In high-school Maths, we were taught that it was possible to manipulate ratios as fractions. For example, $$ 1 : 7 = 3 : x \\ \frac{1}{7} = \frac{3}{x} \\ \frac{x}{7} = 3\\ x = 3 \times 7\\ x = 21\\ $$ My question relates to the second line of working, where we represent $1:7$ as $\frac{1}{7}$. As I understand it, $1:7$ is better represented as one part of 8 total parts, i.e. $\frac{1}{8}$. My question, then, is this:
Is there a simple way of showing (or better still, a rigorous proof!) that a ratio-equation of the form $a:n = b:m$ can in all cases be rewritten as $\frac{a}{n} = \frac{b}{m}$ ?
It does not actually matter; if you want to do this algebraically, consider writing $$a:n=b:m$$ as a fraction in the terms that you're thinking in, i.e.: $$\frac{a}{a+n}=\frac{b}{b+m}$$ then, so long as neither $a$ nor $b$ is $0$, we can apply the function $x\mapsto \frac{1}x$ to both sides of the equation without affecting its truth, yielding $$\frac{a+n}a=\frac{b+m}{b}$$ which reduces to $$1+\frac{n}a=1+\frac{m}b$$ $$\frac{n}a=\frac{m}b$$ and applying $x\mapsto\frac{1}x$ again to both sides, assuming neither $n$ nor $m$ is zero, gives: $$\frac{a}n=\frac{b}m.$$ So, assuming none of the parts of the ratios are $0$, the fraction representation will work just the same.
You would expect this to be true since exactly what a ratio means is ambiguous; the ratio $a:n$ certainly is best interpreted as "$a$ parts of one thing per $b$ parts of another" - and you can uniquely associate a few rational numbers to such an idea. For instance, $\frac{a}b$ is the "conversion" ratio from $b$ to $a$ - i.e. if you multiply how much of the second thing you have by $\frac{a}b$, you retrieve how much of the first thing you have. Your fraction, $\frac{a}{a+b}$ represents how much of the total stuff between $a$ and $b$ is of the first thing. So, of course, you'd expect either representation to be valid where well-defined (since we're only concerned that the two fractions are equal if and only if the ratios were).