Prove that

Solution:

Put $a=1, b=1\left|1^{2}-1^{2}\right| \leq 5,(1,1)$ is an ordered pair.

Put $a=1, b=2\left|1^{2}-2^{2}\right| \leq 5,(1,2)$ is an ordered pair.

Put $a=1, b=3\left|1^{2}-3^{2}\right|>5,(1,3)$ is not an ordered pair.

Put $a=2, b=1\left|2^{2}-1^{2}\right| \leq 5,(2,1)$ is an ordered pair.

Put $a=2, b=2\left|2^{2}-2^{2}\right| \leq 5,(2,2)$ is an ordered pair.

Put $a=2, b=3\left|2^{2}-3^{2}\right| \leq 5,(2,3)$ is an ordered pair.

Put $a=3, b=1\left|3^{2}-1^{2}\right|>5,(3,1)$ is not an ordered pair.

Put $a=3, b=2\left|3^{2}-2^{2}\right| \leq 5,(3,2)$ is an ordered pair.

Put $a=3, b=3\left|3^{2}-3^{2}\right| \leq 5,(3,3)$ is an ordered pair.

$R=\{(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3)\}$

(i) For $(a, a) \in R$

$\left|a^{2}-a^{2}\right|=0 \leq 5$. Thus, it is reflexive.

(ii) Let $(a, b) \in R$

$(a, b) \in R$ è $\left|a^{2}-b^{2}\right| \leq 5$

$\left|b^{2}-a^{2}\right| \leq 5$

$(b, a) \in R$

Hence, it is symmetric

(iii) Put $a=1, b=2, c=3$.

$\left|1^{2}-2^{2}\right| \leq 5$

$\left|2^{2}-3^{2}\right| \leq 5$

But $\left|1^{2}-3^{2}\right|>5$

Thus, it is not transitive.