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

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- September 17th 2006, 11:04 AMMKLyonAnalysis
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 - September 17th 2006, 12:07 PMPlato
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. - September 17th 2006, 12:30 PMMKLyon
I still don't see where the contradiction occurs.

- September 17th 2006, 12:38 PMThePerfectHacker
You have,

a<b+e

then,

a-b<e

Make a contradiction,

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. - September 17th 2006, 12:44 PMtopsquark
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 - September 17th 2006, 01:25 PMPlato
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.