# Thread: Induction Proof

1. ## Induction Proof

Prove by induction that, for all natural numbers, if 0 < x < y, then x^n < y^n.

2. What have you attempted? I am guessing you're stuck on the k->k+1 step.

3. That is exactly where I am stuck. I have never wrote a proof of this knid using mathematical induction. I just dont know how to go about it.

4. Originally Posted by bearej50
Prove by induction that, for all natural numbers, if 0 < x < y, then x^n < y^n.
I write out the most imporant step in this solution but you need to complete all the steps of the proof thyself. If $\displaystyle 0<x<y$ and $\displaystyle x^n < y^n$ then it means $\displaystyle x\cdot x^n < x\cdot y^n$. Therefore, $\displaystyle x^{n+1} < x\cdot y^n$ however, $\displaystyle x\cdot y^n < y\cdot y^n = y^{n+1}$. Putting this together we find that $\displaystyle x^{n+1} < y^{n+1}$.

5. thank you! you enlightened me