continuous function

**rmpatel5** · October 28th 2008, 10:44 AM

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

**ThePerfectHacker** · October 28th 2008, 11:44 AM

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.

**rmpatel5** · October 28th 2008, 12:05 PM

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??

**ThePerfectHacker** · October 28th 2008, 04:17 PM

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 $r\in \mathbb{R}$ .

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