Chapter 1 Foundations

We begin with some foundational material that forms the basis of all applied mathematics.

1.1 Numbers

We deal with various number systems in mathematics.

The set of natural numbers are the familiar counting numbers: \[1, 2, 3,\dotsc\] We denote these by the symbol $\mathbb{N}$.
The integers are the “whole numbers”, which include zero and negative numbers: \[\dots,-2,-1,0,1,2,\dots\] We denote these by the symbol $\mathbb{Z}$.
We sometimes want to focus on just the positive integers, which is another name for the natural numbers.
The non-negative integers are the numbers: \[0, 1, 2, 3,\dotsc\] (the natural numbers and zero).
The negative integers are the numbers: \[-1, -2, -3,\dotsc\]
The rational numbers are numbers which can be written in the form $\frac{a}{b}$ where $a$ and $b$ are integers, with $b\neq 0$, so numbers like \[\frac{1}{2}, \frac{3}{2}, -\frac{4}{5}, \dotsc\] Note these include the integers since they can be written as $\frac{a}{1}$. Rational numbers are denoted $\mathbb{Q}$.

The ancient Greeks discovered that there are numbers that are not rational – so-called irrational numbers. For example, $\sqrt{2}, \pi$ and $e$ are all irrational numbers. One way of characterising irrational numbers is that they have a decimal representation that never repeats… \[ \pi=3.141592653589793238462643383279502884197169399375105820\dots \] In contrast, all rational numbers have an eventually repeating decimal expansion (after perhaps some initial jumble of digits, there is a repeating pattern), for example:

Rational Number	Repeating pattern
$\frac{1}{3}=0.333\dots$	repeating $3$’s
$\frac{3}{2}=1.5000\dots$	repeating $0$’s after the first decimal digit, which we normally don’t write down!
$\frac{219}{1750}=0.125142857142857\dots$	repeating sequence $142857$ of length $6$ after the first $3$ decimal digits

The real numbers consist of all rational and irrational numbers together. We can think of real numbers as points on a continuous straight line, or as numbers that possibly have a decimal representation requiring an infinite number of digits¹. The set of real numbers is denoted by $\mathbb{R}$.
We learn about all of the above numbers early on in our mathematical education. However, there is a further set of numbers that is very useful in science and engineering that you may not have encountered before: the complex numbers, denoted by $\mathbb{C}$. These are not as intuitive as real numbers and they are explored in section 5.

1.2 Algebra

Algebra is the general name given to the rules for manipulating numerical variables.

We often use letters as placeholders for numerical values. We can then manipulate these letters as if they were numbers using the rules of algebra, and hence re-arraging formulae or fing the general solution to an equation. We can then “plug-in” specific numbers to get the answer.

Being fluent in the notation and manipulations of algebra allows us to progress to higher (more useful and more interesting) areas of mathematics.

1.2.1 BEDMAS

Recall that when an expression contains multiple arithmetic operations, there is a certain precedence (they must be carried out in a certain order), so that there is no ambiguity in evaluating the answer. This is usually remembered by the mneumonic BEDMAS:

B - Brackets

E - Exponents

D - Division

M - Multiplication

A - Addition

S - Subtraction

Division and Multiplication can be carried out in either order, as can Addition and Subtraction. We’ll recap the rules of Brackets and Exponents in the next sections.

Example 1.1 (BEDMAS) We have \[3+2\times 4=11,\] whilst \[(3+2)\times 4 = 20.\] Another example: \[18\div 3^2 = 2\] whilst \[(18\div 3)^2 = 36.\]

1.2.2 Notation for Multiplication and Division

Let $a,b$ and $c$ be numerical variables.

