Section 8.3 Operations With Irrational Numbers
Subsection 8.3.1 Addition and Subtraction with Irrational Numbers
As a reminder, when we add two rational numbers, the result is also a rational number. This is known as the closure property of addition for rational numbers. Formally, if \(a\) and \(b\) are rational numbers, then \(a + b\) and \(a - b\) are also rational numbers.
However, we can show that the sum of a rational number and an irrational number is always irrational. Let’s state this formally as a theorem and prove it by contradiction.
Proof.
Assume for the sake of contradiction that \(r + i\) is a rational number. Since the set of rational numbers is closed under addition and subtraction, we can write:
We can then subtract \(r\) from both sides of the equation:
\(i = q - r\text{.}\)
Since both \(q\) and \(r\) are rational numbers, and the difference of two rational numbers is also rational, \(q - r\) must be rational.
This implies that \(i\) is rational, which contradicts our initial assumption that \(i\) is irrational. Therefore, our assumption that \(r + i\) is rational must be false, and \(r + i\) is indeed irrational.
It is important to note that while the sum of a finite number of rational numbers is rational, the sum of infinitely many rational numbers can be irrational. This can be understood by considering the decimal expansion of certain numbers.
For example, the number \(\pi\) (pi) is known to be irrational. Its decimal expansion is non-repeating and non-terminating:
\(\pi = 3.141592653589793238...\)
If we express \(\pi\) as the sum of its decimal parts, we get an infinite series of rational numbers:
\(\pi = 3 + 0.1 + 0.04 + 0.001 +0.0005 + 0.00009 + 0.000002 + \ldots\)
Each term in this series is a rational number, but their infinite sum converges to the irrational number \(\pi\text{.}\) This demonstrates that the sum of infinitely many rational numbers can indeed be irrational.
When adding two irrational numbers, the result can be either rational or irrational. Let’s explore this with two examples:
Example 8.3.NaN.
Consider the irrational numbers \(\sqrt{2}+1\text{,}\) which we know is irrational by Theorem Theorem 8.3.NaN, and \(-\sqrt{2}\text{.}\) Their sum is:
\(\sqrt{2}+1 + (-\sqrt{2}) = 1\text{.}\)
Of course, \(1\) is a rational number. Thus in this example the sum of two irrational numbers is a rational number.
Example 8.3.NaN.
\(\sqrt{2} + \sqrt{3}\text{.}\)
This sum is still an irrational number because there is no way to express it as a ratio of two integers; we will justify this claim in the following section.
We can summarize the possible results of adding or subtracting rational and irrational numbers in the following table:
| + | Rational | Irrational |
|---|---|---|
| Rational | Rational | Irrational |
| Irrational | Irrational | Either |
Checkpoint 8.3.NaN.
Consider the number \(\sqrt{2}\text{.}\) Show that \(\sqrt{2} + 1\) and \(\sqrt{2} - 1\) are both irrational.
Solution.
Assume, for the sake of contradiction, that \(\sqrt{2} + 1\) is rational. Then we can write:
Subtracting 1 from both sides, we get:
\(\sqrt{2} = r - 1\text{.}\)
Since \(r\) is rational, \(r - 1\) is also rational, implying that \(\sqrt{2}\) is rational, which is a contradiction. Hence, \(\sqrt{2} + 1\) is irrational.
A similar argument shows that \(\sqrt{2} - 1\) is irrational by assuming it is rational and arriving at a contradiction that \(\sqrt{2}\) would be rational.
Subsection 8.3.1 Multiplication and Division with Irrational Numbers
Multiplication and division of irrational numbers shares many of the same properties as addition does, as explored above. The biggest difference is keeping in mind that multiplying by 0 is a special case; we always know that multiplying by 0 gives us 0. Also, remember that division by 0 is something that is not defined.
As a reminder, when we multiply two rational numbers, the result is also a rational number. This is known as the closure property of multiplication for rational numbers. Formally, if \(a\) and \(b\) are rational numbers, then \(a \times b\) is also a rational number.
However, similar to addition, we can show that the product of a non-zero rational number and an irrational number is always irrational. Let’s state this formally as a theorem and prove it by contradiction.
Theorem 8.3.5.
Let \(r\) be a non-zero rational number and \(i\) be an irrational number. Then \(r \times i\) is irrational.
Proof.
Assume for the sake of contradiction that \(r \times i\) is a rational number. Since the set of rational numbers is closed under multiplication and division, we can write:
\(i = \frac{q}{r}\text{.}\)
Since both \(q\) and \(r\) are rational numbers, and the quotient of two rational numbers is also rational, \(\frac{q}{r}\) must be rational.
This implies that \(i\) is rational, which contradicts our initial assumption that \(i\) is irrational. Therefore, our assumption that \(r \times i\) is rational must be false, and \(r \times i\) is indeed irrational.
Note that multiplying any number by zero always results in zero, which is rational. Thus, the product of zero and any irrational number is zero.
