# Spivak Calculus Problem 1.iii

1. Mar 18, 2012

### mishima

1. The problem statement, all variables and given/known data
Prove that if x2=y2, then x=y or x=-y.

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

2. Relevant 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.

3. 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...

2. Mar 18, 2012

### Office_Shredder

Staff Emeritus
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

3. Mar 18, 2012

### 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.