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How would I prove that there is a rational number between every two rational numbers ?

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    Have you tried googling this question ? – 2017-01-12

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Since a rational number can be converted to a fraction in simplest form, we can just work with fractions. So Suppose $\dfrac{a}{b} < \dfrac{c}{d}$ are two positive fractions in simplest form and that $a,b,c,d \in \mathbb{N}$, consider the fraction $\dfrac{a+c}{b+d}$. You can check that this fraction is between the other two fractions. For the case where you could have negative fractions, it can be done similarly by working on the sign.

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    What do you mean by working on the sign and why not just say that if $a, b$ are rational, then so is $\frac 12 (a+b)$? – 2017-01-12
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You prove that by taking two arbitrary rational numbers $phalfway between them (which you need to prove is rational by expressing specifically using $p$ and $q$ and algebraic / arithmetic operations, and pointing out that these operations do not break rationality).

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Let $p , q \in \mathbb{Q}$ such that $p < q$. Given $n \in \mathbb{N}$, with $n \geq 0$, then $$ p < p + \frac{q - p}{2^{n + 1}} < q\mbox{.} $$ If you assume that $r_1 + r_2 \in \mathbb{Q}$ and $2^k r \in \mathbb{Q}$ for all $r_1 , r_2 , r \in \mathbb{Q}$ and for all $k \in \mathbb{Z}$, the proof is trivial.