Rules for modulus and multiplication

Question

My question is rather simple as I'm interested about modulus and multiplication, specifically whether it holds that $(a*b)\,mod\,n=(a\,mod\,n)*(b\,mod\,n)$?

Do you know the definition of "$\text{mod} ,, n$"? What have you tried so far? Please share your thoughts. — Krish, Sep 04 '17 at 06:54
Since $mod,n$ produces the remainder from division by $n$, it is obviously true for the cases where either $a$ or $b$ are perfectly divisible by $n$ as the product would also be by nature of multiplication. I'm not quite sure about the other cases, although it seems to hold. I'm practically searching for a proof of why this works(or doesn't). — , Sep 04 '17 at 07:02
It is true that $(a\cdot b) \textrm{mod}n =((a \textrm{mod}n)\cdot (b \textrm{mod}n))\textrm{mod}n$. Try $a=5$, $b=8$, $n=3$ in your equation. — TheSimpliFire, Sep 04 '17 at 07:05

score 27 · Accepted Answer · 2017-09-04T13:34:59.920

27

Not quite, the correct rule is

$$(ab\bmod m)=((a\bmod m)(b\bmod m))\bmod m$$ because the product $(a\bmod m)(b\bmod m)$ can very well exceed $m$ (it actually lies in range $[0,(m-1)^2]$).

edited Sep 04 '17 at 13:34

answered Sep 04 '17 at 07:16

Yep, many people write $=$ for $\equiv$ and forget the additional modulo. – zwim Sep 04 '17 at 08:18
How to proof this? – Clemens Bartholdy Jun 14 '21 at 15:29
@Buraian: to prove what ? That many people forget ? – Jun 14 '21 at 15:53
3

The result, I've figured out how to do it now $a=p_1 m + r_1$ and $b=p_2 m + r_2$, multiplying and taking mod:
$$ ab mod m \equiv r_1 r_2 \mod m$$

Then $r_1 = a \mod m $ , $r_2 = b \mod m $
– Clemens Bartholdy Jun 14 '21 at 15:53

zwim · Answer 2 · 2021-11-19T15:50:34.767

8

$\begin{array}{}a\pmod n\equiv \tilde a\iff a=np+\tilde a\\ b\pmod n\equiv \tilde b\iff b=nq+\tilde b\end{array}$

$ab=n(npq+q\tilde a+p\tilde b)+\tilde a\tilde b\implies ab\equiv \tilde a\tilde b\pmod{n}$

edited Nov 19 '21 at 15:50

answered Sep 04 '17 at 07:12

zwim

29,833

score 2 · Answer 3 · answered Sep 04 '17 at 10:02

2

In modular arithmetic there are two equivalent expressions: $$a \bmod n = a_1 \iff a \equiv a_1 (\bmod n)$$ $$b \bmod n = b_1 \iff b \equiv b_1 (\bmod n)$$ $$(ab) \bmod n = c_1 \iff (ab) \equiv c_1 (\bmod n)$$ According to the multiplication property (see: here): $$(ab) \equiv (a_1b_1) \equiv c_1 (\bmod n)$$ For example: $a=5,b=8,n=3$: $$5 \equiv 2 (\bmod 3)$$ $$8 \equiv 2 (\bmod 3)$$ $$(5\cdot 8) \equiv (2\cdot2) \equiv 1 (\bmod 3)$$

answered Sep 04 '17 at 10:02

farruhota

32,168

$$\because a \equiv a_1 (\bmod n) \iff a \bmod n = a_1$$ $$(ab) \equiv (a_1b_1) (\bmod n) \iff (ab) \bmod n = (a_1b_1) = (a \bmod n)(b \bmod n)$$ But, how do we get, $$(ab) \bmod n = (a_1b_1) \bmod n$$ – KGhatak Jun 05 '22 at 06:48

Krish · Answer 4 · 2017-09-04T07:52:36.043

1

Hint: $a=nr_1+s_1, b=nr_2+s_2$ where $r_1, r_2, s_1, s_2 \in \mathbb{Z}$ and $0\leq s_1< n, 0 \leq s_2 < n$. Then $$a\,\,(mod\,\, n)=?$$ $$b\,\,(mod \,\, n)=?$$ $$ab\,\, (mod\,\, n) = ?$$ Can you conclude from here?

edited Sep 04 '17 at 07:52

answered Sep 04 '17 at 07:09

Krish

7,262

score 0 · Answer 5 · answered Sep 04 '17 at 08:35

The statement

for all integers $a$ and $b$, $(ab)\bmod n=(a\bmod n)(b\bmod n)$

only holds for $n=1$ or $n=2$. The case $n=1$ is trivial, as $a\bmod 1=0$ for every integer $a$. For $n=2$ it's easy as well, because $a\bmod 2=0$ if $a$ is even and $a\bmod 2=1$ if $a$ is odd (just check the four cases).

There is already a counterexample for $n=3$: indeed $$ (2\cdot2)\bmod 3=1 \qquad (2\bmod 3)(2\bmod 3)=4 $$ For general $n>2$ the counterexample is similar: take $a=b=n-1$; then $$ (ab)\bmod n=(n^2-2n+1)\bmod n=1 $$ whereas $$ (a\bmod n)(b\bmod n)=(n-1)^2=n^2-2n+1=n(n-2)+1>1 $$

Rules for modulus and multiplication

5 Answers5