# Straightedge and compass constructions

1) Prove that 45 degrees can be trisected with straightedge and compass.

My attempt:
60 deg constructible since equilateral triangle constructible
and 45 deg constructible since 90 deg constructible and we can bisect any angle.
=>(60-15)=15 deg constructible
Then copy this angle 3 times to trisect 45 deg (fact: any angle can be copied with straightedge and compass)
Did I get the right idea?

2) Let F={a+b√3 | a,b E Q(√2)} where Q(√2)={c+d√2 | c,d E Q}. Show that every element of F is the root of a polynomial of degree 4 with rational coefficients.

No clue...how to begin?

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2) In class, I've learnt about the concepts of number fields, surds, trisectibility of angles, constructible numbers, angles, and polygons. But I still can't figure out how to solve this problem.

For example, I've learnt the following theorems:

Theorem: If a cubic equation with rational coefficients has a constructible root, then it has a rational root.

Theorem:
Let (a+b√r) E F(√r) (i.e. in some tower of number fields).
Suppose p is a polynomail with rational coefficients, if p(a+b√r)=0, then p(a-b√r)=0.

Can anyone help me with Question 2, please? I am feeling desperate on this question...

morphism
Homework Helper
Notice that every element in F can be written as $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}$, where $a,b,c,d \in \mathbb{Q}$. Do you know anything about degrees of extension fields?

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Notice that every element in F can be written as $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}$, where $a,b,c,d \in \mathbb{Q}$. Do you know anything about degrees of extension fields?
I know about "the externsion of F by √r", but not about the degree.

Is it possible to do it without this concept??

x =a + b √ 3, with a, b in Q(√ 2)
then a = u + v √2 and b = m + n √2 for some u,v,m,n in Q

x =a + b √3
x-a = b √3
x^2 - 2ax + a^2 = 3 b^2
x^2 - 2(u+v √2) x + (u+v √2)^2 = 3 (m+n √2)^2
x^2 - 2ux + u^2 + 2v^2 - 3m^2 - 6n^2 = (2vx - 2uv + 6mn) √2
squaring both sides => we get a polynomial equation of degree 4 with rational coefficients

Is this a valid proof??

That looks right assuming the algebra is correct and the question isn't asking you to show that the minimal polynomial is of degree 4. As long as you've constructed a degree 4 polynomial with rational coefficients and x is the root, you'll be fine. Though as morphism says, this would be easier using a "degree of the field extension" argument.

hey..it is not impossible to trisect an angle using compass n a straight edge..ive proved it possible...