This preview shows pages 1–2. Sign up to view the full content.
More on Constructible Numbers and Angles
Original Notes adopted from March 5, 2002 (Week 21)
© P. Rosenthal , MAT246Y1, University of Toronto, Department of Mathematics typed by A. Ku Ong
Lemma
: If x
0
is a root of a polynomial with coefficients in F(r), then x
0
is a root of a polynomial with
coefficients in F.
Proof:
Given (a
n
+ b
n
√
r) x
0
n
+ (a
n1
+ b
n1
√
r) x
0
n1
+ .
.. +(a
1
+ b
1
√
r)x
0
+ a
0
+ b
0
√
r =0 with a
i
, b
i
∈
F.
By Dividing
through by a
n
+ b
n
√
r, we can assume it is poly monic. (ie. coefficients of x
0
n
is 1).
x
0
n
(a
n1
+ b
n1
√
r) x
0
n1
+
... +(a
1
+ b
1
√
r)x
0
+ a
0
+ b
0
√
r =0
Then,
x
0
n
+ a
n1
x
0
n1
+....
+ a
1
x
0
+ a
0
=
√
r (b
n1
x
0
n1
+
... + b
1
x
0
+ b
0
)
Square both Sides
(x
0
n
+ a
n1
x
0
n1
+....
+ a
1
x
0
+ a
0
)
2
 r (b
n1
x
0
n1
+
... + b
1
x
0
+ b
0
)
2
=0
Recall: x
0
algebraic if p(x
0
)= 0 some polynomial p with rational coefficients.
Theorem: Every constructible number is algebraic.
Proof:
Suppose x
0
is constructible.
There exists Q = F
0
⊂
F
1
⊂
F
2
⊂
......
⊂
F
k
with F
i
= F
i1
(
√
r
i
) some r
i
∈
F
i1
, r
i
= 0,
√
r
i
∉
F
i1
such that x
0
∈
F
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '10
 Applebaugh
 Math, Angles

Click to edit the document details