(a) Let A = {0,1, 2, 3, 4, 5}, B = {1, 2, 3, 4, 5, 6}, and consider the relation R = {(a, b) э A x B | a2+b2 < 30}. (i) List all the elements of the relation R and give its cardinality |R|. (3 marks) (ii) Find the domain and range of the relation R. (2 marks) (iii) Find the inverse relation R-1 (2 marks)

(i)Since ${0^2} + {1^2} = 1 < 30$ then $(0;1) \in R$ .

Since ${1^2} + {1^2} = 2 < 30$ then $(1;1) \in R$ .

Arguing in a similar way, we get:

$\begin{array}{l} R = \{ \left( {0;1} \right),\,\left( {1;1} \right),\,\left( {2;1} \right),\,\left( {3;1} \right),\,\left( {4;1} \right),\,\left( {5;1} \right),\,\left( {0;2} \right),\,\left( {1;2} \right),\,\\ \left( {2;2} \right),\,\left( {3;2} \right),\,\left( {4;2} \right),\,\left( {5;2} \right),\,\left( {0;3} \right),\,\left( {1;3} \right),\,\left( {2;3} \right),\,\left( {3;3} \right),\,\left( {4;3} \right),\,\\ \left( {0;4} \right),\,\left( {1;4} \right),\,\left( {2;4} \right),\,\left( {3;4} \right),\,\left( {0;5} \right),\,\left( {1;5} \right),\,\left( {2;5} \right)\} \end{array}$

The cardinality of a finite set is equal to the number of elements of the set, therefore

$|R| = 24$

(ii) Let's find  the domain of thr relation:

$DomR = \{ x|(x,y) \in R\} = \{ 0;1;2;3;4;5\} = A$

Let's find  the range of thr relation:

$ImR = \{ y|(x,y) \in R\} = \{ 1;2;3;4;5\}$

(iii) Let's find   the inverse relation:

$\begin{array}{l} {R^{ - 1}} = \{ (y,x)|(x,y) \in R\} = \{ \left( {1;0} \right),\,\left( {1;1} \right),\,\left( {1;2} \right),\,\left( {1;3} \right),\,\left( {1;4} \right),\,\\ \left( {1;5} \right),\,\left( {2;0} \right),\,\left( {2;1} \right),\,\left( {2;2} \right),\,\left( {2;3} \right),\,\left( {2;4} \right),\,\left( {2;5} \right),\,\left( {3;0} \right),\,\left( {3;1} \right),\,\\ \left( {3;2} \right),\,\left( {3;3} \right),\,\left( {3;4} \right),\,\left( {4;0} \right),\,\left( {4;1} \right),\,\left( {4;2} \right),\,\left( {5;0} \right),\,\left( {5;1} \right),\,\left( {5;2} \right)\} \end{array}$