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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?

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  • $\begingroup$ your title and body don't match $\endgroup$
    – RE60K
    Oct 16 '14 at 12:07
  • $\begingroup$ What do you mean? $\endgroup$
    – Tim H UK
    Oct 16 '14 at 12:08
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    $\begingroup$ If you can show $\dfrac{a}{b} < \dfrac{ad+bc}{2bd}$ and $\dfrac{ad+bc}{2bd} < \dfrac{c}{d}$ you have a proof but you have to prove both statements. $\endgroup$ Oct 16 '14 at 12:18
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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 $b, d>0$.

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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.

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  • $\begingroup$ Well I wrote down what I would do to find a fraction between two other fractions. So I need to show why it is always true? $\endgroup$
    – Tim H UK
    Oct 16 '14 at 12:13
  • $\begingroup$ That would be the main part of a proof by construction, showing why what you had built would work. $\endgroup$
    – kitbeard
    Oct 16 '14 at 12:14
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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.

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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 $\frac{1}{2}(r+s)\in\mathbb Q$?

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