# Check my proof of an algebraic statement about fractions

## I tried to prove the part c) of "Problem 42" from the book "Algebra" by Gelfand.

Fractions $$\frac{a}{b}$$ and $$\frac{c}{d}$$ are called neighbor fractions if their difference $$\frac{cb-ad}{db}$$ has numerator ±1, that is, $$cb-ad = ±1$$. Prove that:

b) If $$\frac{a}{b}$$ and $$\frac{c}{d}$$ are neighbor fractions, then $$\frac{a+c}{b+d}$$ is between them and is a neighbor fraction for both $$\frac{a}{b}$$ and $$\frac{c}{d}$$.

Me: It is easy to prove.

c) No fraction $$\frac{e}{f}$$ with positive integer $$e$$ and $$f$$ such that $$f < b+d$$ is between $$\frac{a}{b}$$ and $$\frac{c}{d}$$.

Me: We know that $$\frac{a+c}{b+d}$$ is between $$\frac{a}{b}$$ and $$\frac{c}{d}$$. The statement says that if we make the denominator smaller than $$b+d$$, the fraction can't be between $$\frac{a}{b}$$ and $$\frac{c}{d}$$ with any numerator.

Let's prove it:

1. Assume that $$\frac{a}{b}$$ < $$\frac{c}{d}$$, and $$cb-ad = 1$$, ($$cb = ad + 1$$). I also assume that $$\frac{a}{b}$$ and $$\frac{c}{d}$$ are positive.

2. Start with the fraction $$\frac{a+c}{b+d}$$, let $$n$$ and $$m$$ denote the changes of the numerator and denominator, so we get $$\frac{a+c+n}{b+d+m}$$ ($$n$$ and $$m$$ may be negative). We want it to be between the two fractions:$$\frac{a}{b} < \frac{a+c+n}{b+d+m} < \frac{c}{d}$$

3. Let's see what the consequences will be if the new fraction is bigger than $$\frac{a}{b}$$: \begin{align*} \frac{a+c+n}{b+d+m} & > \frac{a}{b}\\ b(a+c+n) & > a(b+d+m)\\ ba+bc+bn & > ba+ad+am\\ bc+bn & > ad+am \end{align*}

But $$bc = ad + 1$$ by the definition, so \begin{align*} (ad + 1) + bn & > ad + am\\ bn - am & > -1 \end{align*}

All the variables denote the natural numbers, so if a natural number is bigger than $$-1$$, it implies that it is greater or equal to $$0$$:$$bn - am \geq 0$$

1. Let's see what the consequences will be if the new fraction is less than $$\frac{c}{d}$$: \begin{align*} \frac{a+c+n}{b+d+m} & < \frac{c}{d}\\ ...\\ cm - dn & \ge 0 \end{align*}

2. We've got two equations, I will call them p-equations, because they will be the base for our proof (they both have to be right):$$\begin{cases} bn - am \ge 0\\ cm - dn \ge 0\end{cases}$$

3. Suppose $$\frac{a}{b} < \frac{a+c+n}{b+d+m} < \frac{c}{d}$$. What $$n$$ and $$m$$ have to be? It was conjectured that if $$m$$ is negative, so for any $$n$$ this equation would not be right. Actually if $$m$$ is negative, $$n$$ can be only less or equal $$0$$, because when the denominator is getting smaller, the fraction is getting bigger.

4. Suppose that $$m$$ is negative and $$n = 0$$. Then the second p-equation can't be true:$$-cm - d\cdot 0 \ge 0 \implies -cm \ge 0$$

5. If both $$n$$ and $$m$$ are negative, the p-equations can't both be true. I will get rid of the negative signs so we can treat $$n$$ and $$m$$ as positive: $$\begin{gather*} \begin{cases}(-bn) - (-am) \ge 0\\ (-cm) - (-dn) \ge 0\end{cases}\\ \begin{cases}am - bn \ge 0\\ dn - cm \ge 0\end{cases} \end{gather*}$$

If something is greater or equal $$0$$ then we can multiply it by a positive number and it still will be greater or equal $$0$$, so multiply by $$d$$ and $$b$$: $$\begin{gather*} \begin{cases}d(am - bn) \ge 0\\ b(dn - cm) \ge 0\end{cases}\\ \begin{cases}da\cdot m - dbn \ge 0\\ dbn - bc\cdot m \ge 0\end{cases} \end{gather*}$$

But $$bc$$ is greater than $$da$$ by the definition. You can already see that the equations can't both be true, but I will show it algebraicly:

By the definition $$bc = da + 1$$, then $$\begin{gather*} \begin{cases}dam - dbn \ge 0\\ dbn - (da + 1)m \ge 0\end{cases}\\ \begin{cases}dam - dbn \ge 0\\ dbn - dam - m \ge 0\end{cases} \end{gather*}$$

If two equations are greater or equal $$0$$, than if we add them together, the sum will still be greater than or equal to $$0$$. \begin{align*} (dam - dbn) + (dbn - dam - m) & \ge 0\\ -m & \ge 0 \end{align*}

It is impossible (I changed $$n$$ and $$m$$ from negative to positive before by playing with negative signs).

QED.

If $$n$$ and $$m$$ are positive, the p-equations can both be true, I won't go through it here because it is irrelevant to our problem. But it is the common sense that I can choose such big $$n$$ and $$m$$ to situate $$\frac{a+c+n}{b+d+m}$$ between any two fractions.

PS: Maybe my proof is too cumbersome, but I want to know is it right or not. Also advices how to make it simpler are highly appreciated.

Tiny detail, maybe even a typo: $n$ and $m$ are integers, not necessarily naturals, so the statement at the end of (2) needs to reflect that. But for integers, also $x>-1$ implies $x\geq 0$, so it's not a big deal.
You didn't really need to make $n$ and $m$ positive since the only place you use positivity is at the very, very end you need $m>0$ to derive the contradiction. You don't even use it when you multiply by $d$ since that relied on the expressions being positive and not the individual numbers themselves. This is the only place I can imagine really see simplifying the proof.
As it stands, it would make the reader more comfortable if you named the positive versions of $(n,m)$ as $(N,M)$ or $(n',m')$ or something.