It is important to note that while the product of a finite number of rational numbers is rational, the product of infinitely many rational numbers can be irrational. This can be understood by considering the Wallis product for \(\pi\text{.}\)
The Wallis product for \(\pi\) is an infinite product of rational numbers that converges to the irrational number \(\pi\text{.}\) Showing that the product below converges to \(\pi\) is slightly beyond the scope of this book, but you can explore this yourself by looking at "partial" products and continuing to multiply by successive terms. You should notice that these partial products get closer and closer to \(\pi\text{:}\)
\(\frac{\pi}{2} = \prod_{n=1}^{\infty} \frac{2n}{2n-1}\times \frac{2n}{2n+1} = \left(\frac{2}{1} \times \frac{2}{3}\right) \times \left(\frac{4}{3} \times \frac{4}{5}\right) \times \left(\frac{6}{5} \times \frac{6}{7}\right) \times \ldots\)
Each factor in this product is a rational number, but the infinite product converges to the irrational number \(\pi\text{.}\)
When multiplying two non-zero irrational numbers, the result can be either rational or irrational. Let’s explore this with two examples:
Example 8.3.6.
Consider the irrational numbers \(\sqrt{2}\) and \(\frac{1}{\sqrt{2}}\text{.}\) We know the second of these numbers is irrational by Theorem Theorem 8.3.5. Their product is:
\(\sqrt{2} \times \frac{1}{\sqrt{2}} = 1\text{.}\)
Here, the product of two irrational numbers is a rational number (1).
Example 8.3.7.
\(\sqrt{2} \times \sqrt{5} = \sqrt{10}\text{.}\)
This product is still an irrational number which we know is irrational since it is a square root of a non-square number.
We can summarize the possible results of multiplying or dividing rational and irrational numbers in the following table:
| \(\times \) | Rational | Irrational |
|---|---|---|
| Rational | Rational | Irrational |
| Irrational | Irrational | Either |
Checkpoint 8.3.8.
Consider the number \(\sqrt{2}\text{.}\) Show that \(\sqrt{2} \times 2\) and \(\sqrt{2} \times \sqrt{3}\) are both irrational.
Solution.
Assume, for the sake of contradiction, that \(\sqrt{2} \times 2\) is rational. Then we can write:
Dividing both sides by 2, we get:
\(\sqrt{2} = \frac{r}{2}\text{.}\)
Since \(r\) is rational, \(\frac{r}{2}\) is also rational, implying that \(\sqrt{2}\) is rational, which is a contradiction. Hence, \(\sqrt{2} \times 2\) is irrational.
A similar argument shows that \(\sqrt{2} \times \sqrt{3}\) is irrational by assuming it is rational and arriving at a contradiction that \(\sqrt{6}\) would be rational, which it is not.
Subsection 8.3.2 More Elaborate Examples of Irrationality
Sometimes, demonstrating that a number is irrational can be more challenging and require more sophisticated techniques. In this subsection, we will explore some harder examples to illustrate these methods.
Let’s begin by showing that \(\sqrt{2} + \sqrt{3}\) is irrational. We will do this by assuming that it is rational and then squaring the expression to arrive at a contradiction.
Example 8.3.9.
Assume, for the sake of contradiction, that \(\sqrt{2} + \sqrt{3}\) is rational. Then we can write:
\(\sqrt{2} + \sqrt{3} = r\) for some rational number \(r\).
Squaring both sides of the equation, we get:
\((\sqrt{2} + \sqrt{3})^2 = r^2\).
Expanding the left-hand side, we have:
\(\sqrt{2}^2 + 2\sqrt{2}\sqrt{3} + \sqrt{3}^2 = r^2\).
Simplifying, we get:
\(2 + 2\sqrt{6} + 3 = r^2\).
Combining like terms, we obtain:
\(5 + 2\sqrt{6} = r^2\).
Since \(r\) is rational, \(r^2\) is also rational. Therefore, \(5 + 2\sqrt{6}\) must be rational. However, we know that \(2\sqrt{6}\) is irrational because \(\sqrt{6}\) is irrational (as 6 is not a square number).
Thus, \(5 + 2\sqrt{6}\) is a sum of a rational number (5) and an irrational number (\(2\sqrt{6}\)), which is irrational. This contradicts our assumption that \(5 + 2\sqrt{6}\) is rational. Therefore, \(\sqrt{2} + \sqrt{3}\) must be irrational.
Next, we will show that \(\sqrt[3]{2} \times \sqrt{3}\) is irrational by assuming it is rational and then solving for 2 to reach a contradiction.
Example 8.3.10.
Assume, for the sake of contradiction, that \(\sqrt[3]{2} \times \sqrt{3}\) is rational. Then we can write:
\(\sqrt[3]{2} \times \sqrt{3} = r\) for some rational number \(r\).
We can cube both sides of the equation to remove the cube root:
\((\sqrt[3]{2} \times \sqrt{3})^3 = r^3\).
Expanding the left-hand side, we have:
\(\sqrt[3]{2}^3 \times \sqrt{3}^3 = r^3\).
Simplifying, we get:
\(2 \times 3\sqrt{3} = r^3\).
Combining the constants, we obtain:
\(6\sqrt{3} = r^3\).
Since \(r\) is rational, \(r^3\) is also rational. Therefore, \(6\sqrt{3}\) must be rational. However, we know that \(\sqrt{3}\) is irrational.
Thus, \(6\sqrt{3}\) is a product of a rational number (6) and an irrational number (\(\sqrt{3}\)), which is irrational. This contradicts our assumption that \(6\sqrt{3}\) is rational. Therefore, \(\sqrt[3]{2} \times \sqrt{3}\) must be irrational.
