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Suppose $\epsilon$ > 0 and $\delta$ = min {1,$\frac{\epsilon}{10}$}

Is it true that if 0 < |x -1 |< $\delta$, then this implies 0 < |x -2 |< $\epsilon$?

I am saying yes and here is why. We are given that 0 < |x -1 |< $\delta$, so -1 < |x -2 |< $\frac{\epsilon}{10}$-1, which is
-1< |x -2 |< $\frac{\epsilon}{10}$-1

Now $\frac{\epsilon}{10}$-1 is always less than $\epsilon$. And that is why I say it is true.

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  • $\begingroup$ I would like to introduce you to the MathJaX tutorial page, because when I look at your past questions I get the feeling no one did before. MatJaX is what this site uses to format mathematics. This way you can type your own maths, properly. $\endgroup$ – gebruiker Apr 15 '16 at 8:38
  • $\begingroup$ @NoahSchweber Yup... I was thinking $2/10$. Your example works much better. $\endgroup$ – TokenToucan Apr 15 '16 at 17:37
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Your argument is wrong - here's a counterexample. Take $\epsilon={1\over 2}$, so $\delta={1\over 20}$. Let $x=1+{1\over 100}$. Then $0<\vert x-1\vert<\delta$, but $\vert x-2\vert={99\over 100}>\epsilon$.

Specifically, your error is that your "... so ..." is not justified at all.

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  • $\begingroup$ Is there a way to prove it is false without taking counterexamples? As in, assume it is true, try ti prove it and realize we have reached a contradiction. $\endgroup$ – stackdsewew Apr 15 '16 at 5:21
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    $\begingroup$ @anna_xox Not that I know of - but that said, keep in mind that in some sense this is essentially what a counterexample is: the realization that the statement can't be true, because it contradicts a known fact! $\endgroup$ – Noah Schweber Apr 15 '16 at 5:23
  • $\begingroup$ So we are basically doing trial and error? It just feels as though there would be some methodical way to prove it true or false. $\endgroup$ – stackdsewew Apr 15 '16 at 5:24
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    $\begingroup$ @anna_xox I think that's not really fair - there are in fact methods for finding counterexamples. For instance, in this case we're looking for an $x$ such that $\vert x-1\vert$ is "small" but $\vert x-2\vert$ is "big." That means we want an $x$ which is much closer to $1$ than to $2$. (Also, keep in mind that most methods are heuristic - even if you know what you're doing, you're going to have to try a bunch of things to get everything to work properly. But I think it's wrong to conflate this with "trial and error.") $\endgroup$ – Noah Schweber Apr 15 '16 at 5:28

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