We can already define abs(x) as sqrt(x^2) = (x^2)^(1/2).

sign(x) follows naturally as abs(x)/x.

The step function is a bit trickier, we can solve this with 0^0=1.

Begin by noting, that to construct an arbitrary step function, all we need is a function f such that:

    f(x) = -1 | x<0
    f(x) = 1  | x>= 0

By composing this basic step function, we can construct an arbitrary step function. We can also note, that sign(x) comes very close to this definition. The only problem is that sign(0)=0.

To work around this, we need a function, g, such that:

    g(x) = 1 | x=0
    g(x) = 0 | x/=0

0^x comes close, but is not defined for x<0. To work around this limitation, we can take g(x)=0^(x^2).

With this, we find that f(x) = sign(x) + 0^(x^2), satisfies our initial criteria. By composing functions of the form `a * f(x + b)`, we can construct an arbitrary step function.

"Is it possible to define a step-function in terms of this result?"

It might be possible to define it in terms of the semi-arbitrary choice of what 0^0 is, but that's generally mathematically unsatisfactory, almost certainly leading to no non-equally-arbitrary insights.