I encountered the question given below in I.M. Gelfand's book.

Is it possible that numbers $\frac{1}{2}, \frac{1}{3}, \frac{1}{5}$ are (not necessarily adjacent) terms of the same arithmetic progression? Hint: Yes. Try $\frac{1}{30}$ as a difference.

The question has already been asked here.

I was going through the answers and found this one very interesting.

According to the above answer we have to find the G.C.D in order to get the d.

Now my question is: -

We can always find the G.C.D of any given number of numbers. So whenever we encounter such question (mentioned above) we can blindly say that they are in A.P. ? or does an example exist which proves my theory wrong? [the theory being that any given random numbers (not necessarily being adjacent terms of A.P.) shall always be in A.P.]

  • 3
    $\begingroup$ Try $\pi, e$, and $1$. Unless you are referring to only using rational numbers, then in which case changes things $\endgroup$
    – WaveX
    Feb 2, 2020 at 16:55
  • $\begingroup$ @WaveX I meant rational numbers only, but now you mention about it, is it possible for irrational numbers also? $\endgroup$ Feb 2, 2020 at 16:57
  • $\begingroup$ if we have something along the lines of $\sqrt{2}, 3\sqrt{2}, 5\sqrt{2}$ then yes they can be members of an A.P. but in other cases no they cannot together be members of an A.P, it just depends on which irrational we are dealing with. $\endgroup$
    – WaveX
    Feb 2, 2020 at 17:01
  • $\begingroup$ @WaveX I would love to see a proof that this cannot be done with $\pi,e$ and $1$ :) $\endgroup$
    – Wojowu
    Feb 3, 2020 at 8:30
  • 1
    $\begingroup$ @Wojowu I believe if we are using lulu's strategy, we subtract off $1$ from each and we see the ratio $$\frac{\pi - 1}{e-1}$$. The proof comes down now to showing this ratio is irrational, which is beyond my knowledge and I suppose is something that hasn't been proven yet $\endgroup$
    – WaveX
    Feb 3, 2020 at 11:25

2 Answers 2


Say your three numbers are $a<b<c$. Subtracting $a$ from each we see, without loss of generality, that we could assume we are dealing with $0<b<c$.

Suppose these are in an arithmetic progression, and let $d$ be the associated period. Then $b=md, c=nd$ for some integers $m,n$. It follows that $\frac cb=\frac nm\in \mathbb Q$. Thus, being in progression means that the ratio is rational.

Conversely, if $\frac cb=\frac nm\in \mathbb Q$ we can take $d=\frac bm$ to obtain an arithmetic progression containing $b,c$. (since then $b=dm$ and $c=dn$).


Your theory always holds when the chosen numbers are rational. Irrational numbers will work under very specific cases. This can be demonstrated as follows-

1) Represent the three given numbers as fractions.

2) Multiply all of them with the LCM of all denominators.

3) The sequence will be in an AP with common difference as $\frac 1{LCM}$.

The thing to understand here is that the difference between the two distinct pairs of numbers will always have a common factor. This situation will not occur when you use different irrational numbers. However, if the three numbers are multiples of the same irrational number, it will work. For example $(\pi,2\pi,5\pi)$ do work out.

  • $\begingroup$ I think you messed in the example. diffs are $1+2\pi$, and $3+\pi$ which ratio is not in $\mathbb{Q}$ $\endgroup$
    – RiaD
    Feb 3, 2020 at 1:05
  • $\begingroup$ @RiaD Sorry, now I don't follow that nonsensical (deleted) comment either. $\endgroup$
    – L. F.
    Feb 3, 2020 at 3:14

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