TITLE: Addition, Subtraction, Multiplication and Division on both sides of congruence relation
QUESTION [5 upvotes]: This post is to clear some of the doubts related to modulo arithmetic.
Addition on both sides

$a \equiv b \pmod p \iff$ $(a+k) \equiv (b+k) \pmod p$

Subtraction on both sides

$a \equiv b \pmod p \iff$ $(a-k) \equiv (b-k) \pmod p$

Multiplication on both sides

$a \equiv b \pmod p \iff$ $ak \equiv bk \pmod p$

But for division,

$a \equiv b \pmod p \iff$ $a/k \equiv b/k \pmod {p/k}$
  (if $a/k, b/k$ and $p/k$ are integers)

Is my understanding right on this? Is there an alternate form exists for division or why the general form is different for division? please help to understand this

REPLY [2 votes]: Try some numbers for division and see if you can see why.  20 = 8 mod 12 so 10 = 4 mod 12.  why does that not work? 
Note $a = b \mod n$ means $a = m*n + b$ so $a/k = (m*n)/k + b/k$ assuming $k|a; k|m*n; k|b$.  Well, we want to say that means $a/k = (m/k)*n + b/k$.  But what if $k \not  \mid m$ but $k | nm$
That's what happens with $20 = 1*12 + 8$ and $20/2 = 1*12/2 = 8/2$.  $2 \not \mid 1$ so instead we get $20/2 = 12/2 + 8/2$ so $10 = 6 + 4$ so $10 \equiv 4 \mod 6$ and not $10 \equiv 4 \mod 12$.
So division is more complicated.  Is does work sometimes and sometimes it doesn't.
For example $24 \equiv 4 \mod 5$ and $12 \equiv 2 \mod 5$.  In this case $\gcd(5,2) = 1$ so if $2|24$ and $2|4$ then $2\not \mid 5$ so if $24 = 5*k + 4$ then $\gcd(2,5) = 1$ so $2|k$ so we do get $12 = 5*(k/2) + 2$.