Stuck on equivalence relation question

Question

I have been stuck on this question for a while. I was wondering for a set $A={1,2,3,4,5,6}$, given that its distinct equivalence classes are $\{1,4,5\},\{2,6\},\{3\}$, what is the equivalence relation R on A?

I have tried everything, such as powers, modulo, parity arguments, sum, difference, product, etc.

Any help would be appreciated.

Answer 1

What you need to do is simply define $R$ by explicitly stating which ordered pairs of numbers belong to $R$. There's no need to discover some clever operation(s) on elements that define of a binary relation between elements.

All we need to know is:

• Every element in an equivalence class is related to itself and all other elements in that class, and is NOT related to any element outside the class.

Because $R$ is an equivalence relation, we need to also ensure that we include pairs in $R$ satisfying:

• reflexivity $(a, a) \forall a\in A$,
• symmetry: $(a, b) \in R \implies (b, a) \in R$,
• transitivity: $(a, b), (b, c)\in R \implies (a, c)\in R$

We simply list those ordered pairs:

$$R = \{\underbrace{(1, 1), (1, 4), (1, 5), (4, 1), (4, 4), (4, 5), (5, 1), (5, 4), (5, 5),}_{\text{first class}} \\\underbrace{(2, 2), (2, 6), (6, 2), (6, 6),}_{\text{second class}} \underbrace{(3, 3)}_{3^\text{rd}\text{class}}\}$$

Answer 2

There need not be any nice qualitative description. It’s entirely possible that you’re simply being asked for the set of ordered pairs that make up the equivalence relation. This consists of all of the pairs $\langle x,x\rangle$ for $x\in A$ together with the pairs $\langle 1,4\rangle,\langle 4,1\rangle,\langle 1,5\rangle,\langle 5,1\rangle,\langle 4,5\rangle,\langle 5,4\rangle$, $\langle 2,6\rangle$, and $\langle 6,2\rangle$.

Answer 3

You're not going to like the answer, but it's correct. The equivalence relation is the following set:

$ \{ (1, 4) (4, 1) (1, 5) (5, 1), (4, 5), (5, 4), (2, 6), (6, 2), (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)\}. $

Why? Because a "relation on $A$" is a subset $S$ of $A \times A$; an equivalence relation is one where $S$ satisfies three properties (symmetry, reflexivity, transitivity).

You were probably hoping for some algebraic expression whose zero-set, over the integer grid, consists of exactly those 14 pairs. (Or something like that) You can build such a thing. Here's how you start:

You build an expression of the form

$ e(x, y) = e_1(x, y) e_2(x, y) \cdot \ldots \cdot e_{14}(x, y) $

where each factor is nonzero for every $(x, y)$ except the one you like. For instance, you could let

$ e_1(x, y) = (x-1)^2 + (y - 4)^2. $ That's nonzero except at $x = 1, y = 4$. Then you let

$ e_2(x, y) = (x-4)^2 + (y - 1)^2. $

And you keep doing that. When you're all done, you'll have 14 quadratic expressions, whose product is a degree 28 polynomial on the set ${1, 2, 3, 4, 5, 6}$ whose zeroes are exactly the items in the "relation" above. But even if you multiply it out and simplify, it won't be very useful. Sometimes "algebraic expressions" for things just aren't the best way to express them.

Answer 4

My (lazy) answer would have been:

$xRy\iff\left[\left\{ x,y\right\} \subset\left\{ 1,4,5\right\} \vee\left\{ x,y\right\} \subset\left\{ 2,6\right\} \vee\left\{ x,y\right\} \subset\left\{ 3\right\} \right]$.