Numbers

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 12 February 2012

Numbers

At some point humankind wanted to count things and so we discovered the positive integers,

1, 2, 3, 4, 5, 6, ...

GREAT for counting something,

BUT what if you don't have anything? ... How do we talk about nothing, null, zilch?

... and so we discovered the nonnegative integers,

0, 1, 2, 3, 4, 5, 6, ... GREAT for adding

5 + 3 = 8, 0 + 10 = 10, 21 + 37 = 48,

BUT not so great for subtraction,

5 - 3 = 2, 2 - 0 = 2, 12 - 34 = ??? .

...and so we discovered the integers ...,

−3, −2, −1, 0, 1, 2, 3, ...

GREAT for adding, subtracting, and multiplying,

3 \cdot 6 = 18, -3 \cdot 2 = -6, 0 \cdot 7 = 0, BUT not so great if you only want part of the sausage, ...and so we
discovered the

rational numbers,

$\frac{a}{b}$ , $a$ an integer, $b$ an integer, $b \neq 0$ .

GREAT for addition, subtraction, multiplication, and division,

BUT not so great for finding

\sqrt{2} = ????

,

... and so we discovered the real numbers

all decimal expansions.

Examples:

π = 3.1415926..., e = 2.71828..., \sqrt{2} = 1.414 \dots, \frac{1}{3} = 0.333...,

\frac{1}{8} = 0.125 = 0.125000000000..., 10 = 10.0000...,

GREAT for addition, subtraction, multiplication and division,

BUT not so great for finding $\sqrt{- 9} = ????,$

...and so we discovered the complex numbers,

$a + b i$ , where $a$ and $b$ are real numbers and $i = \sqrt{- 1} .$

Examples of complex numbers:

\begin{array}{l} 3 + 4 i, & 7 + 9 i, & 3.2 + 6.7 i, \\ 5 + 0 i = 5, & 0 + 10 i = 10 i, & π + 0 i = π, \end{array}

\frac{1}{3} + \frac{2}{6} i = \frac{1}{3} + \frac{1}{3} i, 7 + \sqrt{2} i,

and

{(3 i)}^{2} = 3^{2} i^{2} = 9 i^{2} = - 9

. So

\sqrt{- 9} = 3 i

GREAT! We now have

Addition

(3 + 4 i) + (7 + 9 i) = 10 + 13 i .

Subtraction

(3 + 4 i) - (7 + 9 i) = 3 - 7 + 4 i - 9 i = - 4 - 5 i .

Multiplication

\begin{array}{l} (3 + 4 i) - (7 + 9 i) = 3 (7 + 9 i) + 4 i (7 + 9 i) \\ = 21 + 27 i + 28 i + 36 i^{2} \\ = 21 + 27 i + 28 i - 36 \\ = - 15 + 55 i . \end{array}

Division: $\frac{3 + 4 i}{7 + 9 i} = \frac{(3 + 4 i)}{(7 + 9 i)} \frac{(7 - 9 i)}{(7 - 9 i)} = \frac{21 - 27 i + 28 i + 36}{49 - 63 i + 63 i + 81} = \frac{57}{130} = \frac{1}{130} i .$
Square roots: We want $\sqrt{- 3 + 4 i}$ to be some $a + b i$ . $If \sqrt{- 3 + 4 i} = a + b i$ then $\begin{array}{l} - 3 + 4 i = {(a + b i)}^{2} = a^{2} + a b i + a b i + b^{2} i^{2} \\ = a^{2} + - b^{2} + 2 a b i . \end{array}$ So $a^{2} - b^{2} = - 3 and 2 a b = 4 .$ Solve for $a$ and $b$ , $\begin{array}{l} b = \frac{4}{2 a} = \frac{2}{a} . & So a^{2} - {(\frac{2}{a})}^{2} = - 3 . \\ So a^{2} - \frac{4}{a^{2}} = - 3 . \\ So a^{4} - 4 = - 3 a^{2} . \\ So a^{4} + 3 a^{2} - 4 = 0 . \\ So (a^{2} + 4) (a^{2} - 1) = 0 . \end{array}$ So $a^{2} = - 4$ or $a^{2} = 1$ .

So $a = \pm 1$ , and $b = \frac{2}{\pm 1} = 2 or - 2$ .

So $a + b i = 1 + 2 i$ or $a + b i = - 1 - 2 i$ .

So $\sqrt{- 3 + 4 i} = \pm (1 + 2 i)$ .
Graphing:
Factoring:

x^{2} + 5 = (x + \sqrt{5} i) (x - \sqrt{5} i),

x^{2} + x + 1 = (x - (\frac{1}{2} + \frac{\sqrt{3}}{2} i)) (x - (\frac{1}{2} - \frac{\sqrt{3}}{2} i)) .

This is REALLY why we like the complex numbers. The fundamental theorem of algebra says that ANY POLYNOMIAL (for example, $x^{12673} + 2563 x^{159} + π x^{121} + \sqrt{7} x^{23} + 9621 \frac{1}{2}$ ) can be factored completely as $(x - u_{1}) (x - u_{2}) \dots (x - u_{n})$ where $u_{1}, u_{2}, \dots u_{n}$ are complex numbers.

Notes and References

These notes are a retyped version of notes of Arun Ram from http://researchers.ms.unimelb.edu.au/~aram@unimelb/Notespre2005/nums8.13.03.pdf. Very likely this was originally an introductory lecture for a calculus class. It has been used by Arun Ram in various edited forms in numerous research and teaching presentations since 1999. This can be viewed as motivation and introduction for the pages: The integers ℤ, The Rationals ℚ, The Real numbers ℝ, The complex numbers ℂ.

References

[Ram] Lecture 1: September 6, 2000.

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