To simplify our mathematical writing, we shorten the multiplication $a\times b$ to $ab$

We also write division $a\div c$ in the “fraction” form $\dfrac{a}{c}$

Since division and multiplication can be carried out in either order, the following are different ways of writing the same thing: \[\dfrac{a}{c}b=\dfrac{ab}{c}=a\dfrac{b}{c}\]

and since multiplication of two numbers can be carried out in either order, there are even more ways of writing this expression: \[\frac{b}{c}a=\frac{ba}{c}=b\frac{a}{c}\]

1.2.3 Rules of Exponents

In their most basic form, we use exponents as a shorthand for multiplication of a number by itself a given number of times. For example:

$5\times 5 = 5^2$ spoken as “5 squared” or “5 to the power 2”

$5\times 5\times 5=5^3$ spoken as “5 cubed” or “5 to the power 3”

and more generally, for any numbers $x$ and $a$,

$\underbrace{x\times \dotsb \times x}_{\text{with }a\, x\text{'s}} = x^a$ spoken as “x to the power $a$”

The number $x$ is known as the base and the number $n$ is known as the power, index or exponent.

Next, we look at rules for combining and simplifying exponents. We start by assuming the base $x$ is a positive real number and the exponent $a$ is a postive integer.

Now for any positive integers $a$ and $b$, it should be clear that the following rule holds:

Theorem 1.1 (Rule 1) Multiplying powers with the same base: \[x^ax^b=x^{a+b}\]

since we are multiplying $a$ $x$’s by a further $b$ $x$’s to give a total of $a+b$ $x$’s multiplied together

\[\underbrace{x\times \dotsb \times x}_{\text{with }a\, x\text{'s}}\times\underbrace{x\times \dotsb \times x}_{\text{with }b\, x\text{'s}}=\underbrace{x^{a+b}}_{\text{with }a+b\, x\text{'s}}\]

Example 1.2 (Multiplying powers) Example:

$2^3 2^4 = 2^7$

It is also worth seeing a non-example:

$3^3 4^2=?$

There is in general no simple way to combine exponents with different bases (3 and 4 here):

\[3^3 4^2=(3\times 3 \times 3) \times (4 \times 4) = 27 \times 16=432\]

and this number cannot be written as a power of any whole number.

Extending this idea, we also have:

Theorem 1.2 (Rule 2) Power of a Power \[(x^a)^b=x^{ab}\]

This follows from Rule 1 since $$(x^a)^b=\underbrace{x^a\times \dotsb \times x^a}_{b\, \text{times}}=x^{a+\dotsb+a}=x^{ab}$

Example 1.3 (Power of a power) $(5^3)^4=5^{12}$

Exercise 1.1 Show that $(x^a)^b=(x^b)^a$.

The following is a straightforward rule applying when we have powers of products of two different numbers $x$ and $y$.

Theorem 1.3 (Rule 3) Power of a product: \[(xy)^a=x^ay^a\]

Example 1.4 (Power of a product) $(2\times 3)^4=(2\times 3)\times(2\times 3)\times(2\times 3)\times(2\times 3) = 2^4\times 3^4$

And we have a similar rule for powers of quotients.

Theorem 1.4 (Rule 4) Power of a quotient:

\[\left(\frac{x}{y}\right)^a=\frac{x^a}{y^a}\]

This follows from the previous rule:

$\left(\frac{x}{y}\right)^a=\left(x\frac{1}{y}\right)^a=x^a\left(\frac{1}{y}\right)^a=x^a\frac{1}{y^a}=\frac{x^a}{y^a}.$

So far we have allowed the exponents to be positive integers. Can we make sense of negative integers as exponents? Consider for $a$ and $b$ as positive integers the product

\[x^a\frac{1}{x^b}=\frac{x^a}{x^b}\]

Let’s start with a concrete example and for arguments sake let’s take $a>b$ with $a=5, b=2$. Then:

\[\frac{x^5}{x^2}=\frac{x\times x \times x\times x \times x}{x \times x}=x\times x\times x=x^3\]

where we have cancelled the 2 common factors of $x$, leaving us with a product of $5-2=3$ $x$’s.

More generally, taking $\frac{x^a}{x^b}$ for $a>b$ we have $a$ $x$’s on the top and $b$ $x$’s on the bottom, so we can cancel the common factors to leave $a-b$ $x$’s. So,

\[\begin{equation} \frac{x^a}{x^b}=x^{a-b} \tag{1.1} \end{equation}\]

and following Rule 1 we could think of this as $x^{a-b}=x^{a+(-b)}=x^ax^{-b}$, and therefore using a negative exponent $-b$ gives the reciprocal of $x^b$.

