Proving Equality of Real Numbers: Hints & Solutions

[n1person]

Homework Statement

Prove that the real numbers a and b are equal if and only if for each positive real e, the absolute value of a-b satisfies abs(a-b)<e

Homework Equations

The main one I am thinking about is the fact that if a<=b and b<=a then a=b, also the whole sigma thing might mean the archimedean property might come into effect (if r and s are positive ration numbers then there exists a positive integer N such that rN<s).

The Attempt at a Solution

I am tried a proof by contradiction... but generally it falls apart or requires too much non-rigorous work :-/

Any ideas/hints/help?
Thanks!

 

[n1person]

okay; so I have a new idea after another half hour of work, i sort of feel though the last step is flawed;

proof by contradiction

a=b iff abs(a-b)>e
for some e>0

abs(a-b)>e means either
a-b>e or -(a-b)>e
if a-b>e then a>e+b
and if -(a-b)>e then b>e+a

however from these two points it is "obvious" that either a>b or b>a, but is there some theorem which actually proofs that?

 

[snipez90]

First of all, the iff means you have to show that both implications are true.

The forward implication is that if a = b, then for every e > 0, |a-b| < e. This should be evident.

The reverse implication is that if for every e > 0, |a-b| < e, then a = b. To do a proof by contradiction, you have to negate this last statement first. The negation of 'if p then q' for statements p and q is 'p and not q'. Assume the 'p and not q' condition and see if you can find a contradiction.