- #1
alfred_Tarski
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Homework Statement
Prove that x < y and n is odd implies x^n < y^n
This is from chapter 1 problem 6 (b), so I am sure a lot of you are familiar with this. One can only use Spivak's 12 axioms, theorems derived from said axioms, and mathematical induction.
Homework Equations
x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + ... + xy^{n-2} + y^{n-1})
Sorry I do not know how the scripting language works here.
The Attempt at a Solution
I have tried doing this problem for about five hours, *shame*, and I have looked up hints on the internet about it but have been still unsuccessful. I think my problem with this exercise is my accounting.
Here is one of my attempts:
If x < y [itex]\wedge[/itex] n is odd [itex]\Rightarrow[/itex] x^n < y^n
There are three cases to consider:
(1) 0 [itex]\leq[/itex] x < y
(2) x < 0 [itex]\leq[/itex] y
(3) x < y [itex]\leq[/itex] 0
Case (1) is a special case of the previous theorem we proved in problem 6 (a), so it's trivial.
Case (2) is a negative number to an odd power is less than a positive number to an odd power, so again trivial. But since I am so new to proofs, it would be nice if one of you nice gentlemen or ladies would help me show this. (I am about 1/4 the way done with the book How To Prove It and my school has never taught me anything about proofs in my 1 and 1/2 years of attendance)
Case (3) is the case which I have been stuck on.
x < y [itex]\leq[/itex] 0
x - y < 0 < -y
∴ x - y < 0
Notice all three cases can produce this inequality so there must be one logical way to prove them all at once WLOG, but I do not see it?
I've also tried to prove this other ways, namely induction and contrapositive, but they did not get me any closer. Thanks for your time, and I really appreciate any and all responses.