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Consider the binary operation $\ast :\; R \times R \rightarrow R$ and $o :\; R \times R \rightarrow R$ defined as $a \ast b = | a \text{-b}|$ and \(\;a\;o\;b=a, \forall a,\;b \in R.\) Show that \(\ast\) is commutative but not associative, \(o\) is associative but not commutative. Further, show that \(\forall\; a,\; b,\; c \in R,\; a\; \ast\; (b\; o\; c) = (a \ast b) \;o\; (a \ast c)\). [If it is so, we say that the operation $\ast$ distributes over $o$]. Does $o$ distribute over? Justify your answer.

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1 Answer

  • An operation $* : A \to B$ is commutative if $a*b=b*a$ for $a,b \in A$
  • An operation $*:A \to B$ is associative if $ (a*b)*c=a*(b*c) \qquad a,b,c \in A$
  • An operation * is distributive over another operation o if $a*(boc)=(a*b) o (a*c)$
Given $*:R \to R$ defined by $a*b=|a-b| \qquad a,b \in R$
$\textbf {Step 1: Checking if the operation is commutative}$:
An operation $\ast$ on $A$ is commutative if $a\ast b = b \ast a\; \forall \; a, b \in A$
$\Rightarrow a \ast b = |a-b|$
$\Rightarrow b*a=|b-a|=|-(a-b)|$ $=|a-b|$
Therefore $ a*b=b*a$. Hence operation * is commutative
$\textbf {Step 2: Checking if the operation is associative}$:
An operation $*:A \to B$ is associative if $(a*b)*c=a*(b*c) \qquad a,b,c \in A$
$\Rightarrow (a*(b*c)) = a*|b-c| = |a - |b-c||$
$\Rightarrow (a*b)*c = |a-b|*c = ||a-b|-c|$
We can show that they are not equal using a simple example, where $a=1,b=2,c=3$.
$(a*(b*c)) = |1-|2-3|| = |1-|-1|| = |1-1| = 0.$
$(a*b)*c = ||1-2| -3| = ||-1|-3| = |1-3| = |-2| = 2.$
Since $ (a*(b*c)) \neq (a*b)*c$, the * operation is not associative.
$\textbf {Step 3: Showing that a*(boc) = (aob)*(aoc)}$:
Let $a,b,c \in R$, $a*(boc)= a*(b) = |a-b|$
$(a*b) = |a-b|$ and $a*c = |a-c| \rightarrow (a*b) o (a*c) = |a-b| o |a-c| = |a-b|$
Therefore $a*(boc) = (aob)*(aoc)$:
$\textbf {Step 4: Checking to see if o distributes over *}$:
For $o$ to distribute over $\ast$, the following must be true: $ao(b*c) = (aob)*(boc)$
Consider $ao(b*c) \rightarrow ao(|b-c|) = a$
Consider $(aob) = a$ and $(boc) = b \rightarrow (aob)*(boc) = a*b = |a-b|$
Therefore, $ao(b*c) \neq (aob)*(boc) \rightarrow o$ does not distribute over $\ast$.
answered Feb 28, 2013 by meena.p
edited Mar 20, 2013 by balaji.thirumalai

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