Why can't I separate the fraction inside a fraction to the outside

Question

I know when I have something of this form

\begin{equation} {a} \frac{1}{\alpha} \frac{s + \frac{1}{T}}{s + \frac{1}{\alpha T}} \end{equation} I multiply out the fractional part, α with the bottom to get

\begin{equation} {a} \frac{s + \frac{1}{T}}{\alpha s + \frac{1}{T}} \end{equation}

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But, if I instead have

\begin{equation} {a} \frac{\frac{1}{\alpha} ( s + \frac{1}{T} )}{s + \frac{1}{\alpha T}} \end{equation}

with the fractional part

\begin{equation}\frac{1}{\alpha} \end{equation} in the numerator instead.

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I know I can't take the fractional part out of the equation in this second instance. But, I don't know why that is. I know this is a fairly basic rule, but I haven't done algebra in quite a while. So, I wanted to ask for clarification.

Answer 1

I know I can't take the fractional part out of the equation in this second instance

What? Yes you can. (Technically it's "expression" and not "equation.")

In general, $\dfrac{AB}C = A\dfrac BC$, as long as $C \ne 0$ since otherwise these expressions are undefined.

In your case:

$$ a \ \frac{\frac1\alpha ( s + \frac1T )}{s + \frac1{\alpha T}} = a \ \frac1\alpha \frac{ s + \frac1T}{s + \frac1{\alpha T}} $$

Answer 2

if you have this \begin{equation} {a} \frac{\frac{1}{\alpha} ( s + \frac{1}{T} )}{s + \frac{1}{\alpha T}} \end{equation}

then it is exactly the same as \begin{equation} {a} \frac{1}{\alpha} \frac{s + \frac{1}{T}}{s + \frac{1}{\alpha T}} \end{equation}

And it is a matter of asociation: Lets call $b:= \frac{1}{\alpha},\ \ c:=s + \frac{1}{T},\ \ d:=s + \frac{1}{\alpha T}$. Then, rewriting the fractions as divisions, we have that

\begin{equation} \frac{\frac{1}{\alpha} ( s + \frac{1}{T} )}{s + \frac{1}{\alpha T}} = (b.c):d =b.(c:d) = \frac{1}{\alpha} \frac{s + \frac{1}{T}}{s + \frac{1}{\alpha T}} \end{equation}