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): enter image description here

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.

  • 5
    $\begingroup$ Isn't it $\frac{a+c}{b+d}$ instead of $\frac{a+b}{c+d}$ ? $\endgroup$
    – Peter
    Jan 3, 2017 at 0:50
  • 5
    $\begingroup$ Guys, are you saying there is a typo in the book? My goodness and I thought I was stupid for not being able to prove it. $\endgroup$
    – mark
    Jan 3, 2017 at 1:30
  • 1
    $\begingroup$ Please convert the image to actual text. $\endgroup$
    – jpmc26
    Jan 3, 2017 at 11:27
  • $\begingroup$ Not easy. I have the actual hard copy book, this one is from a pdf I found on the Internet, but it is a pdf capturing the picture of the text, so it is not selectable and copyable. $\endgroup$
    – mark
    Jan 3, 2017 at 14:19
  • $\begingroup$ BTW, I checked the Russian origin - no typo there. So this is a translation issue. $\endgroup$
    – mark
    Jan 11, 2017 at 4:12

4 Answers 4


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.

  • 2
    $\begingroup$ Or I can post it on SO and thanks to all the good Samaritans out there the truth came out very quickly. $\endgroup$
    – mark
    Jan 3, 2017 at 5:11

Just so you know, it helps a good deal if at least one parent goes through the same exercises as the child. It may be hard; however, purchasing a book may not be enough. Given your profile, be aware that mathematics and computer science are really not the same thing. The example that comes up over and over again on this site is that of equivalence relations, especially the integers modulo $m$ for some $m.$ If the student is only able to think in terms of the computer command n % m, important nuances are missed.

I can add this: given neighbor fractions, and a positive integer $k,$ the fraction $$ \frac{a + kc}{b + kd} $$ is a neighbor to at least one of them (cannot remember, probably the second) and lies between them. In fact, if you also had $m,$ similar outcome for $$ \frac{ma + kc}{mb + kd} $$

The main examples of neighboring fractions are sequences called Farey fractions. https://en.wikipedia.org/wiki/Farey_sequence

Another example, not quite the same, is the convergents of simple continued fractions. Continued fractions: convergence of fraction expansion

$$ \begin{array}{ccccccccccccccccccccccc} & & 1 & & 1 & & 2 & & 1 & & 1 & & 4 & & 1 & & 1 & & 6 & \\ \frac{0}{1} & \frac{1}{0} & & \frac{1}{1} & & \frac{2}{1} & & \frac{5}{3} & & \frac{7}{4} & & \frac{12}{7} & & \frac{55}{32} & & \frac{67}{39} & & \frac{122}{71} & & \frac{799}{465} \\ \end{array} $$


We have to prove that for positive integers $a,b,c,d$, we either have




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.

  • $\begingroup$ This seems unnecessarily complicated. I have not done it, but I expect the calculation showing the new fraction to be a neighbour of the original ones to also yield signs and hence show that it between them. $\endgroup$
    – Carsten S
    Jan 3, 2017 at 12:38

There appears to be a slight misunderstanding in the above discussion of betweenness in the Gelfand exercise.

The fact that $a+c\over b+d$ is between $a\over b$ and $c\over d$ does not depend on the two fractions being neighbor fractions. It is true for any (positive) fractions (as stated by Gelfand himself on the previous page) and is easily proven algebraically without assuming the fractions are neighbors (a stipulation which only complicates the proof).

An outline of such an algebraic proof would be:

If ${a\over b}\lt {c\over d}$, then $ad\lt bc$. Add $ab$ to both sides, and it follows that ${a\over b}\lt{a+c\over b+d}$. The other half of the betweenness follows similarly.

This betweenness property also follows immediately from a logical view of averaging. If a group of $b$ people has $a$ apples, and a group of $d$ people has $c$ apples, merging the groups will give each person in the new merged group an average number of apples which benefits the people in the "poorer" group and hurts the people in the "richer" group. [Note that Gelfand makes this very argument on the previous page, to explain the betweenness property.]

So in Gelfand's part (b), the fact of betweenness is true always, while the fact that the resulting fraction is a neighbor to both original fractions depends, of course, from the assumption that the original fractions were neighboring.



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