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I was looking over some properties of modular arithmetic and noticed the following one:

If $a \equiv a'$ (mod $n$) and $b \equiv b'$ (mod $n$), then $ab \equiv a'b'$ (mod n)

Now this is ok and I understand the proofs that prove that property, but I have also noticed that reducing only one variable modulo $n$ the congruence will also be satisfied. Namely:

$ab \equiv ab'$ (mod $n$)

Can anyone tell me why this also works? I am not very good with proofs, but I think it is just a corollary of the first property.

Michael Munta
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    Yes, it's just the special case of the first where $,a' = a,$ (so $,a'\equiv a).,$ Both versions of the Congruence Product Rule are equivalent. – Bill Dubuque Jan 11 '19 at 16:33
  • $a \equiv a \pmod n$, so of course it's fine. – Randall Jan 11 '19 at 16:41
  • @BillDubuque Can I also think of it in terms of congruence being an equivalence relation? So analogously to the $=$ relation; in modulo $N$ two numbers being congruent to one another means that they are essentially one and the same and can be swapped without changing the equivalence? – Michael Munta Jan 15 '19 at 08:51
  • @MichaelMunta Yes, by definition congruence relations are equivalence relations that satisfy the additional property of being compatible with the ambient arithmetic operations (in a ring this means that are compatible with addition and multiplication, i.e. they satisfy the Sum and Product Rules in the linked post in my prior comment). Once you learn about quotient structures then you can replace the equivalence relation by equality (of congruence classes in the quotient ring, i.e. $11\equiv 1\pmod {10}$ becomes $[11]{10} = [1]{10},,$ i.e. $, 11+10\Bbb Z = 1+10\Bbb Z)\ \ $ – Bill Dubuque Jan 15 '19 at 15:27

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\begin{align} b\equiv b' \pmod{n} &\Leftrightarrow b-b'\equiv 0 \pmod{n}\\ &\Leftrightarrow b-b'=nq, q\in\mathbb{Z}\\ &\Leftrightarrow a\left(b-b'\right)=a(nq)=n(aq)\\ &\Leftrightarrow a(b-b')\equiv 0 \pmod{n}\\ &\Leftrightarrow ab-ab'\equiv 0 \pmod{n}\\ &\Leftrightarrow ab\equiv ab' \pmod{n}\\ \end{align}

Platt
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