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.

  • $\begingroup$ 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

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.

| cite | improve this answer | |
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.