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Chapter 0: Problem 13

Give an example of two irrational numbers whose product is a rational number.

Short Answer

Expert verified
The two irrational numbers whose product is a rational number are \(\sqrt{2}\) and \(\dfrac{1}{\sqrt{6}}\). Their product is \(\dfrac{1}{3}\), which is a rational number.

Step by step solution

01

Recall the Properties of Irrational Numbers

Irrational numbers are numbers that cannot be expressed as a fraction (i.e., as the quotient of two integers). Examples include square roots of non-perfect square numbers, 蟺 (pi), and e (Euler's number).
02

Choose an Irrational Number

Let's start by choosing a well-known irrational number. We'll use the square root of 2 (\(\sqrt{2}\)), which is a known irrational number.
03

Create Another Irrational Number from the Chosen Number

Now, we'll create another irrational number using the chosen number. Let's multiply \(\sqrt{2}\) by another irrational number, say \(\sqrt{3}\). The product of these two irrational numbers will also be an irrational number. So we have our second irrational number: \(\sqrt{2} \cdot \sqrt{3} = \sqrt{6}\).
04

Create a Product of the Two Irrational Numbers that is a Rational Number

Now let's find the product of our two irrational numbers (\(\sqrt{2}\) and \(\sqrt{6}\)) in such a way that it results in a rational number. To do this, we can multiply the first irrational number by the reciprocal of the second irrational number. The reciprocal of \(\sqrt{6}\) is \(\dfrac{1}{\sqrt{6}}\). Now, let's multiply \(\sqrt{2}\) by the reciprocal of \(\sqrt{6}\): \(\sqrt{2} \cdot \dfrac{1}{\sqrt{6}}\)
05

Simplify the Product

To simplify the product, let's rationalize the denominator by multiplying the numerator and denominator by \(\sqrt{6}\): \(\sqrt{2} \cdot \dfrac{1}{\sqrt{6}} = \dfrac{\sqrt{2} \cdot \sqrt{6}}{6}\) This simplifies to: \(\dfrac{\sqrt{12}}{6}\)
06

Confirm the Product is Rational

We can further simplify the product: \(\dfrac{\sqrt{12}}{6} = \dfrac{2\sqrt{3}}{6} = \dfrac{1}{3}\) The product of our two irrational numbers (\(\sqrt{2}\) and \(\dfrac{1}{\sqrt{6}}\)) is \(\dfrac{1}{3}\), which is a rational number. Thus, we have found two irrational numbers whose product is a rational number: \(\sqrt{2}\) and \(\dfrac{1}{\sqrt{6}}\)

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Most popular questions from this chapter

The sales tax in San Francisco is \(8.5 \% .\) Diners in San Francisco often compute a \(17 \%\) tip on their before-tax restaurant bill by simply doubling the sales tax. For example, a \$64 dollar food and drink bill would come with a sales tax of \$5.44; doubling that amount would lead to a \(17 \%\) tip of \(\$ 10.88\) (which might be rounded up to \(\$ 11\) ). Explain why this technique is an application of the associativity of multiplication.

Expand the given expression $$ (x+y+z)^{2} $$

Give an example to show that division does not satisfy the associative property.

Simplify the given expression as much as possible. $$ \frac{m}{3}+\frac{5}{n-2} $$

Simplify the given expression as much as possible. $$ \frac{2}{3} \cdot \frac{4}{5}+\frac{3}{4} \cdot 2 $$
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