# Is this minus in the proof by substitution a typo?

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?

• Did they mention what are $k$ and $j$? Given congrunces, it depends on how you relate $a,a'$ and $b,b'$ Sep 11, 2021 at 12:58
• I updated the post. The proof has defined: $a = mj + a'$ and $b=mk + b'$ same as I have
– Jim
Sep 11, 2021 at 13:10
• You have expanded $-(a'-b')$ as $-a'-b'$. Oops! Sep 11, 2021 at 13:20
• @GerryMyerson: That was a typo, Corrected it
– Jim
Sep 11, 2021 at 13:26
• The mysterious minus sign is just a typo in the book. It should be +. Sep 12, 2021 at 18:52