Theorem 1.3 (Rule 5) \[x^{-a}=\frac{1}{x^a}\]

Example 1.5 (Negative Exponents)

Note in particular $x^{-1}=\frac{1}{x}$, the reciprocal of $x$
Generally, $\frac{x^a}{x^b}=x^a\frac{1}{x^b}=x^ax^{-b}=x^{a-b}$

so, for example,

\[\frac{5^6}{5^2}=5^4\]

and

\[\frac{7^2}{7^5}=7^{-3}\] (which is $\dfrac{1}{7^3}$)

How do negative exponents work in a “power of a power”?

\[(x^2)^{-3}=\frac{1}{(x^2)^{3}}=\frac{1}{x^6}=x^{-6}\]

which is compatible with the “multiply exponents” rule

\[(x^2)^{-3}=x^{(2\times -3)}=x^{-6}\].

So far we have assumed $x$ is positive. What happens when $x=0$? Then $0^a$ is multiplying $0$ by itself $a$ times, so the result is obviously $0$. Note this does not hold when $a=0$ since we have $0^0$, which is an indeterminate form. Also when $-a<0$, have $0^{-a}=\frac{1}{0^a}=\frac{1}{0}$ which is also an indeterminate form. So we need $a>0$ for this result.

Theorem 1.3 (Rule 6) \[0^{a}=0, \text{ for }a>0\]

Next, we can consider the the exponent $a=0$, that is the value of $x^0$. In (1.1) with the particular case $b=a$, we have

\[1=\frac{x^a}{x^a}=x^{a-a}=x^0\]

so we get that for any non-zero² value of $x$:

Theorem 1.5 (Rule 7) \[x^0=1, \text{ for } x\neq 0\]

So far we have only considered exponents that are integers and values of the base that are non-negative real numbers. It turns out that the previous rules actually apply to any real number exponent and non-negative real number base. We’ll start by thinking about fractional exponents, and in particular exponents of the form

\[x^{\frac{1}{a}}\]

where $a$ is a non-zero integer.

Example 1.6 Take $(9^\frac{1}{2})^2=9^{(\frac{1}{2}\times 2)}=9^1=9$.

This implies $9^{\frac{1}{2}}=3$, the square root of $9$.

More generally, take $(x^\frac{1}{a})^a=x^{\frac{1}{a}a}=x^1=x$. This means that raising $x^\frac{1}{a}$ to the power $a$ gets us back to $x$, hence the number $x^\frac{1}{a}$ is the $a^\text{th}$ root of $x$. In particular the square root is $x^{\frac{1}{2}}=\sqrt{x}$ and the cube root is $x^{\frac{1}{3}}=\sqrt[3]{x}$; we tend to use the root notation for expressions containing square or cube roots on their own, and use the power notation when combining different roots.

Theorem 1.6 (Rule 8) Reciprocal powers are roots: \[x^\frac{1}{a}=\sqrt[a]{x}.\]

Example 1.7 $8^\frac{1}{3}=\sqrt[3]{8}=2$

Now lets try to make sense of a power of the form $x^{\frac{a}{b}}$ where $a$ is an integer and $b$ is a positive integer. That is, we now allow for any fractional (otherwise known as rational) exponent. Since we have $\frac{a}{b}=a\frac{1}{b}$ we can write (using Rule 2):

\[x^{\frac{a}{b}}=x^{a\frac{1}{b}}=(x^a)^{\frac{1}{b}}=\sqrt[b]{x^a},\]

or, equaivalently

\[x^{\frac{a}{b}}=x^{\frac{1}{b}a}=(x^{\frac{1}{b}})^a=(\sqrt[b]{x})^a.\]

Theorem 1.7 (Rule 9) Fractional exponents: \[x^{\frac{a}{b}}=\sqrt[b]{x^a}=(\sqrt[b]{x})^a\]

1.2.4 More on exponents

As we mentioned, an exponent can be any real number. Not all numbers can be written as fractions i.e. rational numbers, such as $\sqrt{2}, \pi, e$: these are called irrational numbers, and the above rules still work with irrational exponents. Proving this is beyond the scope of these notes, but your calculator can happily handle calculations with any real number exponent.

