the Problem
To solve our fourthdegree cyclotomic polynomial:
1 + x + x2 + x3 + x4 = 0
We will proceed in the usual, brash algebraic way: we will pretend that we already have the solutions.
Then we’ll work to discover what they are. The Fundamental Theorem of Algebra tells us that this
fourth degree equation has four solutions, which we will designate (as in the picture)
ζ1 , ζ2 , ζ3 and ζ4 . Roots ζ1 and ζ4 are one pair of complex conjugates; ζ2 and ζ3 are another
pair.
TwoStage Solution
Take the sums of ζ1 , ζ4 and of ζ2 , ζ3 , like this:
η1 = ζ 1 + ζ 4
η2 = ζ 2 + ζ 3
When added together, η1 and η2 sum up to 1 (because all the fifth roots of unity together sum to zero,
and η1 and η2 include all the roots except (+1 + 0 i):
η1 + η2 = ζ 1 + ζ 4 + ζ 2 + ζ 3 = 1
Also, the product of η1 and η2 works out like this:
ζ 1 + ζ 4 ζ 2 + ζ 3 = ζ 3 + ζ 4 + ζ 6 + ζ 7
Restate this result with the exponents taken Modulo 5, because, on the unit circle in the complex
plane, ζ 5 = ζ 0 = 1. Thus, we have
ζ6 = ζ5 ζ1 = ζ1
ζ7 = ζ5 ζ2 = ζ2
Substitute:
�12
Restate
thisand
result
with the exponents
4 The
Pentagon
the Heptagon.nb
5
taken Modulo 5, because, on the unit circle in the complex
0
plane, ζ = ζ = 1. Thus, we have
ζ6 = ζ5 ζ1 = ζ1
ζ7 = ζ5 ζ2 = ζ2
Substitute:
ζ3 + ζ4 + ζ6 + ζ7 = ζ3 + ζ4 + ζ1 + ζ2 = 1
Presto! We have the sum of the four nonreal roots of the equations x 5  1 = 0. We know that these
sum to 1.
Building a Quadratic Equation for η1, η2
Great! We have two terms, η1 and η2 . We don’t know what they are, but we do know that their sum is
1 and their product is also 1. Does that sound like a familiar situation? When we know that when we
know the sum and product of two terms, we can construct a quadratic equation that has these terms as
roots. In this case, we have:
x2 + x  1 = 0
whose roots are given by the quadratic formula:
η1 and η2 =
1 ±
1+4
2
=
1
2
1 +
5 and
1
2
1 
5 . (The approximate values of these are
0.61803 and 1.61803.)
Behold! Now It Factors!
Remember that we said that the expression x 4 + x 3 + x 2 + x + 1 = 0 could not be factored over the
rationals? Now it can be factored in an extended field when we append 12 1 +
append
5  or even if we just
5  to the rationals.
We have:
( x  ζ1 ) ( x  ζ4 ) = x 2  ζ1 x  ζ4 x + ζ1 ζ4
( x  ζ1 ) ( x  ζ4 ) = x 2  (ζ1 + ζ4 ) x + ζ1 ζ4 = x 2  (ζ1 + ζ4 ) x + 1
= η1
This expression, x
2
 (ζ1 + ζ4 ) x + ζ1 ζ4 , has coefficients that are in the extended field. The
= η1
coefficient of x is the sum of the two roots ζ1 + ζ4 ; we don’t know them individually yet, but we know
that they sum to η1 , which is in the extended field. The constant term is ζ1 ζ4 ; we know right away that
the product of these two is 1 (product of their moduli, sum of their arguments).
�4 The Pentagon and the Heptagon.nb
13
The Four Singletons
Now look at the four roots individually:
ζ1 , ζ2 , ζ3 , ζ4
We know how they sum in pairs:
η1 = ζ 1 + ζ 4
η2 = ζ 2 + ζ 3
We also know the products of the same pairs :
ζ1 ζ4 = ζ5 = 1
ζ2 ζ3 = ζ5 = 1
So we can make two more quadratic equations:
w2  η1 w + 1 = 0
whose roots are ζ1 and ζ4 , which are solved as w =
η1 ±
y 2  η2 y + 1 = 0
whose roots are ζ2 and ζ3 which are solved as y =
η2 ±
η1 2  4
2
η2 2  4
2
We now have enough information to solve for the four roots:
ζ
1
ζ
4
ζ
2
ζ
3
=
=
=
=
η1 +
η1 2  4
2
η1 
η1 2  4
2
η2 +
η2 2  4
2
η2 
η2 2  4
2
1
=
2
2
2
1+ 5 
1
1 5 +
1
2
1 5 
= 0.809017 + 0.587785 ⅈ
2
2 1 5  4
2
= 0.309017  0.951057 ⅈ
2
2 1 5  4
1
= 0.309017 + 0.951057 ⅈ
2
2 1+ 5  4
2
1
=
2
2
1
=
1
2 1+ 5  4
2
1
=
1+ 5 +
= 0.809017  0.587785 ⅈ
You can see that these solutions contain radicals of radicals. These expressions are not in the first
extended field, but we can extend that field again (in a finite quadratic algebraic field extension) so that
it includes these four solutions.
�14
4 The Pentagon and the Heptagon.nb
You can see that these solutions contain radicals of radicals. These expressions are not in the first
extended field, but we can extend that field again (in a finite quadratic algebraic field extension) so that
it includes these four solutions.
These can be plotted on the complex plane:
0.31 + 0.95 ⅈ
0.81 + 0.59 ⅈ
0.81  0.59 ⅈ
0.31  0.95 ⅈ
Voila.
More Important Than the Answer
To summarize and review.
More important that getting the answer or than drawing the pentagon is to notice how the field extensions were built. Beginning with the rationals, which are all constructible, we first got the values for η1
and η2 , which were the sums of ζ 1 + ζ 4 and ζ 2 + ζ 3 respectively, the two pairs of complex conjugates. These values were
1
2
1 ±
5 , and thus required that we move into an extended field:
Q ⟶ Q(η1, 2 )
This is a quadratic extension and is thus constructible. Then, getting the four roots themselves
required another field extension. The four roots are
η1,2 ±
η1,2 2  4
2
, and each will require one more
quadratic field extension.
Q ⟶ Q(η1, 2 ) ⟶ Q(η1, 2 ,
η1,2 ±
η1,2 2  4
2
)
Sequences of quadratic field extensions are constructible.
Look again at what is happening here. At the outset, we knew that we had a fourth degree equation with all complex roots.
1 + x + x2 + x3 + x4 = (1  ζ1 ) (1  ζ2 ) (1  ζ3 ) (1  ζ4 )
By segregating out the pairs of complex conjugates, we separated the factors on the right into two pairs
�4 The Pentagon and the Heptagon.nb
15
Look again at what is happening here. At the outset, we knew that we had a fourth degree equation with all complex roots.
1 + x + x2 + x3 + x4 = (1  ζ1 ) (1  ζ2 ) (1  ζ3 ) (1  ζ4 )
By segregating out the pairs of complex conjugates, we separated the factors on the right into two pairs
:
1 + x + x2 + x3 + x4 = {(x  ζ1 ) (x  ζ4 )} × {(x  ζ2 ) (x  ζ3 )}
1 + x + x2 + x3 + x4 = x2  (ζ4 + ζ1 ) x + ζ1 ζ4 × x2  (ζ2 + ζ3 ) x + ζ2 ζ3
1 + x + x2 + x3 + x4 = x2  η1 x + 1 × x2  η2 x + 1
Is this interesting? Yes! If we confine ourselves to rational numbers, then our original equation could
not be factored. If we admit η1 and η2 , it could be factored into two factors. If we admit all the complex numbers  really, we needed go so far; a finite field extension adding
η1,2 ±
η1,2 2  4
2
to the mix would
be enough  then it factors into four factors:
In Q
1 + x + x2 + x3 + x4
In Q(η1, 2 )
"" factors to x2  η1 x + 1 × x2  η2 x + 1
In Q(η1, 2 ,
η1,2 ±
η1,2 2  4
2
)
“”
is irreducible
factors to (1  ζ1 ) (1  ζ2 ) (1  ζ3 ) (1  ζ4 )
The procedure we have followed does exactly what is required for specifying constructible figures: it
has made a sequence of finite field extensions, starting with the rationals, Q, and proceeding by
quadratic field extensions until the polynomial with our desired points as roots is completely factored.
This stepwise factorization works for the pentagon because at each step it was possible to subdivide
the roots into two groups, each of which could be shown to be a quadratic expression of the preceding
group. That is not always possible.
Conclusion
We have seen here an application of the technique of algebraic decomposition. The equation we are
trying to solve is broken into simpler and simpler parts as the field in which we operate is expanded
stepbystep until we arrive at a final field, the “splitting field,” in which the polynomial can be completely decomposed into linear factors.
Unlike Euclid’s way of working, this methodical procedure provides a framework for evaluating which
polygons are constructible and which are not.
We will see this method play out on a larger stage next week with the construction of the hep
�16
We have seen here an application of the technique of algebraic decomposition. The equation we are
trying
to solve
is broken
into simpler and simpler parts as the field in which we operate is expanded
4 The
Pentagon
and the
Heptagon.nb
stepbystep until we arrive at a final field, the “splitting field,” in which the polynomial can be completely decomposed into linear factors.
Unlike Euclid’s way of working, this methodical procedure provides a framework for evaluating which
polygons are constructible and which are not.
We will see this method play out on a larger stage next week with the construction of the heptadecagon.
Thank you.
�
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https://s3.useast1.amazonaws.com/sjcdigitalarchives/original/475e77346471eb8b8d17111a59ad06cb.pdf
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Text
Constructible Numbers
Grant Franks
June 3, 2019, revised September 21, 2019
Introduction
Recapitulation
Two weeks ago, we saw that Rafael Bombelli confronted the possibility that the square root of negative
one might have some positive significance when he solved the cubic equation
x 3  15 x  4 = 0
and, using the formula Cardano stole from Tartaglia, got
x=
3
2 + 11
1
+
3
2  11
1
which he was then able to solve by intuiting that
2 + 11
1 = 2 +
3
1 .
Last week, we followed Caspar Wessel in developing a graphic understanding of the arithmetic of
complex numbers. In particular, we saw that
(i) complex numbers can be expressed in polar coordinates by giving a distance (modulus) and an
angle (argument);
(ii) multiplication of complex numbers amounts to
(a) multiplication of their distances (moduli) and
(b) adding their angles (arguments); and
(iii) the solutions of equations of the form x n  1 = 0, known as the “Roots of Unity,” appear
graphically as the vertices of an equilateral ngon in the unit circle on the complex plane.
So far, so good.
