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Inequality proof

  1. Sep 24, 2010 #1
    1. The problem statement, all variables and given/known data

    Let x and y be real numbers. Prove that if x =< y + k for every positive real number k, then x =< y

    3. The attempt at a solution

    x =< y + k
    -y + x =< k
    since k is positive, the lowest value it can take doesn't include 0: -y + x < 0
    x < y

    So I get x < y from x =< y + k and not the required x =< y. Am I right or I'm screwing up somewhere? Thanks for your help.
  2. jcsd
  3. Sep 24, 2010 #2
    If x and y are equal, then [itex] x \leq y + k [/itex] for all [itex] k>0 \in \mathbb R [/tex], right? So your conclusion must not be right.

    Try this. Assume there are an x and y such that [itex] x \leq y + k [/itex] for all [itex]k>0[/itex] but [itex] x > y[/itex] and show this leads to a contradiction.
  4. Sep 24, 2010 #3
    Out of x > y I got -y + x > 0.
    So that 0 < -y + x =< k

    But that's not quite a contradiction? Or is it?
  5. Sep 24, 2010 #4


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    I haven't studied this at all, but intuitively I would say that lynchu is correct.

    If x and y are equal, then we only have x=y+k in the case that k=0, now if k>0 then x<y+k. Right?
  6. Sep 24, 2010 #5
    Well yes, but that's not the statement that's being evaluated. The theorem to be proved is:

    IF [itex] x \leq y + k [/itex] for all real [itex]k>0[/itex], THEN [itex] x \leq y [/itex].

    lynchu's claim was stronger than this; he claimed that in fact

    IF [itex] x \leq y + k [/itex] for all real [itex]k>0[/itex], THEN [itex] x < y [/itex].

    But this isn't true, because if x and y are equal, then the condition holds, but the conclusion is false.

    What you said was:

    IF [itex]x=y[/itex], THEN [itex]x<y+k [/itex] for all real [itex]k>0[/itex].

    No, not really. All you need to do is show me ONE k value so that if [itex] x \leq y + k [/itex] and [itex] x > y [/itex] you get a contradiction.
  7. Sep 24, 2010 #6


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    Oh I see, yeah now it makes sense why it should be [tex]x\leq y[/tex] :smile:
  8. Sep 25, 2010 #7
    Appreciate the help so far.
    I've spent way too much time on this and yet I just don't see it. :cry:
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