Est. 1998 — 27 years of woodworking knowledge
Separating each integer from its double.
Alex Y
The positive integers are to be partitioned into several sets such that two times an integer and the integer itself do not belong to the same set, and every positive integer belongs to exactly one set. What is the minimum number of such sets required?
Re: Separating each integer from its double.
Bill Earl
Too much negativity from last week. Getting things back on a positive track I see. 
As a prime number, the answer is odd - even though it is not.
correct
Re: Separating each integer from its double.
Larry Barrett
Put 1 in Set A.
Since 2 = 2x1, 2 can not be in Set A, so put 2 in Set B.
Then, for n greater than 2, if n=2m, put n in the set that does not contain m. If n not = 2m, put n in either Set A or Set B.
So 3 can go in either Set A or B,
4 = 2x2, and 2 is in Set B, so 4 goes in Set A
and so on.
Seems like 2 sets will do it, which is what Bill says, I think.
Correct
Alex Y
Another way of looking at it: If the power of 2 in the prime factorization of n is odd, put n in set A. If the power of 2 is even (including zero), put it in Set B.
And yes, that is what Bill was saying. I love Bill's obfuscated answers, which are often a fun puzzle in and of themselves. 