Today’s Agenda
�2 ���
3 Constructible Numbers.nb
Today, we’re going in diﬀerent direction. Today, we will talk about constructing numbers and the
emptiness of the Euclidean plane.
The Geometrical Plane
The Awkward Question
“Are there holes in the geometric plane?” This is a question that rarely gets asked when doing geometry. There is no reason to raise such a question until some awkward questions get asked, and there is
no way to answer the question without being able to step back from intuitively given geometrical space
by thinking algebraically.
Construction
In Euclid’s geometry, one begins with five postulates, three of which authorize constructions:
Postulate One: to draw a line between any point and any other point;
Postulate Two: to continue a line indefinitely; and
Postulate Three: to draw a circle with any center and radius.
The references to “any point,” “any center” and “any radius” give the impression that the Euclidean
plane contains all possible points. But that isn’t quite true or, to be more precises, there are many
points in a plane to which Euclidean geometry provides no access.
For instance, suppose you wanted to “square the circle.” This is a classical problem that amounts to
drawing a rectangle with an area equal to that of a given circle. Archimedes shows that such a rectangle would have a height equal to the radius of the circle and base equal to half its circumference. If you
can draw a circle with any center and radius, could you draw one whose radius is equal to the semicircumference of the circle you are trying to square? If you could do that just by saying it, your problem
would be solved.
That would not satisfy a mathematician. She or he would want to know how to get that radius length,
that is, how to construct it. Beginning with the one length that is given  the radius of the circle to be
squared  and using only permissible manipulations, how can we construct the desired straight line
with a length equal to the semicircumference of the given circle?
The Meno Problem
This problem sounds a little like the geometrical problem of the Meno: Socrates asks the slave boy, “If
you are given this square with these sides, can you show the side of a square with double the area?”
The slave boy is stumped. Some modern students think they have a better answer than the slave boy
and say, “That’s easy! It’s the square root of two!” They don’t realize that they are not giving an
�3 Constructible Numbers.nb
���
3
answer at all. The phrase “square root of two” is just a slightly shortened version of “the magnitude
which, when multiplied by itself, gives two.” So, to Socrates’ question, “What is the magnitude which,
when multiplied by itself, yields two?” they have answered, “The magnitude which, when multiplied by
itself, gives two.” True, no doubt, as any tautology is true, but it doesn’t really advance our knowledge
of, well, anything.
Socrates provides what is really needed: a geometrical construction for finding the square root of two
by looking at the diagonal of a square with sides equal to one. (Really, he gives the construction for 2
2 because his original square had sides equal to two, but that is minor detail.)
For the squaring of the circle, we want a construction for a straight line whose length is equal to the
semidiameter of the given circle. Alas, not all that humankind desires does it obtain! There is no
Euclidean construction for that length. The story that leads to that result culminates in the late nineteenth century. It involves Ferdinand Lindemann’s demonstration of that π is not just an irrational
magnitude  that had been known for over a century (Johann Lambert 1761)  but that it is a particular
kind of irrational magnitude. Getting to that result, if anyone here is interested in it, would require
work that goes beyond what this series of talks will cover. If it is any consolation, however, the ideas
we will cover tonight are necessary preliminaries to that work.
The Delian Problem
Once upon a time, a long time ago, a great plague aﬀlicted the city of Delos. The citizens consulted
Apollo’s oracle at Delphi who told them that the god was dissatisfied with the altar of his temple. The
altar was made in the shape of a cube, and the oracle said that the god wanted an altar twice as big.
The citizens, eager to be rid of the plague, got a great piece of marble and built an altar twice as long,
twice as deep and twice as tall as the one that was there. The plague continued. The priest of the
oracle corrected the people saying the god wanted an altar with twice the volume of the present one.
The newly built altar had eight times the volume, and was not what was wanted. The people were
understandably annoyed, but everyone in the ancient world knew that gods love to mess with people
by issuing weirdly misleading oracular pronouncements. Gods are cruel, that’s all there is to it.
Geometers realized immediately that what was needed was, in eﬀect, an altar whose side was 2
times bigger than the present one. When they set out to design it, however, they found that the god
had been even crueler than expected. They couldn’t figure out how to construct the altar. Finding a
construction for 2 was easy, but finding one for the 2 was surprisingly diﬀicult.
3
3
To get to the bottom of the problem that they faced, we have to sketch out a new kind of algebraic
operation, finite field extensions. Ordinarily this would be a semesterlong study, but since this is St.
John’s College, I will try to compress it into about fi�een minutes.
�4 ���
3 Constructible Numbers.nb
Rational Operations
What is a Field?
To begin, we need to define “a field.”
For our purposes, a “field” is a collection of objects  we’re going to be talking about numbers  that
are closed under the operations of addition (and its inverse, subtraction) and multiplication (and its
inverse, division).
Consider first, then, the whole numbers: 1, 2, 3, …. These are closed under addition, that is, if you add
any two whole numbers you get a whole number. But they are not closed under subtraction. Although
you can subtract 5 from 7 to get 2, you cannot subtract 7 from 5.
So set aside the whole numbers and take up the integers: … 3, 2, 1, 0, +1, +2, +3, …. Now you have
closure under addition and subtraction. You also have closure under multiplication but not division. 10
divided by 5 is 2, but 10 divided by 3 is not among the integers.
So set aside the integers and take up the rational numbers: …
1 1 2
, 5 , 11 ,
3
…0…
1
, 2, 5
10 7 3
…. That is,
all the numbers made up by ratios of integers with one another (forbidding division by zero). Now we
have it: this is a field. It is closed under addition and subtraction, it is closed under multiplication and
division.
For convenience’ sake, we will give the rational numbers a symbolic name, Q. (Why Q and not R?
Because R is reserved for the real numbers. Alas.)
Euclid’s Operations Allow Us to Form a Field
The operations of Euclid’s geometry allow us to construct lengths on a line that correspond to the field
of rational numbers. If we begin with a given length that we will call the “unit,” we can with straightedge and compass easily make a double length, a triple length, etc. If we define “negative” to be mean
motion in one direction from an arbitrary starting point and “positive” to mean going in the opposite
direction, we can construct lengths corresponding to all integers. By an easy construction, we can also
divide our given unit length into equal parts corresponding to any whole number. Thus we can make
lengths corresponding to any positive proper fraction; by multiplying these we can make any proper or
improper fraction, and by directing them toward the negative side of our arbitrary zero point, we can
identify places corresponding to any rational number. The lengths from zero to these points can be
added, subtracted, multiplied and divided at will and the result will always be another rational length.
We have a field.
If we erect two such lines at right angles to one another, we can locate and label any point on a plane
�3 Constructible Numbers.nb
���
5
that corresponds to (a, b), where a and b are rational numbers. As my grandfather used to say, “Now
we’re cookin’ with gas!”
Other Lengths
All points with rational coordinates is a lot of points, but we know that there are other lengths that can
be found in Euclidean geometry. There is, for example, 2 , which is the diagonal of the square with
sides of unit length. In fact, we can construct lengths equal to the square roots of any lengths we can
find through other means.
D
A
C
B
If you want to find a length equal to r , draw line AB in length equal to r + 1. Here, let Ac = r and let CB
= 1. Erect a semicircle on line AB and a perpendicular at C meeting the semicircle at D. Join AD and DB.
Triangle ADC is similar to triangle DCB and to the combined triangle ADB. Therefore:
AC : CD :: CD : CB
AC × CB = CD2
But AC = r and CB = 1; therefore:
r × 1 = 2 = CD2
r = CD.
Combinations
A little examination will show that these are all the operations that are available to us. We have addition, subtraction, multiplication, division  these are suﬀicient to find any rational lengths. In addition
to this, we can take the square root of any length that we can find. Not only that: we can do so as many
times as we please. So, can construct
2 , or
5 , or
17
3
, or of any rational length. And that’s not
�6 ���
3 Constructible Numbers.nb
all! We can construct
2 . Or
2+
2 . Or
17
3
+
2+
7
3
… or any sequence or combina
tion of the rational operations and repeated extraction of square roots.
With these techniques in hand, someone might easily jump to the conclusion that these Euclidean
operations can construct any length whatsoever. That’s probably what I would have said if anyone had
asked me back when I was a Johnnie Freshman more years ago than I care to think about. But I would
have been wrong.
Jumping to Conclusions
When I learned about the Pythagorean theorem and irrational numbers (or their equivalents, incommensurable lengths), I didn’t take time to think about these new numbers as carefully as, in retrospect,
I should have. Looking back, I think my understanding ran something like this:
“We had the whole numbers, but they weren’t enough to do subtraction so we added the negative
numbers and got the integers. But the integers weren’t enough to do division, so we added the fractions and got the rational numbers. But even the rationals weren’t enough to account for all the
lengths we could find in geometry  the 2 is irrational! (Hey! I was just as surprised by this as the
Greeks were!)  so we added the irrational numbers to the rational numbers and now we have all the
real numbers, which is all that there are!”
That understanding didn’t get challenged for decades until I began working on a preceptorial on Galois
Theory and Professor Charles Hadlock, author of Field Theory and Its Classical Problems, introduced me
to finite field extensions. It was here that I learned the humbling lesson that not all irrational numbers
are the same. Some numbers are more irrational than others, and lumping them all together blurred
together distinctions that are best kept separate.
Baby Steps
Let’s start over. Go back to when we had just the rational lengths and could find any point with rational
coordinates. Now, we read the Meno and find out about 2 . Instead of pretending that we are now
able to generate all possible irrational lengths, look carefully at what we have. We can make any
rational length, and we can make the square root of two. If we continue now to use just with our
rational operations (+, , x, ÷) on the two lengths we have at hand, 1 and 2 , we can make any number
that looks like this:
a+b
2
where a and b are rational numbers. Notice something important about these numbers: we can add
them, subtract them, multiply them, and divide them any way we please and we always get other
�3 Constructible Numbers.nb
���
7
numbers of this same kind. So, if we have:
3 + 5
2 + 1 + 7
2 = 2 + 12
3 + 5
2 × 1 + 7
2 = 3 + 21
2.
or
2 5
2
2 + 35 2 = 67 + 16
2
Division is a bit more complicated, but it works as well. The upshot is this: the numbers a + b 2 form
a field of their own. This new field is called an extension field. Because we added a finite number of
elements to form it (in this case, just one), it is called a finite field extension. And because the element
we added was a solution of a polynomial with elements of the original field as coeﬀicients  in this
case, x 2  2 = 0  it is called a finite algebraic field extension. And because it was made by adding an
element that is the squareroot of a member of the original field, it is called a quadratic finite algebraic
field extension. Let’s call it F1 and write F = Q( 2 ) to signify that F was formed by appending 2 to Q
and making all the numbers of the form a + b 2 where a and b are elements of Q.
