Question

Consider the binary operations $\ast :R\times R\to R$ and $o:R\times R\to R$ defined as $a\ast b=|a-b|$ and $aob=a,\forall a,b\in R$. Show that $\ast$ is commutative but not associative, o is associative but not commutative. Further, show that $\forall a,b,c\in R,a\ast \left(boc\right)=(a\ast b)o(a\ast c)$. [If it is so, we say that the operation $\ast$ distributes over the operation o]. Dose o distribute over $\ast$? Justify your answer

Answer 1

Check commutative for $\ast$

$\ast$ is commutative if

$a\ast b=b\ast aa\ast b=|a-b|;b\ast a=|b-a|=|a-b|$

Since,

$a\ast b=b\ast a\forall a,bϵR$

$\therefore \ast$ is commutative.

Check associative for $\ast$

$\ast$ is associative if

$(a\ast b)\ast c=(a\ast b)\ast c$

$(a\ast b)\ast c=\left(|a-b|\right)\ast c=||a-b|-c|$

$a\ast (b\ast c)=a\ast \left(|b-c|\right)=|a-|b-c||$

Since $(a\ast b)\ast c\ne a\ast (b\ast c)$

$\ast$ is not associative.

$aob=a$

Check commutative for $0$

$0$ is commutative if, $a0b=b0a$

$a0b=aandb0a=b$

Since $a0b\ne b0a$

$0$ is not commutative.

Check associative for $0$

$0$ is associative if

$\left(a0b\right)0c=a0\left(b0c\right)$

$\left(a0b\right)0c=a0c=a$

$a0(b\ast c)=a0b=a$

Since, $\left(a0b\right)0c=a0\left(b0c\right)$

$0$ is not associative.

$a\ast b=|a-b|anda0b=a$

$0$ distributes over $\ast$

$Ifa0(b\ast c)=\left(a0b\right)\ast \left(a0c\right),\forall a,b,cϵR$

$0$ distributes over $\ast$

$a0(b\ast c)=a0|b-c|$

$\left(a0b\right)\ast \left(a0c\right)=a\ast a=|a-a|=\left|0\right|=0$

Since

$a0(b\ast c)\ne \left(a0b\right)\ast \left(a0c\right)$

$0$ does not distributes over $\ast$

$\ast$ distributes over 0

$Ifa\ast \left(b0c\right)=(a\ast b)0(a\ast c),\forall a,b,cϵR$

$\ast$ distributes over $0$.

$a\ast \left(b0c\right)=a\ast b=|a-b|$

$(a\ast b)0(a\ast c)=|a-b|0|a-c|=|a-b|$

Since

$a\ast \left(b0c\right)=(a\ast b)0(a\ast c),\forall a,b,cϵR$

$\ast$ distributes over $0$.