Get Ready: The World of Real Numbers
Our journey with numbers often starts simply, but have you ever heard of a special number called the Golden Ratio? It's found everywhere, from nature to famous art.
The Golden Ratio, represented by the Greek letter Phi (\(\Phi\)), is equal to \(\frac{1+\sqrt{5}}{2}\). It's a key part of what makes designs like Leonardo da Vinci's Mona Lisa so visually appealing.
But this raises a question: Why isn't this famous number considered a rational number?
In this lesson, we will explore the concept of irrational numbers and discover the new, larger set they belong to: The Set of Real Numbers (R). This will give us the tools to answer questions just like that.
Learn: Irrational Numbers (Q') and Rational Numbers (Q)
What are Irrational Numbers (Q')?
An Irrational Number (Q') is a number that cannot be written in the form \(\frac{a}{b}\), where \(a\) and \(b\) are integers and \(b \neq 0\). The set of irrational numbers is denoted by the symbol \(Q'\).
- Examples of irrational numbers:
- The square root of any non-perfect square, such as: \(\sqrt{2}\)
where \(\sqrt{2} \approx 1.41421356\dots\) - The cube root of any non-perfect cube, such as: \(\sqrt[3]{5}\)
where \(\sqrt[3]{5} \approx 1.709975947\dots\)
- The square root of any non-perfect square, such as: \(\sqrt{2}\)
Note: Calculators provide approximate values for irrational numbers like \(\sqrt{2}\) and \(\sqrt[3]{5}\). These numbers have decimal expansions that are non-terminating and non-repeating, which is why they cannot be written as a rational number.
Review: Rational Numbers (Q)
A Rational Number (Q) is any number that can be written in the form \(\frac{a}{b}\), where \(a\) and \(b\) are integers and \(b \neq 0\). The set of rational numbers is denoted by the symbol \(Q\).
- Examples: \(0.5 = \frac{1}{2}\), \(3 = \frac{3}{1}\), \(-0.75 = -\frac{3}{4}\), \(0.333\dots = \frac{1}{3}\)
Enrichment Information
Not all irrational numbers are just roots; there are other irrational numbers such as:
- The number \(\pi\), where \(\pi \approx 3.141592654\dots\). It is the ratio between the circumference of a circle and its diameter.
- The Golden Ratio \(\Phi\), where \(\Phi \approx 1.618033989\dots\). It is a distinctive ratio related to the beauty and harmony of designs.
Example 1
Which of the following numbers is a rational number, and which is an irrational number? State the reason.
\(-12\)
\(\sqrt{13}\)
\(2\frac{2}{5}\)
\(0.125\)
\(\sqrt{25}\)
\(\sqrt[3]{-27}\)
\(-\sqrt[3]{10}\)
\(3.\overline{6}\)
Solution
All of the following numbers are rational numbers because each can be written in the form \(\frac{a}{b}\), where \(a\) and \(b\) are integers and \(b \neq 0\):
\(\sqrt{25} = 5 = \frac{5}{1}\)
\(-12 = \frac{-12}{1}\)
\(2\frac{2}{5} = \frac{12}{5}\)
\(\sqrt[3]{-27} = -3 = \frac{-3}{1}\)
\(0.125 = \frac{125}{1000} = \frac{1}{8}\)
Remember that:
The number \(3.\overline{6}\) is a non-terminating repeating decimal, and it is a rational number.
\(3.\overline{6} = 3.6666\dots = \frac{11}{3}\)
Both \(\sqrt{13}\) and \(-\sqrt[3]{10}\) are irrational numbers.
This is because \(13\) is not a perfect square, and \(10\) is not a perfect cube. Therefore, neither can be written in the form \(\frac{a}{b}\), where \(a\) and \(b\) are integers and \(b \neq 0\).
Finding an Approximate Value for an Irrational Number
Any irrational number lies between two consecutive integers, and it can be represented by a point on the number line between these two integers.
Example 2
Find two consecutive integers between which each of the following numbers lies, and determine its approximate location on the number line.
\(\sqrt{11}\)
\(\sqrt[3]{20}\)
Solution
\(\sqrt{11}\)
We choose two consecutive integers, each of which is a perfect square, between which the number 11 lies:
∵ \(9 < 11 < 16\)
∴ \(\sqrt{9} < \sqrt{11} < \sqrt{16}\)
∴ \(3 < \sqrt{11} < 4\)
∴ \(\sqrt{11}\) lies between the numbers 3 and 4.
Since the number 11 is closer to 9 than to 16, then \(\sqrt{11}\) is closer to 3.
