Inequality proof

  • Thread starter lynchu
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  • #1
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Homework Statement



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

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.
 

Answers and Replies

  • #2
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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.
 
  • #3
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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?
 
  • #4
Mentallic
Homework Helper
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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 [itex] x \leq y + k [/itex] for all [itex] k>0 \in \mathbb R [/tex], right?

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?
 
  • #5
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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?

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].


lynchu said:
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?

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.
 
  • #6
Mentallic
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Oh I see, yeah now it makes sense why it should be [tex]x\leq y[/tex] :smile:
 
  • #7
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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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