- Van der Waerden's theorem
**Van der Waerden's theorem**is a theorem of the branch ofmathematics calledRamsey theory . The theorem is about the basic structure of theinteger s. It is named for Dutch mathematician B. L. van der Waerden. [*B.L. van der Waerden, "Beweis einer Baudetschen Vermutung", Nieuw. Arch. Wisk. 15 (1927), 212–216.*]Van der Waerden's theorem states that for any given positive integers "r" and "k", there is some number "N" such that if the integers {1, 2, ..., "N"} are colored, each with one of "r" different colors, then there are at least "k" integers in

arithmetic progression all of the same color. The least such "N" is theVan der Waerden number "V"("r", "k").For example, when "r" = 2, you have two colors, say red and blue. "V"(2, 3) is bigger than 8, because you can color the integers from {1, ..., 8} like this:

1 2 3 4 5 6 7 8 B R R B B R R B

and no three integers of the same color form an arithmetic progression. But you can't add a ninth integer to the end without creating such a progression.

It is an open problem to determine the values of "V"("r", "k") for most values of "r" and "k". The proof of the theorem provides only an upper bound. For the case of "r"=2 and "k" = 3, for example, the argument given below shows that it is sufficient to color the integers {1, ..., 325} with two colors to guarantee there will be a single-colored arithmetic progression of length 3. But in fact, the bound of 325 is very loose; the minimum required number of integers is only 9. Any coloring of the integers {1, ..., 9} will have three evenly spaced integers of one color.

For "r" = 3 and "k" = 3, the bound given by the theorem is 7(2·3

^{7}+ 1)(2·3^{7·(2·37 + 1)}+ 1), or approximately 4.22·10^{14616}. But actually, you don't need that many integers to guarantee a single-colored progression of length 3; you only need 27. (And it is possible to color {1, ..., 26} with three colors so that there is no single-colored arithmetic progression of length 3; for example, RRYYRRYBYBBRBRRYRYYBRBBYBY.)Anyone who can reduce the general upper bound to any 'reasonable' function can win a large cash prize.

Ronald Graham has offered a prize ofUS$ 1000 for showing "V"(2,"k")<2^{"k"2}. [*Ron Graham, "Some of My Favorite Problems in Ramsey Theory", INTEGERS (The Electronic Journal of Combinatorial Number Theory), [*] The current record is due to*http://www.integers-ejcnt.org/vol7-2.html 7(2)*] , 2007, #A2.Timothy Gowers , [] who establishesTimothy Gowers , "A new proof of Szemerédi's theorem", Geom. Funct. Anal., 11(3):465-588, 2001, Preprint available at http://www.dpmms.cam.ac.uk/~wtg10/papers.html.: $V(r,k)\; leq\; 2^\{2^\{r^\{2^\{2^\{k\; +\; 9\},$

by first establishing a similar result for

Szemerédi's theorem , which is a stronger version of Van der Waerden's theorem. The previously best known bound was due toSaharon Shelah and proceeded via first proving a result for theHales-Jewett theorem , which is another strengthening of Van der Waerden's theorem.The best-known lower bound for $V(2,\; k)$ is that $V(2,\; k)\; >\; 2^k/k^epsilon$ for all positive ε. [

*cite book |title=Discrete Mathematics And Its Applications |editor=M. Sethumadhavan |last=Brown | first=Tom C. | pages=80 | chapter=A partition of the non-negative integers, with applications to Ramsey theory |authorlink= |coauthors= |year=2006 |publisher=Alpha Science Int'l Ltd. |location= |isbn=8173197318*]**Proof of Van der Waerden's theorem (in a special case)**The following proof is due to Ron Graham and B.L. Rothschild. [

*R.L. Graham and B.L. Rothschild, "A short proof of van der Waerden's theorem on arithmetic progressions", Proc. American Math. Soc. 42(2) 1974, 385-386.*]We will prove the special case mentioned above, that "V"(2, 3) ≤ 325. Let "c"("n") be a coloring of the integers {1, ..., 325}. We will find three elements of {1, ..., 325} in arithmetic progression that are the same color.

