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I need to prove that absolute value of any real number is greater than or equal to that real number, where $|a| = a ; a\ge0 , |a| = -a ; a<0 $

I came across this on real analysis. I need this proven Filed and Order Axioms and basic definitions.

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    Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? – 2017-02-07
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    Well, all numbers are either positive 0 or negative while absolute value is positive or zero so..... – 2017-02-07
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    What is your definition of absolute value? – 2017-02-07

4 Answers 4

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  • $a\ge0\implies -a\le0\le a\implies -a,a\le a=|a|$.
  • $a\le0\implies a\le0\le-a\implies -a,a\le-a=|a|$.

Alternatively, you can establish that the defintion of the absolute value is equivalent to

$$|a|=\max(a,-a),$$ from which the claim follows.

Alternatively,

$$a^2-a^2=|a|^2-a^2\ge0$$

then

$$(|a|-a)(|a|+a)\ge0$$ which is equivalent to

$$(|a|\ge a\land |a|\ge-a)\lor(|a|\le a\land |a|\le-a).$$

but the second clause of the or is false as the absolute value cannot be negative.

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What is $|a|$? It is either $a$ itself, when $a\geq0$, or $-a$, when $a\leq0$. Just separate in cases, $a\leq0$ and $a\geq0$, and substitute the absolute value by these.

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Definition of absolute value is $|a|=a $ if $a \ge 0$. $|a|=-a $ if $a <0$.

If $a\ge 0$ then $a \le a =|a|$.

If $a < 0$ then we must prove $a \le -a =|a|$

$a < 0$. By axiom $x

That should do it for you. (Let $x=a;y=0; $ and $w =???? $)

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$$|a|-a= \begin{cases} 0 & a\ge0\\ 2|a| & a\le0 \end{cases} $$

So $2|a|\ge|a|-a\ge 0$ and $|a|\ge-a$ from the left inequality, and $|a|\ge a$ from the right inequality.