0

On the set $N\times N$ define $(m,n)\simeq(k,l)$ if $m+l=n+k$.

Draw a sketch of $N\times N$ that shows several equivalence classes. (hint: sketch points on graph paper).

I'm not quite sure how to sketch the equivalence classes. I know that the following is in the relation, but do I just graph the points? Also, how do I find out how many equivalence classes there are and how do I distinguish them? Are there $4$ different equivalence classes ($1$ for $m$, $1$ for $n$, $1$ for $k$, and $1$ for $l$)? I know that if my relation were $\mod p$, there would be $p$ equivalence classes.

hjhjhj57
  • 4,125
mylasthope
  • 241

2 Answers2

1

You can start out by simply using the points on a coordinate grid for $\mathbb{N}\times\mathbb{N}$. Every point $(m,n)$ represents the pair $(m,n)\in\mathbb{N}^2$. So far, so obvious. For the actual equivalence classes, you could ask yourself: what is it that characterizes every equivalence class? For instance, take the point $(3,2)$. We have $3+2=5$. Where do you find other points with $m+n=5?$ Try points nearby first and try to find some kind of pattern.

A different approach would be the following: I am going to assume you are familiar with the normal vector form of a line in $\mathbb{R}^2$.

Every point in your class fulfills $m+n=c$. Note that you could also write this as $(m,n)\cdot (1,1)=c$. Can you find a geometric interpretation of that? I have already mentioned normal vectors, so maybe that will help you to find an actual proof.

Some Math Student
  • 1,502
1

Points on the same line belong to the same equivalence class and each different line represent a different equivalence class.

image

mfl
  • 29,399
  • Thank you! Just to make sure I understand this correctly, we plot (1,1) because we see that (m, n) ~ (k, l) is (1, 1) ~ (1, 1), right? Since (0,1) ~ (2,3) (0 + 3 = 1 + 2) also satisfies the relation, why is there not a line from those two points? Does satisfying the relation simply mean that a point is marked on the graph? Also, am I right in thinking that a line is an equivalence class because (0, 2) ~ (0, 2) => 2 = 2 and (1, 1) ~ (1, 1) => 2 = 2? – mylasthope Oct 05 '14 at 20:59
  • I have not considered $0\in \mathbb{N}.$ If you do that, then you must include two lines, one through $(0,1)$ and $(1,0)$ and other through $(0,0).$ Also, you must include any point of the form $(m,0)$ and $(0,n)$ in the graph. – mfl Oct 05 '14 at 21:01