0
$\begingroup$

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?

  • 0
    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 $\left|\frac{1}{x}\right| = \frac{|1|}{|x|} = \frac{1}{x}$.2012-10-16
  • 0
    I guess he meant $|n|-n=2|n|$ for $n<0$.2012-10-16
  • 0
    @Graphth, sorry `a` was supposed to be `n`. corrected that (typo)2012-10-16

2 Answers 2