Solve this

Question:

Let $R=\left\{(a, b): a=b^{2}\right\}$ for $a l l a, b \in N$.

Show that R satisfies none of reflexivity, symmetry and transitivity.

 

Solution:

We have, $R=\left\{(a, b): a=b^{2}\right\}$ relation defined on $N$.

Now,

We observe that, any element a $\in \mathrm{N}$ cannot be equal to its square except 1 .

$\Rightarrow(a, a) \notin R \forall a \in N$

For e.g. $(2,2) \notin R \because 2 \neq 2^{2}$

$\Rightarrow \mathrm{R}$ is not reflexive.

Let $(a, b) \in R \forall a, b \in N$

$\Rightarrow a=b^{2}$

But b cannot be equal to square of $a$ if $a$ is equal to square of $b$.

$\Rightarrow(b, a) \notin R$

For e.g., we observe that $(4,2) \in R$ i.e $4=2^{2}$ but $2 \neq 4^{2} \Rightarrow(2,4) \notin R$

$\Rightarrow \mathrm{R}$ is not symmetric

Let $(a, b) \in R$ and $(b, c) \in R \forall a, b, c \in N$

$\Rightarrow \mathrm{a}=\mathrm{b}^{2}$ and $\mathrm{b}=\mathrm{c}^{2}$

$\Rightarrow \mathrm{a} \neq \mathrm{c}^{2}$

$\Rightarrow(\mathrm{a}, \mathrm{c}) \notin \mathrm{R}$

For e.g., we observe that

$(16,4) \in R \Rightarrow 16=4^{2}$ and $(4,2) \in R \Rightarrow 4=2^{2}$

But $16 \neq 2^{2}$

$\Rightarrow(16,2) \notin \mathrm{R}$

$\Rightarrow \mathrm{R}$ is not transitive.

Thus, $R$ is neither reflexive nor symmetric nor transitive.

Leave a comment

None
Free Study Material