Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Relations and Functions
0 votes

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.

Can you answer this question?

1 Answer

0 votes
  • 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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App