F is big. It’s bigger than Q, the rational numbers. It includes Q as a subset, so we write:
F1 ⊃ Q.
But F does not include all the numbers (lengths) that we can construct because we can adjoin other
elements if we wish. We can even take the square root of some squirrely element of F1 that already has
a square root of two, say:
3+7
2
We can append this element to F1 and form the numbers:
c+d
3+7
2
where c and d are elements of F. This is a quadratic finite algebraic field extension of F1 . Let’s call it F2
and write F2 = F1 (
3+7
2 ) to signify that F2 was formed by appending
ing all the numbers of the form c + d
3+7
3+7
2 to F1 and mak
2 where c and d are elements of F1 .
F2 is big. It’s bigger than F1 and much bigger than Q. It includes F1 as a subset, so we write:
F2 ⊃ F1 ⊃ Q.
Do you see where this is going? We can continue this process as long as we wish.
�8 ���
3 Constructible Numbers.nb
… F5 ⊃ F 4 ⊃ F 3 ⊃ F 2 ⊃ F 1 ⊃ Q
When I put them all together, I have a tower of finite quadratic field extensions. Every number that
corresponds to every possible constructible length is somewhere in that tower. Altogether, they are
called the constructible numbers. The set of constructible numbers is very big.
But it’s not everything.
There are NonConstructible Numbers
There are, as it turns out, nonconstructible numbers, as can be shown in several ways. For instance,
2 (the real cube root of two) is not a constructible number. It does not belong to any tower of
quadratic field extensions over the rationals.
3
For, proceeding in the time honored way of reductio ad absurdum, suppose that 2 were constructible. Also remember that, since in the real numbers y = x 3 is strictly increasing, there is only one
real cube root of two. Now, if 2 were constructible, it would belong to some quadratic field extension
of a field that was itself part of a tower of quadratic field extensions leading back to the rationals.
3
3
2 ∈ Fn
3
That means that
include 2 .
3
where
2 =a+b
Fn ⊃ Fn1 ⊃ Fn2 ⊃ Fn3 ⊃ … F1 ⊃ Q
c , where a, b and c are all parts of Fn1 , but where Fn1 does not itself
3
Cube both sides of this equation.
3
2 = a + b
3
3
c = a3 + 3 a 2 b
2 = (a3 + 3 a b2 c ) + ( 3 a2 b + b3 c )
c + 3 a b 2 c + b3 c
c
c
The number 2 is a part of Fn  1 ; we know this because it is a member of Q. Therefore, it has no component multiplied by c , which means that ( 3 a2 b + b3 c ) = 0.
Next consider (a3 + 3 a b2 c )  ( 3 a2 b + b3 c ) c (notice the minus sign). Since 3 a2 b + b3 c = 0, this
3
has the same value as (a3 + 3 a b2 c ) + ( 3 a2 b + b3 c ) c . But it unpacks into a  b c . So we
have two cube roots of 2:
a+b
c
ab
c
But there is only one value ; therefore b = 0. That means that a + b
c is really just a, and
3
2 is in
�3 Constructible Numbers.nb
���
9
Fn  1 . By the same reasoning, it is a member of Fn  2 and Fn  3 and so on until we discover it is a member
of Q, that is, that it is rational.
But
3
2 is NOT rational. (Those who doubt this can see Appendix 1.)
Thus, 2 is NOT in ANY tower of quadratic field extensions beginning with the rationals. It is not
constructible.
3
The Delian Problem is Thus Solved
At this point, the architects and engineers at Delos should despair: if 2 is not constructible, then
they cannot double the size of the altar of Apollo with Euclidean mathematics. The gods are cruel, but
they are are mathematically wellinformed.
3
There are Lots of NonConstructible Numbers.
The demonstration that the 2 is not constructible is all well and good, but it is rather ad hoc. It
doesn’t immediately produce any broad conclusions about constructible vs. nonconstructible numbers.
3
A slightly more detailed investigation of field theory allows broader conclusions. It is possible to
characterize the size or “degree” of one algebraic field extension over another. Compare, for example,
the quadratic extension that results from adjoining 2 to the rationals with what would be called the
“cubic” extension that occurs when you adjoin 2 . In the first case, we can express any number in the
extended field by an expression that looks like this:
3
2.
a+b
These numbers form a field. You can add, subtract, multiply and divide to your heart’s content and
never leave the field. If you try this with 2 , however, a problem arises. Form a number like:
3
a+b
3
2.
You can add and subtract alright, but as soon as you start multiplying, you’ll find yourself running into:
3
2 ×
3
2 =
3
4
The cube root of four is not the same as the cube root of two. It can’t be expressed by combinations of
rational numbers and the cube root of two. It is outside of the (purported) field. This problem did not
arise with 2 because
2 ×
2 =2
�10 ���
3 Constructible Numbers.nb
which is within the field defined by a + b 2 . With the cube root, it is not enough to add one term; you
need to add two. To get a field that includes 2 , you need numbers of this form:
3
a+b
3
2 +c
3
4.
A little experimentation will persuade you that these numbers do form a field.
Notice that when you formed a extended field with 2 , your new numbers had two terms, a + b 2 .
With 2 of two, your new numbers have three terms. The quadratic extension is of “degree two,”
while the cubic extension is of “degree three.” Field extensions can be compounded: an extension of
degree two followed by an extension of degree three will yield an extended field of degree six over the
original field. Field extensions can get remarkably complex, but for our purposes it will be enough to
focus on relatively simple extensions of relatively small degrees.
3
3
The degree of an extension is measured by complexity of the minimal polynomial needed to produce
the new elements whose addition to the original field leads to the extension. Determining whether a
polynomial is “minimal” poses some problems, but this approach can produce sweeping knowledge
about whole classes of extensions. So, for instance, every quadratic field extension over the one before
it, so that a tower of quadratic field extensions  that is, the collection leading to any constructible
numbers  will have powers 2, 4, 8, 16 … 2n over the rationals. At the same time, it can be shown that
for equations of the form:
xn  2 = 0
the number n gives the degree of the extension resulting from appending one of the solutions of the
equation to a field. This result allows is to know that
Solutions of
Appended to Q are of degree
And thus are generally
2
2
Constructible
2
3
Not Constructible
 2
4
2
4
Constructible
 2
5
2
5
Not Constructible
x6  2
6
2
6
Not Constructible
x7
7
2
7
Not Constructible
x2
I.e.
 2
2
x3  2
3
x4
x5
 2
Also, it can be show that the solutions of the equations for the nth roots of unity, a�er factoring out (x 1), are minimal when n is prime. So
Roots of
structible
Roots of
structible
x3  1 = 0
appended to Q are of degree
2
over the rationals, so con
x5  1 = 0
appended to Q are of degree
4
over the rationals, so con
�3 Constructible Numbers.nb
Roots of
structible
Roots of
structible
Roots of
structible
Roots of
structible
Roots of
structible
etc.
���
x7  1 = 0
appended to Q are of degree
6
over the rationals, so NOT con
x 11  1 = 0
appended to Q are of degree
10
over the rationals, so NOT con
x 13  1 = 0
appended to Q are of degree
12
over the rationals, so NOT con
x 17  1 = 0
appended to Q are of degree
16
over the rationals, so con
x 19  1 = 0
appended to Q are of degree
18
over the rationals, so NOT con
11
Many  indeed, most  of these create extensions whose degrees are not powers of two over the
rationals. Thus they create field extensions filled with numbers that are not constructible. Literally
infinite fields of nonconstructible numbers emerge.
Hierarchy of Irrationals
Viewing all the irrationals as an undiﬀerentiated mob is a mistake. We can distinguish between those
irrationals that are constructible and those that are not. The constructible numbers are built by successive quadratic field extensions starting from the rationals.
NonConstructibles
Constructibles
Rationals
The distinction between the constructible and nonconstructible numbers is interesting enough, but
situation is even stranger than that.
As we have seen, the Constructibles are made from towers of field extensions all of degree two. We
�12 ���
3 Constructible Numbers.nb
have just seen that many polynomials have solutions which, when appended to a field, give an extension of degree other than two; the equation x 3  2 = 0, for instance, has solutions of degree 3, which
takes us away from the powersoftwo towers of constructible numbers.
Suppose we toss away that restriction. Suppose we consider all algebraic field extensions of any
degree. What if we allow ourselves to build towers in which each step can be of any degree  that is, to
append to a field the solutions of a polynomial of any degree. In this way, we could make towers of
fields that include the solutions of any finite polynomial equations. That would include all the constructible numbers and much, much more. This immense collection is known as the algebraic numbers. It includes the rational numbers and the constructible numbers and much, much more.
It does not, however, include everything. There are numbers that are not included among the algebraic
numbers. You know a few: π is not an algebraic number. Neither is e, the base of the natural logarithm
system. Leibniz and later Euler called these nonalgebraic numbers “transcendental numbers,” a
wonderfully mystical “woowoo” name that stuck and is in common use today.
Transcendentals
Algebraics
Constructibles
Rationals
There are LOTS of Transcendental Numbers
When I name π and e as transcendental numbers, you may be misled into thinking that there are only a
few such numbers and that each of them is a precious rarity, much treasured by mathematicians like
these two specimens.
Au contraire! Far from being scarce, the transcendental numbers not only surpass all other numbers in
�3 Constructible Numbers.nb
���
13
quantity, they do so by an infinite amount. Of course, to characterize one infinity as greater or lesser
than another is a controversial project first pioneered by nineteenth century mathematician Georg
Cantor. According to Cantor, the smallest sort of infinity is like that of the natural numbers which can
be ordered in such a way that one can count oﬀ the members of an infinite set sequentially and be sure
eventually to encounter every member. The natural numbers are obviously countable in this way, as
are the integers if we number them like this:
0
+1 1 +2 2 +3 3 …
1st 2nd 3rd 4th 5th 6th 7th …
It takes a little more work to see that the rationals can be placed in countable order (they can), and a
bit more still to figure out that the algebraic numbers can also be ordered and counted. But they can.
Cantor designates this infinity by the symbol ℵ0 .
The complete collection of real numbers  and also the collection of all complex number a + b i where a
and b can be any real number  cannot be so ordered. These numbers, according to Canto, form a
higher degree of infinity, the infinity of the continuum, ℵ1 which is widely take to be equivalent to 2 ℵ0 ,
a quantity distinctly diﬀerent, and distinctly bigger than ℵ0  insofar as “bigger” is a concept applicable to infinities.
