# Math Help - proof

1. ## proof

Prove if $0 < a < b\,$ then $0 < a^n < b^n\,$ for all $n \in \mathbb{N}\,$

2. Just use the fact that if,
0<a<b
0<c<d
Then,
0<ac<bd.

Thus, consider "n" inequalities,
0<a<b
0<a<b
.........
0<a<b
Multiply them out,
0<aa...a<bb...b
Thus,
0<a^n<b^n

Or you can prove it by induction (basically what I did).

3. observe that this is true for n=1, and then suppose it
true for some positive integer n=k.

Then we have:

0<a<b, and 0<a^k<b^k.

hence as a>0,

0 < a^{k+1} < a b^k

and as b>a:

0 < a^{k+1} < b^{k+1}

Hence by induction if 0<a<b, 0<a^n<b^n for all positive integers n.

RonL