# verification- epsilon-delta

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.

• 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. – gebruiker Apr 15 '16 at 8:38
• @NoahSchweber Yup... I was thinking $2/10$. Your example works much better. – 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 $\vert x-2\vert={99\over 100}>\epsilon$.
• @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.") – Noah Schweber Apr 15 '16 at 5:28