\(\sqrt[3]{20}\)
We choose two consecutive integers, each of which is a perfect cube, between which the number 20 lies:
∵ \(8 < 20 < 27\)
∴ \(\sqrt[3]{8} < \sqrt[3]{20} < \sqrt[3]{27}\)
∴ \(2 < \sqrt[3]{20} < 3\)
Therefore, \(\sqrt[3]{20}\) lies between the numbers 2 and 3.
Since the number 20 is closer to 27 than to 8, then \(\sqrt[3]{20}\) is closer to 3.
Example 3
Nature: The Golden Rectangle is found in the shell of a marine creature, and the ratio of its length to its width is equal to \(\frac{1+\sqrt{5}}{2}\). Estimate this value.
Solution
First, estimate the value of \(\sqrt{5}\) as follows:
∵ The number 5 lies between the two perfect squares 4 and 9:
∵ \(4 < 5 < 9\)
∴ \(\sqrt{4} < \sqrt{5} < \sqrt{9}\)
∴ \(2 < \sqrt{5} < 3\)
∵ The number 5 is closer to 4 than to 9.
∴ \(\sqrt{5}\) is closer to 2.
Use "\(\sqrt{5} \approx 2.2\)" for a better approximation to estimate the value of the expression \(\frac{1+\sqrt{5}}{2}\):
∴ \(\frac{1+\sqrt{5}}{2} \approx \frac{1+2.2}{2} = \frac{3.2}{2}\)
∴ \(\frac{1+\sqrt{5}}{2} \approx 1.6\)
Real Numbers (R)
The set of Real Numbers is the set that consists of the union of the set of Rational Numbers and the set of Irrational Numbers, and it is denoted by the symbol R.
That is, \(R = Q \cup Q'\) where \(Q \cap Q' = \emptyset\)
The following Venn diagram illustrates the relationship between number sets:
\(N \subset Z \subset Q \subset R\)
\(Q' \subset R\)
Example 4
Place the appropriate symbol \(\in\) or \(\notin\) in the blank square:
\(0.\overline{63}\) \(Q'\)
\(\sqrt[3]{3\frac{3}{8}}\) \(Z\)
\(\sqrt{5}\) \(R\)
\(\sqrt[3]{3}\) \(Q\)
\(|-4|\) \(N\)
\(\sqrt{-1}\) \(R\)
Solution
Previous Information:
There is no square root for a negative real number.
Example 5
Find the solution set for each of the following equations:
\(3x^2 - 7 = 8\), where \(x \in Q'\)
\(4x^3 + 15 = -17\), where \(x \in Q'\)
\(\frac{1}{2}x^2 + 1 = -4\), where \(x \in R\)
\(5x^3 - \frac{1}{4} = 4x^3 + 8\frac{3}{4}\), where \(x \in R\)
Solution
∵ \(3x^2 - 7 = 8\)
∴ \(3x^2 = 8 + 7 = 15\)
∴ \(x^2 = \frac{15}{3} = 5\)
∴ \(x = \pm\sqrt{5} \in Q'\)
∴ Solution Set = \(\{\sqrt{5}, -\sqrt{5}\}\)
∵ \(4x^3 + 15 = -17\)
∴ \(4x^3 = -17 - 15 = -32\)
∴ \(x^3 = \frac{-32}{4} = -8\)
∴ \(x = \sqrt[3]{-8} = -2 \notin Q'\)
∴ Solution Set = \(\emptyset\)
∵ \(\frac{1}{2}x^2 + 1 = -4\)
∴ \(\frac{1}{2}x^2 = -4 - 1 = -5\)
∴ \(x^2 = -10\)
∴ \(x = \pm\sqrt{-10}\)
∵ \(\pm\sqrt{-10} \notin \mathbb{R}\) (Not a real number)
∴ Solution Set = \(\emptyset\)
∵ \(5x^3 - \frac{1}{4} = 4x^3 + 8\frac{3}{4}\)
∴ \(5x^3 - 4x^3 = 8\frac{3}{4} + \frac{1}{4}\)
∴ \(x^3 = 9\)
∴ \(x = \sqrt[3]{9} \in \mathbb{R}\)
∴ Solution Set = \(\{\sqrt[3]{9}\}\)
Example 6
Find the solution set for each of the following equations:
\(\frac{1}{3}x^3 + 9 = 0\)
\(\frac{3}{4}x^2 = \frac{9}{2}\)
\(2(x^2 - 1) = -10\)
If the substitution set is:
- First: \(Q'\)
- Second: \(R\)
Solution
Case 1: If the substitution set is \(Q'\) (Irrational Numbers)
∵ \(\frac{1}{3}x^3 + 9 = 0 \implies x^3 = -27 \implies x = -3\)
