Let [A=0,3,4,6,7,8,9,10] and R be the relation defined on A such that [R=(x, y) A A: x-y] …

Question

Let $$A=\{0,3,4,6,7,8,9,10\}$$ and $R$ be the relation defined on $A$ such that $$R=\{(x, y) \in A \times A: x-y$$ is odd positive integer or $$x-y=2\}$$. The minimum number of elements that must be added to the relation $R$, so that ...

Answer: 19

Accepted Answer

We have, $A={0,3,4,6,7,8,9,10}$

Case I : $x-y$ is odd, if one is odd and one is even and $x>y$.

$	herefore $ Possibilites are ${(3,0),(4,3),(6,3),(7,6),(7,4)$, $(7,0),(8,7),(8,3),(9,8),(9,6),(9,4),(9,0),(10,9),(10$, $7),(10,3)}$

No. of cases $=15$

Case II : $x-y=2$

$	herefore $ Possibilities are ${(6,4),(8,6),(9,7),(10,8)}$

$	herefore $ No. of cases $=4$

So, minimum ordered pair to be added $=15+4=19$

Solutions · 1