From this perspective, the relation of the algebraic numbers to the entire set of complex numbers is
pretty much what Ptolemy would call “the ratio of a point to a line.” The whole realm of all our manipulations, geometric and algebraic, occur in a vanishingly small subset of the totality of real numbers. Yet
though we speak of infinities, do not imagine that these transcendentals are far away. They are not far
away in heaven, so that you have to ask, “Who will ascend into heaven to get them?” Nor are they
beyond the sea, so that you have to ask, “Who will cross the sea to get them?” No, they are very near to
you always on every side, crowding about with incredible density. And the net of constructible numbers seems now to spread across the Euclidean plane like everthinning gossamer network of barely
perceptible points, each separated from the next by gulfs teeming full with inaccessible points.
While we all contemplate the miserable smallness of all our endeavors, let us take a brief break and
then there will be time for questions.
Appendix 1: Irrationality of 2
3
This result follows from Euclid, Book X, proposition 9. It can also be shown from arithmetic principles
as follows:
Suppose that
3
2 is rational. Then it can be expressed as a fraction
numbers. We may assume also that the fraction
a
b
a
b
where a and b are finite whole
is expressed in lowest terms, so that a and b have no
�14 ���
3 Constructible Numbers.nb
factors in common.
We have:
3
a
b
2 =
2=
a3
b3
2 b3 = a 3
That means that a3 is even, which means that a is even; thus a3 is divisible by 8. Let it be 8 c3 .
2 b3 = 8 c 3
b3 = 4 c 3
That means that b3 is divisible by 4, which means that b is even (and that b3 is in fact divisible by 64).
But we began with the hypothesis ab was a fraction where a and b have no factors in common.
Therefore,
3
2 cannot be expressed as a fraction ab .
�
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Trigonometric Interpretation of Complex Numbers
Grant Franks
June 3, 2019, revised September 16, 2019
Dedication
Caspar Wessel (1745  1818)
Let us pause for a moment to remember and give thanks for Caspar Wessel, Norwegian mathematician
and cartographer, who conceived the idea that complex numbers might be usefully portrayed on a
map.
Introduction
We ended the last talk with Rafael Bombelli staring at Cardano’s Formula as applied to the cubic
equation
x 3  15 x  4 = 0
The formula gives for a solution:
x=
3
2 + 11
1
+
3
2  11
1
which at first glance appears to be nonsensical since it consists of two terms both containing the cuberoot of expressions involving the squareroot of negative one, which mathematicians in other contexts
�2 ���
2 Trignometry and Complex Numbers.nb
agreed meant that no solution was possible. But Bombelli knew that there was a solution to this
equation. In fact, he knew what at least one of the solutions was: the integer + 4 solves the equation.
43  (15) (4)  4 = 64  60  4 = 0
Bombelli figured out by some combination of guesswork, deduction and just inspired staring that there
are expressions which, when cubed, give 2 + 11 1 and 2  11 1 . They are 2 + 1 and
2 + 1 , respectively. Hard as it is to find them, it is easy to confirm that they work. All you need do
is to multiply them by themselves three times, keeping in mind the one rule we know about 1 ,
namely, that when multiplied by itself it gives 1.
1 = 4 + 4
2 +
1 2 +
2 +
1 = 2 +
2 
1 2 
2 
1 = 2 
3
1  1 = 3 + 4
2
1 2 +
1 = 3 + 4
1
1 2 +
1 = 2 + 11
1
1 = 2  11
1
And
1 = 4  4
3
2
1 2 
Knowing the cube roots of 2 ± 11
x=
3
2 + 11
1
1  1 = 3  4
+
3
1 = 3  4
1
1 2 
1 allowed Bombelli to solve the particular problem facing him:
2  11
1 = 2 +
1 + 2 
1 = 4.
However knowing the answer to this problem doesn’t show us how to deal more generally with other
numbers involving 1 .
The answer to that problem is the subject of tonight’s talk.
Arithmetic of Complex Numbers
Their Real and Imaginary Parts of a Complex Number
The most evident problem with 1 is that it doesn’t stand in a relation of “more” or “less” with
regard to other numbers we have come across. That feature more than any other makes 1 seem
especially weird. When one takes the step from the whole numbers to fractions (that is, to positive
rational numbers), things like “onehalf” or “five and a quarter” could be related as greater or less than
whole numbers we were already familiar with. Later, for all its undefinable strangeness, an irrational
like 2 at least sat snugly between rational numbers, greater than some and less than others. (That,
in fact, is how Dedekind defined irrational numbers, namely, by identifying which rationals each was
�2 Trignometry and Complex Numbers.nb
���
3
greater than and which it was less than.) Even the very strange negative numbers aren’t as peculiar as
1 . If your idea of a number is that it should respond to counting something or measuring something, negative numbers are nonsense because there is less than nothing there to count or to measure.
But if you can get over that problem, at least negative numbers still stand in greaterandlesser relations to one another.
Not so 1 . It is neither greater than nor less than any real number. That much is pretty clear: the
square of every real number is positive, or at least “nonnegative.” The 1 is not anywhere on the
real number line that stretches from enormous negatives to enormous positives. So, if we are to
imagine it at all, we have to picture it being “somewhere else.”
Caspar Wessel set the imaginary numbers apart from the reals on an axis of their own at right angles to
the real number line. He thus established a complex number plane. One axis represents the real
numbers, the other the numbers that include 1 . A real number and an imaginary number together
form a twopart entity called a “complex number.” Each point on the complex number plane represents a single complex number. A complex number can look like:
2+
1 or
3 + 9
1
or
0 +5
1
or
4 + 0
1 .
The first and second examples have both a real and imaginary part. The third has only an imaginary
part; the real part is zero. The last example has only a real part; the imaginary part has a zero coeﬀicient.
At the risk of seeming overly pedantic, I want to note here that “having an imaginary part with a zero
coeﬀicient” is not quite the same thing as “being a real number.” The complex number “4 + 0 1 ” is
not quite the same thing as the real number “4.” The reason for this hypertechnicality and squeamishness about nomenclature is not at all clear at this point and it won’t become clear until the last lecture.
It’s not unusual to overlook this distinction and, for now, doing so won’t cause any problems. It is
common, even convenient, to skip over the zero terms and to write “4 + 0 i” as just “4” I mention this
notyetdeveloped distinction only so that, when it comes back again in the final lecture, I can say “As I
have already said …”, and you will all nod sagely in agreement.
The real and imaginary parts of a complex number stand in diﬀerent orders and, when they are added
or subtracted, they act independently of one other. In modern parlance, one might say that a complex
number can be represented as a vector on a plane with a real axis in one direction (generally, le�right)
and an imaginary axis orthogonal to it (updown). That’s not how Caspar Wessel spoke because the
term of a “vector” wasn’t introduced until the middle of the 19th century, decades a�er Wessel died.
But the fundamental idea is there: a complex number is a twopart object whose parts add independently of one another. That idea had been around for years, at least since Isaac Newton had analysed
motions into components towards and parallel to the sides of a parallelogram.
�4 ���
2 Trignometry and Complex Numbers.nb
5i
4i
3+4i
3i
2i
3 + 2 i
i
5
4
3
2
1
0
1
2
3
4
5
i
4  i
2 i
3 i
4 i
24i
5 i
Here, then, is the representation of four complex numbers on a complex number plane: 3 + 4 i, 2  4 i, 4
 i and 3 + 2 i. So far, this is just a picture. Its value appears as we see how it is used.
Addition and Subtraction of Complex Numbers
In addition, the real and imaginary parts act separately. So, if one adds
3 + 2 i
to
4+2i
one gets
(3 + 4) + (2 + 2) i
The real parts add ordinarily, and the imaginary parts do too, thanks to the (formerly implicit, now
explicit) understanding that “distribution of multiplication over addition” works for the number i as it
does for other numbers, so that we have:
2 i + 2 i = (2 + 2) i = 4 i.
This procedure is just what one would do with components of a vector or of a decomposed Newtonian
force or velocity. Graphically, as Wessel proposes envisioning complex numbers, the result looks like
this:
�2 Trignometry and Complex Numbers.nb
���
5i
4i
1 + 4i
4+2i
3i
2i
3 + 2 i
i
4
3
2
1
0
1
2
3
4
i
Multiplication by a real number (or real part of a complex number) is like ordinary
multiplication
Multiplying a complex number by a real number amounts to multiplying each of the real and complex
parts of the complex number as you would expect. For multiplication by positive integers, the result is
just like repeated addition of the vector representing the complex number. Multiplication by negative
numbers is like repeated subtraction.
5
�6 ���
2 Trignometry and Complex Numbers.nb
5i
4i
Multiplication of 3 + i by 3 + 0 i
3i
2i
i
2
1
3+i
0
1
2
3
4
5
6
7
8
9
10
i
2 i
So far, so good. The graphic representation hasn’t yet shown us anything novel about complex numbers or given us new, but there is more and better yet to come.
The Crux of the Problem: Imaginary Multiplication
Next we have to deal with complex numbers times other complex numbers. This is where things get
interesting. It’s not immediately clear what that means graphically, but we do have an algebraic
understanding. The one thing we know for sure about 1 is that when you multiply it by itself, it
gives 1.
Let’s go back to the example we have already seen: Bombelli’s discovery that 2 + i is the cube root of 2
+ 11 i. As we showed already, we can multiply 2 + i times itself:
(2 + i) (2 + i) = (2 ⨯ 2 )+ (2 ⨯ i )+ (2 i ⨯ 2) + ( i ⨯ i) = 4 + 2 i + 2 i  1 = 3 + 4 i.
�2 Trignometry and Complex Numbers.nb
5i
(2 + i)(2 + i) = 3 + i
4i
3i
2i
i
2
1
2+ i
0
1
2
3
4
i
2 i
So far, this is not too revealing. Multiply the product by 2 + i again:
5
���
7
�8 ���
2 Trignometry and Complex Numbers.nb
12 i
(2 + i)3 = 2 + 11 i
11 i
10 i
9i
8i
7i
6i
5i
4i
(2 + i)(2 + i) = 3 + i
3i
2i
i
2
1
2+ i
0
1
2
3
4
5
i
2 i
What sense does that make?
The meaning appears more easily with Polar Coordinates
So far, we’ve been writing complex numbers like points on a plane using Cartesian coordinates. For
some purposes, it is a LOT easier to understand what is going on if you use polar coordinates. (Trust
me.)
To start, consider a circle with a radius of one centered on the origin. This is the “unit circle in the
complex plane.”
�2 Trignometry and Complex Numbers.nb
���
�
������
1.51 = A
1.11841 = θ
A (cos θ + i sin θ)
Now if you choose any angle θ, the point (Cosine(θ) + i Sine (θ)) will necessarily fall on the unit circle.
