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I was given the following task to solve

Let $n > 1$ be a natural number. Define on $Z$ the following relation: $x \equiv y$ mod $n$ if $x-y$ can be divided by $n$. Show that operations of addition and multiplication on $Z$ are "agreed" with this equivalence relation.

I don't understand this task. Could you please explain what we have to show ?

  1. We have a relation $x \equiv y $ mod $n$ with 3 variables now, i.e $x$~$y$~n ?
  2. Show that operations of addition and multiplication on $Z$ are "agreed" with this equivalence relation. What does it mean ?

UPD. Let's prove that operation of addition is compatible with equivalence relation.

Suppose that $x\equiv y\pmod n$ and $w\equiv v\pmod n$. Then $y -x = n *i$ and $v -w = n * k$. Adding up both equations we get $(y+v) - (x+w) = n*(k*i)$.Hence, $x+w\equiv y+v\pmod n$. Hence addition is compatible with this relation.

Proof that multiplication is compatible.
Consider $y = x + ni$ and $w = z + nl$ we have to show that $yw-xz = n * k$.
Multiplying both equations we get $yw = xz + xnl + zni + n^2li \to yw - xz = n(xl + zi + nli) $

Rustem Sadykov
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1 Answers1

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For any natural number $n$ we have a relation $\equiv_n$ defined as $x\equiv_n y\iff n\mid y-x$. We write $x\equiv_n y$ as $x\equiv y\pmod n$ as well.

That the operations of addition and multiplication are compatible with the equivalence relation means that you should show that, when $x\equiv y\pmod n$ and $z\equiv w\pmod n$, then $x+z\equiv y+w\pmod n$, and $xz\equiv yw\pmod n$.

Hope this helps.

awllower
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