Algebra is the metaphysics of arithmetic.

John Ray

These are not letters, they are numbers.
Me, just there

## Introduction: What is$x$??

Consider an equation: a mathematical statement expressing that two objects — e.g. numbers — are equal, equivalent and one and the same. As an example;

$x^2+1=5x$.

In mathematics there are a number of uses for this = sign. There is the common;

$1+2=3$

which merely asserts that the sum of 1 and 2 is the same as 3. Also there is the definition-type :=;

$3^3:=3\times3\times 3$

which defines $3^3$ for example, and by extension all these positive integer powers. Finally there is the equation or formula type =, the most famous of which is probably

Energy=(mass)$\times$ (speed of light) $\times$ (speed of
light)

$E=mc^2$.

An equation of this type is a statement that one object — e.g. a number — is equal to another.

Suppose now I told you that the difference between two numbers squared was the product of their respective sum and difference, and I stacked up the evidence in front of you as so you would believe me:

$5^2-2^2=25-4=3\times 7=(5-2)(5+2)$

$10^2-11^2=100-121=(-1)\times (21)=(10-11)(10+11)$

etc.

in particular negatives aren’t a problem because for example $(-4)^2=(+4)^2$; the square of a negative number is positive. You could say though how do you know this is true of every number, how can you prove this is true? By going away and thinking I can demonstrate why it’s true and this is when $x$ kicks in.

I do know that real numbers — elements of the set $\mathbb{R}$ — have certain properties. Perhaps I can use the properties of real numbers to show that the difference between two numbers squared is the product of their respective sum and difference — even if it is a picture which gives us the “why”.

These properties of real numbers are listed as an appendix to this post. These properties are ones that we assume without proof. Special properties like this that we assume rather than prove are called axioms.

### Laws of Algebra

Five important facts about numbers (four axioms and one theorem).Recall that a real number is one on the numberline:

• Subtraction: For all real numbers $x$:

there exists a $-x$ such that $x+(-x):=x-x=0$.

• Identity: There is a special number $1$ such that

for all real numbers $x$, $x\times 1=x$

• Distributivity: For any three real numbers $x,\,y,\,z$:

$x\times(y+z)=x\times y+x\times z$.

Note that this works both ways and explains how to group terms and factorise. One way says how to write a product as a sum; the other how to write a sum as a product/factorise.

• Division: For all $x\neq0$,

there exists $x^{-1}=:\frac{1}{x}$, such that $x\times x^{-1}=x\times \frac{1}{x}=:\frac{x}{x}=1$.

• No Zero Divisors: Suppose that $a,b$ are real numbers such that $a\times b=0$

then $a=0$ or $b=0$.

The use of $x$ is a sort of ‘every-number’. It can now be said that the difference between two real numbers squared is the product of their respective sum and difference because the theorem can be proved algebraically. The proof below is incredibly thorough but shows the use of the axioms. Note the notation:

$x\times y=x(y)=xy.$

As you can imagine this can lead to all kinds of confusion. Instead we use $xy$ to signify ‘$x$ multiplied by $y$. This is exactly why I don’t like so-called ‘mixed fractions‘.

#### Theorem

If $x,\,y\in\mathbb{R}$; then

$x^2-y^2=(x-y)(x+y).$

Proof:

$\displaystyle \begin{array}{ccc} (x-y)(x+y)&=(x+(-y))(x+y)&\,\,\,\,\,\text{(subtraction)} \\ &=x(x+y)+(-y)(x+y)&\,\,\,\,\,\text{(distribution)} \\ &=x^2+xy+(-yx)+(-y)(y)&\,\,\,\,\,\text{(distribution)} \\ &=x^2+xy+(-xy)+(-1)(y)(y)&\,\,\,\,\,\text{(commutativity)} \\ & =x^2+(xy+(-xy))+(-1)(y^2)&\,\,\,\,\,\text{(associativity)} \\ &=x^2-y^2&\,\,\,\,\,\text{(additive identity)} \end{array}$

Hence using $x$ and letting it stand for some real number, and manipulating it according to the axioms of the real numbers, a result that is true of all real numbers is proven. Again this is more than anything an exposition of proof from axioms.

