A question about neighboring fractions. I have purchased I.M. Gelfand's Algebra for my soon-to-be high school student son, but I am embarrassed to admit that I am unable to answer seemingly simple questions myself.
For example, this one:

Problem 42. Fractions $\dfrac{a}{b}$ and $\dfrac{c}{d}$ are called neighbor fractions if their difference $\dfrac{ad - bc}{bd}$ has numerator $\pm1$, that is, $ad - bc = \pm 1$.
Prove that
    (a.) in this case neither fraction can be simplified (that is, neither has any common factors in numerator and denominator);
    (b.) if $\dfrac{a}{b}$ and $\dfrac{c}{d}$ are neighbor fractions then $\dfrac{a + b}{c + d}$ is between them and is a niehgbor fraction for both $\dfrac{a}{b}$ and $\dfrac{c}{d}$; moreover, ...

Here is the snapshot from the book online (click on Look Inside on the Amazon page):

So, (a) is simple, but I have no idea how to prove (b). It just does not seem right to me. Embarrassing. Any help is appreciated.

We have to prove that for positive integers $a,b,c,d$, we either have
$$\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$$
or
$$\frac{c}{d}<\frac{a+c}{b+d}<\frac{a}{b}$$
First of all, $\frac{a}{b}=\frac{a+c}{b+d}$ is equivalent to $ab+ad=ab+bc$, hence $ad=bc$, which contradicts the assumption $ad-bc=\pm 1$. We can
disprove $\frac{a+c}{b+d}=\frac{c}{d}$ in the same manner.
In the case of $\frac{a}{b}<\frac{a+c}{b+d}$, we get $ad<bc$, hence $ad-bc=-1$. The condition $\frac{a+c}{b+d}<\frac{c}{d}$ is equivalent to $ad+cd<bc+cd$, hende $ad<bc$, which implies $ad-bc=-1$ again. So, $\frac{a}{b}<\frac{a+c}{b+d}$ is equivalent to $\frac{a+c}{b+d}<\frac{c}{d}$.
So, if we have $\frac{a}{b}>\frac{a+c}{b+d}$, we must have $\frac{a+c}{b+d}>\frac{c}{d}$ (which can also be proven directly analogue to the calculation above)
To show that the middle fraction is a neighbor-fraction to both fractions, just use the definition of neighbor-fractions.