As the angle θ goes through the complete cycle from 0 to 2 π  we measure angles in radians, which is
easier for all sorts of reasons once you get used to it; if you are thinking in degrees, say “0° to 360°” the point (Cosine(θ) + i Sine (θ)) goes around the circle. If the angle continues to grow, the point spins
endlessly around the unit circle.
If you want a point, that is to say “a complex number,” inside or outside the unit circle, multiply the
result by some constant A. If A is greater than one, the corresponding point (complex number) will be
outside the unit circle; if it is between zero and one, the point (complex number) will be inside the
circle. Any point on the complex plane can be designated with a pair of numbers A (for length) and θ
(for angle).
In complexnumberspeak, the angle of the complex number expressed in polar coordinate form is
called the “argument”; the length is called the “modulus” of the number.
9
�10 ���
2 Trignometry and Complex Numbers.nb
Multiplication of Two Arbitrary Complex Numbers
Try multiplication again with two arbitrary complex numbers, this time expressed in polar form. Let
the two numbers be:
A B (cos(θ) cos(ϕ)  sin(θ) sin(ϕ) + i (cos(θ) sin(ϕ) + cos(ϕ) sin(θ))
“Okay,” you say. “How has this helped me?” The answer to that would be clear if you had been careful
about memorizing trigonometric identities, in particular, the identities for the sine and cosine of the
sum of two angles. On the oﬀ chance that you don’t have those identities burned into the forefront of
your minds, let me show you what you need to “remember” or, as Socrates might say, “recollect.”
Digression: Trigonometric Identities for Sine and Cosine of the Sum of Two
Angles.
Consider a portion of a unit circle with center at O. From center, draw a line OA at any (acute) angle;
call the angle ϕ. Drop a perpendicular AF to the horizontal diameter of the circle. The right triangle
formed as lengths that represent cos ϕ (horizontal OF) and sin ϕ (vertical AF). Now draw a line OB,
creating another angle, θ, on top of the first one. Drop a perpendicular BC to OA, the hypotenuse of the
first triangle. The segments OC and BC represent cos θ and sin θ, respectively. Drop perpendicular CH
to the original diameter OA. Also, drop a perpendicular from BD at the top of angle θ down onto the
original diameter. The segments thus created, OD and BD, represent cos (ϕ + θ) and sin (ϕ + θ) respectively.
Note draw a horizontal CE from C to the line BD. In triangle BEC notice that angle EBC is equal to ϕ.
Since segment BC is equal to sin ϕ, we conclude that BE = sin θ cos ϕ and that EC = sin θ sin ϕ.
Meanwhile, since OC = cos ϕ, we conclude that CH = cos θ sin ϕ and that OH = cos θ cos ϕ.
Examination will show that:
BD = sin (θ + ϕ) = BE+ EC = sin θ cos ϕ + cos θ sin ϕ ; and
CD = cos (θ + ϕ) = OH  DH = OH  EC = cos θ cos ϕ  sin θ sin ϕ.
�2 Trignometry and Complex Numbers.nb
���
11
������ ����� ϕ
��� ����� θ
������
������ ������
�������� ���� ���
����������
B
Sin
θ
Sin θ Cos ϕ
ϕ
A
G
θ
C
Sin ϕ
sθ
Co
Sin θ Sin ϕ
Cos θ Sin ϕ
E
ϕ
O
D
Cos θ Cos ϕ
Cos ϕ
H
F
Now look back at the product that we just obtained in multiplying two complex numbers.
A B (cos (θ) cos (ϕ)  sin (θ) sin (ϕ) + i (cos (θ) sin (ϕ) + cos (ϕ) sin (θ))
cos (θ + ϕ )
sin(θ + ϕ )
The collection of trigonometric terms associated with the real portion of the expression is cos (θ + ϕ).
The collection of trigonometric terms associated with the imaginary portion of the expression is sin (θ +
ϕ). The numbers associated with the lengths (modulus) are multiplied; the angles (arguments) are
added.
The significance of the imaginary multiplication is now visible:
In multiplying two complex numbers, whether written as A (cos(θ) + i sin(θ)) and B (cos(ϕ) + i
sin(ϕ)) or as a + b i and c + d i, graphically speaking what happens is that one
�12 ���
2 Trignometry and Complex Numbers.nb
(i) multiplies the distances of each number from the origin of the plane (the moduli), and
(ii) add the angles (arguments) made between the positive real axis and the line from the
origin to the point representing the number.
In short, again: in complex multiplication, distances from the center (moduli) multiply; angles
from the center add.
All sorts of neat things follow from this observation.
Raising Complex Numbers to Powers Causes Them to Spin!
If you raise a complex number to a (real) power, the argument (angle) of the result will grow continually
as the distance from the center grows (if it begins outside the unit circle) or shrinks (if it begins inside
the unit circle). Raising complex numbers to real powers therefore causes the results to trace spirals in
the complex plane. Here is the exponentiation of a complex number represented by a point a little bit
outside the unit circle:
�������
{Modulus =, 1.0435}
If we reduce the modulus (the “length”) so that the point falls inside the unit circle, the spiral will go
inwards because increasing powers of a length (modulus) less than one will shrink.
�2 Trignometry and Complex Numbers.nb
���
13
Between these two cases is the balanced point, where the modulus is one and the point lies on the unit
circle. Then, increasing powers of the complex numbers will result in a representative point that spins
forever around the circumference of the unit circle.
The investigation of complex numbers is a vast field. Thick textbooks are devoted to “functions of a
complex variable.” The Mandelbrot set, which lies at the beginning of complexity studies, exists in the
complex field. (It is defined as the set of complex numbers c that do not diverge when the function
fc (z) = z2 + c is iterated from z = 0.)
All this would be subject matter for an immense study. However, for the present , I want only to point
to two results that are relevant to the particular path that these talks are taking toward their goal,
constructing the heptadecagon.
Taking Integral Roots
First, now that we understand how complex numbers are multiplied and raised to powers, we can
easily find how to find integral roots of any complex number and thereby develop a general solution to
the problem that faced Rafael Bombelli. His great triumph, recall, was finding the cube root of one
complex number, 2 + 11 i, which he did by a combination of great genius, immense labor and fabulous
luck. (Almost any other complex number would have been much harder for him to deal with.)
However now we can see how easily to take the cube root of any complex number. Remember, to cube
a complex number, you cube the real number that is its modulus and triple the angle (argument). So,
�14 ���
2 Trignometry and Complex Numbers.nb
to take the cube root of a number, all you need do is to (i) take the cube root of the length (the
“modulus”) and (ii) and divide the angle (the “argument”) by three.
The Cube Root of 2 + 11 i
The particular problem that Bombelli faced was finding the cube root of 2 + 11 i. To take its cube root
the new way, first calculate its modulus (length) and argument (angle). The length of the vector from
the origin to (2 + 11i) we can get with the Pythagorean Theorem:
length (modulus) =
22 + 112 =
4 + 121 =
125
If we allow ourselves some trigonometry, the angle is easy enough, too:
= 1.39094 radians (79.7 degrees).
angle (argument) = ArcTan 11
2
To take the cube root, take the cube root of the length (modulus). In this case, we are assisted because
125 = 53 :
3
125 =
3
125 =
Take the angle and divide by three:
5.
1.39094
3
= 0.463648 radians.
So we get:
5 (Cos(0.463648) + i Sin (0.463648) )
= (2.236) (0.894427 + i 0.447214)
=2+i
Just the result that Bombelli arrived at by genius, sweat and divine guesswork.
Here, for comparison, are the values Bombelli worked on plotted atop the graph of the spiral
z = (2 + i)n
�2 Trignometry and Complex Numbers.nb
���
15
12 i
(2 + i)3 = 2 + 11 i
11 i
10 i
9i
8i
7i
6i
(2 + i)n
5i
4i
(2 + i)(2 + i) = 3 + i
3i
2i
i
2
1
2+ i
0
1
2
3
4
5
i
2 i
The Roots of Unity
When the Modulus Equals One
We have seen that when complex numbers whose representative points lie outside the unit circle spiral
outward when squared, cubed, or generally raised to powers greater than one. Those that lie inside the
unit circle spiral inward.
Those that lie on the unit circle  those with a modulus that is exactly equal to one  spin around the
unit circle with out moving inward or outward. These are very interesting, very handy numbers.
Because the cosine of a given angle and the sine of the same angle can form the sides of a right triangle
whose hypotenuse is equal to one, we can write these complex numbers with modulus one in the form:
z = cos θ + i sin θ
�16 ���
2 Trignometry and Complex Numbers.nb
We have seen that multiplying two complex numbers adds their angles (arguments) and multiplies
their lengths (moduli). In the case of these numbers, the modulus is one, so multiplying it any number
of times leaves it unchanged. For these numbers, multiplying means just adding the angles. So, if we
take a number and multiply it by itself, we get:
z2 = (cos θ + i sin θ) (cos θ + i sin θ) = (cos 2 θ + i sin 2 θ)
If we do it again, we get:
z3 = (cos θ + i sin θ) (cos θ + i sin θ) (cos θ + i sin θ) = (cos 3 θ + i sin 3 θ)
And in general,
zn = (cos θ + i sin θ)n = (cos n θ + i sin n θ).
If two diﬀerent modulus one numbers are multiplied, we get:
z1 z2 = (cos θ + i sin θ) (cos ϕ + i sin ϕ) = cos (θ + ϕ) + i sin (θ + ϕ)
�2 Trignometry and Complex Numbers.nb
���
θ
ϕ
Cos ϕ + i Sin ϕ
Cos θ + i Sin θ
Cos (θ + ϕ) + i Sin (θ + ϕ)
In this operation, multiplication of the complex numbers is tightly bound up with addition of the
angles. Such tight linkage of multiplication and addition is characteristic of exponentiation and logarithms, and in fact it is a very short step from what we have seen here to a formula expounded by
Leonhard Euler in his work Introduction to the Analysis of the Infinite that identifies the two:
ei θ = cos θ + i sin θ.
(A few years ago I gave a whole lecture on this identity; I’ll see about having it available on the library
website alongside this one.)
For now, we will be especially interested in a subset of these numbers that bear the intriguing and
evocative name, the “Roots of Unity.”
Roots of Unity
The “Roots of Unity” sounds like a New Age metaphysical treatise or the name of a theologically
inclined folkrock ensemble, but in our present context it means something rather diﬀerent and more
precise. It refers to numbers that, when raised to integral powers come to the result 1. Numbers like:
17
�18 ���
2 Trignometry and Complex Numbers.nb
2
1,
3
1,
4
1,
5
1 … etc.
