Question

Let $R$ be a relation from $N\to N$ defined by

$R=\left\{\right(a,b):a,b\in N\text{and}a=b^2\}$.
Are the following true ?

(i) $(a,a)\in R\text{for all}a\in N$
(ii) $(a,b)\in R$ implies $(b,a)\in R$
(iii) $(a,b)\in R,(b,c)\in R$ implies $(a,c)\in R$

Answer 1

(i) $(a,a)\in R\text{for all}a\in N$
Given relation: $R=\left\{\right(a,b):a,b\in N\text{and}a=b^2\}$
Let $(3,3)\in R\Rightarrow 3=3^2$ which is not true
$\therefore (a,a)\in R$, for all $a\in N$ is not true.

(ii) $(a,b)\in R$ implies $(b,a)\in R$
Given relation: $R=\left\{\right(a,b):a,b\in N\text{and}a=b^2\}$
We know, $(4,2)\in R$ since $4=2^2$
But $(2,4)\notin R$ Since $2=4^2$ is not true.
$\therefore (a,b)\in R$ implies $(b,a)\in R$ is not true.

(iii) $(a,b)\in R,(b,c)\in R$ implies $(a,c)\in R$
Given relation: $R=\left\{\right(a,b):a,b\in N\text{and}a=b^2\}$
It is clearly seen that $(16,4)\in R,(4,2)\in R$
Because $16,4,2\in N\text{and}16=4^2\text{and}4=2^2$
Now,
$16\ne 2^2=4$ therefore, $(16,2)\in N$
Thus, the statement ${}^{\prime \prime }(a,b)\in R,(b,c)\in R$
implies $(a,c)\in R^{\prime \prime }$ is not True.