### Why Bother?

Consider the following two problems:

1. What real number, when added to its square, equals $0$?
2. Show that $n^2+n+41$ is always a prime number when $n$ is a natural number.

Here there are two types of problem. One is a puzzle — what is $x$? Another is asking us to prove that a statement is true for each of the infinitely many natural numbers.

Solution of .:  Now first off, many people can guess one solution. How do we know that there are no more solutions? This is how we do algebra: suppose that $x$ has the property that

$x^2+x=0$.

Now we use some of our facts. Well $x^2=x\times x$ and $x=x\times 1$. Also using distributivity we can write this sum as a product:

$x(x+1)=0$.

Now using No Zero Divisors’:

$x=0$ or $x+1=0$.

Now considering $x+1=0$ separately. I’m happy enough that $x=-1$ but if we still weren’t convinced we could add $-1$ to both sides:

$x+1-1=0-1\Rightarrow x+0=-1\Rightarrow x=-1$.

We set up $x$ as the fall guy — and by a series of logical deductions, show what $x$ must be.

Solution to 2: There is great numerical evidence for this ‘theorem’:

$1^2+1+41=43$, prime

$2^2+2+41=47$, prime

….

$39^2+39+41=1601$, prime.

However it is not true; finite evidence is nothing in an infinite world:

$40^2+40+41=41^2$.

#### Some More Facts (!)

• Multiplication by Zero For all $x\in\mathbb{R}$, $0\times x=0$.
• Division by Zero is Contradictory There does not equal a real number equal to $1/0$. (Division by any number $y$ may is realised as multiplication by $1/y$.)
• For all real numbers $x\in\mathbb{R}$, $-1\times x=-x$.
• $-1\times -1=+1$.
• $(-a)\times (-b)=ab$.
• $ax+bx=(a+b)x$.

Before doing any serious mathematics, it is important that this knowledge is understood to ensure competency at algebra, and prevent many of the common pitfalls faced by students who are not algebraically aware. Now whenever $x$ or $y$ is seen it is understood it stands for a real number.

One of the biggest mistakes that people make is to assume that everything is linear. A linear function is a function $T:A\rightarrow B$ which has the property that

$T(a_1+\lambda a_2)=T(a_1)+\lambda T(a_2)$,

for all $a_1,\,a_2\in A$ and $\lambda\in\mathbb{R}$. The only linear real functions are functions of the form $f(x)=ax$ for a constant $a\in\mathbb{R}$.

Everything else is not linear.

#### Squaring is NOT Linear

Is $(x+y)^2= x^2+y^2$? NO:

$(2+3)^2=5^2=25\neq 13=4+9=2^2+3^2$.

The correct thing is to know what squaring means: $(x+y)^2=(x+y)(x+y)=x^2+2xy+y^2$ when we multiply out (later).

#### The Square Root is NOT Linear

Is $\sqrt{x+y}=\sqrt{x}+\sqrt{y}$? NO:

$\sqrt{16+9}=\sqrt{25}=5\neq 7=4+3=\sqrt{4}+\sqrt{3}$.

You cannot simplify $\sqrt{x+y}$ any further without knowing what $x$ and $y$ are (although we can show that $\sqrt{xy}=\sqrt{x}\cdot\sqrt{y}$ for $x,y>0$).

#### Inverting/Taking the Reciprocal is NOT Linear

Is $\displaystyle \frac{1}{x+y}=\frac{1}{x}+\frac{1}{y}$? No:

$\displaystyle \frac{1}{2+3}=\frac{1}{5}\neq \frac{5}{6}=\frac{1}{2}+\frac{1}{3}$.

Again there is nothing which you can do to simplify $\displaystyle \frac{1}{x+y}$ without knowing more about $x$ and $y$.

## Algebraic Manipulation & Simplification

### Simplifying Expressions

An expression in real numbers $x$s, $y$s, $z$s, etc. may often be simplified greatly. Because of commutativity and the distributive law of multiplication over addition; if $a$, $b$ are real constants;

$ax+bx=x(a+b)=(a+b)x$

This means common terms can be added together to simplify an expression.

#### Example

Simplify $3x-2+2x-4$.

Solution:

$3x-2+2x-4=5x-6$.

