Prove x^2 = y^2 if x = y or x = -y

[DavidSnider]

I'm trying to Prove x^2 = y^2 if x = y or x = -y and I'm getting stuck.

Some different things I think are relevant but can't seem to connect together to form a proof. Am I on the right path?

Squares are non-negative. 0 ≤ a^2

x^2 - y^2 = 0

x^2 - y^2 = (x-y)(x+y)
= (x-y) * x + (x-y) * y : Distributive Law
= x^2 - xy + xy - y2 : Distributive Law
= x^2 - y^2 : Additive Inverse

 

[berkeman]

Can you use the fact that (-1)^2 = 1 ?

 

[DavidSnider]

Ah! Yes I can. Thank you!

 

[D H]

x^2 - y^2 = 0

x^2 - y^2 = (x-y)(x+y)

Stop here. Don't go any further. What does (x-y)(x+y) = 0 tell you?

 

[DavidSnider]

Stop here. Don't go any further. What does (x-y)(x+y) = 0 tell you?

That the difference between x and y multiplied by the sum of x and y is equal to zero

So.. let's say that (x-y) is M and (X+Y) is N then M * N = 0.

The only way for this to happen is if one or both of those is equal to 0.
(x-y) can only be zero if X = Y. X+Y can only be zero if X = -Y.

Is there a better way I should be expressing that?

 

[D H]

You got it.