# For every $\epsilon >0$ , if $a+\epsilon >b$ , can we conclude $a>b$?

If $a+\epsilon > b$ for each $\epsilon >0$, can we conclude that $a>b$?

No, you cannot: for every $\epsilon>0$ it’s true that $1+\epsilon>1$, but it’s not true that $1>1$. What you can conclude is that $a\ge b$. To see this, suppose that $a<b$, and let $\epsilon=b-a$; then $\epsilon>0$, but $a+\epsilon=a+(b-a)=b\not>b$, contradicting the hypothesis. Thus, it must be the case that $a\ge b$.
• Suppose $a = 5$ and $b = 10$. Now what? Sep 29, 2013 at 5:17
• @Don: Then obviously it’s not the case that $a+\epsilon>b$ for each $\epsilon>0$, so the example is irrelevant. Sep 29, 2013 at 5:19
• @alu: It’s a proof by contradiction. We are given that $a+\epsilon>a$ for each positive $\epsilon$. To get a contradiction, we suppose that $a<b$. Then for $\epsilon=b-a$, which is positive, we have a+\epsilon=b$. This contradicts what we were given. Thus, the supposition that$a<b$must be false, since it is what led us to the contradiction, and therefore$a\ge b$. Jan 15, 2021 at 18:57 • @alu: It was supposed to say: To get a contradiction, we suppose that$a<b$. then for$\epsilon=b-a$, which is positive, we have$a+\epsilon=b$. This contradicts what we were given. That is, we were told that$a+\epsilon$is always greater than$b$when$\epsilon>0\$, and now we’ve found a counterexample. Jan 18, 2021 at 7:11