# Inequality proof (Spivak 1.6-b)

1. Aug 18, 2011

### carlosbgois

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

Prove that if x < y, and n is odd, then x$^{n}$< y$^{n}$

3. The attempt at a solution

My attempt was to solve three different cases:

Case 1: If 0 $\leq$ x < y, we have

y-x > 0
y*y*...*y > 0 (closure of the positive numbers under multiplication)
x*x*...*x $\geq$ 0

y$^{n}$-x$^{n}$ = (y-x)(y$^{n-1}$ + y$^{n-2}$x +...+ yx$^{n-2}$ + x$^{n-1}$)

So, as every piece of the second member of this last equation is positive, their sums and multiplications are also positive, hence proving that y$^{n}$ > x$^{n}$

Case 2: If x$\leq$ 0 < y, we have: x$^{j}$ $\leq$ 0 (j is odd), and also y$^{j}$ > 0, which is the same as -y$^{j}$ < 0. Now, as we have closure under sum, then x$^{n}$ + (-y$^{n}$) < 0, so y$^{n}$ > x$^{n}$

Case 3: If x < y $\leq$ 0 ... ?

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Are my proofs of case 1 and 2 ok? What should I do in case 3?
Thanks

2. Aug 18, 2011

### HallsofIvy

Staff Emeritus
For (2), if x< 0< y, for odd n, if $x^n< 0$ and $0< y^n$. That's all you need to say.

For (3), if x< y< 0, then -x> -y> 0. And, of course, for n odd, $(-x)^n= -x^n$. Use (1) that you have already proved.

3. Aug 18, 2011

### carlosbgois

Thanks! It's actually simple, but sometimes I find hard to have these insights.