Let $\ast$ be a binary operation on the set $Q$ of rational numbers as follows:
(i) $a\ast b = a - b$ (ii) $a\ast b = a^{2} + b^{2}$
(iii) $a\ast b = a + ab$ (iv) $a\ast b = (a - b)^{2}$
(v) $a\ast b = \dfrac {ab}{4}$ (vi) $a\ast b = ab^{2}$
Find which of the binary operations are commutative and which are associative
- $ii, iv, v$ are commutative and $v$ associative
- $ii, iv, v$ are not commutative and $v$ associative
- $iii, iv, v$are commutative and $v$ associative
- $vi, iv, v$are commutative and $v$ associative
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A
Correct answer
Explanation
$(i)$ $a\ast b=a-b$
Check commutative is
$a\ast b=b\ast a$
$a\ast b=a-b$
$b\ast a=b-a$
Since, $a\ast b\neq b\ast a$
$\ast$ is not commutative.
Check associative
$\ast$ is associative if
$(a\ast b)\ast c=a\ast (b\ast c)\ (a\ast b)\ast c={ (a-b) }^{ \ast }c=(a-b)-c=a-b-c\ a\ast (b\ast c)=a\ast (b-c)=a-(b-c)=a-b+c$
Since $ (a\ast b)\ast c\neq a\ast (b\ast c)$
$\ast$ is not an associative binary operation.
$(ii)$ $a\ast b={ a }^{ 2 }+{ b }^{ 2 }$
Check commutative
$\ast$ is commutative if $a\ast b=b\ast a$
$a\ast b={ a }^{ 2 }+{ b }^{ 2 }\ b\ast a={ b }^{ 2 }+{ a }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }$
Since $ a\ast b=b\ast a\quad \forall\quad a,b\epsilon Q$
$\ast$ is commutative.
Check associative
$\ast$ is associative if
$(a\ast b)\ast c=a\ast (b\ast c)\ (a\ast b)\ast c=({ a }^{ 2 }+{ b }^{ 2 })\ast c={ ({ a }^{ 2 }+{ b }^{ 2 }) }^{ 2 }+{ c }^{ 2 }\ a\ast (b\ast c)=a\ast ({ b }^{ 2 }+{ c }^{ 2 })={ a }^{ 2 }+{ ({ b }^{ 2 }+{ c }^{ 2 }) }^{ 2 }$
Since $ (a\ast b)\ast c\neq a\ast (b\ast c)$
$\ast$ is not an associative binary operation.
$(iii)$ $a\ast b=a+b$
Check commutative
$\ast$ is commutative is $a\ast b=b\ast a$
$ a\ast b=a+ab;\quad b\ast a=b+ba$
Since $ a\ast b\neq b\ast a$
$\ast$ is not commutative.
$(iv)$ $a\ast b={ (a-b) }^{ 2 }$
Check commutative
$\ast$ is commutative if $a\ast b=b\ast a$
$ a\ast b={ (a-b) }^{ 2 }\quad ;\quad b\ast a={ (b-a) }^{ 2 }={ (a-b) }^{ 2 }$
Since $ a\ast b=b\ast a\quad \forall\quad a,b\epsilon Q$
$\ast$ is commutative.
Check associative
$\ast$ if
$(a\ast b)\ast c=a\ast (b\ast c)\ (a\ast b)\ast c={ (a-b) }^{ 2 }\ast c={ [{ (a-b) }^{ 2 }-c] }^{ 2 }\ a\ast (b\ast c)=a\ast { (b-c) }^{ 2 }={ [a-{ (b-c) }^{ 2 }] }^{ 2 }$
Since $ (a\ast b)\ast c\neq a\ast (b\ast c)$
$\ast$ is not an associative binary operation.
$(v)$ $a\ast b=\cfrac { ab }{ 4 } $
Check commutative.
$\ast$ is commutative if $a\ast b=b\ast a$
$ a\ast b=\cfrac { ab }{ 4 } \quad ;\quad b\ast a=\cfrac { ba }{ 4 } =\cfrac { ab }{ 4 } $
Since $ a\ast b=b\ast a\quad \forall\quad a,b\epsilon Q$
$\ast$ is commutative.
Check associative.
$\ast$ is association if $(a\ast b)\ast c=a\ast (b\ast c)$
$(a\ast b)\ast c=(\cfrac { \cfrac { ab }{ 4 } \ast c }{ 4 } )=\cfrac { abc }{ 16 } \ a\ast (b\ast c)=a\ast (\cfrac { bc }{ 4 } )=\cfrac { a\times \cfrac { bc }{ 4 } }{ 4 } =\cfrac { abc }{ 16 } $
Since $ (a\ast b)\ast c=a\ast (b\ast c)\quad \forall\quad a,b,c\epsilon Q$
$\ast$ is an associative binary operation.
$(vi)$ $a\ast b={ ab }^{ 2 }$
check commutative.
$\ast$ is commutative if $a\ast b=b\ast a$
$ a\ast b={ ab }^{ 2 }\quad ;\quad b\ast a={ ba }^{ 2 }$
Since $ a\ast b\neq b\ast a$
$\ast$ is not commutative.
Check associative
$\ast$ is associative if $(a\ast b)\ast c=a\ast (b\ast c)$
$(a\ast b)\ast c={ ab }^{ 2 }\ast c=({ ab }^{ 2 }){ c }^{ 2 }=a{ b }^{ 2 }{ c }^{ 2 }.\ a\ast (b\ast c)=a\ast { bc }^{ 2 }=a{ ({ bc }^{ 2 }) }^{ 2 }=a{ b }^{ 2 }{ c }^{ 4 }$
Since $ (a\ast b)\ast c\neq a\ast (b\ast c)$
$\ast$ is not an associate binary operation.