To put the matter slightly diﬀerently, we are talking about numbers that are the solutions of equations
like:
x2
x3
x4
x5

1
1
1
1
=
=
=
=
0
0
0
0
or in general,
xn  1 = 0
Based on what I learned in high school, these equations are not hard to solve. For x 2  1 = 0, I know
that there are two solutions, + 1 and 1. For x 3  1 = 0, there is only one solution, +1, because
(1)3 = 1. That pattern continues down the line, with even numbered powers having two solutions
and odd numbered powers having only one. That understanding works so long as one considers only
the real numbers. But in the complex number field the answer is more complete, more interesting and
in some ways more satisfying.
Take x 3  1 = 0 for example. We are looking here for a number which, when cubed, is equal to one,
that is, the cubed root of one. Easy! One, when cubed, is equal to one. That’s fine, but it’s not the full
story. Consider the number on the unit circle whose angle is 120°: when squared it is still on the unit
circle and its angle is 120° × 2 = 240°; when cubed, it is still on the unit circle and its angle is 120° × 2 =
360° = 0°. That number is +1 + 0 i. Thus, the complex number at 120° on the unit circle is also a
cubed root of one! So, for that matter, is the number on the unit circle at 240°: squared, its angle is
480° = 120°; cubed, its angle is 360° = 0°. There are, in fact, three cube roots of one, and the points that
represent them form an equilateral triangle in the unit circle.
�2 Trignometry and Complex Numbers.nb
���
2i
The Cube Roots of Unity
i
1
 , + i ,
2
2
3
2
0
1
1
3
 , + i ,
2
2
{1, + i , 0}
1
2
i
2 i
The Algebraic Approach
The graphical approach to the cube root of unity is simple: take the 360° of the circle and divide them
by three. One can also take a strictly algebraic approach which is a little more intricate but which
reaches the same result. Begin with the equation:
x 3  1 = 0.
As you noticed at first, the integer 1 (or, better, the complex number, 1 + 0 i) is solution. Therefore, we
expect that this polynomial will be divisible by the linear factor (x  1), as indeed it is:
x 3  1 = (x  1 ) (x 2 + x + 1) = 0.
The new factor, (x 2 + x + 1), can easily be broken down into two linear factors by applying the
quadratic formula to the equation x 2 + x + 1 = 0 :
x=
1 ±
1  4 (1)
2
=
1
2
±i
3
2
.
So, the complete breakdown of the equation x 3  1 = 0 into linear factors is:
(x  1), x  12
+ i
3
2
, x  12
 i
3
2
19
�20 ���
2 Trignometry and Complex Numbers.nb
You can verify this result by multiplying any of the solutions  1, 12
3
2
+ i
or 12
3
2
 i
 by itself three times and seeing that you get the result 1 + 0 i.
More Roots of Unity
It should not surprise you to learn that the equation x 4  1 = 0 gives four fourth roots of unity: +1, 1, +i
and i. And the equation x 5  1 = 0 gives five fi�h roots of unity, like this:
2i
2i
i
2
1
i
0
1
2
2
i
2 i
1
0
1
2
i
2 i
And so forth. Generally speaking, there are always n nth roots of unity. This tidy fact is a special case of
a more general result proved by Gauss and called the “Fundamental Theorem of Algebra” which states
that in the complex number field a polynomial equation of the nth degree always has n solutions. That
is a wonderful result, but we don’t need its full generality for our taskathand.
Look again at the polynomials that define the n roots of unity. We can see from the graphic representations that the number 1 + 0 i is a solution of each of these “roots of unity” equations. Consequently, we
can divide any of them by the factor (x  1), just as we did with the cuberoot of unity equation:
x 4  1 = (x  1) (x 3 + x 2 + x + 1) = 0
x 5  1 = (x  1) (x 4 + x 3 + x 2 + x + 1) = 0
and generally:
x n  1 = (x  1) (x n  1 + x n  2 + … + x 2 + x + 1) = 0
The increasingly lengthy remainder terms are of special interest to us. The solutions to the corresponding polynomial equations
�2 Trignometry and Complex Numbers.nb
���
21
x3 + x2 + x + 1 = 0
x4 + x3 + x2 + x + 1 = 0
and generally:
xn  1 + xn  2 + … + x2 + x + 1 = 0
are precisely what we need in order to find the vertices of regular polygons inside a unit circle. For
fairly evident reasons, these equations are called collectively the cyclotomic (that is, “circlecutting”)
polynomials.
They will be the subject, not of the next talk, but the one a�er that.
Oh, By the Way … One More Thing to Note About Roots of Unity!
Before closing, I want to note one more feature about the roots of unity that will show up in a later talk.
It is this: for any whole number n, the sum of the nth roots of unity comes to zero.
This can be seen pretty easily by looking at the case of the four fourth roots of unity:
2i
i
2
1
0
1
2
i
2 i
The four roots are +1, +i, 1 and i. It is evident (isn’t it?) that when these four are added together, the
sum is zero. A�er all, the pair + 1 and 1 add to zero, as do the pair +i and i.
�22 ���
2 Trignometry and Complex Numbers.nb
Only a little less evident is what happens with the three third roots of unity:
2i
i
2
1
0
1
2
i
2 i
The two red vectors are parallel and equal to the blue vectors to the two complex third roots of unity.
Placing the three vectors endtoend in the usual way for vector addition gives a closed triangle, beginning and ending at (0, 0).
Similarly for all nth roots of unity: their sum always comes out to zero.
As a quick corollary, if one takes all nth the roots of unity for any n, the whole collection excluding the
number +1 sum up to 1. This follows easily from the fact that all the roots of unity sum to zero; if one
excludes +1, the rest must sum to 1 so that all of them together come to zero.
These facts will be used repeatedly in what follows. If you don’t remember them, I’ll remind you of
them when they come up again.
Conclusion
So these are the fundamentals of the arithmetic of complex numbers. The next talk will concern itself
with another topic altogether, the algebraic diﬀerence between points that can be constructed and
those that can’t. In the fourth lecture, these two topic will come together to demonstrate how algebra
can decide whether a construction is possible or not; we’ll look at two classical problems  the trisection of an angle and the doubling of the cube  and then at a new problem: the construction of the
sevengon. The fourth Tuesday lecture will bring all that has been said to bear on Gauss’s surprise, the
construction of the seventeengon.
�
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Text
Lecture 1: Meet Your New Best Friend: the Square Root of
Negative One
Grant Franks
June 19, 2019, rev Sept. 9, 2019
Introduction
This is the first of a series of six lectures on algebra. A�er today’s lecture, there will be four more talks
on successive Tuesday evenings, each unpacking a step along the way to the final result that I want to
share with you, Gauss’s demonstration of the Constructibility of the SeventeenSided Polygon or
“Heptadecagon.” (The schedule is available on the handouts.) A�erwards, on October 16, there will be
a final Wednesday a�ernoon lecture that very briefly recaps the contents of the Tuesday evening
technical talks and concludes with some reflections on the significance of the square root of negative
one to the foundations of arithmetic and the relation of ordinary experience to mathematics.
This series of talks grew out of a remark that a former dean of the College made to me years ago. This
person, whom I respect and admire greatly, said something I thought was seriously questionable.
“Algebra is boring,” he opined. “Our students love geometry,” he said, “because its beauty strikes
them immediately. But algebra is just a tool, a technique. Nobody wants to spend any class time
studying it. It’s just dull.”
I took those words as a challenge. It didn’t seem plausible that sane people would devote countless
hours of intense intellectual eﬀort to something that is inherently dull, at least without being paid a lot
of money. (Algebraists generally are not highly paid.) Do they really enjoy tedium?
No, they don’t. Algebraic structures have a real beauty, even if it takes a little work to notice and
appreciate it.
For better or for worse, algebraic beauty is invisible beauty, and the taste for it is an acquired taste, like
that for single malt scotch, twelvetone music or the wordplay of Finnegans Wake. My onagain, oﬀagain quest, therefore, for many years has therefore been to find the some entryway into what algebra
has to oﬀer, something more appealing than exercises in factoring polynomials. I wanted to find the
algebraic equivalent of the Pythagorean Theorem, some result that would make someone stand still in
wonder and say, “Whoa! Really?,” as Thomas Hobbes reportedly did when he saw a copy of Euclid
open to proposition 47 of book one. Legend has it that he stood transfixed at a library table for hours
reading the entire first book of the Elements BACKWARD until he arrived at the postulates. That
encounter reportedly made him “in love with geometrie.” (John Aubrey, Brief Lives (c. 1700))
What, then, would make someone “in love with algebra?” If anything could do it, I think it would be
Carl F. Gauss’s construction of the heptadecagon. It’s a beautiful result, and the pathway to it, while
�2 ���
1 Square Root of Negative one.nb
not perfectly smooth, requires only a few hours of preliminary work, not years. Also, for this audience,
it speaks directly to geometrical demonstrations that all St. John’s students have encountered by the
middle of the first semester of their freshman year, namely, the propositions of Book Four of Euclid’s
Elements where we see the construction of regularsided polygons in circles. Euclid shows how to
construct the equilateral triangle, the square, the pentagon  that’s a hard one!  and the hexagon
inside a given circle. Then, without explanation, he skips ahead to the fi�eengon (Elements IV, 16).
Then he stops. Why? Euclid, characteristically laconic, says nothing.
Two thousand years later, a very young Carl Friedrich Gauss provided the answer.
However, in order reach that result, you need to come to terms with [horror suspense sound eﬀect] the
square root of negative one. [mad scientist laugh sound eﬀect]
ζ5
ζ4
ζ6
ζ3
ζ2
ζ7
ζ
ζ8
1
ζ9
ζ 10
ζ0
ζ 16
ζ 15
ζ 11
ζ 12
ζ 13
ζ
14
Some people have diﬀiculty accepting this number. They say, for instance, that they are put oﬀ by the
fact that it doesn’t exist. Which is ridiculous! You shouldn’t let so trivial a problem prevent you from
embracing this concept, for the square root of negative one, also denoted by the single letter i, is the
gateway to an algebraic realm of amazing results. Through it we come to the Complex Number Field.
This is a spectacular realm where we see hidden machinery that links algebra and trigonometry,
wonderfully expressed in DeMoivre’s formula
(cos θ + i sin θ)n = cos n θ + i sin n θ ;
where we find Euler’s famous identity “ei π = 1,” where all polynomial equations can be completely
decomposed into linear factors and where we hear
the buzzin’ of the bees in the cigarette trees
‘round the soda water fountains!
No, wait. That’s the Big Rock Candy Mountains. Never mind. The Complex Number Field is still an
algebraic paradise. It is here that you must go to find the result that I most want to show you, the
construction of the heptadecagon.
