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Subfields of C (Hoffman/Kunze LA)

  1. Oct 8, 2011 #1
    I'm taking my first proof heavy class, linear algebra. Unfortunately, I'm taking a long time picking up proofs in general, so I'm going to try and work through some of the material in Hoffman/Kunze.
    1. The problem statement, all variables and given/known data
    Prove that the set of all complex numbers in the form of x + y√2, where x and y are rational, is a subfield of C.


    2. Relevant equations
    The nine rules of algebra...
    x+y=y+x
    .
    .
    .

    3. The attempt at a solution
    In general, a subfield has to obey the nine rules of algebra (addition and multiplication) and if x and y are elements of the field F, then so are (x+y), -x, xy, and x^-1.

    I'm not going to spend time writing a proof in this particular post. I just want some general guidance on how to approach certain proofs.

    For this one, my first idea is to define two numbers, (x1+y1√2) and (x2+y2√2), and then show that "(x+y), -x, xy, and x^-1" are all complex numbers. I'll do this by simple addition/multiplication and, which stem from the nine rules.

    Would this work? Sorry for the lack of LaTeX. I just want to get this question out before dinner.
     
  2. jcsd
  3. Oct 8, 2011 #2

    D H

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    That won't prove the conjecture. Denoting F={x+y√2:x,y ∈ Q}, you need to show that (for example), (x+y√2)-1 is a member of F. Showing that the multiplicative inverse is a member of C doesn't help prove the conjecture.
     
  4. Oct 8, 2011 #3
    Can I use the previously mentioned method for the other conditions?

    For example:
    (x1+y1√2)+(x2+y2√2)=(x1+x2)+(y1+y2)√2
    The sum must be in F since the sum of rational numbers are rational.

    Likewise, a rational number multiplied by negative one will still be rational, thus they still fit under the conditions of F. Following that train of thought, I would just foil (x*y) and use the same reasoning.

    As for (x+y√2)-1, the denominator must not be equal to zero. Then I reason that [(x+y√2)-1]-1=(x+y√2) is a member of F, which it is.

    Thanks for the help.
     
  5. Oct 8, 2011 #4

    D H

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    That's good. You used a constructive proof to show that the addition is closed in F.

    That's not good. Your proof assumes that for 1/z is a member of F if z is in F. It doesn't prove it. You can either use a constructive approach as you did for closure of addition, or non-constructive approach such as proof by contradiction.
     
  6. Oct 9, 2011 #5
    Ok, so I think I have the constructive proof:

    (x+y√2)-1=(x+y√2)-1*[(x-y√2)/(x-y√2)]=(x-y√2)/(x2-2y2)=x/(x2-2y2) + (-y)√2/(x2-2y2)

    where both x/(x2-2y2) and (-y)/(x2-2y2) are rational.

    For the proof by contradiction, do I start by asserting (x+y√2)-1 is not a member of F? So x and/or y are irrational numbers?

    Thanks.
     
  7. Oct 9, 2011 #6

    D H

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    Very good, but there is one minor problem here: How do you know that both of these are rational? Just because both the numerator and denominator are integers is not enough; x/0 is not rational. In other words, how do you know that the denominator is not zero?

    First off, you have a constructive proof, and constructive proofs are almost always preferred over non-constructive proofs.

    To prove this by contradiction, you would start by assuming that there exists some non-zero z∈F such that the multiplicative inverse of z is not an element of F. All it takes is one counterexample to disprove closure. From this you would have to arrive at a contradiction. This will not be easy. The constructive proof is much easier.
     
  8. Oct 9, 2011 #7
    If the denominator were zero, then either (x+y√2) or (x-y√2) would have to equal zero, but this can't be since division by zero is impossible. (x+y√2)-1 or (x-y√2)/(x-y√2) would provide an undefined answer.

    Should I have stated beforehand that both terms are non-zero?

    If you wouldn't mind, how would I structure the proof in a "professional manner"? I've just started reading a logic/proof book (Velleman's How to Approach It), so hopefully this will become easier.
     
  9. Oct 9, 2011 #8

    Dick

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    The missing ingredient that DH is talking about is that in finding inverses, you can only assume x+sqrt(2)*y is nonzero. If x-sqrt(2)*y were zero then that would mean x+sqrt(2)*y would have an undefined inverse. Can you show that doesn't ever happen? I.e. can you show x^2-2*y^2 with x and y rational is only zero if x and y are both zero??
     
  10. Oct 9, 2011 #9
    I'm not sure where I would fit this detail in the proof, but:

    x2-2y2 would only equal zero if x-y√2 were zero.

    I'll first let x-y√2=0
    so, x=y√2.
    Because x and y are rational, this can only occur if x=y=0. This means that when both x and y aren't zero, x2-2y2 will give a non-zero rational number.
     
  11. Oct 9, 2011 #10

    Dick

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    Sure. The important thing is that sqrt(2) is irrational. So your denominators are never zero unless x=y=0.
     
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