If every real number is coloured either black or white, prove that there exist distinct real numbers $a,b,c$ all of the same colour such that $a-b=b-c$.

  • $\begingroup$ @rand al'thor But what if 1 is the only black number? :( Now the "distinct" is cramping my style. $\endgroup$
    – Curmudgeon
    Jul 15, 2015 at 11:54
  • $\begingroup$ If every real number is involved somewhere in that infinity there must be three that fit the rule. $\endgroup$
    – Bob
    Jul 15, 2015 at 20:23
  • $\begingroup$ @Bob Yep, that's restating the problem. Proof? :-) $\endgroup$ Jul 15, 2015 at 22:42
  • $\begingroup$ I dunno infinity is pretty big :P Maybe if I just looked at the first ten integers ;) BTW does this work for imaginary and complex numbers? $\endgroup$
    – Bob
    Jul 15, 2015 at 22:53
  • $\begingroup$ @Bob - You need only look at the first nine integers, and regardless of whether you start with 0 or 1. Any nine complex numbers in arithmetic progression will also show the property. $\endgroup$
    – h34
    Jul 15, 2015 at 23:07

5 Answers 5


WLOG, 2 and 4 are both black. (This is WLOG because we can scale and translate the real line freely, and can also swap the colors.)

If 0 is black, then

$(a, b, c) = (4, 2, 0)$ are all black.
This satisfies $4-2=2-0$.

If 3 is black, then

$(a, b, c) = (4, 3, 2)$ are all black.
This satisfies $4-3=3-2$.

If 6 is black, then

$(a, b, c) = (6, 4, 2)$ are all black.
This satisfies $6-4=4-2$.

If none of ${0,3,6}$ is black, then

$(a, b, c) = (6, 3, 0)$ are all white.
This satisfies $6-3=3-0$.

  • 1
    $\begingroup$ You're a quick solver, Mr White-square :-) $\endgroup$ Jul 15, 2015 at 12:05
  • 1
    $\begingroup$ For completeness, the proof that scaling and translating does not affect the answer should probably be included. $\endgroup$ Jul 15, 2015 at 19:03
  • 2
    $\begingroup$ Please do not write at that level of completeness. It makes it too hard for the reader. $\endgroup$
    – Brian
    Jul 15, 2015 at 21:21

Say that $a$ and $b$ are black, with $a<b$. Then, there are three points $x$ such that $\{a,b,x\}$ in some order is an arithmetic progression:

  • $\bf 2a-b$, $a$, $b$
  • $a$, $\bf \frac{a+b}{2}$, $b$
  • $a$, $b$, $\bf 2b-a$

If any of the three is also colored black, we're done. If all are colored white, we're also done because they themselves form an arithmetic progression $2a-b, \frac{a+b}{2}, 2b-a$.

If there are not two black points, do the same argument with white points. In fact, because this argument applies to every pair of same-colored points, there are an uncountable number of solutions.

  • $\begingroup$ I like this answer as it is a better generalization than simply say "without loss of generality" without providing proof that it doesn't lose generality. $\endgroup$ Jul 15, 2015 at 20:20
  • $\begingroup$ Shouldn't one of the 2b-a at the end of the second paragraph be 2a-b? $\endgroup$
    – Togashi
    Jul 15, 2015 at 20:38
  • $\begingroup$ @Togashi Good catch, fixed it. $\endgroup$
    – xnor
    Jul 15, 2015 at 20:43

Van der Waerden's theorem tells us that if we color a sufficiently long interval of integers with $c$ colors, there will exist a same-color arithmetic progression of length $l$. Since the reals contain a copy of the integers, which contain every finite interval of integers, they contain the required arithmetic progression.

The problem asks us for the special case $c=2, l=3$.


Not the first answer, but a bit of a different spin:

For it to not happen,

  • 3 numbers with the same difference between them must not have the same color (in every sequence of 3 consecutive numbers, one color should be seen twice and the other once).

  • Numbers can't be colored abab... all the time, so we have to include aas on occasion. Starting with an aa in the middle, we get axxbaabxxa, then axbbaabbxa and finally aabbaabbaa (no escape). If it's impossible even for those countable painted points, then it's true for all real numbers by extension.

    Edit: Before downvoting it, note that the unit difference can be any real number and two real numbers don't have to be located "next to" each other (so to speak) to be consecutive or for the question to be solved.

  • $\begingroup$ How do you define "consecutive" for the reals? Given any two real numbers, there are an infinite number of reals between them, so you can't have consecutive reals. $\endgroup$ Jul 15, 2015 at 17:52
  • $\begingroup$ Doesn't have to be "consecutive reals", the difference can be any real number. Let's say we're starting with multiples of r (which can be any positive or negative real number). Even then it applies. If that satisfies you, can you please take back your downvote? $\endgroup$
    – Nautilus
    Jul 15, 2015 at 18:07
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    $\begingroup$ @GentlePurpleRain I think Nautilus's idea is to prove that there exist three integers $a,b,c$ as required, which is sufficient and which f'''s answer also proves. I was also about to downvote, before realising that this is a sensible answer! $\endgroup$ Jul 15, 2015 at 18:17
  • $\begingroup$ Thanks! I shouldn't have used the word "consecutive" at all though. $\endgroup$
    – Nautilus
    Jul 15, 2015 at 18:21

It is sufficient to colour the integers between 1 and 9 inclusive. If each one is coloured white or black, there will be three of them $a,b,c$ all of the same colour such that $a−b=b−c$.

This is because Van der Waerden's $W(2,3)$ is 9.

The result holds for every set of nine real (or complex) numbers in arithmetic progression.

  • $\begingroup$ I proved by brute force that nine consecutive integers were sufficient, but then saw @xnor's answer above. $\endgroup$
    – h34
    Jul 15, 2015 at 22:00
  • $\begingroup$ Your first paragraph is covered by the accepted answer, your second by xnor's second answer. What does this post add to the rest of the page? $\endgroup$ Jul 15, 2015 at 22:21
  • $\begingroup$ I doubt that everyone who reads the accepted answer or all of the other answers will deduce that the result holds for every colouring of nine consecutive integers but not for every colouring of eight, although it is true they could work it out or look it up. $\endgroup$
    – h34
    Jul 15, 2015 at 23:02

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