1. ## continuous function

Let h: be continuous on satisfying =0 for all m within integers and n within naturals. Show that h(x)=0 for all x within reals

2. Originally Posted by rmpatel5
Let h: be continuous on satisfying =0 for all m within integers and n within naturals. Show that h(x)=0 for all x within reals
Let $r\in \mathbb{R}$ with $h(r) > 0$ WLOG. Take $\epsilon > 0$ so that $h(r) - \epsilon > 0$. Then it means there is $\delta > 0$ such that if $x\in (r-\delta,r+\delta) \implies h(x) > h(r) - \epsilon > 0$. The problem is that on any interval we can find a number having the form $\tfrac{m}{2^n}$. And $h( \tfrac{m}{2^n} ) = 0 \not > 0$. This is a contradiction.

3. Originally Posted by ThePerfectHacker
Let $r\in \mathbb{R}$ with $h(r) > 0$ WLOG. Take $\epsilon > 0$ so that $h(r) - \epsilon > 0$. Then it means there is $\delta > 0$ such that if $x\in (r-\delta,r+\delta) \implies h(x) > h(r) - \epsilon > 0$. The problem is that on any interval we can find a number having the form $\tfrac{m}{2^n}$. And $h( \tfrac{m}{2^n} ) = 0 \not > 0$. This is a contradiction.
would u not have to prove h(r)<0. If so how would i go about doing that??

4. Originally Posted by rmpatel5
would u not have to prove h(r)<0. If so how would i go about doing that??
If $h(r) < 0$ choose $\epsilon > 0$ so that $h(r) + \epsilon < 0$. Then there is $\delta > 0$ so that $x\in (r - \delta,r+\delta) \implies h(x) < h(r) + \epsilon < 0$. But since we can find $\tfrac{n}{2^m}$ in this interval it would mean $0=h(\tfrac{n}{2^m}) < 0$. A contradiction.

Therefore, the only possibility is that $h(r) = 0$ for all $r\in \mathbb{R}$.