# Simple Proof

1. Sep 20, 2009

### evry190

0<x<y, prove x^n<y^n for n = natural numbers

I know its obvious, but I don't really know what to write to proove it...

2. Sep 20, 2009

### VeeEight

You should show your attempt at a solution. Perhaps try induction

3. Sep 20, 2009

### evry190

well

P(1) is true (x<y)

and P(n) is true (x^n<y^n)

and p(n+1) is true (x^n+1 < y^n+1)

4. Sep 20, 2009

### tara123

Induction is the best way to show this proof.
the way you should start would be:
Let P(n) be the statement 0<x<y then x^n<y^n where n=natural numbers

Then your base case is where x and y are the smallest natural numbers such that they apply to your restrictions, then of course you assume that P(n) is true for some n like you have and then just go through the steps to show that P(n+1) is true. You just need to beef up your proof. Your on the right track!=)

5. Sep 20, 2009

### evry190

but x and y don't have to be natural numbers
im a bit confused : (

6. Sep 20, 2009

### Gregg

This is easily proved with axioms of the real number system. You can use multiplicative and order axioms to show that for z>0, xz<yx since, z(x-y)<0 for example. Then you can show that for 0<x<y and 0<u<v, 0<xu<yv. In the case that 0<x<y and u=x and y=v, x^2<y^2...x^n<y^n.

7. Sep 20, 2009

### jgens

Induction seems like the best way to give a formal proof of the fact, but a sloppy and informal proof could show that $x^n < x^{n-1}y < \dots < xy^{n-1} < y^n$.

8. Sep 21, 2009

### Gregg

I don't see how that would work, I did not give a good way to prove this but I think the only way to do this properly (and formally) would be with axioms of the real number system. With induction it seems much less accessible since you have not only prove it for all n, but for real numbers x and y. I'd be interested to see how this works, since the choice of inequality at the start is comletely arbitrary isn't it? Maybe I'm missing something.