∵ \(-3 \notin Q'\)
∴ Solution Set = \(\emptyset\)
∵ \(\frac{3}{4}x^2 = \frac{9}{2} \implies x^2 = 6 \implies x = \pm\sqrt{6}\)
∵ \(\pm\sqrt{6} \in Q'\)
∴ Solution Set = \(\{\sqrt{6}, -\sqrt{6}\}\)
∵ \(2(x^2 - 1) = -10 \implies x^2 - 1 = -5 \implies x^2 = -4\)
∵ \(\pm\sqrt{-4} \notin Q'\) (Not a real number)
∴ Solution Set = \(\emptyset\)
Case 2: If the substitution set is \(R\) (Real Numbers)
∵ \(x = -3\)
∵ \(-3 \in \mathbb{R}\)
∴ Solution Set = \(\{-3\}\)
∵ \(x = \pm\sqrt{6}\)
∵ \(\pm\sqrt{6} \in \mathbb{R}\)
∴ Solution Set = \(\{\sqrt{6}, -\sqrt{6}\}\)
∵ \(x^2 = -4\)
∵ \(\pm\sqrt{-4} \notin \mathbb{R}\) (Not a real number)
∴ Solution Set = \(\emptyset\)
Example 7
Write all possible sets from the number sets (N, Z, Q, Q', R) to which each of the following numbers belongs:
\(\sqrt{10}\)
\(\sqrt{100}\)
\(0.\overline{25}\)
\(-\sqrt[3]{8}\)
\(\sqrt[3]{9}\)
Solution
∵ \(\sqrt{10}\) is an irrational number (10 is not a perfect square).
∴ \(\sqrt{10} \in Q'\) and \(\sqrt{10} \in R\)
∵ \(\sqrt{100} = 10\)
∴ \(10 \in N\), \(10 \in Z\), \(10 \in Q\), \(10 \in R\)
∵ \(0.\overline{25}\) is a repeating decimal, so it is a rational number (\(0.\overline{25} = \frac{25}{99}\)).
∴ \(0.\overline{25} \in Q\) and \(0.\overline{25} \in R\)
∵ \(- \sqrt[3]{8} = -2\)
∴ \(-2 \in Z\), \(-2 \in Q\), \(-2 \in R\)
∵ \(\sqrt[3]{9}\) is an irrational number (9 is not a perfect cube).
∴ \(\sqrt[3]{9} \in Q'\) and \(\sqrt[3]{9} \in R\)
Example 8
A square room has a floor area of \(15 \text{ m}^2\). Ahmed wants to surround its floor with a decorative strip. Estimate the length of this strip.
Solution
∵ The area of the square room's floor = \(15 \text{ m}^2\)
∴ The side length of the room = \(\sqrt{15}\) meters
The length of the decorative strip = Perimeter of the square room
Perimeter of a square = Side length \(\times\) 4
∴ The length of the decorative strip = \(4 \times \sqrt{15} = 4\sqrt{15}\) meters
Example 9
Find an irrational number between the two numbers \(0.\overline{3}\) and \(0.\overline{6}\).
Solution
First, convert the repeating decimals to fractions:
∵ \(0.\overline{3} = \frac{3}{9} = \frac{1}{3}\)
∵ \(0.\overline{6} = \frac{6}{9} = \frac{2}{3}\)
We need to find an irrational number \(y\) such that \(\frac{1}{3} < y < \frac{2}{3}\).
Let's square the boundaries to find a range for \(y^2\):
\(\left(\frac{1}{3}\right)^2 = \frac{1}{9}\) and \(\left(\frac{2}{3}\right)^2 = \frac{4}{9}\)
So, we need to find a number \(x\) such that \(\frac{1}{9} < x < \frac{4}{9}\), where \(x\) is not a perfect square of a rational number. Then \(y = \sqrt{x}\) will be our irrational number.
Let's choose a number between \(\frac{1}{9}\) and \(\frac{4}{9}\), for example, \(\frac{2}{9}\).
Let \(y = \sqrt{\frac{2}{9}} = \frac{\sqrt{2}}{3}\). This number is irrational.
Since \(1 < 2 < 4\), taking the square root gives \(1 < \sqrt{2} < 2\). Dividing by 3 gives \(\frac{1}{3} < \frac{\sqrt{2}}{3} < \frac{2}{3}\).
∴ \(\frac{\sqrt{2}}{3}\) is an irrational number between \(0.\overline{3}\) and \(0.\overline{6}\).