1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Analysis Equality

  1. Sep 7, 2010 #1
    1. The problem statement, all variables and given/known data

    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

    2. Relevant 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).

    3. 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?
  2. jcsd
  3. Sep 7, 2010 #2
    okay; so I have a new idea after another half hour of work, i sorta 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?
  4. Sep 7, 2010 #3
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook