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Math Help - continuous function

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by rmpatel5 View Post
    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.
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    Quote Originally Posted by ThePerfectHacker View Post
    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??
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  4. #4
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    Quote Originally Posted by rmpatel5 View Post
    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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