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(i) On ${Z^ + }$ , define $ * $ by $a * b = a - b$

(ii) On ${Z^ + }$ , define $ * $ by $a * b = ab$

(iii) On $R$ , define $ * $ by $a * b = a{b^2}$

(iv) On ${Z^ + }$ , define $ * $ by $a * b = \left| {a - b} \right|$

(v) On ${Z^ + }$ , define $ * $ by $a * b = a$

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Hint: Using definition of binary operation.

(i) On ${Z^ + }$ , the binary operation$ * $ defined by $a * b = a - b$ is not a binary operation

Because if the points are taken as $\left( {1,2} \right)$ , then by applying binary operation, it becomes $1 - 2 = - 1$ and $ - 1$ does not belong to${Z^ + }$ .

(ii) On ${Z^ + }$ , the binary operation$ * $ defined by$a * b = ab$ is a binary operation because each element in ${Z^ + }$ has a unique element in ${Z^ + }$ .

(iii) On $R$ , the binary operation$ * $ defined by $a * b = a{b^2}$ is a binary operation because each element in $R$ has a unique element in $R$ .

(iv) On ${Z^ + }$ , the binary operation $ * $ defined by $a * b = \left| {a - b} \right|$ is a binary operation because each element in ${Z^ + }$ has a unique element in ${Z^ + }$ .

(v) On ${Z^ + }$ , the binary operation $ * $ defined by $a * b = a$ is a binary operation because each element in \[{Z^ + }\] has a unique element in \[{Z^ + }\] .

Note: - In order to prove that a given operation is not a binary operation just as in case I, we just need to show an example satisfying that the operation is not binary. But in all other cases, or to show that the given operation is binary we need to consider all the different possibilities and also some exceptional cases.

(i) On ${Z^ + }$ , the binary operation$ * $ defined by $a * b = a - b$ is not a binary operation

Because if the points are taken as $\left( {1,2} \right)$ , then by applying binary operation, it becomes $1 - 2 = - 1$ and $ - 1$ does not belong to${Z^ + }$ .

(ii) On ${Z^ + }$ , the binary operation$ * $ defined by$a * b = ab$ is a binary operation because each element in ${Z^ + }$ has a unique element in ${Z^ + }$ .

(iii) On $R$ , the binary operation$ * $ defined by $a * b = a{b^2}$ is a binary operation because each element in $R$ has a unique element in $R$ .

(iv) On ${Z^ + }$ , the binary operation $ * $ defined by $a * b = \left| {a - b} \right|$ is a binary operation because each element in ${Z^ + }$ has a unique element in ${Z^ + }$ .

(v) On ${Z^ + }$ , the binary operation $ * $ defined by $a * b = a$ is a binary operation because each element in \[{Z^ + }\] has a unique element in \[{Z^ + }\] .

Note: - In order to prove that a given operation is not a binary operation just as in case I, we just need to show an example satisfying that the operation is not binary. But in all other cases, or to show that the given operation is binary we need to consider all the different possibilities and also some exceptional cases.

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