Question

Show that addition and multiplication are associative binary operation on $R$. But subtraction is not associative on $R$. Division is not associative on $R_{\ast }$ .

Answer 1

Checking associativity of Addition on $R$.
$\ast$ is associative if $(a\ast b)\ast c=a\ast (b\ast c)$
Now checking $(a\ast b)\ast c$ and $a\ast (b\ast c)$
$(a\ast b)\ast c=(a+b)\ast c$
$=(a+b)+c$
$=a+b+c$
$a\ast (b\ast c)=a\ast (b+c)$
$=a+(b+c)$
$=a+b+c$
Since $(a\ast b)\ast c=a\ast (b\ast c)\forall a,b,cϵR,+$ is an associative binary operation.

Checking associativity of Multiplication on $R$.
$\ast$ is associative if
$(a\ast b)\ast c=a\ast (b\ast c)$
Now checking $(a\ast b)\ast c$ and $a\ast (b\ast c)$
$(a\ast b)\ast c=\left(ab\right)\ast c$
$=\left(ab\right)c$
$=abc$
$a\ast (b\ast c)=a\ast \left(bc\right)$
$=a\left(bc\right)$
$=abc$
Since $(a\ast b)\ast c=a\ast (b\ast c)\forall a,b,cϵR,\times$ is an associative binary operation.

Checking associativity of Subtraction on $R$.
${}^{\ast }$ is associative if
$(a\ast b)\ast c=a\ast (b\ast c)$
Now checking $(a\ast b)\ast c$ and $a\ast (b\ast c)$
$(a\ast b)\ast c=(a-b)\ast c$
$=(a-b)-c$
$=a-b-c$
$a\ast (b\ast c)=a\ast (b-c)$
$=a-(b-c)$
$=a-b+c$
Since $(a\ast b)\ast c\ne a\ast (b\ast c)\forall a,b,cϵR-$ is not an associative binary operation.

Checking associativity of Division on $R_{\ast }$.
$\ast$ is associative if,
$(a\ast b)\ast c=a\ast (b\ast c)$
Now checking $(a\ast b)\ast c$ \taxt{and} $a\ast (b\ast c)$
$(a\ast b)\ast c=\left(\frac{a}{b}\right)\ast c$
$=\left(\frac{a}{b}\right)÷c=\left(\frac{a}{b}\right)\times \frac{1}{c}=\frac{a}{bc}$
$a\ast (b\ast c)=a\ast \left(\frac{b}{c}\right)$
$=a÷\left(\frac{b}{c}\right)=a\times \frac{c}{b}=\frac{ac}{b}$
Since $(a\ast b)\ast c\ne a\ast (b\ast c)\forall a,b,cϵR_{\ast },÷$ is not an associative binary operation.