Divide {1, ..., 325} into the 65 blocks {1, ..., 5}, {6, ..., 10}, ... {321, ..., 325}, thus each block is of the form {"b" ·5 + 1, ..., "b" ·5 + 5} for some "b" in {0, ..., 64}. Since each integer is colored either red or blue, each block is colored in one of 32 different ways. By the

pigeonhole principle , there are two blocks among the first 33 blocks that are colored identically. That is, there are two integers "b"_{1}and "b"_{2}, both in {0,...,32}, such that: "c"("b"

_{1}·5 + "k") = "c"("b"_{2}·5 + "k")for all "k" in {1, ..., 5}. Among the three integers "b"

_{1}·5 + 1, "b"_{1}·5 + 2, "b"_{1}·5 + 3, there must be at least two that are the same color. (Thepigeonhole principle again.) Call these "b"_{1}·5 + "a"_{1}and "b"_{1}·5 + "a"_{2}, where the "a"_{"i"}are in {1,2,3} and "a"_{1}< "a"_{2}. Suppose (without loss of generality) that these two integers are both red. (If they are both blue, just exchange 'red' and 'blue' in what follows.)Let "a"

_{3}= 2·"a"_{2}− "a"_{1}. If "b"_{1}·5 + "a"_{3}is red, then we have found our arithmetic progression: "b"_{1}·5 + "a"_{"i"}are all red.Otherwise, "b"

_{1}·5 + "a"_{3}is blue. Since "a"_{3}≤ 5, "b"_{1}·5 + "a"_{3}is in the "b"_{1}block, and since the "b"_{2}block is colored identically, "b"_{2}·5 + "a"_{3}is also blue.Now let "b"

_{3}= 2·"b"_{2}- "b"_{1}. Then "b"_{3}≤ 64. Consider the integer "b"_{3}·5 + "a"_{3}, which must be ≤ 325. What color is it?If it is red, then "b"

_{1}·5 + "a"_{1}, "b"_{2}·5 + "a"_{2}, and "b"_{3}·5 + "a"_{3}form a red arithmetic progression. But if it is blue, then "b"_{1}·5 + "a"_{3}, "b"_{2}·5 + "a"_{3}, and "b"_{3}·5 + "a"_{3}form a blue arithmetic progression. Either way, we are done.A similar argument can be advanced to show that "V"(3, 3) ≤ 7(2·3

^{7}+1)(2·3^{7·(2·37+1)}+1). One begins by dividing the integers into 2·3^{7·(2·37 + 1)}+ 1 groups of 7(2·3^{7}+ 1) integers each; of the first 3^{7·(2·37 + 1)}+ 1 groups, two must be colored identically.Divide each of these two groups into 2·3

^{7}+1 subgroups of 7 integers each; of the first 3^{7}+ 1 subgroups in each group, two of the subgroups must be colored identically. Within each of these identical subgroups, two of the first four integers must be the same color, say red; this implies either a red progression or an element of a different color, say blue, in the same subgroup.Since we have two identically-colored subgroups, there is a third subgroup, still in the same group that contains an element which, if either red or blue, would complete a red or blue progression, by a construction analogous to the one for "V"(2, 3). Suppose that this element is yellow. Since there is a group that is colored identically, it must contain copies of the red, blue, and yellow elements we have identified; we can now find a pair of red elements, a pair of blue elements, and a pair of yellow elements that 'focus' on the same integer, so that whatever color it is, it must complete a progression.

It should be noted that the proof for "V"(2, 3) depends essentially on proving that "V"(32, 2) ≤ 33. We divide the integers {1,...,325} into 65 'blocks', each of which can be colored in 32 different ways, and then show that two blocks of the first 33 must be the same color, and there is a block coloured the opposite way. Similarly, the proof for "V"(3, 3) depends on proving that

: $V(3^\{7(2\; cdot\; 3^7+1)\},2)\; leq\; 3^\{7(2\; cdot\; 3^7+1)\}+1$

By a double induction on the number of colors and the length of the progression, the theorem is proved in general.

**References****External links*** [

*http://www.math.uga.edu/~lyall/REU/ramsey.pdf Proof of Van der Waerden's theorem*]

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