In "Fundamentals of Electrical Engineering" we learned about piecewise functions for the "unit-step" and "ramp" which are represented by $f(x)= \begin{cases}0, & \text{if }x< 0 \\ 1, & \text{if }x>0\end{cases}$ and $f(x)= \begin{cases}0, & \text{if }x< 0 \\ x, & \text{if }x\ge 0\end{cases}$ respectively. I was bored in calculous class and determined these functions could be represented in traditional algebra by $f(x)= \frac{|x|}{x} \cdot \frac12 + \frac12$ and $f(x)= \frac{x + |x|}2$ So this got me thinking. Can any piecewise function be represented as a traditional equation? Just as an example, how about this one: $$f(x)= \begin{cases} 5, & \text{if }x=0 \\ x^2, & \text{if }x<0 \\ \sqrt{x} & \text{if }x>0 \end{cases}$$
edit: For the sake of the question, substitute $\sqrt{x^2}$ for $|x|$
The crucial step is to come up with an acceptable way to describe indicator functions, i.e. for certain subsets $S\subseteq \mathbb R$ to replace the piecewise definition $$1_S(x)=\begin{cases}1&\text{if }x\in S\\0&\text{if }x\notin S\end{cases} $$ with something not involving piecewise, but only "traditional" definitions. Provided that taking limits is allowed as "traditional", we should accept the functions $$\begin{align}\max\{x,y\} &=\frac{x+2}{2}+\left|\frac{x-y}{2}\right|\\ \min\{x,y\} &=x+y-\max\{x,y\}\\ 1_{[0,\infty)}(x)&=\lim_{n\to\infty}\min\{e^{nx},1\} \\ 1_{(-\infty,0]}(x)&=1_{[0,\infty)}(-x)\\ 1_{[a,b)}(x) &=1_{[0,\infty)}(x-a)-1_{[0,\infty)}(x-b)\\ 1_{[a,b]}(x) &=1_{[0,\infty)}(x-a)\cdot1_{[0,\infty)}(b-x)\\ 1_{\{a\}}(x) &=1_{[0,\infty)}(x-a)\cdot 1_{[0,\infty)}(a-x)\end{align}$$ and similar combinations for arbitrary intervals $\subseteq \mathbb R$. With these you get for example $$\begin{align}f(x)&= \begin{cases} 5, & \text{if }x=0 \\ x^2, & \text{if }x<0 \\ \sqrt{x} & \text{if }x>0 \end{cases}\\& = 1_{\{0\}}(x)\cdot 5+1_{(-\infty,0)}(x)\cdot x^2+1_{(0,\infty)}(x)\cdot\sqrt x.\end{align} $$ Or, just to make sure, you may want to replace $\sqrt {x}$ with $\sqrt{1_{[0,\infty)}(x)\cdot x}$ (otherwise you'd need a convention that $0$ times undefined is $0$).