0

I am currently stuck trying to prove that LHS = RHS,

such that for LHS:

$$ab \equiv (b\pmod m)\cdot(a\pmod m)$$

and RHS:

$$[(a\pmod m)\cdot(b\pmod m)] \equiv ab\pmod m$$

I understand that:

$a\equiv b\pmod m$ is by definition equivalent to $m\mid(a−b)$.

$a\equiv b\pmod m$; $a,b\in\mathbb{Z},m\in\mathbb{Z}^+$ iff $a\pmod m \equiv b\pmod m$.

and using:

$a = mq_1 + r_1$

$b = mq_2 + r_2$.

homosapien
  • 4,157
DrMichaelMorbius
  • 51

1 Answers1

1

$\textbf{Hint:}$ Since $(b \mod m )\cdot (a \mod m) \equiv ba \mod m$, use this and the fact that $m \mid 0$.

homosapien
  • 4,157
  • wont there be an error when m| 0 or am i thinking this the wrong way? – DrMichaelMorbius Jun 28 '22 at 23:07
  • 1
    Makes sense, so once i show that m | 0, how do I then prove that LHS = RHS to be equivalent? – DrMichaelMorbius Jun 28 '22 at 23:11
  • you've just shown $ab \equiv ab \mod m \equiv a \mod m \cdot b \mod m$ since $m \mid ab - ab$. – homosapien Jun 28 '22 at 23:11
  • You're done by transitivity of $\equiv$ since you know $a \mod m \cdot b \mod m \equiv ab \mod m \equiv ab$ and $m \mid ab - ab$, this shows the two sides are equivalent. – homosapien Jun 28 '22 at 23:13
  • OK, but im still concern of m|0 because any number divides by zero is undefined. – DrMichaelMorbius Jun 28 '22 at 23:24
  • See here : https://math.stackexchange.com/questions/1591697/why-zero-is-a-multiple-of-every-integer-but-not-a-divisor-of-zero – homosapien Jun 28 '22 at 23:26
  • 1
    $\textit{divides}$ does not mean we are dividing it by that number, it means that number times some integer is your given number. I.e., $a$ divides $b$ means there exists some integer $k$ such that $ak=b$. Every integer divides $0$ because $0$ times any integer is $0$. And $0$ is an integer. – homosapien Jun 28 '22 at 23:26
  • see here also : https://math.stackexchange.com/questions/2419900/prove-that-every-number-divides-0-while-the-only-number-that-0-divides-is-itself – homosapien Jun 28 '22 at 23:32