Proving abs(x-y) < ε for all ε>0 in Real Analysis

[mb55113]

Prove that abs(x-y) < ε for all ε>0, then x=y.

I really do not know how to start this... I have tried to do the contra positive which would be If x does not equal y, then there exist a ε>0 such that abs(x-y) >= ε. Can someone help me and lead me to the right direction.

 

[jgens]

Here's how I might approach the problem. Given any two real numbers $x,y \in \mathbb{R}$ we clearly have that $x-y \in \mathbb{R}$ and similarly that $|x-y| \in \mathbb{R}$. Now, suppose that $|x-y| > 0$, can you see how this imediately results in a contradiction? What does this contradiction suggest about the value of $|x-y|$?

 

[jgens]

However, if you want to prove the contrapositive, you could do it this way. Pick any two real numbers $x,y$ such that $x-y \neq 0$. Then clearly $|x-y| > 0$. Now, what happens if you choose something like $\varepsilon = \frac{|x-y|}{2}$?

 

[Hurkyl]

If x does not equal y, then there exist a ε>0 such that abs(x-y) >= ε. Can someone help me and lead me to the right direction.

Quick: without thinking, tell me a positive number that is either equal to or less than 5.

 

[mb55113]

Ok...I tried the contrapositive...I think that i got it so suppose that abs(x-y)=m>0. Therefore we make ε=m/2 which makes abs(x-y)=m>(m/2)=ε and therefore we have found an ε>o and which makes abs(x-y)>= ε...is that right?