S denotes the set of pairs (x,y) with x ≤ y, 0 < x ≤ 1 and 0 < y ≤ 1, and having the restriction that:

Each of floor (Log_{2}(1/x) ) and floor (Log_{2}(1/y) ) is even.

Find P+Q, given that that the area of the graph of S is P/Q, where P and Q are relatively prime positive integers.

Let's just do one dimension first.

What is the length of the line segment such that 0 < x <= 1?

Well, 1/x is between 1 and 2, or between 4 and 8, or between 16 and 32, etc,

So 1/x is between 1 and 1/2, or between 1/4 and 1/8, or between 1/16 and 1/32, etc.

Total length = 1/2 + 1/8 + 1/32 + ... = 2/3

Then the total area would be (2/3)^2 = 4/9 if x and y were independent.

However, x <= y, so one half the area qualifies.

Area = 2/9 = P/Q, so P+Q = 11