Gauss’s result is not overly complicated, but it does involve several separate stages, so I thought it best
not to try to jam all of it into a single talk. Doing that would generate more confusion than understand
�1 Square Root of Negative one.nb
���
3
ing. So, all I want to do today is to set up the issue of the square root of negative one by describing how
it first appeared, how it was at first dismissed, and how an algebraic triumph led an Italian algebraist to
reconsider the possibility that it might not be gibberish but an important numerical results.
Start Simple: The Linear Equation
We will be concerned with polynomial formulas and their solutions. The simplest polynomial formula
is an equation in the first degree, that is, one where the variable  we’ll call it x  appears only to the
first power:
x  a = 0,
Here “x” is an unknown quantity and “a” is something given. To be more particular, we might have:
x  3 = 0.
What is x? In this case, x is 3. Is that obvious? You don’t need to do the explicit manipulation of adding
3 to both sides of the equation, although you could:
x3+3=0+3
x = 3.
Already there are, in fact, subtleties and complexities that could be discussed at length. What are these
quantities “x” and “a”? Are they lengths? areas? numbers? magnitudes? Do we know? Do we care?
Are they particular lengths or numbers? It is possible to imagine a “general” quantity that is nothing in
particular? Even if we cannot imagine it, can we conceive of it? What is a “variable” like x? How is a
variable similar to or diﬀerent from a “constant term” like a? In a situation in which we don’t know
what “x” and “a” are, are there diﬀerences in the manner in which we “don’t know x” and in which we
“don’t know a?” Can we “add the same thing to both sides of an equation” if we have no idea what
those things are?
All these questions are interesting and important. However, I’m going to pass by all of them because
the real concern of this lecture lies further down the road in the direction of more complex equations.
The Next Step: the quadratic, x2 + b x + c = 0
I hope you found the linear equation x  3 = 0 easy to solve. Things get more complicated quickly.
In the realm of polynomials, the next step in complexity takes us to the quadratic equation:
x2 + b x + c = 0
�4 ���
1 Square Root of Negative one.nb
where “b” and “c” are rational numbers. (Why start with “b” and not “a”? Because the really simple
form is “a x 2 + b x + c = 0” where “a,” “b,” and “c” are all integers. But it is convenient to divide out
the lead term “a” and redefine b and c so that one has a polynomial whose leading coeﬀicient is 1.
Such an expression is called a “monic polynomial.”)
There is a path to solving this equation quickly and reliably. It’s called “completing the square.” I
suspect it may be familiar to many of you. In case it isn’t, I will review it here. Stepping through the
derivation of the quadratic formula will be useful when we turn in a moment to the next step, the cubic.
Consider: what does a “perfect square” polynomial look like in algebra? That is, what do we get when
we multiply a single factor by itself. Try making one:
(x  b)(x  b) = x 2  2 b x + b2
That’s what a “perfect square” polynomial looks like, one whose two solutions are both b. If someone
posed that polynomial for us to solve,
x 2  2 b x + b2 = 0
life would be easy! We would just take the square root of both sides:
x 2  2 b x + b2 =
0 =0
(x  b)2 = 0
x  b =0
x = b.
Sadly, x 2 + b x + c = 0 is not perfect square. Happily, however, we can make it into a perfect square, or
at least get close enough. Look again at the general equation for a perfect square:
x 2  2 b x + b2 = 0
Notice the relation between the coeﬀicient of x ( 2 b) and the constant term (b2 ). If you take the
coeﬀicient of x, divide by 2 then square it, you get the constant term. Now look at the equation we
actually have:
x2 + b x + c = 0
�1 Square Root of Negative one.nb
By brute force, let’s make a perfect square. First, subtract c from both sides of the equation, just to
“clear the decks.”
x2 + b x =  c
Take half the middle term ( b2 ) and square it: you get
x2 + b x +
b2
4
b2
4
=
c =
b2
.
4
Add that to both sides of our equation:
b2  4 c
4
Now, on the le�, you have a perfect square: x 2 + b x +
b2
4
2
= x  b2 . This is great! Take the square
root of both sides:
x2 + b x +
b2
4
b2  4 c
4
=
On the le�, we have the square root of a perfect square, a situation that we deliberately contrived:
x + b2
2
± x + b2 =
b2  4 c
4
=
b2  4 c
2
(We need to say “±” because both x +
b
2
and x +
b
when
2
squared give the same result.)
x =
b
2
±
b2  4 c
2
This is the standard form of the quadratic formula for a monic, quadratic polynomial. Plug in the
coeﬀicients b and c, turn the crank and out pop two values for x.
A Cute Quadratic Trick
Before we go on, however, there’s a clever little trick involving quadratics that I want to show you. It
will be used many times in what follows.
Suppose you have two diﬀerent factors, (x  a) and (x  b) of a polynomial equation.
(x  a) (x  b) = 0
Multiply and expand:
���
5
�6 ���
1 Square Root of Negative one.nb
x 2  (a + b) x + a b = 0
Look at the coeﬀicient of x and at the constant term: a + b and a times b. The first is the sum of a and b,
the second is the product, where a and b are the two roots of the polynomial. This observation can
easily be generalized. If we had three terms:
(x  a) (x  b)(x  c)= 0
The expanded version would look like this:
x 3  (a + b + c) x 2 + (a b + a c + b c) x + a b c = 0
For the fourth degree:
(x  a) (x  b)(x  c)(x  d)= 0
The expanded version would look like this:
x 4  (a + b + c + d) x 3 + (a b + a c + a d + b c + b d + c d) x 2 + (a b c + a b d + b c d) x + a b c d = 0
Maybe you see where this is going: the first nonzero coeﬀicient is always the sum of all the roots. The
second is the sum of all the roots taken two at a time. The third is the sum of all the roots taken three at a
time. The constant term, when you get to it, is always the product of all the roots. These patterns are
very interesting and very useful. The patterns that you see here can be described by saying that the
coeﬀicients are “symmetric functions” of the roots because, as you can see, each of a, b, c play the
same role in each expression. Exploration of symmetric functions is fascinating, but for us right now it
is beside the point.
Look back at the second degree equation, and particularly at the coeﬀicients of the equation: the
coeﬀicient of x is the sum of the solutions a and b. The constant term, is the product of the solutions a
and b. Pause over that for a second: just looking at the polynomial may not tell us the two solutions
immediately, but even a glance at the coeﬀicients give us the sum and the product of the solutions.
Now, turn that observation inside out. Suppose you have two unknown numbers, call them r1 and r2 .
Suppose further that you don’t know what these two numbers are, but you DO know their sum and their
product. In that case, you can make a quadratic equation that has these two numbers as its solutions:
x 2  (r1 + r2 ) x + r1 r2 = 0
The solutions of this equation can be found with the quadratic formula:
�1 Square Root of Negative one.nb
(r1 + r2 ) ±
(r1 + r2 )2  4 r1 r2
2
���
7
=x
You may say to yourself, “That’s great, Mr. Franks. But how o�en, really, does it happen that I come to
know the sum and the product of two numbers without knowing what those numbers are individually?”
Well, in your daily life, maybe not so o�en. However, in the algebraic journey that lies before us, this
little trick is going to show up more o�en than you may imagine.
What Happens When Things Go Wrong
So far, we have constructed the Quadratic Formula, which solves a general quadratic equation:
x2 + b x + c = 0
x =
⟹
b
2
±
b2  4 c
2
You notice that the solution contains a radical, b2  4 c . If the quantity under this radical, which is
called “the discriminant,” is positive, all is well. But if that quantity becomes negative, that is, if 4c is
greater than b2 , then we have a problem.
Here is a graph of the equation y = x 2 + 2 x + c. We have arranged matters so that we can vary the
value of the constant term c:
�
y = x2 + 2 x + c,
b2  4 c = , 9.16
10
5
3
2
1
1
2
3
5
The “solutions”  that is, the x values where the function equals zero, are given by:
x=
2 ±
44c
2
=
2 ± 2
2
1 c
= 1 ±
1 c
�8 ���
1 Square Root of Negative one.nb
When c is less than one, the expression is positive and there are two solutions. When c is equal to 1, the
expression under the radical is zero and there is one solution. When it is greater than one, the expression under the radical is less than one and, as you can see graphically, there appears to be no solution.
From this example, we might conclude that the square root of a negative number means “impossible.”
Another Example
Descartes came to this conclusion looking at a slightly diﬀerent example. He considered the expression:
x=
1  y2
He could picture this by an illustration like the one below. Consider a semicircle with radius 1. Draw a
line parallel to the diameter with a variable height, y.
�
������
y
x
So long as y lies between zero and one, the quantity under the radical is positive and there are two
solutions, a negative one and a positive one, indicated by the intersection of the horizontal blue line
with the semicircle. When y reaches the value 1, then the horizontal line is tangent to the semicircle
and there is one solution only. When y is greater than 1, the horizontal line misses the semicircle. It
looks as if then there are no solutions at all.
Tentative conclusion: when a radical contains a negative sign, the formula is meaningless. There is no
solution to the equation, and the expression should be rejected as absurd.
That conclusion was generally accepted before Rafael Bombelli began to work with the cubic equation.
�1 Square Root of Negative one.nb
���
9
The Cubic
The cubic equation poses greater challenges than the quadratic. The quadratic equation is not exactly
simple, but its solution has been known for a long time. Babylonians had techniques that were more or
less equivalent of solving a quadratic equation. Some of Euclid’s geometrical manipulations in Book II
of the Elements also answer questions that are closely analogous to finding the solution of the general
quadratic equation.
By contrast, the solution for the cubic equation was not published until the sixteenth century. The
formula for the cubic is known as “Cardano’s Formula,” named a�er Girolamo Cardano who succeeded
where many others had failed … by stealing it from the man who invented it, Niccolo Tartaglia. Cardano published this formula in his [Cardano’s] book, The Great Art or the Rules of Algebra (1545). Let
this be a lesson to you: if you want to make a name for yourself in mathematics, steal freely and publish early. (This is an example of “Stigler’s Law of Eponomy” which states that no result in science or
mathematics is named a�er the person who first discovered it. See, Tom Lehrer’s song, Lobachevsky.)
Because Cardano’s formula is important, I propose to walk through its derivation here. I am aware,
however, that it is hard to follow a sequence of algebraic steps in a lecture format; that’s why I have
printed copies for anyone who wishes to review the derivation later at her or his leisure. The lecture
will also be posted on some part of the College’s web page. Of course, you are also free (if you wish)
just to accept the result on faith.
One begins, not with the full form of the cubic, but with the “depressed cubic,” a formula that has been
manipulated so that the x 2 term disappears. There is a routine procedure for making this happen, so it
does not limit the generality of the demonstration. (The procedure for “depressing the cubic” is
included as an appendix to the printed version of this lecture for anyone to examine at leisure.) Thus,
we begin with:
x3 + c x + d = 0
We want to know what x is. To solve this equation, Cardano  or Tartaglia, really  came up with a
special trick: reconceive x as divided into two parts, p and q so that
x = p + q.
