Question

For any two real numbers, an operation $\ast$ defined by $a\ast b=1+ab$ is

• A. commutative but not associative
• B. associative but not commutative
• C. neither commutative nor associative
• D. both commutative and associative

Answer 1

Answer: (A)

For Commutative
Given, $a^{\ast }b=1+ab$
Now, $b^{\ast }a=1+ba=1+ab$
$\therefore a^{\ast }b=b^{\ast }a$
So, it is commutative.
For Associative
${\left(a^{\ast }b\right)}^{\ast }c=(1+ab{)}^{\ast }c$
$=1+(1+ab)c$
$=1+c+abc$
and $a^{\ast }\left(b^{\ast }c\right)=a^{\ast }(1+bc)$
$=1+a(1+bc)$
$=1+a+abc$
$\therefore {\left(a^{\ast }b\right)}^{\ast }c\ne a^{\ast }\left(b^{\ast }c\right)$
So, it is not associative.