Proof of Modulo Equation

Question

I'm taking a cryptography class and we are going over proofs and modulus equations for hashing. On our homework assignment, one of the questions is to prove the following...

(ab) mod m = [(a mod m)(b mod m)] mod m

I understand that a = b mod m means that m | (a - b) and that m | (a - b) by definition means that a = b + mk but I'm struggling to make the connection on how to prove the above equation and the professor's slides and notes aren't very helpful as this is far more difficult than the examples he's done in class.

Any help or guidance is much appreciated!

That's the usual definition of multiplication mod $m$! What definition do you have? — Bernard, Feb 10 '17 at 23:46

score 2 · Accepted Answer · answered Feb 10 '17 at 23:54

2

This is a just do it type question.

Let $a \equiv a' \mod m$ and $b \equiv b' \mod m$. That means $a = a' + k m$ for some integer $k$ and that $b = b' + jm$ for some integer $j$.

So just do it.

$ab = (a' + km)(b'+jm) = a'b' + b'km + a'jm + jkm^2$

$= a'b' + m(b'k + a'j + jkm)$ so

$ab = a'b' + m(b'k + a'j + jkm)$ "for some integer $b'k + a'j + jkm$".

So $ab \equiv a'b' \mod m$

or $ab \equiv (a \mod m)(b \mod m) \mod m$

answered Feb 10 '17 at 23:54

fleablood

130,341

so the entire statement of b′k+a′j+jkm boils down to a single constant? – m_callens Feb 11 '17 at 00:00
A single integer, yes, I'm not sure "constant' is the right word but yeah. – fleablood Feb 11 '17 at 00:05
@fleablood, I had the same question. Why does it deduce to an integer? – Math and ML Jun 18 '19 at 01:22
Because each of $b',k,a',j$ and $m$ are all integers. Products and sums of integers are integers. – fleablood Jun 18 '19 at 19:18

Adren · Answer 2 · 2017-02-10T23:56:59.320

1

if

$a=mp+r$ with $0\le r<m$

and

$b=mq+s$ with $0\le s<m$

then

$ab=m(mpq+ps+qr) + rs$

so that

$(ab)$ mod $m$ = $rs$ mod $m$ = [($a$ mod $m$)($b$ mod $m$)] mod $m$

edited Feb 10 '17 at 23:56

answered Feb 10 '17 at 23:48

Adren

8,184

score 1 · Answer 3 · answered Feb 10 '17 at 23:57

There's a math mod and then there's a mod used in computer programs, a function of two integer variables which returns an integer, sometimes called the remainder.

In math terms, $x=y \bmod m$ means there is some integer $z$ such that $x=y+zm$, i.e., equality mod $m$ is a relation rather than a function. Thus, you want to prove is that if $u=a\bmod m$ and $v=b\bmod m$, then $ab = uv \bmod m$. Easy: write $u=a+km$, $v=b+lm$, so $uv=ab+vkm+ulm+klm^2$. Because $vkm=0\bmod m$, $ulm=0\bmod m$, and $klm^2=0 \bmod m$, you get $uv=ab\bmod m$ as desired.

Proof of Modulo Equation

3 Answers3