We have $0\le x \lt h$ for all $h \gt 0$. Prove that $x=0$. I know it is obvious but I can't prove it.
My try : I think it has some connections with concept of limit but I can't prove it.
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Approach 1:
Suppose that $x\ne 0$ then take $h=\frac{x}{2}$ and then
$$h<x$$
what is a contradiction.
Approach 2:
Another approach is use the sequence $h_n=\frac{1}{n}$, $0_n=0$ and use Squeeze Theorem:
$$0_n\le x<h_n$$
because $h_n\to 0$ and $0_n\to0$.
Arnaldo
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Very nice answer ! Can you provide a different answer without using contradiction method ? – S.H.W Apr 11 '17 at 14:52
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4You can even take $h=x$ and that'll still be a contradiction – Apr 11 '17 at 14:54
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@S.H.W: You can, for example, use a sequence $h_n=1/n$ and use the squeeza theorem – Arnaldo Apr 11 '17 at 14:54
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@Arnaldo, could you please help me with this question :: http://math.stackexchange.com/questions/2228579/if-x-be-the-a-m-between-y-andz?? – pi-π Apr 11 '17 at 15:48
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Suppose $x$ is an infinitesimally small positive number close to $0$, e.g., $x = 2^{-(2^{127} - 1)}$. Then $0 < x$. But now $x < h$ is now true only for almost all $h > 0$, but not all $h > 0$. So the condition $h > 0$ essentially pushes $x$ all the way down to $0$.
Mr. Brooks
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