<s>[INST] Why is $\frac{1}{4/3} - \frac{1}{3/2}$ not the same as $\bigl(\frac{4}{3} - \frac{3}{2}\bigr)^{-1}$ If you have the problem:$$\frac{1}{\frac{4}{3}} - \frac{1}{\frac{3}{2}} =?$$
Why can't you change the problem to $(\frac{4}{3} - \frac{3}{2})^{-1}$   and get the same answer?
In the first scenario, the answer is $1/12$
In the second scenario, the answer is $-6$
Why doesn't this work?
 [/INST] You are basically asking why $\frac{1}{a} + \frac{1}{b} \neq \frac{1}{a+b}$. Take $ a = b = 2$, for example. You would expect that adding a half plus a half would get you back to one, i.e. $\frac{1}{2}+\frac{1}{2} = 1$, and certainly not a quarter, $\frac{1}{2+2} = \frac{1}{4}$.
Let's say you have some function $f(x)$ that takes some number $x$ and spits out a different number $y$. Let's take $f(x) = 3\cdot x$. If we put in a number that is a sum of two other numbers, for example, $x = a+b$, then we have that $f(x) = f(a+b) = f(a) + f(b)$. Same if $x$ is some multiple of a different number, $x = 4\cdot a$, namely $f(x) = f(4\cdot a) = 4f(a)$. These properties only hold because we choose our $f(x)$ to be very specific, i.e. multiplication by $3$. In general they do not have to hold however! For example if $f(x) = x^2$, or $f(x) = \frac{1}{x}$.
You are probably too used to functions behaving in the way $f(x) = 3\cdot x$ did, but if you look carefully almost everything that you can come up with will not have these nice properties.
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