Question

Define a relation R on Z by R={(n,n+3k)| k€Z}.
Check whether R is an equivalence relation or not. If it is, find all the distinct equivalence classes. If R is not an equivalence relation, define an equivalence relation on Z

Answer 1

Answer on Question #76812 – Math – Abstract Algebra

Question

Define a relation $R$ on $Z$ by $R = \{(n, n + 3k) | k \in Z\}$. Check whether $R$ is an equivalence relation or not. If it is, find all the distinct equivalence classes. If $R$ is not an equivalence relation, define an equivalence relation on $Z$.

Solution

The relation $R$ is an equivalence relation, if it is reflexive, symmetric and transitive relation. Let's check it.

For any integer $n \in Z$, we have $(n, n) = (n, n + 3 \cdot 0) \in R$, where $k = 0$. This means that $R$ is reflexive.

If $n \in Z$ and $(n, n + 3k) \in R$ for some $k \in Z$, then $(n + 3k, n) = (n + 3k, n + 3k + 3(-k)) \in R$. This means that $R$ is symmetric.

If $(n, m) \in R$ and $(m, l) \in R$, then there exists integers $k_1$ and $k_1$ in $Z$, such that $m = n + 3k_1$ and $l = m + 3k_2$. Therefore, $l = m + 3(k_1 + k_2)$, i.e. $(m, l) \in R$. The last statement means that $R$ is transitive.

Hence, the relation $R$ on $Z$ is an equivalence relation.

Now find all distinct equivalence classes of the relation $R$.

$$[0] = \{n \in Z: (0, n) \in R, \text{ i.e. } n = 3k, k \in Z\} = \{3k: k \in Z\} = 3Z;$$ $$[1] = \{n \in Z: (1, n) \in R, \text{ i.e. } n = 1 + 3k, k \in Z\} = \{3k + 1: k \in Z\} = 3Z + 1;$$ $$[2] = \{n \in Z: (2, n) \in R, \text{ i.e. } n = 2 + 3k, k \in Z\} = \{3k + 2: k \in Z\} = 3Z + 2;$$

**Answer**: The relation $R$ is an equivalence relation on $Z$ and there is only three distinct equivalence classes of $R$: $[0] = \{3k: k \in Z\}$; $[1] = \{3k + 1: k \in Z\}$; $[2] = \{3k + 2: k \in Z\}$.

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