### Expanding Linear Products

Suppose $x\in\mathbb{R}$ and $a$, $b$, $c$, $d$ are real constants. Then using the distributive law twice:

$(ax+b)(cx+d) =ax(cx+d)+b(cx+d)$
$=acx^2+adx+bcx+bd$
$=acx^2+(ad+bc)x+bd.$

#### Examples

• Expand $(2x+4)(x+3)$

Solution:

$=2x(x+3)+4(x+3)$

$2x^2+6x+4x+12=2x^2+10x+12$.

• Multiply out $(x+1)(3x^2+x+1)$.

Solution:

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

$=3x^3+x^2+x+3x^2+x+1$

$3x^3+4x^2+2x+1$.

Suppose that we have real numbers $a$ and $b\neq 0$ and form the fraction $\displaystyle\frac{a}{b}$. How do we add fractions like this? Recall that we can add fractions using the lowest common multiple. Take

$\displaystyle \frac{5}{12}+\frac{3}{8}$

$\displaystyle =\frac{5}{12}\cdot \frac{2}{2}+\frac{3}{8}\cdot \frac{3}{3}$

$\displaystyle =\frac{10}{24}+\frac{9}{24}=\frac{19}{24}$.

but this is not strictly necessary:

$\displaystyle \frac{5}{12}+\frac{3}{8}$

$\displaystyle =\frac{5}{12}\cdot \frac{8}{8}+\frac{3}{8}\cdot \frac{12}{12}$

$\displaystyle =\frac{40}{96}+\frac{36}{96}=\frac{76}{96}=\frac{19}{24}$.

All that we need is a common denominator. The second approach to adding $5/12$ and $3/8$ can be used here. The general method of writing a general $\displaystyle \frac{a}{b}+\frac{c}{d}$ as a single fraction is the same:

$\displaystyle \frac{a}{b}\cdot \frac{d}{d}+\frac{c}{d}\cdot \frac{b}{b}$

$\displaystyle =\frac{ad}{bd}+\frac{bc}{bd}=\frac{ad+bc}{bd}$.

After practise some might do it in one line

$\displaystyle \frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$.

#### but you are probably better off taking your time.Examples

Write as a single fraction:

• $\displaystyle \frac{x+2}{3}+\frac{x+5}{4}$

Solution:

$\displaystyle \frac{x+2}{3}\cdot \frac44+\frac{x+5}{4}\cdot \frac33$

$\displaystyle \frac{4(x+2)}{12}+\frac{3(x+5)}{12}$

$\displaystyle =\frac{4x+8+3x+15}{12}=\frac{7x+23}{12}$.

• $\displaystyle \frac{1}{x+2}+\frac{1}{x+3}$.

Solution:

$\displaystyle \frac{1}{x+2}\cdot \frac{x+3}{x+3}+\frac{1}{x+3}\cdot \frac{x+2}{x+2}$

$\displaystyle =\frac{x+3}{(x+2)(x+3)}+\frac{x+2}{(x+3)(x+2)}$

$\displaystyle =\frac{x+3+x+2}{(x+2)(x+3)}=\frac{2x+5}{(x+2)(x+3)}$.

### Factorising

Factorising is taking a sum or difference and writing it as a product. There are a multitude of reasons for factoring. For example,

•  Simplifying expressions. Often after a factorisation a simplification of a supposedly complicated object may be made.

Example: Recalling the difference of two squares:

$\displaystyle \frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3$

$\text{ if }x\neq 3$.

• Solving quadratic equations. To solve a quadratic equation ($a\neq 0$, $b$ and $c$ are constants/actual numbers):

$a\cdot x^2+b\cdot x+c=0$.

means to find numbers that when plugged in for $x$ make the equation true. We use the no-zero divisors theorem to do this: if $a$ and $b$ are real numbers and $a\cdot b=0$ then either $a=0$ or $b=0$. i.e. we simplify by writing the quadratic in the form:

$ax^2+bx+c=a(x-\alpha)(x-\beta)$,

$ax^2+bx+c=0\Rightarrow \underbrace{a=0}_{a\neq 0\text{ by assumption}} \text{ or }x-\alpha=0\text{ or }x-\beta=0$

$x=\alpha\text{ or }x=\beta$.

