If a set N
_{9} = {1, 2, 3, 4, 5, 6, 7, 8, 9} of 9 numbers is split into two subsets, then at least one of them contains three terms in arithmetic progression.
The statement is not true for a set N
_{8} of 8 integers.
Seems obvious?
Prove it.
(In reply to
By hand by Jer)
Good reasoning.
..there are four possibilities where this subset has no APs. In each case the other subset has an AP in its 4 numbers instead...
Absolutely correct, however the last digit in the 4th line should be 7, not 9.