p, q are be rational numbers with p < q and q ≠ 0. Prove that ${\frac{3}{p - q}}$ is rational.
By definition: "n is a rational number if and only if n = a/b for some integers a, b, with b ≠ 0:
So it doesn't matter what p - q are as long as it isn't 3? I'm not sure how to translate the p-q part into an argument for the proof.
On
To prove that it is a rational number you must prove that it can be presented as a fraction of two integers. That’s trivial. I mean you already have a fraction there. Just simplify a bit. Say, if $p$ is rational, we can write it as $m/n$. Because $q$ is also rational we can write it as $t/s$. Here $m,n,t,s$ are all integers and $n,s\neq0$. That can be done by definition (a rational number is defined as a number that can be written as a fraction of two integers, where a denominator is not zero). So you have $3/(m/n-t/s)=3ns/(ms-tn)$ So we arrived at a more complicated fraction whose numerator and denominator are also integers (denominator $\neq0$ from the get-go). That concludes the proof. It's trivial. For me just saying it's rational is obvious. You can do it as Jonah recommended -- it's all trivial in my opinion. It's better to focus on more difficult proofs.
As we know, the difference of two rational numbers must be rational. If we therefore assign $p-q$ as a rational number $r$, we simplify this to $$ \frac{3}{r} $$ Which satisfies the condition that a number is rational if it can be written as the quotient of rationals, as $r$ was established to be rational.