Prove that s/t is rational where s and t are rational

[RM86Z]

TL;DR

given the rational numbers s and t, with t != 0, prove that s/t is rational.

Is my proof correct? The steps from hypothesis to conclusion are in order below:

1) given rational numbers s,t with t != 0
2) take s = p/1 and t = q/1 where p,q are integers
3) (p/1)/(q/1) = p/q is rational
4) therefore by substitution s/t is rational

 

[Office_Shredder]

Why are the denominators of s and and t 1? Aren't they arbitrary rational numbers?

 

[RM86Z]

Ah yes of course thanks for pointing that out - is this better?

1) given rational numbers s,t with t != 0
2) s = a/b, t = c/d for integers a,b,c,d with b,c,d != 0
3) (a/b) / (c/d) = (a/b) * (d/c) = ad / bc is a rational number as ad and bc are integers (product of integers is an integer)
4) s = a/b and 1/t = d/c and so by substitution,
5) s/t is a rational number

 

[FactChecker]

A slight improvement:
Start 3) with s/t = ... and the proof is done. Line 4) is not needed and what it says is distracting.

 

[RM86Z]

Awesome thank you

 

[PeroK]

Awesome thank you

Here's an alternative approach. First, we ask what is meant by $\frac s t$ in the first place? The most logical answer is that $$\frac s t \equiv st^{-1}$$ where $t^{-1}$ is the rational such that $tt^{-1} = 1$. And, we see that if $t = \frac c d$, then $t^{-1} = \frac d c$. And then $$\frac s t = st^{-1} = \frac{ad}{bc}$$ is rational.

This means that showing that $\frac s t$ is rational is equivalent to showing that every non-zero rational has a rational multiplicative inverse,

 

[RM86Z]

Nice one thank you PeroK