# Math Help - Prove: Let x >=0, x is real number such that any(Epsilon) e > 0, x <= e. Prove x = 0

1. ## Prove: Let x >=0, x is real number such that any(Epsilon) e > 0, x <= e. Prove x = 0

Would someone please give me some help with my case 2 below?

The problem: Let x $\ge$ 0, x $\in$ $\Re$ such that for any $\varepsilon$ > 0, x $\le$ $\varepsilon$. Prove that x = 0

Solution:

Case 1: Let x > 0 and $\varepsilon$ = $\frac{x}{2}$

then

$\varepsilon$ = $\frac{x}{2}$ > 0 because x > 0

since $\varepsilon$ > 0 by given, x $\le$ $\varepsilon$

Case 2: Let x = 0 and $\varepsilon$ = $\frac{x}{2}$

then

$\varepsilon$ = 0/2 = 0

since $\varepsilon$ = 0 = x, I cannot conclude x $\le$ $\varepsilon$

unless I let $\varepsilon$ = x + y (where y $\in$ ${\Re ^ + }$). But can I do that? Some thing does not look right in my opinion, but I am stuck and unsure.

Thanks,

2. ## Re: Prove: Let x >=0, x is real number such that any(Epsilon) e > 0, x <= e. Prove x

Originally Posted by Mathlv
Case 2: Let x = 0 and $\varepsilon$ = $\frac{x}{2}$

then

$\varepsilon$ = 0/2 = 0

since $\varepsilon$ = 0 = x, I cannot conclude x $\le$ $\varepsilon$

unless I let $\varepsilon$ = x + y (where y $\in$ ${\Re ^ + }$). But can I do that? Some thing does not look right in my opinion, but I am stuck and unsure.

Thanks,
Why do you have a case 2? There are two possibilities. Either x=0 or x>0. First, you showed that if x>0, you arrive at a logical contradiction because there exists an $\varepsilon>0$ such that $x>\varepsilon>0$. Hence x is not greater than 0. That only leaves one possibility. If x=0 and $\varepsilon>0$, then $x\le \varepsilon$ is true.

3. ## Re: Prove: Let x >=0, x is real number such that any(Epsilon) e > 0, x <= e. Prove x

SlipEternal,

Thank you so much for your help. I am understood now. It was a logical contradiction for the case of x > 0; therefore, that case was false. That left x = 0 is true for x <= espilon.