Question

For each binary operation $\ast$ defined below, determine whether $\ast$ is binary, commutative or associative.
(i) On Z, define a $\ast$ b =a-b

(ii) On Q, define $a\ast b=ab+1$

(iii) On Q, define $a\ast b=\frac{ab}{2}$

(iv) On $Z^{+}$, define $a\ast b=2^{ab}$

(v) On $Z^{+}$, define $a\ast b=a^b$

(vi) On (R-{-1} ,define $a\ast b=\frac{a}{b+1}$

Answer 1

On Z, define a $\ast$ b =a-b
$a-b\in Z$, so the operation $\ast$ is binary
It can be observed that $1\ast 2=1-2=-1$ and $2\ast 1=2-1=1.$
Therefore, $1\ast 2\ne 2\ast 1$ where $1,2\in Z$
Hence, the operation $\ast$ is not commutative.
Also, we have $(1\ast 2)\ast 3=(1-2)\ast 3=-1\ast 3=-1-3=-4$
$1\ast (2\ast 3)=1\ast (2-3)=1\ast -1=1-(-1)=2$
Therefore, $(1\ast 2)\ast 3\ne 1\ast (2\ast 3)$, where $1,2,3\in Z$
Hence, the operation $\ast$ is not associative.

On Q, define $a\ast b=ab+1$
$ab+1\in Q$, so operation $\ast$ is binary
It is known that
ab =ba for $a,b\in Q$
Therefore, ab +1 =ba +1 for $a,b\in Q$
Therefore, $a\ast b=b\ast$ a for $a,b\in Q$
Therefore, the opeartion $\ast$ is commutative. It can be ovserved that
$(1\ast 2)\ast 3=(1\times 2+1)\ast 3=3\ast 3=3\times 3+1=101\ast (2\ast 3)=1\ast (2\times 3+1)=1\ast 7=1\times 7+1=8$
Therefore, $(1\ast 2)\ast 3\ne 1\ast (2\ast 3)$, where $1,2,3\in Q$
Therefore, the operation $\ast$ is not associative.

On Q, define $a\ast b=\frac{ab}{2}$
$\frac{ab}{2}\in Q$, so the operation $\ast$ is binary. It is known that
ab =ba for $a,b\in Q$
Therefore, $\frac{ab}{2}=\frac{ba}{2}$ for $a,b\in Q$
Therefore, $a\ast b=b\ast a$ for a, $b\in Q$
Therefore, the operation $\ast$ is commutative.
For all $a,b,c\in Q$, we have
$(a\ast b)\ast c=\left(\frac{ab}{2}\right)\ast c=\frac{\left(\frac{ab}{2}\right)c}{2}=\frac{abc}{4}a\ast (b\ast c)=a\ast \left(\frac{bc}{2}\right)=\frac{a\left(\frac{bc}{2}\right)}{2}=\frac{abc}{4}$
Therefore,$(a\ast b)\ast c=a\ast (b\ast c)$
Therefore, the operation $\ast$ is associative.

On $Z^{+}$, define $a\ast b=2^{ab}$
$2^{ab}\in Z^{+}$, so the operation $\ast$ is binary opeartion
It is known that
ab=ba for $a,b\in Z^{+}$
Therefore, $2^{ab}=2^{ba}fora,b\in Z^{+}$
Therefore, $a\ast b=b\ast a$ for $a,b\in Z^{+}$
Therefore, the operation $\ast$ is commutative. It can be observed that
$(1\ast 2)\ast 3=2^{(1\times 2)}\ast 3=4\ast 3=2^{4\times 3}=2^{12}$
$1\ast (2\ast 3)=1\ast 2^{2\times 3}=1\ast 2^6=1\ast 64=2^{64}$
Therefore, $(1\ast 2)\ast 3\ne 1\ast (2\ast 3)$, where $1,2,3,\in Z^{+}$
Therefore, the operation $\ast$ is not associative.

On $Z^{+},\ast$ is defined by $a\ast b=a^b,$
$a^b\in Z^{+}$, so the operation $\ast$ is binary operation.
It can be observed that $1\ast 2=1^2=1$ and $2\ast 1=2^1=2$
Therefore, $1\ast 2\ne 2\ast 1$ where $1,2,\in Z^{+}$
Therefore, the operation $\ast$ is is not commutative.

It can also be observed that
$(2\ast 3)\ast 4=2^3\ast 4=8\ast 4=8^4=(2^3{)}^4=2^{12}2\ast (3\ast 4)=2\ast 3^4=2\ast 81=2^{81}$
Therefore, $(2\ast 3)\ast 4\ne 2\ast (3\ast 4);where2,3,4\in Z^{+}$
Therefore, the operation $\ast$ is not associative

On R-{-1},define $a\ast b=\frac{a}{b+1}$
$\frac{a}{b+1}\in R$ for $b\ne -1$ so that operation $\ast$ is binary.
It can be observed that $1\ast 2=\frac{1}{2+1}=\frac{1}{3}$ and $2\ast 1=\frac{2}{1+1}=1$
Therefore, $1\ast 2\ne 2\ast 1$ where $1,2\in R--1$
Therefore, the operation $\ast$ is not commutative.
It can also be obseved that $(1\ast 2)\ast 3=\frac{1}{3}\ast 3=\frac{\frac{1}{3}}{3+1}=\frac{1}{12}$
$1\ast (2\ast 3)=1\ast \frac{2}{3+1}=1\ast \frac{2}{4}=1\ast \frac{1}{2}=\frac{1}{\frac{1}{2}+1}=\frac{1}{\frac{3}{2}}=\frac{2}{3}$
Therefore, $(1\ast 2\ast 3)\ne 1\ast (2\ast 3)$ where $1,2,3\in R--1$
Therefore, the operation $\ast$ is not associative.