Question

Let a binary operation ${}^{\prime }{\forall }^{\prime }$ on Q (set of all rational numbers) be defined by $a\forall b=a+2b$ for all $a,b\in Q$. Then

• A. Q is closed under the given operation
• B. The given operation is commutative
• C. The given operation is associative
• D. Q is not closed under the given operation

Answer 1

Answer: (A)

$a\forall b=a+2b$, $b\forall a=b+2a=>a\forall b\ne b\forall a$
$\therefore \forall$ is not commutative
$(a\forall b)\forall c=(a+2b)\forall c=a+2b+c$
Again, a $\forall (,b\forall c)=a\forall (b+2c)$
= a + 2 (b + 2c)
$\therefore a\forall (b\forall c)\ne (a\forall b)\forall c$
$\therefore$ Ass. law does not hold.
$\forall a,b\in Q,a+2b\in Q\therefore Q$ is closed