Spivak Calculus: Problem 1-(iii)

[Saladsamurai]

Homework Statement

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

Homework Equations

The 12 Properties of numbers

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?

 

[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.

 

[Saladsamurai]

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

Adding -(a*a) to both sides:

(-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!

 

[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.

 

[Dick]

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.

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.

 

[Saladsamurai]

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.

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.

I see. Thanks for the tips! I got a lot of mileage out of this one.