Factoring is based on the distributive law for multiplication over addition; for all $x,y,z\in\mathbb{R}$:

$x\cdot (y+z)=x\cdot y+x\cdot z$.

If there is a sum like the right-hand side of this, with a factor common to both, it can be taken out as shown.

#### Examples

Write as a product (i.e. factorise)

• $5ab+10ac=5a\cdot b+5a\cdot 2c=5a\cdot (b+2c)$.
• $6xy-3y^2=3y\cdot 2x-3y\cdot y=3y\cdot (2x-y)$.
• $pq-pr+p=p\cdot q+p\cdot r+p\cdot 1=p\cdot (q+r+1)$.
• $2a(x+y)+3(x+y)=(x+y)\cdot 2a+(x+y)\cdot 3=(x+y)\cdot (2a+3)$.

• $16ax^2-12a-8x^2+6$.

Solution: It’s not actually clear that we can factorise this. Let us start however by taking out what is common out of the first two terms… and something out of the second two that will leave something common:

$4a\cdot (4x^2-3)-2\cdot (4x^2-3)=(4x^2-3)\cdot (4a-2)$.

### Dividing above and below

Suppose $a,b,c\in\mathbb{R}$, $b,c\neq 0$. Then

$\displaystyle \frac{ac}{bc}=\frac{a}{b}$.

We have done this before! Remember these letters’ are only numbers. Therefore be careful not to manipulate ‘symbolically’!! Let $a,b,c,d\in\mathbb{R}$, $b,c\neq 0$. The following ‘move’ is nonsense:

$\displaystyle \frac{a+bc}{bd}=\frac{a+c}{d}$.

Nowhere does it say that this should make sense. Don’t ever forget that we are dealing with numbers here. Therefore we will try not to use the word cancel.

#### Examples

• If $\displaystyle x=\frac{2t}{1+t^2}$ and $\displaystyle y=\frac{1-t^2}{1+t^2}$, write $\displaystyle \frac{x}{y}$ in terms of $t$.

Solution:

$\displaystyle \frac{x}{y}=\frac{\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}}\cdot\underbrace{\frac{1+t^2}{1+t^2}}_{=1}$

$\displaystyle =\frac{2t}{1-t^2}$.

• Simplify $\displaystyle \frac{10ab}{2b}$.

Solution:

$\displaystyle \frac{10ab}{2b}\cdot \underbrace{\frac{1/2b}{1/2b}}_{=1}=\frac{5a}{1}=5a$.

• Simplify

$\displaystyle \left[\frac{1+x}{1-x}-1\right]\div \frac{1}{1-x}.$

Solution: We can if we want just write $\displaystyle a\div b=\frac{a}{b}$:

$\displaystyle \frac{\frac{1+x}{1-x}-1}{\frac{1}{1-x}}$

Now how about multiply above and below by $(1-x)$?

$\displaystyle =\frac{1+x-(1-x)}{1}=2x$.

### Piece of cake!Factorisation for Simplification

If we have a fraction then it might not be initially clear if a simplification can be found. However because we will need the numerator and denominator to be both products if we want to divide out a common term, the first thing to do should be to factorise above and below to see if these will be possible.

#### Examples

Simplify

• $\displaystyle \frac{4x-8}{x^2-4}$.

Solution: Hmmm, factorise above and below recalling $a^2-b^2=(a-b)(a+b)$:

$\displaystyle \frac{4(x-2)}{(x-2)(x+2)}=\frac{4}{x+2}$; $x\neq 2$.

• $\displaystyle \frac{3x}{3xy+6xz}$.

Solution: Factorise the bottom:

$\displaystyle \frac{3x}{3x(y+2z)}\underset{x\neq 0}{=}\frac{1}{y+2z}.$

• $\displaystyle \frac{a^2-b^2}{5(a-b)+6(a-b)+7(a-b)}.$

Solution: Dealing with the top and bottom separately for now:

$\displaystyle \frac{(a-b)(a+b)}{(a-b)(5+6+7)}=\frac{(a-b)(a+b)}{18(a-b)}$.

If $a\neq b$ we can divide above and below by $a-b$:

$\displaystyle \frac{a+b}{18}$.