Question

Let S be the set of all real numbers. Show that the relation R=\left\\{ \left( a,b \right):{{a}^{2}}+{{b}^{2}}=1 \right\\} is symmetric but neither reflexive nor transitive.

Answer 1

Hint:To show that R is symmetric, prove that if $\left( a,b \right)\in R\ then\ \left( b,a \right)\in R$. To show that R is not reflexive, find an example such that a is a real number, but $\left( a,a \right)\notin R$. To show that R is not transitive, find an example such that $\left( a,b \right)\in R\ and\ \left( b,c \right)\in R\ and\left( a,c \right)\notin R$.

Complete step-by-step answer:
Let ‘A’ be a set, then:
Reflexivity, symmetry and transitive of a relation on set ‘A’ is defined as follows:
Reflexive relation: A relation R on a set ‘A’ is said to be reflexive if every element of A is related to itself. i.e. for every element say (a) in set A, $\left( a,a \right)\in R$.
Thus, R on a set A is not reflexive if there exists an element $a\in A\ such\ that\ \left( a,a \right)\notin R$.
Symmetric Relation: A relation R on a set ‘A’ is said to be a symmetric relation if $\left( a,b \right)\in R\ then\ \left( b,a \right)\$must belong to R.
i.e. $\left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R\ for\ all\ a,b\in A$
Transitive Relation: A relation R on A is said to be a transitive relation if $\begin{aligned} & \left( a,b \right)\in R\ and\ \left( b,c \right)\in R\ then\ \left( a,c \right)\in R \\\ & i.e.\ \ \left( a,b \right)\in R\ and\ \left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R \\\ \end{aligned}$
To show that R is symmetric, we have to prove that if $\left( a,b \right)\in R\ then\ \left( b,a \right)\in R$.
Let us assume $\left( a,b \right)\in R$.
$\begin{aligned} & \Rightarrow {{a}^{2}}+{{b}^{2}}=1 \\\ & \Rightarrow {{b}^{2}}+{{a}^{2}}=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ \text{Rearranging the terms in LHS} \right] \\\ & \Rightarrow \left( b,a \right)\in R \\\ \end{aligned}$
Thus, If $\left( a,b \right)\in R\ then\ \left( b,a \right)\in R$
$\therefore$R is symmetric on a set of real numbers.
Next,
To show that R is not reflexive, we need to find a real number ‘a’ such that $\left( a,a \right)\notin R\$.
Let us take a = 1
$a\in \text{ set of real numbers}$ [As ‘1’ is a real number]
$But\ \left( 1,1 \right)\notin R\ \left[ \text{As }{{\left( 1 \right)}^{2}}+{{\left( 1 \right)}^{2}}\ne 1 \right]$
$\therefore$R is not reflexive on a set of real numbers.
Next, we have to show that R is not transitive. It means we have to find an example such that $\left( a,b \right)\in R\ and\ \left( b,c \right)\in R\ but\ \left( a,c \right)\notin R$.
Let us take a = 0, b = 1 and c = 0.
$\begin{aligned} & a,b,c\in \left( set\ of\ real\ numbers \right)\ \ \ \ \ \ \ \ \left[ As'0'\ \And \ '1'\ are\ real\ numbers \right] \\\ & And\ \left( a,b \right)\in R\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ As\ {{0}^{2}}+{{1}^{2}}=1 \right] \\\ & And\ \left( b,c \right)\in R\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ As\ {{1}^{2}}+{{0}^{2}}=1 \right] \\\ & And\ \left( a,c \right)\notin R\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[ As\ {{0}^{2}}+{{0}^{2}}\ne 1 \right] \\\ \end{aligned}$
$\therefore$R is not transitive on a set of real numbers.
Hence, we have shown that R is symmetric but neither reflexive nor transitive on a set of real numbers.

Note: Be careful that to prove that the given relation is not transitive or not reflexive, we need to just find an example which doesn’t satisfy the condition of symmetric or reflexive. But to prove that it is symmetric, we can’t use an example, instead we need to prove that the condition is symmetric for all possible cases.