Uniform distribution presentations seem incomplete I'm studying the book "A practical guide to quantitative finance interviews." This is how it presents the discrete uniform distribution:

I've seen several other sources present it the same way (http://www.math.wm.edu/~leemis/chart/UDR/PDFs/Discreteuniform.pdf, )
This presentation relies on $\Delta x= 1$. If $\Delta x = 1$, then the number of possible values for $x$ is $b-a+1$, i.e., the difference between the max and min plus 1. But what if $\Delta x \neq 1$, then the number of points is no longer the difference between the max and min plus 1. Wouldn't a better presentation make $P(x)$ and $Var(x)$ a function of $a, b, n$, where n is the number is possible $x$ values? Is there a particular reason why this is not done? 
For example, in this case, say $\Delta x = 0.5$, so $x = a, a+0.5, a+1, \cdots, b$, then $P(x) \neq \frac{1}{b-a+1}$, but instead, $P(x) = \frac{1}{2b-2a+1}$
 A: You are right - there is no particular reason to make this convention.
In general we can derive the expectation and variance (and higher order moments) by using centering and scaling arguments from the uniform distribution on $\{0, \ldots, n\}$. I present this below.
As becomes clear - the calculations do not simplify that well, and likely if the sources you are reading only use the case of unit steps, then presenting the result in this context allows for much simpler formulae.

If $X \sim \text{Unif}(0, \ldots, n)$ then the transformed variable
$$Y = a + \frac{b-a}{n} X$$
will be uniform on $\{a, a + \frac1n\, \ldots, b - \frac1n, b\}$.
One can show that
$$\mathbf E X = \frac{n}{2}, \qquad \text{Var}(X) =  \frac{(n+1)^2 - 1}{12},$$
(both follow from the formulae for $\sum_{k=0}^n k$ and $\sum_{k=0}^n k^2$).
And subsequently centering/scaling of expectations and variances give:
$$ \mathbf E[Y] = a + \frac{b-a}{n} \mathbf E[X] = \frac{b + a}{2},$$
and
$$
\begin{align*}
\text{Var}(Y) & = \left(\frac{b-a}{n}\right)^2 \text{Var}(X) \\
& = \left(b-a\right)^2\frac{(n+1)^2 - 1}{12n^2}
\end{align*}$$
In the case of unit steps, $n = b-a$, and the above simplifies to the original formula you gave.
