I am reading a proof about showing that if $a \equiv a' \pmod m$ and $b\equiv b' \pmod m$ then $a + b \equiv a' + b'\pmod m$ and $ab \equiv a'b'\pmod m$.
The proof uses substitution to show that (for the case of addition) the difference $(a + b) - (a' + b')$ is divisible by $m$. By substitution the proof shows that the above expression is $m(j -k)$ which is a multiple of $m$. With the same approach it shows that for the case of multiplication $ab - a'b'$ is equivalent to $m(ka + kb -jkm)$.
The proof has defined: $a = mj + a'$ and $b=mk + b'$
Now the question I have is if that $-$ (minus) in the expression after the substitution is a typo or it serves some specific convention. Because in the case of the addition we have:
$(a + b) - (a' + b') \Leftrightarrow (mj + a' + mk + b') - a' - b' \Leftrightarrow mj + mk \Leftrightarrow m(j + k)$
So I don't understand why the proof says: $m(j -k)$. I understand that we could just define $k = -z$ and consider they are the same expression but I don't understand why we need to do that. Same for the multiplication.
Is this a typo or am I misunderstanding something?
If a is congruent to a' (mod m) then a= a'+ km for some integer k. If b is congruent to b' (mod m) then b= b'+ jm for some integer j. ab= (a'+ km)(b'+ jm)= a'b'+ kmb'+ a'jm+ kjm^2= a'b'+ (kb'+ a'j+ kjm)m. That is, ab is equal to a'b' plus some integer multiple of m so ab is congrjent to a'b' (mod m).