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Straightedge and compass constructions

  • Thread starter kingwinner
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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?

Can someone please help me? Particularly with Q2. Thanks a lot!
 
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Answers and Replies

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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.
 
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Can anyone help me with Question 2, please? I am feeling desperate on this question...
 
morphism
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Notice that every element in F can be written as [itex]a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}[/itex], where [itex]a,b,c,d \in \mathbb{Q}[/itex]. Do you know anything about degrees of extension fields?
 
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Notice that every element in F can be written as [itex]a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}[/itex], where [itex]a,b,c,d \in \mathbb{Q}[/itex]. 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??
 
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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??
 
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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...
 

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