Proving x=y or x=-y from x2=y2.

[mishima]

Homework Statement

Prove that if x²=y², then x=y or x=-y.

This is from Spivak's Calculus, problem 1(iii).

Homework Equations

Distributive law. If a, b, and c are any numbers, then a(b+c)=a*b+a*c.
Existence of additive inverse. If a is any number, then a+0=0+a.
Existence of multiplicative inverse. For every number a≠0, there is a number a^-1 such that a*a^-1=a^-1*a=1.

The Attempt at a Solution

First by the existence of additive inverse,

x²+(-y²)=y²+(-y²)=0.

Then in an earlier problem I proved x²-y²=(x-y)(x+y) using a triple application of the distributive law and the existence of additive inverse. So,

x²-y²=0=(x-y)(x+y).

Now here's where I get uncomfortable. I want to use the existence of multiplicative inverses to do this

(x-y)^-1(x-y)(x+y)=1*(x+y)=0

But multiplicative inverse has the stipulation that (x-y)≠0. So then when I do the other part

(x+y)^-1(x-y)(x+y)=1*(x-y)=0

I get a contradiction. I must be making a mistake since this only shows that x=-y, right? Or is it just an exclusive "or" in the problem statement or something? Earlier he says to assume inclusive ors...

 

[Office_Shredder]

There are two possibilities - x=y or x is not equal to y. If x=y you are done! That was one of the solutions you had to show is possible. If they're not equal do your division to find the other solution possibility

 

[jgens]

You can also just use the zero product property instead of imitating its proof. Recall that (x-y)(x+y)=0 implies (x-y)=0 or (x+y)=0. The proof is pretty simple from there.