Equivalence classes - on $\mathbb{N}^2$

Question

Let $R$ be the relation on $\mathbb{N}^2$ defined by

$(a,b)R(c,d)$ if $2a + 3b = 2c + 3d$

Write $4$ elements in the equivalence class of $(1,2)$

So I think I need to find all the pairs $(a,b)$ with $2a + 3b = 2(1) + 3(2) = 8$

But no other positive integers other than $(1,2)$ will satisfy that equation so obviously I'm doing something wrong.

Any suggestions?

Answer 1

Your reasoning is spot on. The equivalence class consisting of all pairs $(a, b)$ that are equivalent to $(1, 2)$ under $R$, with the restriction that $a, b \in \mathbb Z^+$ consists only of $(1, 2)$ itself.

Otherwise, the set of pairs of integers (no restriction) which are in the equivalence class $\Big[(1, 2)\Big]$ is exactly $$\Big[( 1,2)\Big] = \{(a, b) \mid a = 1 + 3k \;\land \;b = 2-2 k,\;\; k \in \mathbb Z\} $$

As others have suggested, there must by some misprint involved here in the problem statement you've been given, or the instructor/author of the problem statement was having a very bad day and was careless in thinking the problem through.