1
$\begingroup$

I was reading Algebra of Hungerford, the (sketch of) proof of Theorem 3.6.

Definitions : Let $m$ be a positive integer. The equivalence relation '$\equiv$' modulo $m$ partitions $\mathbb Z$ into $m$ equivalence classes $\bar a$ for $a=0,1,2,\cdots,m-1$. The set $\mathbb Z_m$ of all such equivalence classes is a group under addition, defined by $\bar a+\bar b=\overline{a+b}$. Note that $\bar a=\bar b$ in $\mathbb Z_m$ iff $a\equiv b$ (mod $m$).

Then, is the following true for every integer $n$ and $k$? $$n\bar k=\overline{nk}\quad(\text{in }\mathbb Z_m)$$

my answer : yes

examples : In $\mathbb Z_6$, $$3\bar5=\bar5+\bar5+\bar5=\overline{15}$$ $$0\bar5=\bar0$$ $$(-2)\bar5=-(\bar5+\bar5)=-\overline{10}=\overline{-10}.$$

$\endgroup$

1 Answer 1

2
$\begingroup$

Yes:

$n\bar k=\underbrace{\bar k+\cdots+\bar k}_{\text{n-times}}=\overline {\underbrace{k+\cdots+k}_{\text{n-times}}}=\overline{nk}$, since $\overline{a+b}=\bar a+\bar b$ by definition.

$\endgroup$
1
  • $\begingroup$ Thanks for your answer! $\endgroup$
    – shyzealot
    Commented Dec 26, 2019 at 12:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .