Understanding Rational and Irrational Numbers

Numbers can be classified into different categories, including rational and irrational numbers. A rational number is any number that can be expressed as a fraction \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\). Examples include \(3\) (\(\frac{3}{1}\)), \(-0.5\) (\(\frac{-1}{2}\)), and \(0.333...\) (\(\frac{1}{3}\)).

An irrational number, on the other hand, cannot be expressed as a simple fraction \(\frac{p}{q}\). Their decimal representations are non-terminating and non-repeating. Famous examples include \(\pi\) and \(\sqrt{2}\).

The question asks us to identify which of the given numbers is irrational. The given numbers are different roots of 64. Let's evaluate each option.

Evaluating the Given Options

We will evaluate each option to determine if it results in a rational or irrational number.

Option 1: \(\sqrt[3]{{64}}\)

This is the cube root of 64. We need to find a number that, when multiplied by itself three times, equals 64.

We know that \(4 \times 4 \times 4 = 16 \times 4 = 64\).

So, \(\sqrt[3]{{64}} = 4\).

Since 4 can be written as \(\frac{4}{1}\), which is a fraction of integers, 4 is a rational number.

Option 2: \(\sqrt {64} \)

This is the square root of 64. We need to find a number that, when multiplied by itself, equals 64.

We know that \(8 \times 8 = 64\).

So, \(\sqrt {64} = 8\).

Since 8 can be written as \(\frac{8}{1}\), which is a fraction of integers, 8 is a rational number.

Option 3: \(\sqrt[6]{{64}}\)

This is the sixth root of 64. We need to find a number that, when multiplied by itself six times, equals 64.

We can write 64 as a power of 2: \(2 \times 2 = 4\), \(4 \times 2 = 8\), \(8 \times 2 = 16\), \(16 \times 2 = 32\), \(32 \times 2 = 64\).

So, \(2^6 = 64\).

Therefore, \(\sqrt[6]{{64}} = \sqrt[6]{{2^6}} = 2\).

Since 2 can be written as \(\frac{2}{1}\), which is a fraction of integers, 2 is a rational number.

Option 4: \(\sqrt[4]{{64}}\)

This is the fourth root of 64. We need to find a number that, when multiplied by itself four times, equals 64.

Let's express 64 as a power of a prime number: \(64 = 2^6\).

So, \(\sqrt[4]{{64}} = \sqrt[4]{{2^6}}\).

Using the property of roots and exponents, \(\sqrt[n]{a^m} = a^{m/n}\), we can write:

\(\sqrt[4]{{2^6}} = 2^{6/4} = 2^{3/2}\).

Now, let's simplify \(2^{3/2}\):

\(2^{3/2} = 2^{1 + 1/2} = 2^1 \times 2^{1/2} = 2\sqrt{2}\).

The number is \(2\sqrt{2}\). We know that \(\sqrt{2}\) is an irrational number. When a rational number (2) is multiplied by an irrational number (\(\sqrt{2}\)), the result is an irrational number.

Alternatively, we can see that 64 is not a perfect fourth power of any integer:

• \(1^4 = 1\)
• \(2^4 = 16\)
• \(3^4 = 81\)

Since 64 is not a perfect fourth power, its fourth root, \(\sqrt[4]{64}\), is an irrational number.

Conclusion

Based on the evaluation of each option:

• \(\sqrt[3]{{64}} = 4\) (Rational)
• \(\sqrt {64} = 8\) (Rational)
• \(\sqrt[6]{{64}} = 2\) (Rational)
• \(\sqrt[4]{{64}} = 2\sqrt{2}\) (Irrational)

Therefore, the irrational number among the given options is \(\sqrt[4]{{64}}\).

Revision Table: Roots of 64

Additional Information: Identifying Irrational Numbers

Here are some tips for identifying irrational numbers:

• Square roots (\(\sqrt{n}\)), cube roots (\(\sqrt[3]{n}\)), or any nth root (\(\sqrt[n]{n}\)) of integers that are not perfect squares, perfect cubes, or perfect nth powers (respectively) are generally irrational. For example, \(\sqrt{5}\), \(\sqrt[3]{10}\), \(\sqrt[4]{7}\).
• Numbers like \(\pi\) (pi) and \(e\) (Euler's number) are irrational.
• The sum, difference, product, or quotient of a non-zero rational number and an irrational number is always irrational. For example, \(2 + \sqrt{3}\), \(5\pi\), \(\frac{\sqrt{7}}{3}\).
• The sum, difference, product, or quotient of two irrational numbers can be either rational or irrational. For example, \(\sqrt{2} \times \sqrt{2} = 2\) (rational), but \(\sqrt{2} + \sqrt{3}\) is irrational.

When dealing with roots like \(\sqrt[n]{a}\), check if \(a\) is a perfect nth power. If it is, the root is rational. If it is not, the root is likely irrational (unless \(a\) can be factored in a way that simplifies the root into a rational number, which is not the case here for \(\sqrt[4]{64}\)).