What is the list of one member from each equivalence relation in this context?

Question

Good afternoon, I have a question: Let $X = \{1, 2,\dots, 10\}$. Define a relation $R$ on $X \times X$ by $(a, b)R(c, d)$ if $a + d = b + c$. Show that R is an equivalence relation on $X \times X$ and list one member from each equivalence class of $X \times X$.

So, I was able to demonstrate that R is an equivalence relation but I don't understand the last part, what is a list of one member from each equivalence class of $X \times X$? I searched the definition but I still don't get it... Can someone help me?

Thank you

Answer 1

So, according to definition, the equivalence class of $(a,b) \in X \times X$ is: \begin{equation*} [(a,b)]_{R} = \{(c,d) \in X \times X \hspace{.2cm} | \hspace{.2cm} (a,b)R(c,d) \} \end{equation*} To list a member from each equivalence is no more than the direct aplication of the definition above. As an extra, i would suggest every equivalence class, and then picking one member of each. I will give an example: \begin{align*} [(1,1)]_{R} = \{(1,1),(2,2),(3,3),(4,4),\dots,(10,10)\} \end{align*} I obtained this with the procedure below: \begin{align*} (1,1)R(c,d) \Leftrightarrow 1 + d = 1 + c \Leftrightarrow c = d \end{align*}

What you have to do now is to calculate the remainder equivalence classes and then pick one member of each. Also, note that $$[(1,1)]_{R} = [(2,2)]_{R} = \dots = [(10,10)]_{R}$$ And so, you don't need to calculate those. (This is a general result, appliable in the rest of the equivalence classes aswell)

Answer 2

You are being asked to provide elements such that:

• (a) They are not in the relation with each other, but
• (b) Every element of $X\times X$ is in relation with at least one of those (and then by (a) with exactly one of those).

The answer can be, for example:

$$(10,1), (9,1),(8,1),(7,1),(6,1),(5,1),(4,1),(3,1),(2,1),(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(1,7),(1,8),(1,9)$$

Notice your relation is nothing else but this: $(a,b)R(c,d)\iff a-b=c-d$. The pairs above make up all the differences from $9$ to $-9$. This means no two of them can be equivalent, and any pair $(a,b)$ is equivalent to exactly one of them - the one having the same difference $a-b$. For example, the pair $(5,2)$ is equivalent with the pair above with the difference $5-2=3$, which is $(4,1)$.