You want to show that (x-y) > e implies (x-y) >= 0 is false for any e < 0.
Let e < 0. Let y be a real number, and let x = y + e/2. Then x - y = (y + e/2) - y = e/2 > e, but e/2 < 0. Doesn't this work?
I have a trivial question which is rather obvious but I need a short and concise proof. I need a constructive proof and not just a counterexample. I have constructed a proof by contradiction myself but I think there are shorter ways of doing this. I would be greatful for any help!
I want to show that the implication:
(x-y) > e implies (x-y) => 0
cannot be true for any e<0.
x any y are two scalar with support on the real line
But that says that
(x-y) > e and (x-y) >= 0 together with x = y + e/2 cannot mean that e < 0. I need something more general that doesn't impose any structure on x or y.
I have a proof like:
Assume (x-y)>e and e=>0 but (x-y)<0. Combining inequalities gives -e>(y-x)>0 or equivalently e<0 which is contradiction.
Could you please use >= for "greater than"? The symbol => is usually used for "implies", so it's confusing to have it used for something else. Better yet, use LaTeX.
The very nature of proving an implication false means that you must show that there is a truth assignment of TRUE for the antecedent, as well as a truth assignment of FALSE for the consequent. That is the only way to show that an implication is false, because that's the only row in its truth table with the implication is false. An implication is true if either the antecendent is false, or the consequent is true.
Hence, you must show that there is a situation in which (x-y) > e, but (x-y) < 0. That is, you must show that there exists x and y such that e < (x-y) < 0. But surely if e < 0, this is a possibility. That is, there's enough space in-between e and 0 to insert a number (x-y). However, it will not be a possibility for just any x and y. Therefore, you are simply going to have to use a particular x and a particular y. That's still a constructive proof, and it will show that the implication is false.