Chapter 6.1, Problem 44PPSE

Solution

a.

To determine that given statement is true or not.

The given statement is true for all positive integer values of $n.$

Given:

If $x^n=b,$ then $x$ is an $n^{th}$ root of $b$

Explanation:

For

  $x^n=b$

where $x$ is a real number and $n$ is a positive integer, then the number $x$ is the $n^{th}$ root of $b$ .

For example, the cube root of number $27$ is $3.$ The index $n$ of a cube root is positive integer equal to $3.$ Hence, if $x^n=b,$ the real number $x$ is the $n^{th}$ root of $b$ for all positive integer values of $n.$

Therefore, the given statement is true.

b.

To determine that given statement is true or not.

If $x^n=b,$ then $x=\sqrt[n]{b}$ is true, for odd integers values of $n.$

Given:

If $x^n=b,$ then $x=\sqrt[n]{b}.$

Explanation:

Let consider that $x$ is a real number and the expression $\sqrt[n]{b}$ represents the non-negative $n^{th}$ root of $b$ . When the $n^{th}$ root of a number $b$ is calculated, then there are two possible roots are $\sqrt[n]{b}$ and $-\sqrt[n]{b}.$

For example, the number $2$ and $-2$ are the two real square roots of $4$ as both ${\left(2\right)}^2$ and ${\left(-2\right)}^2$ are equal to $4.$ It means that square root of the number $4$ are both $\sqrt{4}$ and $-\sqrt{4}$ .

Hence, for

  $x^n=b$

then $x=\sqrt[n]{b}$ is not necessarily true. It could be $-\sqrt[n]{b}$ when $n$ is even.Hence, the given statement is not true for even integers value of $n.$

Next, consider an example that the number $2$ is the only cube root of $8$ as $2^3=8.$ Hence, the $n^{th}$ root of $x$ is equal to only $\sqrt[n]{b}$ when $n$ is an odd number. Therefore, the given statement is true for odd integers value of $n.$