Is a < b+ε for all ε>0 true if a is not ≤ b?

[Torshi]

Homework Statement

Supopse that "a" and "b" are two numbers and that a < b+ε for all ε>0. Show that a ≤ b

Homework Equations

None given for problem

The Attempt at a Solution

See attachment of my jpeg for my attempt. I'm not sure, but i feel like it satisfies the statement

 

[haruspex]

The graph does not constitute an argument. I would not even consider drawing a graph for this question (unless instructed to). Or are you just checking you have understood the question?

 

[Torshi]

The graph does not constitute an argument. I would not even consider drawing a graph for this question (unless instructed to). Or are you just checking you have understood the question?

For problems like these, can I use real numbers to try and figure it out?

Assuming a = 5 and b + e = 6. These two values would work because 5 < 6. Since e > 0, it has to be a positive number. To make things easy, we'll pick the smallest positive integer greater than zero: 1. If e becomes equal to 1, then I can say that b is equal to 5. Now I can say that a = b, which satisfies the first part of the second equation.

"b" can be greater than "a" if I pick any number I want "b" to equal, so long as it is greater than "a" by a substantial amount.

Algebraically, I could write that

a < b when e > (b-a), and a = b when e = (b-a) ?

 

[haruspex]

This is not a question about integers, I'm sure. It's about reals. And no argument based on specific numbers could be considered a general proof.
Try reductio ad absurdum: assume a is not <= b and see if you can demonstrate this violates "a < b+ε for all ε>0".