1
$\begingroup$

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.

2 Answers 2

1

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.

  • 0
    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?2012-11-10
  • 0
    I see, one little error in mine: it's suppose to be mod 5, not mod n.2012-11-10
  • 0
    @EMACK: I made a mistake in my answer. I updated the answer. Hopefully it is correct now.2012-11-10
  • 0
    Now, if I wanted to see if those x-values, $\{…,−8,−3,2,7,12,…\}$, that are related to n=2, where would I plug those x-values into?2012-11-10
  • 0
    @EMACK: I am not sure that I understand your question. So the congruence class (for $n=2$) is the set of all those $x$ for which $x-2$ is divisible by $5$. This set is the set $2 + 5\mathbb{Z} = \{\dots, -13, -8, -3, 2, 7, 12, 17, \dots\}$.2012-11-10
  • 0
    Hmm, I am not sure if I understand what I asked either. Umm, so they relation relates two arbitrary values in a set together, and the equivalence class relates a specific value, 2 in this case, to all the other values in the set that meet the condition that $x \equiv 2(mod~5)$? Is that a proper way to think about all of this?2012-11-10
  • 1
    @EMACK: Yes, that is exactly the way to think about it.2012-11-10
  • 1
    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.2012-11-10
  • 0
    I have another quick question. For formality sake, would I have to say that $n,x,y \in Z$?2012-11-10
  • 0
    @EMACK: I would say that it depends on the context. In your problem I would just mention in the beginning of the solution that you are "working" over the integers. It should be pretty obvious that $x,y$, and $z$ are integers.2012-11-11
2

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