Question

Check whether the relation R on the set S = {1, 2, 3} is an equivalent

Answer 1

Answer:

$R$ isn't an equivalence relation. It is reflexive but neither symmetric nor transitive.

Step-by-step explanation:

Let $S$ denote a set of elements. $S \times S$ would denote the set of all ordered pairs of elements of $S\!$.

For example, with $S = \lbrace 1,\, 2,\, 3 \rbrace$, $(3,\, 2)$ and $(2,\, 3)$ are both members of $S \times S$. However, $(3,\, 2) \ne (2,\, 3)$ because the pairs are ordered.

A relation $R$ on $S\!$ is a subset of $S \times S$. For any two elements$a,\, b \in S$, $a \sim b$ if and only if the ordered pair $(a,\, b)$ is in $R\!$.

 

A relation $R$ on set $S$ is an equivalence relation if it satisfies the following:

• Reflexivity: for any $a \in S$, the relation $R$ needs to ensure that $a \sim a$ (that is: $(a,\, a) \in R$.)

• Symmetry: for any $a,\, b \in S$, $a \sim b$ if and only if $b \sim a$. In other words, either both $(a,\, b)$ and $(b,\, a)$ are in $R$, or neither is in $R\!$.

• Transitivity: for any $a,\, b,\, c \in S$, if $a \sim b$ and $b \sim c$, then $a \sim c$. In other words, if $(a,\, b)$ and $(b,\, c)$ are both in $R$, then $(a,\, c)$ also needs to be in $R\!$.

The relation $R$ (on $S = \lbrace 1,\, 2,\, 3 \rbrace$) in this question is indeed reflexive. $(1,\, 1)$, $(2,\, 2)$, and $(3,\, 3)$ (one pair for each element of $S$) are all elements of $R\!$.

$R$ isn't symmetric. $(2,\, 3) \in R$ but $(3,\, 2) \not \in R$ (the pairs in $\! R$ are all ordered.) In other words, $3$ isn't equivalent to $2$ under $R\!$ even though $2 \sim 3$.

Neither is $R$ transitive. $(3,\, 1) \in R$ and $(1,\, 2) \in R$. However, $(3,\, 2) \not \in R$. In other words, under relation $R\!$, $3 \sim 1$ and $1 \sim 2$ does not imply $3 \sim 2$.