How to prove/disprove proof on limits (delta-epsilon)

Question

Prove or disprove: $$ \forall \epsilon > 0, \exists \delta>0, \forall x, y \in \mathbb{R}^+, |x - y| > \delta ⇒ |x + y| > \epsilon $$

I've been trying this for some time now but can't seem to get anywhere. I tried proving it then disproving it. Need help figuring out what to choose for the values of $\epsilon$, $x$, $y$ if disproving or of $\delta$ if proving. Also not sure if i'm following the right steps, some clarity would be really helpful.

score 1 · Answer 1 · answered Nov 27 '14 at 05:57

1

Choose $\delta=\epsilon$, if $|x-y|>\delta$ then $|x+y|>|x-y|>\delta=\epsilon$.

answered Nov 27 '14 at 05:57

Alistair

1,720

Suzu Hirose · Answer 2 · 2014-11-27T02:33:04.707

0

If $y<x$ then $|x+y|=x+y\geq x-y+y\geq|x-y|$ so if $|x-y|>\delta$ then $|x+y|>\delta$. If $x<y$ then $x+y\geq y-x+x\geq|x-y|$.

edited Nov 27 '14 at 02:33

answered Nov 27 '14 at 02:26

Suzu Hirose

11,660

I think there is a typo here somewhere. You say that $x + y = x - y + y$, which is generally false. – user141592 Nov 27 '14 at 02:28
@Johanna The = sign should have been $\geq$. The method still works. – Suzu Hirose Nov 27 '14 at 02:31
That's what I thought :) – user141592 Nov 27 '14 at 02:31

How to prove/disprove proof on limits (delta-epsilon)

2 Answers2