Determine whether or not each of the definition of given below gives a binary operation. In the event that * is not a binary operation, give justification for this.
(i) On $\mathbf{Z}^{+}$, define ${ }^{*}$ by $a{ }^{*} b=a-b$
(ii) $\mathrm{On} \mathbf{Z}^{+}$, define ${ }^{*}$ by $a^{*} b=a b$
(iii) On $\mathbf{R}$, define * ${ }^{*}$ by $a^{*} b=a b^{2}$
(iv) On $\mathbf{Z}^{+}$, define ${ }^{*}$ by $a^{*} b=|a-b|$
(v) On $\mathbf{Z}^{+}$, define ${ }^{*}$ by $a^{*} b=a$
(i) On $\mathbf{Z}^{+}$, ${ }^{*}$ is defined by $a^{*} b=a-b$.
It is not a binary operation as the image of $(1,2)$ under ${ }^{*}$ is $1^{*} 2=1-2$ $=-1 \notin Z^{+}$.
(ii) $\mathrm{On} \mathbf{Z}^{+}{ }^{*}{ }^{*}$ is defined by $a{ }^{*} b=a b$.
It is seen that for each $a, b \in \mathbf{Z}^{+}$, there is a unique element $a b$ in $\mathbf{Z}^{+}$.
This means that * carries each pair $(a, b)$ to a unique element $a^{*} b=a b$ in $\mathbf{Z}^{+}$.
Therefore, ${ }^{*}$ is a binary operation.
(iii) On $\mathbf{R}$, ${ }^{*}$ is defined by $a^{*} b=a b^{2}$.
It is seen that for each $a, b \in \mathbf{R}$, there is a unique element $a b^{2}$ in $\mathbf{R}$.
This means that * carries each pair $(a, b)$ to a unique element $a$ * $b=a b^{2}$ in $\mathbf{R}$.
Therefore, ${ }^{*}$ is a binary operation.
(iv) On $\mathbf{Z}^{+},{ }^{*}$ is defined by $a^{*} b=|a-b|$.
It is seen that for each $a, b \in \mathbf{Z}^{+}$, there is a unique element $|a-b|$ in $\mathbf{Z}^{+}$.
This means that * carries each pair $(a, b)$ to a unique element $a{ }^{*} b=$ $|a-b|$ in $\mathbf{Z}^{+}$.
Therefore, ${ }^{*}$ is a binary operation.
(v) $\mathrm{On} \mathbf{Z}^{+}{ }^{*}{ }^{*}$ is defined by $a^{*} b=a$.
${ }^{*}$ carries each pair $(a, b)$ to a unique element $a^{*} b=a$ in $\mathbf{Z}^{+}$.
Therefore, ${ }^{*}$ is a binary operation.