I've been trying to find a way to simplify $\frac{|x|}{x}$ if $x$ is real and $\neq{0}$. The two possible outcomes to this are $\pm{1}$ but I believe there is one required answer. I've noticed that if x is positive, we will have +1, and if x is negative, we will have -1. However, I'm unable to simplify the equation such that we have such an outcome.
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$\begingroup$I suppose what you're looking for is:
$$\frac{|x|}{x}=\left\{\begin{array}+1&\text{if}&x>0\\-1&\text{if}&x<0\end{array}\right..$$ This function is sometimes written $\text{sgn}(x)$.
$\endgroup$ $\begingroup$An alternative answer of $\quad \text{sgn}(x)\quad$ is $\quad 2\left(H(x)-\frac12\right)$.
$H(x)$ is the Heaviside step function :
$\endgroup$ $\begingroup$I don't quite get what you mean by "required answer". You can just write: $$\frac{|x|}{x} = \begin{cases} 1, \mbox{if } x > 0 \\ -1, \mbox{if } x < 0\end{cases}$$ Or, write $\frac{|x|}{x} = \mathrm{sgn}(x)$, where $\mathrm{sgn}: \Bbb R \setminus \{0\} \to \{-1, 1\}$ is the sign function.
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