Question

Determine whether or not each of the definition of $\ast$ given below gives a binary operation. In the event that $\ast$ is not a binary operation, give justification for this.
(i) On $Z^{+}$, defined $\ast$ by $a\ast b=a-b$

(ii) On $Z^{+}$, defined $\ast$ by $a\ast b=ab$

(iii) On R, defined $\ast$ by $a\ast b=a{b}^2$

(iv) On $Z^{+}$, defined $a\ast b=|a-b|$

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

Answer 1

On $Z^{+},\ast$ is defined by $a\ast b=a-b$
It is not a binary operation as the image of (1,2) under $\ast$ is $1\ast 2=1-2=-1\text{/}\in Z^{+}$

On $Z^{+}$, $\ast$ is defined by $a\ast b=ab.$
It is seen that for each a,b in $Z^{+}$, there is a unique element ab in $Z^{+}.$
This means that as $\ast$ carries each pair (a,b) to a unique element $a\ast b=abin{Z}^{+}.$
Therefore, $\ast$ is a binary operation.

On R, defined $\ast$ by $a\ast b=a{b}^2$
It is seen that for each $a,b\in R$, there is a unique element $a{b}^2$ in R.
This means that $\ast$ carries each pair (a,b)to a unique element $a\ast b=a{b}^2$ in R. Therefore, $\ast$ is binary operation.

On $Z^{+}$, defined $a\ast b=|a-b|$
It is seen that for each a,b, $\in Z^{+}$, there is a unique element $|a-b|in{Z}^{+}.$
This means that $\ast$ carries each pair (a,b) to a unique element $a\ast b=|a-b|in{Z}^{+}$. Therefore, $\ast$ is a binary operation.

On $Z^{+}$, defined $\ast$ by $a\ast b=a$
It is seen that for each $a,b\in Z^{+}$, there is a unique element $a\in Z^{+}.$
This means that $\ast$ carries each pair (a,b) to a unique element $a\ast b$ =a in $Z^{+}.$
Therefore, $\ast$ is a binary operation.