Solve the
$2\frac{1}{5}-\frac{-1}{3}$

$R=\{(a,b):a,b\in N;a=b^2\}$
(i) It can be seen that $2\in N$; however  $2\ne 2^2=4$
Therefore the statement $(a,a)\in R$ for all $a\in N$ is not true

(ii) It can be seen that $(9,3)\in N$ because $9,3\in N$ and $9=3^2$
Now $3\ne 9^2=81$ 
$\therefore (3,9)\notin N$
Therefore the statement  $(a,b)\in R$ implies  $(b,a)\in R$ is not true.

(iii) It can be seen that $(9,3)\in R,(16,4)\in R$ because $9,3,16,4\in N$ and $9=3^2$ and $16=4^2$
Now $9\ne 4^2=16$
$\therefore (9,4)\notin N$
Therefore the statement $\left(a,b\right)\in R,\left(b,c\right)\in R$ implies $\left(a,c\right)\in R$ is not true