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

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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
 
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Why are the denominators of s and and t 1? Aren't they arbitrary rational numbers?
 
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
 
A slight improvement:
Start 3) with s/t = ... and the proof is done. Line 4) is not needed and what it says is distracting.
 
Awesome thank you
 
RM86Z said:
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,
 
Nice one thank you PeroK
 
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