Chapter ISG, Problem 7.1.7P

a.

To determine

If the operation of raising one number to an exponent is commutative.

Answer to Problem 7.1.7P

No, the operation is not commutative.

Explanation of Solution

Given: The operation $\ast$ defined as raising one number to the exponent of another. Thus,

  $a\ast b=a^b$ , is the given operation.

Concept Used: An operation $*$ defined on a set S is said to be commutative if-

  $a\ast b=b\ast a$ , $\forall a,b\in S$

Calculation: The given operation is defined as-

  $a\ast b=a^b$

Consider any two numbers, say 2 and 3. Then,

  $2\ast 3=2^3$

  $3\ast 2=3^2$

Clearly, $2^3\ne 3^2$ and hence $2\ast 3\ne 3\ast 2$.

Since the operation doesn’t follow commutative law for a particular set of numbers, thus the operation can’t be commutative.

b.

To determine

If the operation of raising one number to an exponent is associative.

Answer to Problem 7.1.7P

No, the operation is not associative.

Explanation of Solution

Given: The operation $\ast$ defined as raising one number to the exponent of another. Thus,

  $a\ast b=a^b$ , is the given operation.

Concept Used: An operation $*$ defined on a set S is said to be associative if-

  $a\ast (b\ast c)=(a\ast b)\ast c$ , $\forall a,b,c\in S$

Calculations: The given operation is defined as-

  $a\ast b=a^b$

Consider any three numbers, say 2, 3 and 4. Then,

  $2\ast (3\ast 4)$

  $=2\ast 3^4$

  $=2\ast 81$

  $=2^{81}$

Again,

  $(2\ast 3)\ast 4$

  $=2^3\ast 4$

  $=8\ast 4$

  $=8^4$

  $={\left(2^3\right)}^4$ (Since $2^3=8$ )

  $=2^{12}$ (Since from the laws of indices, ${\left(a^m\right)}^n=a^{mn}$ )

Clearly, $2^{81}\ne 2^{12}$ and hence $2\ast (3\ast 4)\ne (2\ast 3)\ast 4$.

Since the operation doesn’t follow associative law for a particular set of numbers, thus the operation can’t be associative.

c.

To determine

If the operation of raising one number to an exponent is distributive over addition.

Answer to Problem 7.1.7P

No, the operation is not distributive over addition.

Explanation of Solution

Given: The operation $\ast$ defined as raising one number to the exponent of another. Thus,

  $a\ast b=a^b$ , is the given operation.

Concept Used: Consider two operations operation $\ast$ and $\odot$ defined on a set S . Then the operation $\ast$ is said to be distributive over $\odot$ if-

  $a\ast (b\odot c)=(a\ast b)\odot (a\ast c)$ , $\forall a,b,c\in S$

Calculations: The given operation is defined as-

  $a\ast b=a^b$

To show that the operation $\ast$ is distributive over addition, the relation-

  $a\ast (b+c)=(a\ast b)+(a\ast c)$ , $\forall a,b,c\in S$ must be true.

Consider any three numbers, say 2, 3 and 4. Then,

  $2\ast (3+4)$

  $=2\ast 7$

  $=2^7$

Again,

  $(2\ast 3)+(2\ast 4)$

  $=2^3+2^4$

  $=2^3(1+2)$

  $=3\times 2^3$

Clearly, $2^7\ne 3\times 2^3$ and hence $2\ast (3+4)\ne (2\ast 3)+(2\ast 4)$.

Since the operation doesn’t follow distributive law for a particular set of numbers, thus the operation can’t be distributive over addition.

d.

To determine

If the operation of raising one number to an exponent is distributive over multiplication.

Answer to Problem 7.1.7P

No, the operation is not distributive over multiplication.

Explanation of Solution

Given: The operation $\ast$ defined as raising one number to the exponent of another. Thus,

  $a\ast b=a^b$ , is the given operation.

Concept Used: Consider two operations operation $\ast$ and $\odot$ defined on a set S . Then the operation $\ast$ is said to be distributive over $\odot$ if-

  $a\ast (b\odot c)=(a\ast b)\odot (a\ast c)$ , $\forall a,b,c\in S$

Calculations: The given operation is defined as-

  $a\ast b=a^b$

To show that the operation $\ast$ is distributive over multiplication, the relation-

  $a\ast (b\times c)=(a\ast b)\times (a\ast c)$ , $\forall a,b,c\in S$ must be true.

Consider any three numbers, say 2, 3 and 4. Then,

  $2\ast (3\times 4)$

  $=2\ast 12$

  $=2^{12}$

Again,

  $(2\ast 3)\times (2\ast 4)$

  $=2^3\times 2^4$

  $=2^{3+4}$ (Using laws of indices $a^m\times a^n=a^{m+n}$ )

  $=2^7$

Clearly, $2^{12}\ne 2^7$ and hence $2\ast (3\times 4)\ne (2\ast 3)\times (2\ast 4)$.

Since the operation doesn’t follow distributive law for a particular set of numbers, thus the operation can’t be distributive over multiplication.