# Counting the number of equivalence classes given the relation (a,b),(c,d) elements of AXA where (a,b)R(c,d) if and only if a+b=c+d

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 !!

• Welcome to MSE. Please edit and use MathJax to properly format nubmers and math expressions. – Lee David Chung Lin Apr 27 at 6:26

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.
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.