Prove that $a\le|a|$ and $-a\le|a|$?

Question

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.

Answer 1

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

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

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

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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.

Answer 2

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.

Answer 3

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 <y\implies x+w <y+w $

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

Answer 4

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

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