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The question is, "What is the congruence class$[n]_5$ (that is, the equivalence class of $n$ with respect to congruence modulo 5) when n is

a) 2?

b)3?

c) 6?

d)−3?"

I know this is more work than the question is asking for, but I am just trying to get as much practice as I can with relations, and such. They didn't specify the set that I am going to find an equivalence class on, but I believe it is the integers, but am not certain why. But, if that is true, then I would have a relation, $R$, on the set $Z$, $R \subset Z$. And wouldn't we be able to determine the relation on this set, because a equivalence class is taken with respect to the relation? Would the relation be $R=\{(x,y)|x \equiv y~(mod~5)\}$? It just seems wrong for some reasons.

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2 Answers 2

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Yes, that's what "congruence mod 5" means; thus $\rm\,[n]_5 = n + 5\,\Bbb Z,\:$ i.e. all $\rm\:k\:$ such that $\rm\:5\:|\:k\!-\!n.$

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I will try again. The relation that you wrote is correct.

You have the set $\mathbb{Z}$. On that set you have the relation $$ R = \{(x,y)\lvert x\equiv y (\text{mod}\; 5)\} $$

Another way to say this is that $x$ and $y$ are equivalent ($x\sim y$) if $x-y$ is divisible by $5$.

Now for an integer $n$ you have the equivalence class (congruence class). This class is defined as all those $x$ for which $x\sim n$. So for example if $n = 2$, you get $$\begin{align} [2]_5 &= \{x\in \mathbb{Z}\lvert x\sim 2\} \\ &= \{x\in \mathbb{Z} \lvert x\equiv 2 (\text{mod } 5)\}\\ &= \{x\in \mathbb{Z}: 5\lvert x-2 \} \\ &= \{\dots , -8, -3, 2, 7, 12, \dots\} \end{align} $$

You can probably now figure out what happens if $n=3$ or one of the other values in your question.

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  • $\begingroup$ Where does the n come in place, though? Should I have used different variables than x and y? Also, is the relation I wrote correct? $\endgroup$
    – Mack
    Commented Nov 10, 2012 at 17:08
  • $\begingroup$ I see, one little error in mine: it's suppose to be mod 5, not mod n. $\endgroup$
    – Mack
    Commented Nov 10, 2012 at 17:11
  • $\begingroup$ @EMACK: I made a mistake in my answer. I updated the answer. Hopefully it is correct now. $\endgroup$
    – Thomas
    Commented Nov 10, 2012 at 17:13
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    $\begingroup$ @EMACK: Yes, that is exactly the way to think about it. $\endgroup$
    – Thomas
    Commented Nov 10, 2012 at 17:37
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    $\begingroup$ Wow, really! That is a good feeling to know that I have properly formed an idea. Thank you so much for your help! I am going to look at the rest of these questions. $\endgroup$
    – Mack
    Commented Nov 10, 2012 at 17:39

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