# Spivak's Calculus problem 6

1. Aug 12, 2012

### alfred_Tarski

1. The problem statement, all variables and given/known data

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.

2. Relevant 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.

3. 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 $\wedge$ n is odd $\Rightarrow$ x^n < y^n

There are three cases to consider:

(1) 0 $\leq$ x < y
(2) x < 0 $\leq$ y
(3) x < y $\leq$ 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 $\leq$ 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.

2. Aug 12, 2012

### christoff

I know you've said that you tried induction, but I think you should give it another try. For case 3, this is my recommended outline/hint guide:

1. The base case is n=1. This is trivially true.
2. Let $m\geq 1$ be odd. Then m=k+1 for some k even.
3. $x<y \Leftrightarrow -y<-x$. What do you know about even powers and inequalities? (note that -y and -x are both positive).
4. Since $-x>0$, you know that for any $0<a<b$, the following holds: $-xa<-xb$.
5. But then, you know from 3. that $-y<-x$...

That's the gist of it.

3. Aug 13, 2012

### ehild

You can prove that xn<yn for x<y and n odd if both x and y are positive.

If case both x and y negative, x<y means |x|>|y|. Both are positive numbers, so |y|n<|x|n. And you know that |y|=-y, |x|=-x.... Can you proceed?

ehild