For each binary operation $^*$ defined below,determine whether $^*$ is commutative or associative. On $Z$,define $a ^* b = a - b$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
On $Z$,$^*$ is defined by $a ^* b = a - b$.
It can be observed that $1 ^* 2 = 1 - 2 = -1$ and $2 ^* 1 = 2 - 1 = 1$.
$\therefore 1 ^* 2 \neq 2 ^* 1$,where $1, 2 \in Z$.
Hence,the operation $^*$ is not commutative.
Also,we have:
$(1 ^* 2) ^* 3 = (1 - 2) ^* 3 = -1 ^* 3 = -1 - 3 = -4$.
$1 ^* (2 ^* 3) = 1 ^* (2 - 3) = 1 ^* (-1) = 1 - (-1) = 2$.
$\therefore (1 ^* 2) ^* 3 \neq 1 ^* (2 ^* 3)$,where $1, 2, 3 \in Z$.
Hence,the operation $^*$ is not associative.

Explore More

Similar Questions

Is $^*$ defined on the set $\{1, 2, 3, 4, 5\}$ by $a \,^*\, b = \text{L.C.M. of } a \text{ and } b$ a binary operation? Justify your answer.

Let $*$ be a binary operation defined on the set of rational numbers $Q$. Determine whether the binary operation defined by $a * b = a^{2} + b^{2}$ for all $a, b \in Q$ is commutative.

Given a non-empty set $X$,let $^*: P(X) \times P(X) \rightarrow P(X)$ be defined as $A \,^*\, B = (A - B) \cup (B - A)$,$\forall A, B \in P(X)$. Show that the empty set $\Phi$ is the identity for the operation $^*$ and all the elements $A$ of $P(X)$ are invertible with $A^{-1} = A$.

Difficult
View Solution

The set $\{-1, 0, 1\}$ is not a multiplicative group because of the failure of

For each binary operation $^*$ defined below,determine whether $^*$ is commutative or associative. On $Q$,define $a ^* b = ab + 1$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo