Proof Inequality: x ≤ y | Homework Statement

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Homework Help Overview

The discussion revolves around proving the inequality involving two real numbers, x and y, specifically that if x is less than or equal to y plus any positive real number k, then x must also be less than or equal to y. Participants are exploring the implications of this statement and the conditions under which it holds true.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are examining the relationship between x and y under the given conditions, questioning whether the conclusion drawn from the inequality is valid. There are attempts to derive contradictions from assuming x is greater than y, and discussions about the implications of equality between x and y.

Discussion Status

The conversation is ongoing, with some participants expressing confusion about the implications of the inequality and the conditions under which it holds. There is a recognition that the original poster's conclusion may not be correct, and guidance has been offered to explore specific cases that could lead to contradictions.

Contextual Notes

Participants note that the problem involves proving a conditional statement and are grappling with the nuances of equality and inequality in the context of real numbers. There is an emphasis on the need for clarity regarding the assumptions made in the proof.

lynchu
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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.
 
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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.<br /> <br /> 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.[/itex]
 
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?
 
I haven't studied this at all, but intuitively I would say that lynchu is correct.

hgfalling said:
If x and y are equal, then [itex]x \leq y + k[/itex] for all [itex]k>0 \in \mathbb R [/tex], right?[/itex]
[itex] <br /> 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?[/itex]
 
Mentallic said:
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.
 
Oh I see, yeah now it makes sense why it should be [tex]x\leq y[/tex] :smile:
 
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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