The square root of 8 (\(\sqrt{8}\)) is not a rational number. This value belongs to the category of irrational numbers, which are numbers that cannot be represented as a simple fraction of two integers. Understanding this classification requires looking at the fundamental differences between the two main types of real numbers.
Defining Rational and Irrational Numbers
A rational number is defined as any number that can be expressed as the ratio of two integers, \(p\) and \(q\), where \(q\) is not zero. The word “rational” comes from “ratio,” signifying this comparison between two whole numbers. In decimal form, rational numbers either terminate or repeat in a predictable pattern.
Irrational numbers are real numbers that cannot be written as a simple fraction of two integers. Their decimal expansions are non-terminating and non-repeating, meaning the sequence of digits continues forever without any cyclical pattern. Examples include the number pi (\(\pi\)) and square roots of numbers that are not perfect squares.
Analyzing the Square Root of 8
To analyze the square root of 8, it is helpful to simplify the expression by looking for perfect square factors within the number 8. Since 8 can be factored into \(4 \times 2\), where 4 is a perfect square, \(\sqrt{8}\) simplifies to \(2\sqrt{2}\).
The number 2 in the simplified expression \(2\sqrt{2}\) is a rational number. The rationality of the entire expression, however, depends entirely on the nature of \(\sqrt{2}\). The product of any non-zero rational number and an irrational number is always irrational.
If \(2\sqrt{2}\) were rational, dividing it by the rational number 2 would imply that \(\sqrt{2}\) itself is rational. This is a contradiction because \(\sqrt{2}\) is a known irrational number. Therefore, since \(\sqrt{2}\) is irrational, the product \(2\sqrt{2}\), or \(\sqrt{8}\), must also be irrational.
Why Non-Perfect Square Roots Are Irrational
The irrationality of \(\sqrt{2}\) is the key to understanding why \(\sqrt{8}\) is irrational. Historically, the discovery of \(\sqrt{2}\) demonstrated that not all lengths could be expressed as a ratio of two integers. The proof of its irrationality uses a method called proof by contradiction.
This method starts by assuming \(\sqrt{2}\) can be written as a fraction \(p/q\) in simplest terms, where \(p\) and \(q\) are integers with no common factors. Squaring both sides of the equation leads to \(2 = p^2/q^2\), or \(2q^2 = p^2\). This shows that \(p^2\) must be an even number, and if \(p^2\) is even, \(p\) itself must also be even.
If \(p\) is even, it can be written as \(2k\) for some integer \(k\). Substituting \(2k\) back into the equation yields \(2q^2 = (2k)^2\), which simplifies to \(2q^2 = 4k^2\). Dividing by 2 results in \(q^2 = 2k^2\), proving that \(q^2\) is also an even number. Consequently, \(q\) must be even as well.
The logical contradiction arises because the initial assumption was that the fraction \(p/q\) was in its simplest form, meaning \(p\) and \(q\) had no common factors. Since the proof shows that both \(p\) and \(q\) are even, they share a common factor of 2. Because the initial assumption leads to a contradiction, it must be false, confirming that \(\sqrt{2}\) cannot be written as a simple fraction.