Let $a,b \in \mathbb{R}$, I must proof:

$ \forall \epsilon \in \mathbb{R}^{>0}(a< b+\epsilon) \to a \leq b$

Proof by contradiction: I have by negation of thesis "$a>b$ (or $b \leq a \wedge a \neq b$), but if $a>b$ then $a-b>0$ and by hypothesis $a<b+(a-b)=a$ therefore $a<a$ is absurd! Is correct?

Thanks in advance!

P.S.= $ \forall \epsilon \in \mathbb{R}^{>0}(a< b+\epsilon) \to a < b$ is true?


1 Answer 1


Maybe the problem is: You are given two real numbers $a,b$ such that for all $\epsilon > 0$, you have $a < b+ \epsilon$, then prove that $a\leq b$.

In this case your solution is correct. So you can assume by contradiction that $a > b$. Then $a - b > 0$, and so $b + (a-b) = a < a$ is the contradiction.

  • $\begingroup$ I edited my post, is correct? $\endgroup$
    – mle
    Dec 30, 2013 at 17:00
  • 1
    $\begingroup$ @Soviet: Not quite. I think you want $(\forall \epsilon \in \mathbb{R}^{>0}, a< b+\epsilon )\to (a \leq b)$ or something like that. $\endgroup$
    – Thomas
    Dec 30, 2013 at 17:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .