# Inequality proof

1. Sep 24, 2010

### lynchu

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. Sep 24, 2010

### hgfalling

If x and y are equal, then $x \leq y + k$ for all $k>0 \in \mathbb R [/tex], right? So your conclusion must not be right. Try this. Assume there are an x and y such that [itex] x \leq y + k$ for all $k>0$ but $x > y$ and show this leads to a contradiction.

3. Sep 24, 2010

### lynchu

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. Sep 24, 2010

### Mentallic

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?

5. Sep 24, 2010

### hgfalling

Well yes, but that's not the statement that's being evaluated. The theorem to be proved is:

IF $x \leq y + k$ for all real $k>0$, THEN $x \leq y$.

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

IF $x \leq y + k$ for all real $k>0$, THEN $x < y$.

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 $x=y$, THEN $x<y+k$ for all real $k>0$.

No, not really. All you need to do is show me ONE k value so that if $x \leq y + k$ and $x > y$ you get a contradiction.

6. Sep 24, 2010

### Mentallic

Oh I see, yeah now it makes sense why it should be $$x\leq y$$

7. Sep 25, 2010

### lynchu

Appreciate the help so far.
I've spent way too much time on this and yet I just don't see it.