Question

Determine whether each of the following operations define a binary operation on the given set or not :
(i) $\text{'}*\text{'}\mathrm{on}N\mathrm{defined}\mathrm{by}a*b=a^b\mathrm{for}\mathrm{all}a,b\in N.$
(ii) $\text{'}O\text{'}\mathrm{on}Z\mathrm{defined}\mathrm{by}aOb=a^b\mathrm{for}\mathrm{all}a,b\in Z.$
(iii) $\text{'}*\text{'}\mathrm{on}N\mathrm{defined}\mathrm{by}a*b=a+b-2\mathrm{for}\mathrm{all}a,b\in N$.

(iv) $$\begin{gathered} \text{'}{\times }_6\text{'}\mathrm{on}S=\left\{1,2,3,4,5\right\}\mathrm{defined}\mathrm{by} \\ a{\times }_{6}b=\mathrm{Remainder}\mathrm{when}ab\mathrm{is}\mathrm{divided}\mathrm{by}6. \end{gathered}$$

(v) $$\begin{gathered} \text{'}{+}_6\text{'}\mathrm{on}S=\left\{0,1,2,3,4,5\right\}\mathrm{defined}\mathrm{by} \\ a{+}_{6}b=\left\{\begin{array}{ll}a+b& ,\mathrm{if}a+b<6\\ a+b-6& ,\mathrm{if}a+b\ge 6\end{array}\right. \end{gathered}$$
(vi) $\text{'}\odot \text{'}\mathrm{on}N\mathrm{defined}\mathrm{by}a\odot b=a^b+b^a\mathrm{for}\mathrm{all}a,b\in N$
(vii) $\text{'}*\text{'}\mathrm{on}Q\mathrm{defined}\mathrm{by}a*b=\frac{a-1}{b+1}\mathrm{for}\mathrm{all}a,b\in Q.$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\text{Let}a,b\in N.\mathrm{Then}, \\ {a}^b\in N\left[\text{∵}{a}^b\ne 0\text{and}{\mathit{\text{a}}}^b\text{is positive integer}\right] \\ \Rightarrow a*b\in N \\ \text{Therefore,} \\ a*b\in N,\forall a,b\in N \end{gathered}$$

Thus, * is a binary operation on N.

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{Both}a=3\mathrm{and}b=-1\mathrm{belong}\mathrm{to}Z. \\ \Rightarrow a*b=3^{-1} \\ =\frac{1}{3}\notin Z \end{gathered}$$
Thus, * is not a binary operation on Z.

(iii) If a = 1 and b = 1,
a * b = a + b$-$ 2
= 1 + 1$-$ 2
= 0 $\notin N$
Thus, there exist a = 1 and b = 1 such that a * b $\notin N$
So, * is not a binary operation on N.

(iv) Consider the composition table,

${\times }_6$	1	2	3	4	5
1	1	2	3	4	5
2	2	4	0	2	4
3	3	0	3	0	3
4	4	2	0	4	2
5	5	4	3	2	1

Here all the elements of the table are not in S.

$$\begin{gathered} \Rightarrow \mathrm{For}a=2\mathrm{and}\mathrm{b}=3, \\ a{\times }_{6}b=2{\times }_{6}3=\mathrm{remainder}\mathrm{when}6\mathrm{divided}\mathrm{by}6=0\ne S \end{gathered}$$
${\text{Thus, ×}}_6\text{is not a binary operation on}\mathit{\text{S.}}$

(v) Consider the composition table,

${+}_6$	0	1	2	3	4	5
0	0	1	2	3	4	5
1	1	2	3	4	5	0
2	2	3	4	5	0	1
3	3	4	5	0	1	2
4	4	5	0	1	2	3
5	5	0	1	2	3	4

Here all the elements of the table are in S.
$\text{⇒}{\mathit{\text{a+}}}_{\mathit{6}}\mathit{\text{b ∈ S,∀ a, b ∈ S}}$

Thus, ${\times }_6$ is a binary operation on S.

$$\begin{gathered} \left(\mathrm{vi}\right)\text{Let}a,b\in N.\mathrm{Then}, \\ {a}^b,b^a\in N \\ \Rightarrow a^b+b^a\in N\left[\because \mathrm{Addition}\mathrm{is}\mathrm{binary}\mathrm{operation}\mathrm{on}N\right] \\ \Rightarrow a\odot b\in N \\ \text{Thus, ⊙ is a binary operation on}\mathit{\text{N.}} \end{gathered}$$

(vii) If a = 2 and b = $-$1 in Q,
$$\begin{gathered} a*b=\frac{a-1}{b+1} \\ =\frac{2-1}{-1+1} \\ =\frac{1}{0}\left[\text{which is not defined}\right] \\ \text{⇒ For}a=2\text{and}b=-1, \\ a*b\notin Q \end{gathered}$$

So, * is not a binary operation on Q.