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I stumbled upon this question on Yahoo answers here :

Prove that for every positive real number x, l/x is also a positive real number.?

Answering to the question is [now] closed, but I was thinking about the answer, when I sketched this argument below, but am not sure if am right, and whether this could be a part of the required proof (or whether this is a mere echo of obvious statements?)

if $p$ is positive real number then this is true:

$p \ge 0 \implies |p| - p = 0$

A similar argument for negative number $n$ would be:

$n \lt 0 \implies |n| - n = 2|n|$

With the above truths, I can then say:

If $x \ge 0$ then ${1 \over x} \ge 0$ since $|{1 \over x} | - {1 \over x} = {0 \over x} = 0$

Is this right?

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    @Graphth, sorry `a` was supposed to be `n`. corrected that (typo)2012-10-16

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I think perhaps what you are doing could work but there are key details missing. We don't know that $\frac{1}{x}$ is positive, that's what we're trying to prove. So, to say that $\left|\frac{1}{x}\right| - \frac{1}{x} = 0$, which is your definition of $\frac{1}{x}$ being positive, with no other justification, is assuming what you want to prove.

But, if you are allowed to use properties of absolute value, you get $\left|\frac{1}{x}\right| = \frac{|1|}{|x|} = \frac{1}{x},$ which proves that little step, since we already know $x$ is positive and thus $|x| = x$.

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What about this: x . 1/x = 1, so x and 1/x must have the same sign.

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    yes, but only when$x$is positive2012-10-16