# Spivak Calculus: Problem 1-(iii)

1. Jan 31, 2012

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

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

2. Relevant equations

The 12 Properties of numbers

3. The attempt at a solution

I think I should do this case-wise:

Case (a) if x=y then x2 = x*x=y*y=y2. Simple enough.

Case (b) if x = -y then x2 = x*x=(-y)*(-y) = ...

So what is left to show is that (-y)*(-y) = y*y which is where I am getting stuck.

I was thinking of doing something like:

(-y)(-y) = (-y)(-y)*1 = (-y)(-y) * [(-y)(-y)]*[(-y)(-y)]-1

... but I don't see this going anywhere really. Any thoughts?

2. Jan 31, 2012

### Dick

I think you are actually doing the thing a bit backwards. If x^2=y^2 then x^2-y^2=0=(x-y)(x+y). You should be able to prove that from the 12 properties of numbers, whatever they are. You should also have if a*b=0 then a=0 or b=0. Try it that way.

3. Jan 31, 2012

Hmmm..... After some searching around, it seems that my approach might not be the best. The simplest, is to use the given information:

x2 = y2

x2 - y2 = y2 - y2 = 0

so

x2 - y2 = x2 +xy - xy - y2

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

from here it is easy enough to show that since (x+y) or (x-y) is zero, then x = y or x = -y.

***********************************

Does this mean that the original way I was trying never would have panned out? I think it can still be done. I am just having trouble trying to prove that:

(-a)(-a) = aa

in a rigorous way. Though I feel like I have done this. Actually, Spivak does prove this on page 7:

(-a)*a + a*a = a*[(-a) + a] = a*0 = 0

(-a)*a + a*a + -(a*a) = -(a*a) so that

(-a)*a = -(a*a)

now adding (-a)*(-a) to both sides:

(-a)*a + (-a)*(-a) = -(a*a) + (-a)*(-a)

factor:

(-a)*[(-a) + a] = -(a*a) + (-a)*(-a) or

-(a*a) + (-a)*(-a) = 0

Finally, add a*a to both sides:

a*a + -(a*a) + (-a)*(-a) = 0 + a*a

so that:

(-a)*(-a) = a*a.

yay!

Edit: Hi Dick! By "backwards" do you just mean "not the way I would do it" ? Or do you mean wrongwards? Just curious. Either way, I am satisfied with the answer!

4. Feb 1, 2012

### Char. Limit

Well, what you proved using your original method is that if x=y or x=-y, then x^2 = y^2. It's TRUE, but it's the converse of what you're trying to prove.

5. Feb 1, 2012

### Dick

What Char. Limit has pointed out is what I meant by backwards. Proving x=y and x=(-y) are solutions doesn't prove there might not be other solutions. Doing it forwards does.

6. Feb 1, 2012