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: Comparison Proof via axioms, almost done need hints for finish and proof read

  1. Sep 17, 2010 #1

    silvermane

    User Avatar
    Gold Member

    1. The problem statement:

    Prove that
    [tex]0\leq2x^2 - 3xy + 2y^2[/tex]​


    2. These are the axioms we are permitted to use:

    01) Exactly one of these hold: a<b, a=b, or b<a
    02) If a<b, and b<c, then a<c
    03) If a<b, then a+c < b+c for every c
    04) If a<b and 0<c, then ac<bc.


    3. The attempt at a solution

    By axiom 03) we are permitted to add [tex]+-y^2[/tex] to both sides:
    [tex]0 = -y^2 + y^2 \leq 2x^2 -3xy+2y^2-y^2+y^2[/tex]
    [tex]0 \leq 2x^2 -3xy +y^2 + y^2[/tex]

    Again, by axiom 03) we add [tex]+-x^2[/tex] to both sides:
    [tex]0 = x^2 - x^2 \leq 2x^2 - 3xy +y^2 +y^2 + x^2 - x^2 = x^2 - 3xy + y^2 +y^2 + x^2[/tex]

    By axiom 03) again, we add +-xy to both sides:
    [tex]0 = xy - xy \leq 2x^2 - 3xy +y^2 +y^2 + x^2 - x^2 +xy - xy = x^2 - 3xy + y^2 +y^2 +xy +x^2[/tex]

    Case 1: If x=y=0, then our expression is zero

    Case 2: If x doesn't = y, and y=0, then we have [tex]0<x^2+x^2=2x^2[/tex]

    Case 3: If x doesn't = y, and x=0, then we have [tex]0<y^2[/tex]

    Case 4: If x doesn't = y, and [tex]y\neq0[/tex],
    [tex]0 < (x-y)^2 < (x-y)^2 +x^2 +xy +y^2 [/tex] ??
    then I'm stuck and need help :uhh:

    If anyone can help me figure out the last part, it would be greatly appreciated. I've worked very hard trying to figure it out, but that xy we would get could possibly be negative and I just need to prove that last part.

    Thank you in advance for any hints/tips!!!!
     
  2. jcsd
  3. Sep 18, 2010 #2
    Case 1

    [tex] x,y >0[/tex] This is trivial since everything is positive.

    Case 2
    [tex] x >0 ,y <0 \Rightarrow x>y[/tex]

    [tex] -x^{2}< xy \Rightarrow 0< xy + x^{2} [/tex]

    [tex] (x-y)^{2} + y^{2}< (x-y)^{2} +x^{2} + xy +y^{2}[/tex]


    Case 3

    This is similar to case 2.


    How come you can't use the distributive properties ?

    Without them what does xy mean ? And how does it differ from yx ?


    Assuming you could use axioms of multiplication and the distributive property you could divide by 2 and complete the square and the result falls into your hands.

    [tex]x^2 - \frac{3}{2}xy + 2y^2 = (x- \frac{3}{2}y)^{2} + \frac{7}{16}y^{2}[/tex]
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook