Let R be a relation from N to N defined by R = {(a, b): a, b ∈ N and a = b2}.

Question:

Let $R$ be a relation from $N$ to $N$ defined by $R=\left\{(a, b): a, b \in N\right.$ and $\left.a=b^{2}\right\}$. Are the following true $\}$

(i) $(a, a) \in R$, for all $a \in \mathbf{N}$

(ii) $(a, b) \in \mathrm{R}$, implies $(b, a) \in \mathrm{R}$

(iii) $(a, b) \in \mathbf{R},(b, c) \in \mathbf{R}$ implies $(a, c) \in \mathbf{R}$.

Solution:

$\mathrm{R}=\left\{(a, b): a, b \in \mathbf{N}\right.$ and $\left.a=b^{2}\right\}$

(i) It can be seen that $2 \in \mathbf{N}$, however, $2 \neq 2^{2}=4$.

Therefore, the statement " $(a, a) \in R$, for all $a \in \mathbf{N}$ " is not true.

(ii) It can be seen that $(9,3) \in \mathbf{N}$ because $9,3 \in \mathbf{N}$ and $9=3^{2}$.

Now, $3 \neq 9^{2}=81$; therefore, $(3,9) \notin \mathrm{N}$

Therefore, the statement " $(a, b) \in \mathrm{R}$, implies $(b, a) \in \mathrm{R}^{n}$ is not true.

(iii) It can be seen that $(16,4) \in R,(4,2) \in R$ because $16,4,2 \in N$ and $16=4^{2}$ and $4=2^{2}$.

Now, $16 \neq 2^{2}=4$; therefore, $(16,2) \notin \mathbf{N}$

Therefore, the statement " $(a, b) \in \mathrm{R},(b, c) \in \mathrm{R}$ implies $(a, c) \in \mathrm{R}^{\prime \prime}$ is not true.