Spivak Calculus Problem 1.iii

  • Thread starter mishima
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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...
 

Answers and Replies

  • #2
Office_Shredder
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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
jgens
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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.
 

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