Imagine, then, a cube whose whole side is x, a length that has been divided into two parts, p and q:
�10 ���
1 Square Root of Negative one.nb
�
We can see that the cube is broken up into
(i) a cube of side p (red);
(ii) a cube of side q (blue); and
(iii) three “slabs” of volume p q (p + q) (yellow, green and purple).
�1 Square Root of Negative one.nb
�
�
Symbolically, we have:
(p + q)3 = p3 + 3 p2 q + 3 p q2 + q3 = 3 p q (p + q) + (p3 + q3 )
or
(p + q)3  3 p q (p + q)  p3  q3 = 0
which gives us:
(p + q)3 + (3 p q) (p + q) + ( p3  q3 ) = 0
x3
c
x
Compare this to our depressed cubic:
x3 + c x + d = 0
d
���
11
�12 ���
1 Square Root of Negative one.nb
The two are the same PROVIDED:
3pq=c
pq=
c
3
p3 q3 =
and
 (p3 + q3 ) = d
or
and
 (p3 + q3 ) = d
or, cubing the first expression 
 c3
27
(p3 + q3 ) =  d
and
Now look: we have expressions for the the sum and the product of the two quantities, p3 and q3 .
Therefore, we can construct the quadratic equation. (I told you this procedure would be helpful!) I’ll
use “w” as a variable:
w2  (p3 + q3 ) w + p3 q3 = 0
values we just determined:
which will have p3 and q3 as it solutions. Substitute the
3
w2 + d w  3c = 0
whose two solutions are:
w=
d±
c 3
(d)2 + 4 3
2
=
d±
d 2
c 3
4 2 + 4 3
2
=
d± 2
d 2
c 3
2 + 3
2
=
d
2
±
2
d2 + 3c
3
The two solutions of this are p3 and q3 . If we take the cube root of each and add them, we get p + q = x:
x=
3
d
2
+
2
d2 + 3c
p
3
+
3
d
2

2
d2 + 3c
3
q
That is Cardano’s formula  the one he stole from Tartaglia  for solving the cubic equation.
What Happens When Things Go Wrong?
For the Quadratic: Apparent Impossibility
If you have followed so far, we have two formulas, one for second degree (quadratic) equations and
one for third degree (cubic) equations.
The quadratic formula:
�1 Square Root of Negative one.nb
For the equation:
x2 + b x + c = 0
x is given by:
x =
b
2
���
13
b2  4 c
2
±
The cubic formula (“Cardano’s Formula”):
For the equation:
x3 + c x + d = 0
x is given by:
x=
d
2
3
2
d2 + 3c
+
3
+
d
2
3

2
d2 + 3c
3
This is all very well … except if the coeﬀicients are such that the quantities under the squareroot signs
become negative.
We have already seen what happens to the quadratic when the quantity under the radical is negative:
the formula seems to give no answer at all and the expression appears to be meaningless.
A priori, we seem to have no reason to suspect that the cubic will act diﬀerently.
But it does.
You Can Use Cardano’s Formula to Solve Cubics
Sometimes, Cardano’s formula works just fine. Consider, just as an example:
0 = x 3 + 6 x + 20
where c = 6 and d = 20
Apply the formula to get:
x=
x=
x=
x=
3
d
2
3
 20
2
3
3
+
+
2
d2 + 3c
3
2
20
+ 63
2
10 +
100 + (2)3
10 +
108
+
3
+
3
d
2
3
+
+
10 
3
3
2
d2 + 3c

 20
2

10 
108
3
2
20
+ 63
2
100 + (2)3
3
�14 ���
1 Square Root of Negative one.nb
x=
10 + 12
3
3 +
3
10  12
3
x = 0.73205 + (2.73205)
x = 2
… which really is one of the solutions, as you can check. (The other two are imaginary, which you
can find a�er you have factored out (x + 2) from 0 = x 3 + 6 x + 20.
x 3 + 6 x + 20 = (x + 2)(x 2  2 x + 10)
You can solve x 2  2 x + 10 = 0 with the quadratic formula.)
However, Sometimes Things are Very Weird
But things can also go terribly, terribly wrong. Take another example:
x 3  15 x  4 = 0
Here, c =  15 and d =  4. Put the values into Cardano’s Formula:
x=
x=
3
d
2
3
4
2
2
d2 + 3c
+
3
2
4
+ 15
2
3
+
x=
3
2+
4 + 125
x=
3
2+
121
x=
3
2 + 11
1
+
+
+
3
3
+
3
3
+
3
2
2
2
3
d
2

d2 + 3c
4
2

4
+ 15
2
3
3
2
3
4 + 125
121
2  11
1 .
You have an imaginary quantity in each part of the solution. Does result this mean that the equation
has no solution?
No. It obviously does have a solution. Every cubic has at least one real solution. Just look at the graph
of this formula:
�1 Square Root of Negative one.nb
Plotx3  15 x  4, {x,  5, 5}, PlotLabel → "y = x3  15 x  4"
y = x3  15 x  4
40
20
4
2
2
4
20
40
This formula should have three solutions. The solution farthest to the right seems to be +4. If there
were any doubt, you can check and see that one of the solutions is +4.
43  (15) 4  4 = 64  60  4 = 0 .
If the solution of the equation is +4, why did Cardano’s formula give a result with two expressions
involving the square roots of negative numbers? What went wrong with Cardano’s Formula?
This is the question that confronted sixteenth century student of algebra, Rafael Bombelli.
(Rafael Bombelli, 1525  1572)
The short answer is that nothing went wrong. The slightly longer answer is that
2 + 11 1 + 2  11 1 is in fact equal to four. That is far from obvious, which is why we
should be grateful that Rafael Bombelli was a genius.
3
3
���
15
�16 ���
1 Square Root of Negative one.nb
For it was Bombelli who stared at this formula, aware that one of the answers of the equation should
be 4, and finally figured out how to make some sense of what he saw. It occurred to him  we do not
know how (although mathematician have reconstructed some plausible guesses)  that the quantities
2 + 11 1 and 2  11 1 were in fact cubes of other complex quantities:
2+
and 2 
1
1 .
Try it!
2
1 = 2 +
2 +
1 2 +
1 = 4 + 2
1 + 2
1  1 = 3 + 4
1
3
2 +
1 =
2
2 +
1 2 +
1 = 3 + 4
1 2 +
1 = 6 + 8
1 + 3
1  4 = 2 + 11
1
And:
2
1 = 2 
2 
1 2 
1 = 4  2
1  2
1  1 = 3  4
1
3
2 
1 =
2 
2
1 2 
1 = 3  4
1 2 
1 = 6  8
1  3
1  4 = 2  11
1
With this in hand, the strange result of Cardano’s Formula reduces like this:
x=
3
3
2 + 11
1
2 +
1
+
3
+
2  11
3
3
2 
1 =
3
1 = 2 +
1 + 2 
1 = 4
The square roots of negative one cancel out and the simple answer appears. This looks like magic,
especially if you have spent hours staring hopelessly at the problem, increasingly convinced that it was
created by the devil to torment incautious humans.
This strange result seems to suggest that the square root of a negative number can form part of a
meaningful formula, provided the imaginary quantities in the expression cancel one another out
before the final solution appears. We still don’t know quite what 1 means, but its appearance no
longer implies instantly that a formula is meaningless. It can participate in the strange morricedance
of algebra and lead ultimately to verifiable solutions.
Bombelli’s result, wonderful as it is, is far from fully enlightening. Pleased as we might be that Bombelli
could guess what complex number, when cubed, would give the particular quantity he was looking for,
�1 Square Root of Negative one.nb
���
17
his success doesn’t give us much help in solving other problems. Moreover, it doesn’t give us much
help in figuring out what the square root of negative one is. Do the imaginary numbers have any real
meaning? Or are they bizarre brambles that need to be cleared away by ad hoc trickery? Was Bombelli’s result a lucky accident? Or does it represent something valuable?
All that is the subject of the next lecture.
Appendix 1: How to Depress a Cubic Equation
Grant Franks
August 28, 2019
Begin with the general cubic:
x3 + b x2 + c x + d = 0
For x substitute y  b3 :
3
2
y  b3 + b y  b3 + c y  b3 + d = 0
y3  b y2 +
b2
3
y
b3
27
+ b y2 
2 b2
3
y+
y3 + ( b  b) y2 + c 
b2
y
3
+ 227b 
y3 +
b2
y
3
+ 227b 
…
+c
b3
9
+ cy 
3
cb
3
+ d = 0
3
cb
3
+ d = 0
cb
3
+d=0
Et voila, a depressed cubic.
For example, suppose you had x 3 + 9 x 2  3 x + 2 = 0. Substitute x = y  93 :
(y  3)3 + 9 (y  3)2  3 (y  3) + 2 = 0
(y3  9 y2 + 27 t  27) + (9 y2  54 y + 81 )  (3 y + 9) + 2 = 0
y3  24 y + 47 = 0
You can solve this (depressed) equation using Cardano’s formula. Then from the values of y, find the
values of x from the relation:
x = y  b3
which means
x+
b
3
=y
�18 ���
1 Square Root of Negative one.nb
Appendix 2: Lectures in the Series
Wednesday a�ernoon lectures, 3:15 pm in the Junior Common Room
Tuesday evening lectures, 7:30 pm in Room FAB 109
1.
Say Hello to Your New Best Friend:  1
(Wednesday, September 11)
2.
The Bridge between Algebra and Trigonometry
(Tuesday, September 17)
3.
Constructible Numbers
(Tuesday, September 24)
4.
Building the Pentagon with Algebra
(Tuesday, October 1)
5.
The Heptadecagon (17gon)
(Tuesday, October 8)
6.
Numbers and Meaning
(Wednesday, October 16)
�
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St. John's College Lecture Transcripts—Santa Fe
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Meet your new best friend : the square root of negative one
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Transcript of a lecture given on September 11, 2019 by Grant Franks as part of the Dean's Lecture and Concert Series.
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Franks, Grant H.
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20190911
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Algebra
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English
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SF_FranksG_Meet_Your_New_Best_Friend_The_Square_Root_of_Negative_One_20190911

https://s3.useast1.amazonaws.com/sjcdigitalarchives/original/cb3cd922369386f7f500fc5e2c4cabdc.mp3
5fe281203c78e0e19d5b75ce58526b97
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What distinguishes a person from a word? An invitation to the thought of Charles Sanders Peirce
Description
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20171117
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Peirce, Charles S. (Charles Sanders), 18391914
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20014389
Friday night lecture