Chapter 8 — What Numbers Really Are
Topics in this chapter:
8.1 Integers and Rationals
The Number Hierarchy
| Set | Symbol | Description |
|---|---|---|
| Natural numbers | $\mathbb{N}$ | $1, 2, 3, \ldots$ (positive counting numbers) |
| Integers | $\mathbb{Z}$ | $\ldots, -2, -1, 0, 1, 2, \ldots$ (naturals + negatives + zero) |
| Rationals | $\mathbb{Q}$ | Numbers expressible as $\frac{p}{q}$ with $p, q \in \mathbb{Z}$, $q \neq 0$ |
| Reals | $\mathbb{R}$ | All points on the number line |
| Complex | $\mathbb{C}$ | Numbers of the form $a + bi$, $i = \sqrt{-1}$ |
Decimals and Fractions
Every rational number has a decimal representation that either terminates or repeats.
- Terminating: $\frac{3}{8} = 0.375$
- Repeating: $\frac{1}{3} = 0.\overline{3} = 0.333\ldots$
Conversely, every terminating or repeating decimal is rational.
Converting Terminating Decimals to Fractions
Example 8-1: Convert $0.375$ to a fraction.
\[0.375 = \frac{375}{1000} = \frac{3}{8}\]Converting Repeating Decimals to Fractions
Example 8-2: Convert $0.\overline{36} = 0.363636\ldots$ to a fraction.
Let $x = 0.\overline{36}$.
\[100x = 36.\overline{36}\]Subtract: $100x - x = 36$, so $99x = 36$, giving:
\[x = \frac{36}{99} = \frac{4}{11}\]Example 8-3: Convert $0.1\overline{6} = 0.16666\ldots$ to a fraction.
Let $x = 0.1\overline{6}$.
$10x = 1.\overline{6}$ and $100x = 16.\overline{6}$.
Subtract: $100x - 10x = 15$, so $90x = 15$:
\[x = \frac{15}{90} = \frac{1}{6}\]General method: If the repeating block has $k$ digits, multiply by $10^k$ to shift the block, then subtract to eliminate the repeating part.
If there are $m$ non-repeating digits after the decimal before the repeating block starts, multiply by $10^m$ first.
Comparing Fractions
Example 8-4: Which is larger, $\frac{5}{17}$ or $\frac{4}{13}$?
Cross-multiply: Compare $5 \times 13 = 65$ with $4 \times 17 = 68$.
Since $65 < 68$: $\;\frac{5}{17} < \frac{4}{13}$.
Rule: $\frac{a}{b} < \frac{c}{d}$ iff $ad < bc$ (assuming $b, d > 0$).
Exercise 8-1: Convert to fractions: $0.\overline{45}$, $0.2\overline{3}$, $0.\overline{142857}$.
Exercise 8-2: Convert to decimals: $\frac{5}{7}$, $\frac{11}{13}$, $\frac{5}{6}$.
Exercise 8-3: Which is larger: $\frac{7}{11}$ or $\frac{9}{14}$?
Exercise 8-4: Arrange from least to greatest: $\frac{3}{5}, \frac{5}{8}, \frac{7}{11}$.
Exercise 8-5: Find a fraction between $\frac{3}{7}$ and $\frac{4}{9}$.
Exercise 8-6: Show that $0.\overline{9} = 1$.
8.2 Lowest Terms and Irrationals
Lowest Terms
A fraction $\frac{p}{q}$ is in lowest terms if $\gcd(p, q) = 1$ (i.e. $p$ and $q$ share no common factor other than 1).
Proof That $\sqrt{2}$ is Irrational
Theorem: $\sqrt{2}$ is irrational.
Proof (by contradiction): Suppose $\sqrt{2} = \frac{p}{q}$ in lowest terms. Then $2 = \frac{p^2}{q^2}$, so $p^2 = 2q^2$.
This means $p^2$ is even, so $p$ is even (since an odd number squared is odd). Write $p = 2k$.
Then $4k^2 = 2q^2$, so $q^2 = 2k^2$, meaning $q$ is also even.
But if both $p$ and $q$ are even, the fraction $\frac{p}{q}$ was not in lowest terms — contradiction!
Therefore $\sqrt{2}$ is irrational. $\blacksquare$
This same argument generalizes: $\sqrt{n}$ is irrational whenever $n$ is not a perfect square. The key idea is “lowest terms” — the assumption that $\gcd(p,q) = 1$ eventually leads to a contradiction.
Types of Irrational Numbers
- Algebraic irrational: Root of a polynomial with integer coefficients. E.g. $\sqrt{2}$, $\sqrt[3]{5}$.
- Transcendental: NOT a root of any polynomial with integer coefficients. E.g. $\pi$, $e$.
Proving a number is transcendental is extremely difficult. Lindemann proved $\pi$ is transcendental in 1882 (settling the ancient “squaring the circle” problem).
Rational Approximations
Irrational numbers can be approximated by sequences of rationals:
\[\sqrt{2} \approx 1, \; 1.4, \; 1.41, \; 1.414, \; 1.4142, \ldots\]Each successive decimal approximation is rational and converges to $\sqrt{2}$.
Exercise 8-7: Prove that $\sqrt{3}$ is irrational.
Exercise 8-8: Prove that $\sqrt{6}$ is irrational.
Exercise 8-9: Is $\sqrt{2} + \sqrt{3}$ rational or irrational? Prove your answer.
Exercise 8-10: Can the sum of two irrational numbers be rational? Give an example.
8.3 Complex and Beyond
The complex numbers extend the reals by introducing $i = \sqrt{-1}$. Every complex number has the form:
\[z = a + bi \qquad (a, b \in \mathbb{R})\]- $a$ is the real part, $b$ is the imaginary part.
- The conjugate $\bar{z} = a - bi$.
-
The modulus $ z = \sqrt{a^2 + b^2}$.
Fundamental Theorem of Algebra
Theorem: Every polynomial of degree $n \geq 1$ with complex coefficients has exactly $n$ roots in $\mathbb{C}$ (counted with multiplicity).
This means every polynomial can be completely factored over $\mathbb{C}$:
\[a_n x^n + \cdots + a_0 = a_n(x - r_1)(x - r_2)\cdots(x - r_n)\]For real polynomials: Complex roots come in conjugate pairs. If $a + bi$ is a root, so is $a - bi$.
This means any real polynomial can be factored into linear and irreducible quadratic factors over the reals.
Beyond complex numbers: Mathematicians have defined even larger number systems — the quaternions ($\mathbb{H}$, dimension 4) and octonions (dimension 8) — but these sacrifice familiar properties like commutativity.
Exercise 8-11: Factor $x^4 + 1$ completely:
i. Over the reals. ii. Over the complex numbers.
(Hint for the reals: $x^4 + 1 = (x^4 + 2x^2 + 1) - 2x^2$.)
Exercises & Problems
End-of-Chapter Problems
Problem 162. Express $3.0\overline{36}$ as a fraction in lowest terms. (MAΘ 1991)
Problem 163. Prove that between any two distinct rational numbers there is another rational number (i.e. the rationals are dense).
Problem 164. Prove that $\sqrt{p}$ is irrational for any prime $p$.
Problem 165. Show that $\log_2 3$ is irrational.
(Hint: if $\log_2 3 = p/q$, then $2^{p/q} = 3$, so $2^p = 3^q$. Why is this impossible?)
Problem 166. Prove or disprove: the product of two irrational numbers is always irrational.