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

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The discussion centers on proving that the quotient of two rational numbers, s and t (with t not equal to zero), is also rational. The initial proof presented was refined to clarify that both s and t can be expressed as fractions of integers, leading to the conclusion that their division results in a rational number. An alternative approach emphasizes understanding the expression s/t as a multiplication of s by the multiplicative inverse of t, reinforcing that the inverse of a non-zero rational number is also rational. The conversation highlights the importance of clarity in mathematical proofs and the equivalence of different methods to demonstrate the same conclusion. Overall, the proof successfully establishes that the quotient of two rational numbers is 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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