Another thing we have not mentioned so far is what happens when $x$ is a negative number. This is easy to understand when the exponent $a$ is an integer, since we are just multiplying a negative number by itself multiple times (and taking the reciprocal if $a$ is negative). But how do we interpret something like $y=(-1)^\frac{1}{2}$? That is, $y$ must be a number such that when it is multiplied by itself it gives the negative number $-1$, but this is impossible for a real number! Answering this conundrum requires us to invent a new set of numbers, called complex numbers – this is addressed in section 5.

A summary of the rules:

Theorem 1.8 (Rules of Exponents)

$x^ax^b=x^{a+b}$
$(x^a)^b=x^{ab}$
$(xy)^a=x^ay^a$
$\left(\frac{x}{y}\right)^a=\frac{x^a}{y^a}$
$x^{-a}=\frac{1}{x^a}$
$0^a=0$, for $a>0$
$x^0=1$, for $x\neq 0$
$x^{\frac{1}{a}}=\sqrt[a]{x}$
$x^{\frac{a}{b}}=\sqrt[b]{x^a}=(\sqrt[b]{x})^a.$

1.2.5 Brackets

We use brackets to set the precedence (or order) of evaluating the different parts of an expression. We evaluate the innermost brackets first. Note we usually omit the multiplication sign when multiplying a bracketed expression.

Example 1.8 (Evaluating brackets) \[5(((2+4)\div (5-2))+1)=5((6\div 3)+1)=5(2+1)=5\times 3 = 15\]

Sometimes you might find it helpful to add “redundant” brackets to make the calculation clearer. For example, we could write \[3+2\times 4=11\] as \[3+(2\times 4)=11.\] The brackets are not really needed in the second version since we get the same result from following BEDMAS, but adding brackets can make an expression easier to read and evaluate.

1.2.5.1 Expansion

Recall that we can expand expressions involving multiplication of brackets by multiplying each term, such as the following:

\[a(b+c)=ab+ac\] \[(a+b)(c+d)=ac+ad+bc+bd\] \[(a+b)^2=(a+b)(a+b)=a^2+ab+ba+b^2=a^2+b^2+2ab\] \[(ab)^2=(ab)(ab)=abab=a^2b^2.\]

We often want to expand expressions in order to simplify them by grouping together like-terms, for example:

\[(a+b)(b+c)+2ac=ab+ac+b^2+bc+2ac=ab+3ac+b^2+bc.\]

With more than two brackets, we do the expansion in two steps: it does not matter which expansion we perform first. For example, we could start by expanding the first two sets of brackets in the following

\[\underbrace{(a+b)(c+d)}_{\text{first expand these two}}(e+f)=(ac+ad+bc+bd)(e+f)=ace+ade+bce+bde + acf+adf+bcf+bdf\]

but since the order of multiplication is unimportant, we could have started by expanding any two of the three sets of brackets.

In the case of the sum of two numbers (a binomial term) to a postive integer power we have a general formula for the expansion, i.e. something of the form

\[(a+b)^n.\]

which involves multiplying out the brackets $n$ times. We start by looking at a slightly simpler form \[(1+x)^n.\]

Let’s take a look at the expansion for the first few values of $n$.

$n=0:$ $(1+x)^0 = 1$

$n=1:$ $(1+x)^1=1+x$

$n=2:$ $(1+x)^2=(1+x)(1+x)=x^2+2x+1$

$n=3:$ $(1+x)^3=(1+x)(1+x)^2=(1+x)(x^2+2x+1)=x^3+3x^2+3x+1$

The coefficients of $x$ form the $n^\text{th}$ row of Pascal’s triangle.

The general formula is:

\[(1+x)^n = \binom{n}{0}x^0+\binom{n}{1}x^1+\binom{n}{2}x^2+\dotsb+\binom{n}{n-1}x^{n-1}+\binom{n}{n}x^n\]

where $\binom{n}{k}$ is the binomial coefficient

\[\binom{n}{k}=\frac{n!}{k!(n-k)!}\]

and $n!$ is the factorial of $n$:

\[n!=n\times(n-1)\times(n-2)\times\dotsb \times3\times 2\times 1\].

