Binary search on array of size 8 for the first element.

is_number_in_array(0, {0, 1, 2, 3, 4, 5, 6, 7}, 8)
  is_number_in_array(0, {0, 1, 2, 3},           4)
    is_number_in_array(0, {0, 1},               2)
      is_number_in_array(0, {0, 1},             1)

We get an answer only when we reduce the array to size 1 (or 0).

On the Mth invocation of is_number_in_array(…), it gets an array of size
N / 2^(M-1).

N / 2^(M-1) = 1   # that is point where we get an answer, so this M tells us how
                  # many invocations we need to search the array.

N = 2^(M-1)
log₂ N = M - 1
M = log₂ N + 1