Identify whether the following number is either a rational number or irrational number. And give the decimal representation of the rational number \[ - \sqrt {64} \].

Hint: Rational numbers: The numbers which can be expressed as the form \[\dfrac{p}{q}\], provided \[q \ne 0\]. For example, any integers (positive and negative), fraction.
Irrational numbers: The numbers which cannot be expressed as the form \[\dfrac{p}{q}\]. For example, \[13.3333...\].

Complete step by step answer:
To identify whether \[ - \sqrt {64} \] is a rational number or an irrational number, first we need to find the square root of the number.
The given number is \[ - \sqrt {64} \]
The prime factorization of \[64\] gives,
\[64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2\]
Now let us take square root on both sides of the above equation,
\[\sqrt {64} = \sqrt {2 \times 2 \times 2 \times 2 \times 2 \times 2} \]
\[\sqrt {2 \times 2 \times 2 \times 2 \times 2 \times 2} = \sqrt {{2^2} \times {2^2} \times {2^2}} \]
\[\sqrt {{2^2} \times {2^2} \times {2^2}} = 2 \times 2 \times 2 = 8\]
So,
The square root of the number \[\sqrt {64} \] is \[8\],
Therefore we can write the given number as \[ - \sqrt {64} = - 8\]
Here,
The number \[ - 8\] can be written as in the fraction form as \[\dfrac{{ - 8}}{1}\],
Which is the form of \[\dfrac{p}{q}\]. So, from the definition it is clear that the number given is a rational number.
Now, we have to represent the value of \[ - \sqrt {64} \] into decimal form.
The decimal representation of \[ - \sqrt {64} \] is: \[ - 8.0\]

Hence, We have found that the number \[ - \sqrt {64} \] is a rational number. And the decimal representation of \[ - \sqrt {64} \] is: \[ - 8.0\]

Note:
There is no value of \[0\] at the right side of the decimal. So, it will give the same value for an integer ‘a’ to write as \[a\] or \[a.0000\]. While finding the square root of the number \[64\] if we know that \[64\] is nothing but the square of the number \[8\] it can be easily done that \[\sqrt {64} = 8\] and the simplifications are made further easier.