Given the more general form $(a+b)^n$, we can first extract a factor of $a$ to give

\[(a+b)^n=(a(1+\frac{b}{a}))^n=a^n(1+\frac{b}{a})^n\]

so we now just need to evaluate $(1+\frac{b}{a})^n$ using the formula and substituting $x$ with $\frac{b}{a}$. Alternatively, the full formula is:

\[(x+y)^n = \binom{n}{0}x^ny^0+\binom{n}{1}x^{n-1}y^1+\binom{n}{2}x^{n-2}y^2+\dotsb+\binom{n}{n-1}x^1y^{n-1}\binom{n}{n}x^0y^n.\]

1.2.5.2 Factorisation

At other times, we wish to simplify by extracting common factors from an expression, which we call factorisation. In the simplest form, this could just be factorising a single number

\[2x+2y=2(x+y)\]

Or, it could involve extracting more complicated factors, for example

\[2xy+6x^2y=2xy(1+3x)\] \[x^2-3x-10=(x+2)(x-5)\]

The second example here is factorising a quadratic which we can do systematically by finding the roots – we will see how to do this later in section 3.1.

1.2.6 Simplifiying Expressions

In order to simplify an expression we can carry out the following proceedures:

Expand brackets
Collect together any like-terms, e.g. $2a+3b+a=3a+3b$, or $a^3a^2+b^2=a^5+b^2$
Simplify any fractions: use common denominators, e.g. $\frac{a}{b}+\frac{c}{d}=\frac{da+bc}{bd}$; cancel any common factors, e.g. $\frac{ab}{ac}=\frac{b}{c}$
Factorise

Note: these are “rules of thumb”, as there usually does not exist such a thing as a “simplest” form for a given algebraic expression. The meaning of “simple” is dependent on what we want to gain from it. The intention is to put the expression in a form that is easier to understand (in the given context) and which usually uses fewer symbols (but not always).

1.3 Equations

An equation indicates the equality of two mathematical expressions: what is on the left of the $=$ sign is the same as what is on the right of the $=$ sign. When manipulating equations, we must perform the same operation to each side in order to maintain equality – think about a traditional mechanical balance scale, where if we were to change the mass on one side of the balance we would need to do the same to the other side in order to maintain balance. We will use the abreviations l.h.s. for left-hand-side and r.h.s. for right-hand-side. There are a few subtly different types of equations; the main two are formulas and conditional equations.

1.3.1 Formulas

Equations are often regarded as formulas, which can be used to calculate the value of a mathematical or physical quantity in terms of other known quantities. In this case, we write a single dependent variable on the left hand side and the right hand side will contain independent variables and constants. The value of the dependent variable depends on the values of the independent variables and is obtained by evaluating the right hand side when the values of the independent variables are known. For example, the formula for the area $A$ of a circle is

\[A=\pi r^2\]

where the independent variable $r$ is the radius of the cirle and $\pi$ is the mathematical constant $3.141...$. By plugging in a particular value for $r$ we then obtain a value for the area $A$. We often want to re-arrange equations to make a different variable the dependent variable, or subject, of the formula, i.e. to place it alone on the left hand-side. Re-arranging formulae is also known as transposition. For example, if we knew the area of a cirlce and wanted to calculate the corresponding radius, we can re-arrange the equation to make $r$ the subject:

\[r=\sqrt{\frac{A}{\pi}}.\]

In obtaining this, we carry out the “opposite” of the operations on the right hand side to both sides of the equation. These opposites are:

addition and subtraction;
multiplication and division;
powers and roots.

