So i have a discrete math final and I don't know why but the profs decided not to post answer key for this one final and I need to check my understanding of this question.
Let $A = \{1, 2, 3, 4, \ldots, 271\}$. Define the relation $R$ on $A\times A$ for any $(a,b), (c,d) \in A \times A$, we have $(a,b) R (c,d)$ if and only if $a + b = c + d$.
I already proved how it is an equivalence relation so the other 3 parts are:
a) List all the elements of $[(3,3)]$, the equivalence class of $(3,3)$.
For this part I thought the answer was $6$ because $(3,3)$ basically says the sum is $6$ and so it belongs to the equivalence class with a sum of $6$ and so $(1,5),(2,4),(3,3)$ are elements and at first I just put $3$ but then I realized that since its $A\times A$ then $(1,5)$ and $(5,1)$ are elements etc so there would be $6$ elements total which are $(1,5),(2,4),(3,3)(3,3)(5,1)(4,2)$.
Is this right or should it be $3$? Also $(0,6)$ or $(6,0)$ is not included since $0$ is not an element of $A$.
b)How many equivalence classes does $R$ have? Explain.
For this one I thought that the equivalence classes are divided based on the sum of the two numbers $a,b$ right so the lowest number possible is $2$ since there is no $0$ and the highest would be up to $542 = 271+271$ and so that means there are $542-1 - 541$ total equivalence classes.
Am I right?
c) Is there an equivalence class that has exactly 271 elements? Explain.
I said "no" for this because I noticed a pattern: for any equivalence class of an even number $n$, there are $n$ elements (like the $6$ in part a), and if its odd there are $n-1$ ways. Hence $5$ would have $4$ ways and $3$ would have $2$ etc. So then that means that $[271]$ has $270$ elements and $[272]$ has $272$ elements so there would be no equivalence class that has exactly $271$ elements.
Is this thinking correct?
I know this may be a lot, but I would really appreciate any feedback !!
a. You are right that $(1,5)$, $(2,4)$, and $(3,3)$ are elements in the equivalence class. You are also right that $(6,1)$ is an element of $A \times A$. Why don't you just list all the pairs of positive integers that sum to $6$? The question asks you to list all elements in the equivalence class, not just state how many there are.
b. This looks fine to me.
c. If you re-do part a), you should see that the equivalence class of "pairs that sum to $n$" has $n-1$ elements, regardless of whether $n$ is odd or even.