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In our mathematics course, we have defined $\overline{a}=\{a \in \mathbb{Z} \mid k \equiv a \text{ mod } m \}=\frac{a-k}{m}$. We have also defined $\overline{a} \times \overline{b}=\overline{ab}$.

In order to calculate $\overline{ab}$, can I simply multiple together $\frac{a-k}{m}$ and $\frac{b-k}{m}$ to get $\frac{(a-k)(b-k)}{m^2}$?

My apologies if this question is too simplistic.

Kurt Mueller
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    I don't understand the definition. Is $\overline a$ a number or a set? Is $k$ fixed? Is $m$ fixed? Is $\overline a$ just $\lfloor \frac am \rfloor$ for some specified $m$ or is it something else? – lulu Oct 22 '16 at 17:35
  • To @lulu 's point, I think you should reconsider what equality "$=$" actually means and how you're using it – Tom Oct 22 '16 at 17:38
  • One would imagine that he intended to have the standard modular arithmetic definition, in which case $\frac{a-k}{m}$ is wrong and shouldn't be there. $\frac{a-k}{m}$ would be the quotient of $a$ when dividing by $m$, but we usually refer to equivalence classes by their remainder instead. – JMoravitz Oct 22 '16 at 17:42

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Assuming what definitions you've written above are written poorly, and what I am thinking of is indeed what you intended, I will clean up the definitions here below and show what and why the multiplication is well defined.

  • We define $a \equiv k \pmod{m}$ as meaning $(a-k)$ is an integer multiple of $m$. I.e. $a\equiv k\pmod{m}$ iff $\exists z\in\Bbb Z$ such that $a-k=zm$

Recognize then that $\equiv\dots \pmod{m}$ is an equivalence relation on $\Bbb Z$

  • We define $\overline{a}$ to be the equivalence class of $a$ for the relation defined above. That is to say $\overline{a} = \{b~~|~~a\equiv b\pmod{m}\}$ is the set of all integers who are equivalent to $a$ via the equivalence relation defined above.

Recognize then that $\overline{a}=\overline{b}$ if and only if $a\equiv b\pmod{m}$

We define $\overline{a}\times \overline{b}$ to be equal to $\overline{(ab)}$. We wish to prove this is well defined. That is to say, we want to show that regardless how we represent the classes, the result is the same.

Suppose $\overline{a_1}=\overline{a_2}$ and $\overline{b_1}=\overline{b_2}$. We wish to prove that $\overline{(a_1b_1)}=\overline{(a_2b_2)}$

Since $a_1\equiv a_2\pmod{m}$ we have $(a_1-a_2)=xm$ for some integer $x$. Similarly $(b_1-b_2)=ym$ for some integer $y$.

We have $a_1b_1 - a_2b_2 = a_1b_1-\underbrace{(a_1-a_2-xm)}_{\text{equals 0}}b_1-a_2b_2-\underbrace{(b_1-b_2-ym)}_{\text{equals 0}}a_2$

$=a_1b_1-a_1b_1+a_2b_1+xmb_1-a_2b_2-a_2b_1+a_2b_2+yma_2$

$=xmb_1+yma_2 = (xb_1+ya_2)m$ which is an integer multiple of $m$

Therefore we do indeed have $\overline{(a_1b_1)}=\overline{(a_2b_2)}$

You ask how to calculate $\overline{ab}$ given an $a$ and a $b$. You simply multiply $a$ and $b$ together like normal and then put a line over top of it.

For example, suppose we are working in $\Bbb Z_{10}$. One has $\overline{5}\times \overline{82} = \overline{(5\cdot 82)}=\overline{410}$

Granted, this is not necessarily the most useful way of expressing the result, but it is none-the-less a correct way of expressing the result.

More often, however, we are interested in using canonical representatives of the equivalence classes, preferring only to refer to equivalence classes by the unique element in the range $[0,m)$. Here, we would have $\overline{5}\times \overline{82}=\overline{410}=\overline{0}$.

To find the canonical representative for the resulting equivalence class, one can run the quotient-remainder theorem on it. $\overline{ab}=\overline{(ab-m\lfloor\frac{ab}{m}\rfloor)}$

JMoravitz
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