[thedufer]

The laws given in the post were explicitly derived from the equation. Please explain how you used that to determine f(0, -1)?

[Strilanc]

Pick any `b` you want, even infinity, and evaluate:

    sum_n^b 2^n (floor(0/2^n) mod 2) (floor(-1/2^n) mod 2)

    sum_n^b 2^n * 0 * 1

    sum_n^b 0

    0

[thedufer]

Excellent. I'm glad we agree. You'll notice that the stated identity f(x,x) = x fails to hold over x < 0. I hope this is sufficient to make it clear that the equation does not consider the negatives?

[Strilanc]

Seems to work fine for -1:

    f(-1, -1)
    = sum_{n=0}^b 2^n (floor(-1/2^n) mod 2) (floor(-1/2^n) mod 2)
    = sum_{n=0}^b 2^n * 1 * 1
    = 1 + 10 + 100 + 1000 + ... [in binary]
    = ...11111 [in the 2-adics]
    = -1