Math 308 Lecture 6 - Divisibility and Congruence of Integers

Errata:

@ 1:09:00 Division in modular arithmetic is more difficult and nuanced than I said it is, and we cannot assume that the same congruence mod n holds by dividing both sides by 2. We can divide both sides of a congruence ka = kb (mod n) to get a = b (mod n) only if gcd(k,n) = 1. Otherwise, we may have to change the n in order to keep congruence. It is also the case that division by zero ("0") is not the only concern, we also have to worry about dividing by any number that is equivalent to zero mod n. My modular arithmetic is more rusty than I thought :p

@38:43
The remainder should be strictly less than the divisor not the dividend. Thanks to Em that pointed this out in the comments.

[tobiasksr]

Love your classes, professor. Thank you.

[Em-iezg]

@38:43
The remainder should be strictly less than the divisor not the dividend.
Since we are dividing b by n, then ∃! (q_1, r_1 ∈ ℤ) with 0 ≤ r_1 < n, such that, b = nq_1 + r_1.