Epsilon form of inequalities

Stub grade: C

This page is a stub

This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:

This is not an important page. It was created so anyone unsure of what it was supposed to be from the title can see what the statement is. The proof of it is very easy

Statement

For [ilmath]a,b\in\mathbb{R} [/ilmath] then:

• [ilmath]\left(\forall\epsilon>0[a<b+\epsilon]\right)\iff(a\le b)[/ilmath]

Proof

• To prove [ilmath]\implies[/ilmath] it's easiest to use the contrapositive
• To prove [ilmath]\impliedby[/ilmath] is straightforward.

Grade: F

This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.

Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:

Extremely easy

This proof has been marked as an page requiring an easy proof