Let A = \ 2,3,6,8,9,11\ and B = \ 1,4,5,10,15\. Let R be a relation on A x B defined by (a,b)R(c,d) if and only if 3 ad - 7 bc is an even integer. Then the relation R is

(b): We have, (a,b)R(c,d) => 3ad - 7bc is an even integer. For reflexive : (a,b)R(a,b) => 3ab - 7ab = - 4ab, Which is an even integer. For symmetric, (a,b)R(c,d) \ => 3ad - 7bc is an even integer. \ => 3ad - 7bc + 4bc + 4ad is also an even integer ( Because even + even = even number ) \ => 7ad - 3bc is an even integer \ => 3cb - 7da is also an even integer \ => (c,d)R(a,b) For transitive, (a,b)R(c,d) and (c,d)R(e,f) \ => 3ad - 7bc and 3cf - 7de is an even integer. For a = 2,b = 5,c = 6,d = 8,e = 9,f = 1 3af - 7bc = 3 x 2 x 1 - 7 x 5 x 9 = 6 - 315 = - 309 which is not an even integer. \ Therefore Given relation is not transitive.

Let A = { 2,3,6,8,9,11} and B = { 1,4,5,10,15}. Let R ... | Relation (Mathematics)

(b): We have, $(a,b)R(c,d) \Rightarrow 3ad - 7bc$ is an even integer. $\\$ For reflexive : $(a,b)R(a,b) \Rightarrow 3ab - 7ab = - 4ab,\\$ Which is an even integer. For symmetric, $(a,b)R(c,d) \\ \Rightarrow 3ad - 7bc$ is an even integer. $\\ \Rightarrow 3ad - 7bc + 4bc + 4ad$ is also an even integer $\\$ ( $\because$ even + even $=$ even number ) $\\ \Rightarrow 7ad - 3bc$ is an even integer $\\ \Rightarrow 3cb - 7da$ is also an even integer $\\ \Rightarrow (c,d)R(a,b)$ For transitive, $(a,b)R(c,d)$ and $(c,d)R(e,f)$ $\\ \Rightarrow 3ad - 7bc$ and $3cf - 7de$ is an even integer. $\\$ For $a = 2,b = 5,c = 6,d = 8,e = 9,f = 1$ $\\3af - 7bc = 3 \times 2 \times 1 - 7 \times 5 \times 9 \\= 6 - 315 = - 309$ which is not an even integer. $\\ \therefore$ Given relation is not transitive.

Explanation

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