Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Zeros proof math

  1. Sep 4, 2010 #1

    jgens

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    Prove that if x3+x2y+xy2+y3 = 0, then x = y = 0 or x = -y.

    2. Relevant equations

    N/A

    3. The attempt at a solution

    Assume that x3+x2y+xy2+y3 = 0, in which case, it follows that x3+y3 = -(x2y+xy2) or (x+y)(x2-xy+y2) = -xy(x+y). Equality clearly holds if x+y = 0. Now, suppose that x+y =/= 0, and divide through by x+y. This leaves the equality x2-xy+y2 = -xy or x2+y2 = 0, which can only happen if x = y = 0. Therefore, if x3+x2y+xy2+y3 = 0, then x = -y or x = y = 0.

    Does this 'proof' work?
     
  2. jcsd
  3. Sep 4, 2010 #2

    Mentallic

    User Avatar
    Homework Helper

    Re: Zeros

    Yes, of course it does. Your proof is sound and logical.
     
  4. Sep 4, 2010 #3

    jgens

    User Avatar
    Gold Member

    Re: Zeros

    Thanks! I've been second guessing myself a lot lately, so it's always good to get confirmation that I've done something correctly.
     
  5. Sep 4, 2010 #4

    Mentallic

    User Avatar
    Homework Helper

    Re: Zeros

    Yeah, I know the feeling :biggrin:
     
  6. Sep 5, 2010 #5

    HallsofIvy

    User Avatar
    Science Advisor

    Re: Zeros

    Excellent! I would have done it a slightly different way: it appeared to me, looking at the first two terms, that [itex]x^2+ x^2y= x^2(x+ y)[/itex] and then that [itex]xy^2+ y^3= y^2(x+ y)[/itex]. Since (x+ y) appears in both of those we can factor it out and have [itex]x^3+ x^2y+ xy^2+ y^3= (x^2+ y^2)(x+ y)[/itex].

    That will be 0 only if one or the other of those factors is 0. [itex]x^2+ y^2= 0[/itex] only if x= y= 0 and x+ y= 0 only if y= -x.
     
  7. Sep 5, 2010 #6

    jgens

    User Avatar
    Gold Member

    Re: Zeros

    Thanks Halls! I like the way that your proof works out much better. It looks a lot more like what I typically see.
     
  8. Sep 6, 2010 #7

    HallsofIvy

    User Avatar
    Science Advisor

    Re: Zeros

    On the other hand, your proof is much better for you than mine because it is yours!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook