Is what I've done a proof? Proving there is always an rational number between two distinct rational numbers 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?
 A: 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 $b, d>0$.
A: How did you construct that? If you can show clearly why it is always the case, then you have a proof by construction / proof by example. But you'll need to show it more clearly than you have here.
A: I don't agree with kitbeard that you need to show how you constructed the answer. For a proof you need to show that your answer is indeed a solution. How you came up with it, might be curious, but it does not need to be part of the proof.
If you are showing this for a mathematician (not in your exam) this is enough. If you are in a high-school class/exam or explaining it to a person who has difficulty with mathematics, you would need a couple of more sentences to explain why each of the inequalities is true.
A: Hint:
I cannot recognize a proof in what you wrote down in your question.
Are you able to prove that $r,s\in\mathbb Q$ implies that $\frac{1}{2}(r+s)\in\mathbb Q$?
