1. ## Analysis

Have a question:

Suppose a, b can be any real number and for every e greater than 0, a is less than or equal to b +e. Prove that a is less than or equal to b.

It makes sense, just thinking about it, but how do you actually prove it?

MK

2. Say it is false, then a>b.
Then a-b>0. So let e=a-b.
See what happens when you use the e in the given.

3. I still don't see where the contradiction occurs.

4. You have,
a<b+e
then,
a-b<e
If,
a>b then, a-b>0
Say,
c=a-b
then,
c<e
c>0
thus
0<c<e
but for any positive integer c
we can choose a smaller e, i.e. e=c/2
thus,
0<c<e is not true for all e.

5. Originally Posted by MKLyon
I still don't see where the contradiction occurs.
Let e = a - b > 0 (the last step is by assumption of a > b.)

Thus b + e = b + (a - b) = a. (Which holds for the condition in the problem statement.)

Now, e is positive so there is some real number e' such that 0 < e' < e.

The problem statement says that a =< b + e' also, since any positive real number will do.

But e' < e implies b + e' < b + e.

Thus b + e' < a, contrary to the problem statement.

Thus a is not greater than b.

What can you do to show that a is not equal to b?

-Dan

6. We are given that for each e>0, a<b+e.
If it were the case that a>b then let e=a-b>0.
By the given a<b+e=b+(a-b)=a.
The above says a<a. That is the clear contradiction.