When taking cases for let's say $|x|$ we say :
It is $x$ for all $x \geq 0$ And
It is $-x$ for all $x<0$
I don't understand why we have "or equal to" in the first case
Why can't we have $\leq$ for the second case ?
Does this imply zero is positive?
This is something which I have blindly learnt without understanding why, hopefully I can change that.
Thanks in advance guys.
On
When you define a function piecewise, so to say $$f(x)=\begin{cases} g_1(x)&\text{if }\mathcal A(x)\\ g_2(x)&\text{if }\mathcal B(x)\\ g_3(x)&\text{if }\mathcal C(x)\end{cases},$$
it is good practice to divide the cases in such a way that conditions $\mathcal A(x)$, $\mathcal B(x)$ and $\mathcal C(x)$ are mutually exclusive. In principle, there is nothing wrong with having them not be mutuallly exclusive, but then one would need to check that, for instance, $g_1(x)=g_2(x)$ for all $x$ such that $\mathcal A(x)\land \mathcal B(x)$. Otherwise we are not defining a function. This necessity makes the exposition unnecessarily messy, other than being a cause of errors for people whose strong suit is the neglect of details.
In the instance of $\lvert x\rvert$, there is no reason to put the case $x=0$ with $x>0$ rather than with the other: for all that matters, you could even put it by itself and say that $$\lvert x\rvert=\begin{cases}x&\text{if }x>0\\ \arcsin x&\text{if }x=0\\ -x&\text{if }x<0\end{cases},$$ though it's better to have fewer and simpler cases.
Because $0 = -0$, the following ways of writing $|x|$ are all equivalent: \begin{align} |x| &= \begin{cases} -x, &x<0\\ x, & x\ge 0 \end{cases} \tag{1}\\ |x| &= \begin{cases} -x, &x\le 0\\ x, &x \ge 0 \end{cases}\tag{2}\\ |x| &= \begin{cases} -x, &x \le 0\\ x, &x > 0 \end{cases}\tag{3} \end{align}
Some authors/instructors may prefer their domains of definition to be disjoint, and so prefer either $(1)$ or $(3)$, but that's just a matter of preference.