1. Tardigrade
  2. Question
  3. Mathematics
  4. Let a binary operation '∀' on Q (set of all rational numbers) be defined by a ∀ b= a+ 2b for all a, b ∈ Q. Then

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

Relations and Functions - Part 2

Solution:

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