Inequality proof (Spivak 1.6-b)

[carlosbgois]

Homework Statement

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

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 ... ?


------------------------------------------------------------------------


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

 

[HallsofIvy]

For (2), if x< 0< y, for odd n, if x^n&lt; 0 and 0&lt; 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.

 

[carlosbgois]

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