We know that for any two numbers $a$ and $b$, if $a\le b$, then $a=b$ or $a<b$
I want to prove that the converse is true, i.e. if: $$a=b\implies a\le b\tag{1}$$ and in particular if: $$a<b \implies a\le b\tag{2}$$
When the condition (1) or (2) is satisfied, we know that the non-strict inequality holds, since only 1 condition must be satisfied.
How can one prove it in a strict way?
Precisely, I was during the process to prove that if:
$$\lvert a\rvert\le b \space\space\space\text{then}\space\space -b\le a \le b$$
For $a\ge0$, proof works nicely and I have the inequality.
Problem occurs when I consider case, when $a<0$:
If $a<0$, then $\lvert a\rvert=-a\le b$, so $a\ge-b$.
Since $b\ge0$ (because $\lvert a\rvert\ge0$) and $a<0$, then $a<b$
(because if $a<b$ and $b<c$ then $a<c$)
But I need to prove that $a\le b$.
Obviously I can consider the cases for $a\ge 0$ and $a\le 0$, but I would like to arrive at satisfactory answer for $a<0$.
Incredibly, it's the definition. Given an order of type $>$, which is a strict partial order, you can define another one $\geq$, which is a non-strict partial order, as $a\geq b$ if and only if $a > b $ or $a = b $. This way, you ensure that all the expected properties are valid, as the one you are trying to prove, but it's valid by definition.