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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)
6 c) Prove that if [itex]x^{n}=y^{n}[/itex] and n is odd, then [itex]x=y[/itex].
6 c) Prove that if [itex]x^{n}=y^{n}[/itex] and n is even, then [itex]x=y or x=-y[/itex].
12 Properties of Real Numbers
For 6c, I considered using the contrapositive and claiming that:
If y≠x, then either y>x or x>y, which implies [itex]x^{n}<y^{n}[/itex] or [itex]x^{n}>y^{n}[/itex], 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?)
Homework Statement
6 c) Prove that if [itex]x^{n}=y^{n}[/itex] and n is odd, then [itex]x=y[/itex].
6 c) Prove that if [itex]x^{n}=y^{n}[/itex] and n is even, then [itex]x=y or x=-y[/itex].
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 [itex]x^{n}<y^{n}[/itex] or [itex]x^{n}>y^{n}[/itex], 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?)