one time a CS professor took points off an assignment (build interpreting calculator in assembly) and I had to bring in my number theory textbook to demonstrate that it is in fact the C language that does division incorrectly

C either rounds towards 0 on integer division or maybe does something implementation specific if the denominator is negative?

anyway

the "correct" definition of integer / and % is:
a / b = x, a % b = r, where
a = bx + r
0 <= r < |b|

(remainder is always positive, which requires rounding towards negative infinity)

*math is made up and you can define division however you want, unless you want algebra to work, then there's some rules you gotta follow

integer division //

real division has these nice properties: if
a / b = q, then
1: (a+b) / b = q + 1
2: (-a) / b = -(a / b) = a / (-b) = -q

how should integer division handle negative numbers? simple approach is to follow property 2 above; convert both to positive numbers, then negate the result if exactly one of the operands was negative:

5 / 3 = 1 rem 2
(-5) / 3 = -(5/3) = -1 rem -2

the problem is this breaks property 1 above:

(5 + 6) / 3 = 3 rem 2
(5 + 3) / 3 = 2 rem 2
(5 + 0) / 3 = 1 rem 2
(5 - 3) / 3 = 0 rem 2 (so far so good!)
(5 - 6) / 3 = (-1) / 3 = -(1 / 3) = -(0 rem 1) = 0 rem -1
(5 - 9) / 3 = -1 rem -1 (b/c -4/3)

(-1) / 3 == (-1 + 3) / 3 !!

to be consistent, the sequence of quotients should have gone 3 2 1 0 -1 -2, but instead we got 0 twice in a row and the remainder changed from +2 to -1!

property 2 (for real numbers) is a special case of:
a / b = q --> (ka / b) = kq
where k = -1

but that can't work with integer division in the first place!
5 / 3 = 1 rem 2
(2*5) / 3 = 10 / 3 = 3 rem 1

@octopus Quality post and the last part is a good reminder.
Even if you like something, it's good and important to know the limits of it.

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