The exercise I am working on is about proving whether there is always a rational number between two other distinct rational numbers.
I came up with this
$\frac{a}{b} < \frac{ad + bc}{2bd} < \frac c d$
But is what I've written a proof, or is it just an algorithm?
Best Answer
Your observation is not a proof, but can be the central part of a proof. Any proof description depends on the mathematical sophistication of your audience so we might have two different proofs:
Proof 1 (intended for a mathematician). The average of any distinct reals lies strictly between the two and the average of any two rationals is rational.
Proof 2 (intended for my students). First, we show that for any distinct real numbers, their average lies strictly between them. Let $r_1 < r_2$ be real numbers. Then $$ r_1+r_2<r_2+r_2=2\,r_2 \quad\text{ so }\quad \frac{r_1+r_2}{2}< \frac{2\,r_2}{2}=r_2 $$ In a similar way we can show $r_1<(r_1+r_2)/2$ and hence have established the first claim.
Now let $r_1, r_2$ be rational numbers. By definition, there exist integers $a, b, c, d$ with $b, d\ne 0$ for which $r_1=a/b, r_2=c/d$. Then $$ \frac{r_1+r_2}{2}=\left(\frac{a}{b}+\frac{c}{d}\right)\frac{1}{2}=\frac{ad+bc}{2bd} $$ but the rightmost expression is the quotient of two integers with $2bd\ne0$, and so the average of two rationals is by definition rational. This, with the above result, shows there is a rational number strictly between any two distinct rationals.
By the way, a slightly different alternative is to use $$ \text{If }\frac{a}{b}<\frac{c}{d}\text{ then }\frac{a}{b}<\frac{a+c}{b+d}< \frac{c}{d} $$ assuming $b, d>0$.