Spivak Calculus - Chapter 1 Problems

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I have begun in my spare time to work through Spivak's Calculus. Although I have been stuck on many problems, I am most troubled by my general clumsiness and nonelegance of answers (particularly proofs). It always seems that there likely is a much simpler route available, and yet I forgo this route for a long-winded tedious argument. (Perhaps of my lack of proof experience)

Homework Statement



6 c) Prove that if x^{n}=y^{n} and n is odd, then x=y.
6 c) Prove that if x^{n}=y^{n} and n is even, then x=y or x=-y.

Homework Equations



12 Properties of Real Numbers

The Attempt at a Solution



For 6c, I considered using the contrapositive and claiming that:
If y≠x, then either y>x or x>y, which implies x^{n}<y^{n} or x^{n}>y^{n}, which should complete the proof? (This was proven as the first part of the problem.)

However, I tried to use a more direct proof from considering the factorization of the terms. This required a long winded explanation that required several rewritings for different cases, and an absolute value claim that I believe to be non-rigorous.

Is there a simple way to do these proofs directly? (And is the contrapositive proof I provided sound?)
 
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Here is a counter example to your first proof. Let y = 2, x = -2. Now y > x. But x^2 = y^2.How about if x^n = y^n and n odd, then divide both sides by x^n, factor the n, and show x/y = 1 which leads to x = y. Then make a special case for when x is 0.

Similarly for n even.
 
Thank you for your quick reply.

Here is a counter example to your first proof. Let y = 2, x = -2. Now y > x. But x^2 = y^2.

Sorry about my inclarity. The proof I provided was only for problem 6c) where n is odd case. I realize this will not work for the even case.


How about if x^n = y^n and n odd, then divide both sides by x^n, factor the n, and show x/y = 1 which leads to x = y. Then make a special case for when x is 0.

Similarly for n even.

Interesting, thank you for this. I have just been somewhat confused about what facts I am and am not allowed to use. Thus, I was generally hesitant with the general exponent rules. (Should I think this way?)
 
So, is there a particularly clean way to show these through the factorization?

(I am uneasy about using the exponent rules at this point of the book.)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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