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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?

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

1 Answer 1

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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).

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    $\begingroup$ Your answer is not readable, please use MathJax. $\endgroup$ Sep 11, 2021 at 14:26
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    $\begingroup$ This is not related to what I am asking in the post though $\endgroup$
    – Jim
    Sep 11, 2021 at 19:33

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