“Conversion” between dimensionally incompatible physical quantities I am writing a program that takes two terms like 60 mph and mm/s and converts the first to the second, assuming the units are compatible. (It does this by first converting both to pure SI units.) But my question is, what exactly makes units compatible?
Initially, I assumed it was if the unit dimensions were equivalent, but that excludes seemingly valid conversions like m/s to s/m, which still make good physical sense. These types of conversions can be done if the difference in power of the units is applied to the coefficient:
$$ 5 \text{ m s}^{-1} \equiv 0.2 \text{ s m}^{-1}$$
But, following the same sort of logic, should
$$ 5 \text{ m} \equiv 25 \text{ m}^2$$
make good sense, since $ \text{m} = ({\text{m}^2})^{1/2} $, and $ 5 \text{ m} = ({25\text{ m}^2})^{1/2} $ ?
Furthermore, would it make sense to convert $ \text{m/s}^2 $ to $ \text{m}^{1/2}/\text{s} $, or something similar? Should I allow this as a valid conversion?
$$ \begin{split}
9 \text{m/s}^2 &=  (3\text{ m}^{1/2}/\text{s})^2 \\
\therefore 9 \text{m/s}^2 &\equiv 3\text{ m}^{1/2}/\text{s}
\end{split} $$
I guess I'm struggling to see where physical context breaks down and nothing but abstract (meaningless) math is left… where is this 'line'? What counts as a meaningful conversion and what doesn't?
Thanks!
 A: For an equation to make sense, or, in technical term, dimensionally compatible, each term in it has to have the same dimension, or, in your words, signature of units. So your equation 
$$5 \text{ m s}^{-1} \equiv 0.2 \text{ s m}^{-1}$$
does not make sense. Although you can say, "Travelling five meters for each second is the same as spending a fifth of a second for each meter." However, you cannot write it as the above equation. "Equivalent" statement does not always end up with "equation". 
To see whether each term has the same dimension, usually we look at the the quantity in dimensions $M,L,T$. Here $M$ represents mass, $L$ represents length, $T$ represents time. There are more, but these are the often-used ones.
For your above equation, the left hand side has dimension $LT^{-1}$ and the right hand side $TL^{-1}$. This is dimensionally incompatible. 
This is the topic of dimension analysis. 
A: To put it plainly, two measure units are (dimensionally) compatible if they measure the same thing. This is the object of study of dimensional analysis. Quoting the Wikipedia page:

The SI standard recommends the usage of the following dimensions and corresponding symbols: mass $[M]$, length $[L]$, time $[T]$, electrical current $[I]$, absolute temperature $[\Theta]$, amount of substance $[N]$ and luminous intensity $[J]$.

In your question you are asking if
$$
[LT^{-1}] = [L^{-1} T]
$$
which is not the case. On the other hand, you could indeed get the same information from either saying that you are covering $5$ meters every second or that you are taking $0.2$ seconds to cover a meter, since you are measuring two reciprocal quantities.
A nice example of this is that in the US car fuel efficiency is usually measured in miles per gallon, while in Europe it is often measured in litres per $100$ kilometers. You can find a humoristic (though accurate) interpretation of this at the end of this blog post. (note that, again, these are reciprocal measures)
A: Very interesting question. This not-quite-answer is too long for a comment. 
I would interpret your assertion that $5 m s^{-1}$ and $0.2 m^{-1}s$ are equivalent (you carefully and correctly don't call them equal) as meaning that they convey the same information about the underlying physical situation you are modeling. In that situation the variables "distance" and "time" are real valued functions of the physical state once you've chosen units for those quantities. Then since 
$$
 \left( 5 m s^{-1} \right)^{-1} =  0.2 m^{-1}s 
$$
knowing the velocity is the same as knowing its inverse.
You can make a similar argument for your second seemingly wrong example. There the "physical system" is the set of all squares. If you know the numerical value   of the side of a square (in, say, meters) then you know its area (in square meters).
You will certainly enjoy the answer to this question at https://what-if.xkcd.com/11/

If you went outside and lay down on your back with your mouth open,
  how long would you have to wait until a bird pooped in it?

