Proof that one can not Trisect an angle of 60 degrees with Straight Edge
and Compass
Original Notes adopted from February 12, 2002 (Week 19)
© P. Rosenthal , MAT246Y1, University of Toronto, Department of Mathematics typed by A. Ku Ong
F, r
∈
F, r>0 & if
√
r
∈
F,
F(
√
r) = {a + b
√
r: : a,b
∈
F}
Tower of number fields
is a collection
F
0
⊂
F
1
⊂
F
2
⊂
.........
⊂
F
n
of number fields such that
F
0
= Q to
∀
i, there exists r
i
∈
F
i
such that r
> 0,
√
r
i
∉
F
i
i+1
= F
i
(
√
r
i
)
Collection of surds
is U {F
n
: F
n
at top of a tower}
S, S number field ("the surd field")
Set of Constructible Numbers:
We showed: C is a number field
Also, if r
∈
C, r > 0, then
√
r
∈
C.
We showed:
Theorem: C = S.
We'll show: Can't Trisect (with compass and straightedge) an angle of 60 degrees.
Can construct angle of 60 degrees.
If we could trisect that angle of 60 degrees, we could construct an angle of 20 degrees.
We'll show that this
is impossible.
Lemma 1: If angle
θ
(acute) is constructible, then cos
θ
is a constructible number.
cos
θ
= x/1 = x.
Constructible x from
θ
.
∴
cos
θ
constructible.
Suffices for:
Theorem: That 60 degrees not trisectable to show cos20 degrees not constructible.
I.e. Show cos20 degrees is not a surd.
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View Full DocumentRecall: cos(A+B) = cosAcosB – sinAsinB
cos2
θ
= cos
2
θ
 sin
2
θ
= 2 cos
2
θ
1
cos3
θ
= cos(2
θ
+
θ
)
cos3
θ
= cos2
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 Spring '10
 Applebaugh
 Math, Prime number, Rational number, Compass and straightedge constructions, rational coefficients, Constructible number

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