Question

The cube root of a negative integer is
(a) Negative
(b) Complex
(c) Real
(d) Positive

Answer 1

Hint:
Here, we need to find which of the options is correct. Let \[x\] be a positive integer. Thus, \[ - x\] is a negative integer. First, we will write the cube of the negative integer \[ - x\], and simplify the right hand side such that it is a negative number. Then, we will cube root both sides. Finally, using the equation and the definitions of a negative, complex, real, and positive number, we will find which of the given options is correct.

Complete step by step solution:
We need to find which of the given options is true for the cube root of a negative integer.
An integer is a rational number that is not a fraction.
For example: 1, \[ - 1\], 3, \[ - 7\], are integers.
Integers can be positive like 1, 3, etc. or negative like \[ - 1\].
The cube of a number \[x\] is given by \[x \times x \times x\]. It can be written using an exponent in the form \[x \times x \times x = {x^3}\], where \[x\] is the base and 3 is the exponent.
Let \[x\] be a positive integer. Thus, \[ - x\] is a negative integer.
The cube of \[ - x\] can be written as
\[ \Rightarrow {\left( { - x} \right)^3} = \left( { - x} \right) \times \left( { - x} \right) \times \left( { - x} \right)\]
The number \[ - x\] can be written as the product of the negative integer \[ - 1\], and the positive integer \[x\].
Thus, we get
$ \Rightarrow {\left( { - x} \right)^3} = \left( { - 1} \right) \times x \times \left( { - 1} \right) \times x \times \left( { - 1} \right) \times x \\
   \Rightarrow {\left( { - x} \right)^3} = {\left( { - 1} \right)^3} \times x \times x \times x \\ $
We know that \[{\left( { - 1} \right)^n}\] is equal to 1 if \[n\] is an even number, and is equal to \[ - 1\] if \[n\] is an odd number.
Therefore, \[{\left( { - 1} \right)^3} = - 1\].
The equation becomes
\[ \Rightarrow {\left( { - x} \right)^3} = \left( { - 1} \right) \times x \times x \times x\]
Now, the product of the three positive integers \[x\] is positive.
The product of the negative integer \[ - 1\] and the positive product \[x \times x \times x\] will be negative.
Therefore, \[\left( { - 1} \right) \times x \times x \times x\] is a negative integer.
Taking cube root of both sides, we get
\[ \Rightarrow - x = \sqrt[3]{{\left( { - 1} \right) \times x \times x \times x}}\]
Therefore, we can observe that the cube root of the negative integer \[\left( { - 1} \right) \times x \times x \times x\] is the negative integer \[ - x\].
Thus, option (a) is correct.
We will also check the remaining options because there may be more than one answer.
In a number of the form \[a + bi\], where \[i = \sqrt { - 1} \], if \[b = 0\], then the number is not complex.
The cube root of the negative integer \[\left( { - 1} \right) \times x \times x \times x\] is \[ - x\].
The number \[ - x\] can be written as \[ - x + 0i\].
Since \[b = 0\], the number \[ - x\] is not complex.
Therefore, the cube root of the negative integer \[\left( { - 1} \right) \times x \times x \times x\] is not complex.
Thus, option (b) is incorrect.
A real number is any number which is not complex.
We have proved that the cube root of the negative integer \[\left( { - 1} \right) \times x \times x \times x\] is not complex.
Therefore, the cube root of the negative integer \[\left( { - 1} \right) \times x \times x \times x\] is a real number.
Thus, option (c) is correct.
We know that a number is either positive or negative, or 0. Any negative number cannot be positive.
We have proved that the cube root of the negative integer \[\left( { - 1} \right) \times x \times x \times x\] is the negative integer \[ - x\].
Therefore, the cube root of the negative integer \[\left( { - 1} \right) \times x \times x \times x\] is not a positive number.
Thus, option (d) is incorrect.

We get that options (a) and (c) are correct.

Note:
A complex number is a number which can be written in the form \[a + bi\], where \[a\] and \[b\] are real numbers, and \[i\] is the imaginary unit. Here, \[i = \sqrt { - 1} \], which is not real.