TITLE: A question about neighboring fractions.
QUESTION [13 upvotes]: 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.

REPLY [11 votes]: As @Peter noticed, (b) is a typo and should be $\frac{a+c}{b+d}$.  To show that the original isn’t correct, you would just need to find a counter example.  One such example is $a=1$, $b=2$, $c=3$, and $d=5$.  Then $\frac{a}{b}=\frac{1}{2}$ and $\frac{c}{d}=\frac{3}{5}$ are neighbor fractions, since $ad-bc=-1$, but $\frac{a+b}{c+d}=\frac{3}{8}$ is not between the original fractions (it’s less than both) and is not a neighbor fraction to either of the original fractions (the specified differences are 2 and 9).
The take-away message for you is: if you are having trouble proving what seems to be a simple statement (especially if it doesn’t seem right to you), try out a few examples to see if you can see what is happening and why it’s true.  In this case, you likely would have found a counter example to the statement (the above was the first case of neighbor fractions I tried where $b\neq c$), and a counter example would have meant that a proof was impossible.
Once you know the original isn’t correct, finding the correct version is optional and more difficult.  You could try googling for the book’s errata (I was unsuccessful).  You could also try rearranging the $\frac{a+b}{c+d}$ to get a correct statement (on the assumption that the original must have been close to the truth), and then see if you can prove a statement that matches a correct example.  Or you could decide that since you don’t know what the correct statement actually is, it may as well be anything from the trivially easy to the impossibly hard, and you can put down your pencil.