Schur's Theorem (Ramsey Theory)

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Theorem

Let $r$ be a positive integer.

Then there exists a positive integer $S$ such that:

for every partition of the integers $\left\{{1, \ldots, S}\right\}$ into $r$ parts, one of the parts contains integers $x$, $y$ and $z$ such that:
$x + y = z$


Proof

Let:

$n = R \left({3, \ldots, 3}\right)$

where $R \left({3, \ldots, 3}\right)$ denotes the Ramsey number on $r$ colors.

Take $S$ to be $n$.


partition the integers $\left\{{1, \ldots, n}\right\}$ into $r$ parts, which we denote by colors.

That is:

The integers in the first part are said to be colored $c_1$
The integers in the second part are said to be colored $c_2$

... and so on till color $c_r$.

Thus $\left\{{1, \ldots, S}\right\}$ has been $r$-colored.

(This terminology is common in Ramsey theory.)


Now consider the complete graph $K_n$.

Now color the edges of $K_n$ as follows:

An edge $xy$ is given color $c$ if $\left|{x - y}\right|$ was colored $c$ in the partitioning.


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From the definition of $R \left({3, \ldots, 3}\right)$ and Ramsey's Theorem, $K_n$ will definitely contain a monochromatic triangle, say built out of the vertices $i > j > k$.

Let the triangle be colored $c_m$.

Now $i - j$, $i - k$ and $j - k$ will also be colored $c_m$.

That is, $i - j$, $i - k$ and $j - k$ will belong to the same part in the partition.

It only remains to take $x = i - j$, $y = j - k$ and $z = i - k$ to complete the proof.

$\blacksquare$


An extension


The above proving technique allows to obtain a variety of similar and further going results. Here is just a sample:

THEOREM 1

Let $r$ be a positive integer.

Then there exists a positive integer $S$ such that:

for every partition of the integers $\left\{{1, \ldots, S}\right\}$ into $r$ parts, one of the parts contains integers $x$, $y$ and $z$ such that:
$x + y = z$
and:
$x \ne y$


This theorem follows from the following one:

THEOREM 2

Let $r$ be a positive integer.

Then there exists a positive integer $S$ such that: for every partition of the integers $\left\{{1, \ldots, S}\right\}$ into $r$ parts, one of the parts contains integers $a$, $b$, $a + b$, $c$, $b + c$, and $d$ such that:

$a + b + c = d$

PROOF

The proof is nearly the same as of the original Schur's theorem above, except that one uses $R \left({4, \ldots, 4}\right)$.

$\blacksquare$


Source of Name

This entry was named for Issai Schur.