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?

  • $\begingroup$ What is $a$???? Do you mean $|n| - n = 2|n|$? And, how do you know $\left|\frac{1}{x}\right| - \frac{1}{x} = 0$ if you don't know that $\frac{1}{x}$ is positive? You're assuming what you want to prove. If you know properties of absolute value, then since you already know $x$ is positive and thus $|x| = x$, we have $\left|\frac{1}{x}\right| = \frac{|1|}{|x|} = \frac{1}{x}$. $\endgroup$ – Graphth Oct 16 '12 at 12:43
  • $\begingroup$ I guess he meant $|n|-n=2|n|$ for $n<0$. $\endgroup$ – coffeemath Oct 16 '12 at 12:47
  • $\begingroup$ @Graphth, sorry a was supposed to be n. corrected that (typo) $\endgroup$ – nemesisfixx Oct 16 '12 at 13:00

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 $|x| = x$.


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

  • $\begingroup$ note that this would hold even for negative $x$. Our task is to prove $1 \over x$ as positive. $\endgroup$ – nemesisfixx Oct 16 '12 at 13:21
  • 1
    $\begingroup$ yes, but only when x is positive $\endgroup$ – Chris Card Oct 16 '12 at 13:41

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