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SO basically I am learning number theory chapter in the Discrete Maths. Now I really don't understand a proof here. Picture SO basically it is a corollary which says

Let m be a positive integer and let a and b be integers. Then
(a + b) mod m = ((a mod m) + (b mod m)) mod m
and
ab mod m = ((a mod m)(b mod m)) mod m.

Now the proof goes something like this.

Proof: By the definitions of mod m and of congruence modulo m, we know that a ≡
(a mod m) (mod m) and b ≡ (b mod m) (mod m). Hence, Theorem 5 tells us that
a + b ≡ (a mod m) + (b mod m) (mod m)
and
ab ≡ (a mod m)(b mod m) (mod m).

What i really can't get here is how is the

a ≡ (a mod m) (mod m) and b ≡ (b mod m) (mod m)

I know about the reflexive property about mod but how come this two mod. what i know is a ≡ (a mod m) but how they are saying a ≡ (a mod m) (mod m)

Christoph
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DeathNet123
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    There are dozens too many "mod m" appearing in that... I hate the way it is written. – JMoravitz Jan 09 '21 at 16:18
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    The superior way of writing and thinking about this in my opinion from a mathematical standpoint is not that mod m is an operator, but rather a context. The context only needs to be written once to the side or mentioned ahead of time. You have by definition $a \equiv A\pmod{m}$ if and only if $(a-A)$ is a multiple of $m$. The theorems they are trying to share with you here rewritten are that if $a\equiv A\pmod{m}$ and $b\equiv B\pmod{m}$ then $a+b\equiv A+B\pmod{m}$ as well as $a\times b\equiv A\times B\pmod{m}$ – JMoravitz Jan 09 '21 at 16:22
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    The proof? Well, in the case of addition $(a+b)-(A+B)=(a-A)+(b-B)$ is a multiple of $m$ plus another multiple of $m$ and therefore a multiple of $m$, proving the claim by definition. For multiplication, $(a\times b)-(A\times B)=(a\times b)-(a\times B) + (a\times B)-(A\times B) = a\times (b-B) + B\times (a-A)$, which is the sum of multiples of multiples of $m$, and thus also a multiple of $m$. – JMoravitz Jan 09 '21 at 16:27
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    Your expression "$a\equiv (a \operatorname{\mathbf{mod}} m)$" is not well-formed. Maybe you mean "$a\equiv a \pmod m$"? The text uses the two notations "$a\operatorname{\mathbf{mod}}m$" for the unique integer $r\in{0,\dots,m-1}$ such that $a=qm+r$ for some $q\in\mathbb Z$ and the notation "$a\equiv b\pmod{m}$" for the congruence relation given by $m \mid (a-b)$. – Christoph Jan 09 '21 at 16:30

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