This came up in a C++ program I am writing for iterating over a loop, where it was pointed out that the for loop condition a * b > c could overflow. I am rusty with integer math but I think, mathematically ignoring overflow, a * b > c is equivalent to a > c // b, so the second form can be used to avoid overflow in C++ code. For the sake of clarity, I will use the python syntax of / means rational division and // means integer division:
Is this correct? The definition of floor I use is for any real x, floor(x) is the integer satisfying floor(x) <= x < floor(x) + 1.
The same question applies for whether a * b >= c is equivalent to a >= c // b. The forward argument I think holds, but backwards fails for a=2, b=2, c=5, since 2 >= 5 // 2 but NOT 2 * 2 >= 5 as Mark Glisse points out in the comments. In that case, how can a * b >= c be safely tested without overflow?
The > case is obvious if you phrase c//b as being “the largest integer which, when multiplied by b, does not exceed c”. The >= case can be handled by simply rewriting it as a*b > c-1, reducing it to the previous.
All this for positive b, of course, since dividing by a negative changes the direction of the inequalities. (Also, in C++ dividing by negative integers isn’t floor division and dividing by -1 can itself overflow.)