Let’s see this step by step in our example. We start with

\[A=\pi r^2.\]

Then, divide both sides by $\pi$

\[\frac{A}{\pi}=\frac{\pi}{\pi}r^2=r^2\]

so that $r^2$ is no longer multiplied by $\pi$. Now, to get from $r^2$ to $r$ we need to take the square root of both sides

\[\sqrt{\frac{A}{\pi}}=\sqrt{r^2}=r\]

and finally, we put the subject on the left hand side (since we read from left to right it is simply convention to do this, but mathematically it makes no difference since it is still an equality).

\[r=\sqrt{\frac{A}{\pi}}\]

1.3.2 Conditional equations

Another type of equation is a conditional equation. These may not hold true for all values of the variables and we will want to find the values for which equality is true. For example, \[5x+4=19\] is only true when $x=3$. We can solve this equation by making $x$ the subject:

subract $4$ from both sides

\[5x=15\]

divide both sides by $5$

\[x=3.\]

Note that an equation may have more than one solution, or perhaps no solutions. For example,

\[x^2=4\]

has two possible solutions, either $x=2$ or $x=-2$. On the other hand, the equation

\[x+2=x+3\]

has no solutions. To see this, subtract $x+2$ from both sides and we obtain

\[0=1\]

which is false, so no value of $x$ can give us equality.

1.4 Inequalities

We are familiar with the symbols $<, \leq, >$ and $\geq$, (less than, less than or equal to, greater than and greater than or equal to) which are used to define inequalities.

Example 1.9 \[x<5\] denotes all numbers that are strictly less that $5$ (so not including $5$ itself). \[-1< x \leq 1\] denotes all numbers that are strictly greater than $-1$ but less that or equal to $1$ (so $x=1$ is a possibility).

When re-arranging inequalities we need to be a bit more careful with their manipulation than with equalities. If we perform the same operation to both sides in an equality then the equality still holds true, but this is not the case in general for inequalities – in particular we need to be careful with multiplication by negative numbers which “flips” the inequality.

For example, if we take all values $x$ such that $x>10$, then an equivalent expression after multiplying by $-1$ is $-x<-10$ (NOT $-x>-10$).

Let $a,u,v,x,y,z$ be real numbers, then some basic relations that hold true are:

if $x>y$ and $y>z$, then $x>z$;
if $x>y$, then $x+z>y+z$ and $x-z>y-z$;
if $x>y$ and $u>v$, then $x+u>y+v$;
if $x>y$, then $ax>ay$ if $a>0$, and $ax < ay$ if $a<0$;
if $x>y$, then $\dfrac{x}{a}>\dfrac{y}{a}$ if $a>0$, and $\dfrac{x}{a}<\dfrac{y}{a}$ if $a<0$;
if $x>y>0$ and $u>v>0$, then $xu>yv$ and $\dfrac{x}{v}>\dfrac{y}{u}$;
if $x>y>0$, then $\dfrac{1}{x}<\dfrac{1}{y}$.

It should not be necessary to memorise all of these, it just requires a little thought when manipulating inequalities: the important ones to be careful with are multiplying or dividing by negative values.

1.5 Common mistakes!

We finish this section with some common algebraic errors to watch out for! Invent your own examples to convince yourself that these things are true.

Fractions: $\frac{a}{b+c}\neq \frac{a}{b}+\frac{a}{c}$ and $\frac{a+b}{c+b}\neq\frac{a}{c}$ (can’t “cancel” the $b$’s.).

Inequalities: $ab>c$ does not imply that $a>\frac{c}{b}$ (what happens if $b$ is negative?).

Powers: $(a+b)^2\neq a^2+b^2$.

Roots: $\sqrt{a+b}\neq \sqrt{a}+\sqrt{b}$.

Real numbers require quite a bit of work to define in a rigorous mathematical way and we will not do this here; the interested reader should look up “construction of real numbers by Dedekind cuts” or “construction of real numbers by Cauchy sequences”.↩︎
We already know that $0^0$ is an indeterminate form.↩︎

Rational Number	Repeating pattern
\(\frac{1}{3}=0.333\dots\)	repeating \(3\)’s
\(\frac{3}{2}=1.5000\dots\)	repeating \(0\)’s after the first decimal digit, which we normally don’t write down!
\(\frac{219}{1750}=0.125142857142857\dots\)	repeating sequence \(142857\) of length \(6\) after the first \(3\) decimal digits