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Straightedge and compass constructions

  1. Mar 21, 2008 #1
    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!
     
    Last edited: Mar 21, 2008
  2. jcsd
  3. Mar 21, 2008 #2
    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.
     
  4. Mar 22, 2008 #3
    Can anyone help me with Question 2, please? I am feeling desperate on this question...
     
  5. Mar 22, 2008 #4

    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?
     
    Last edited: Mar 22, 2008
  6. Mar 23, 2008 #5
    I know about "the externsion of F by √r", but not about the degree.

    Is it possible to do it without this concept??
     
  7. Mar 24, 2008 #6
    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??
     
  8. Mar 24, 2008 #7
    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.
     
  9. Apr 6, 2008 #8
    hey..it is not impossible to trisect an angle using compass n a straight edge..ive proved it possible...
     
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