Numbers

Basic Ideas

This course isn't the place to deeply probe the philosophy of numbers, so we will take it as a given that they exist1 and we can use them. Most people are familiar with numbers, but let's go over a few basics

Integers are whole numbers, which can be positive, negative or 0. Examples of integers include 0, 1, 3, 5, -9, -4. Let's look over a few rules with these numbers

\[a + 0 = 0 + a = a - 0 = -0 + a = a\]

The above statement in English more or less means "adding or subtracting 0 doesn't have any affect on the value of the other number".

However

\[a \times 0 = 0\]

Zero times any number is 0.

\[1 \times a = a\]

One, like 0, can be defined by its behaviour. One is the number that when it multiples any other number, you get the other number back. We now have two fixed reference points: a number that makes no difference when adding and a number that makes no difference when multiplying. The difference between these numbers is 1. Once 0 and 1 are fixed, you can continue to add, or take away, this same value to get all the integers.

Negative Numbers

Negative numbers have some interesting properties, let's explore those

\[-a + a = 0\]

In English this roughly translates to "taking a number away from itself gives 0", which is another equally valid definition of 0.

\[- (-a) = a\]

In English the above more or less means "a double negative is a positive". The way to think about this is that you are taking away negative stuff. Imagine a room with hot (positive) and cold (negative) air. If you took away the cold air, it would get hotter.

Addition

There are a few standard addition rules you should keep in mind.

\[a + b = b + a\]

This is commutativity, which means flipping the order doesn't matter.

\[(a + b) + c = a + (b + c)\]

Now we introduce brackets. Brackets are a way of telling in what order to do things. So, on the left hand side (LHS) of the above, this is like saying "do a plus b first, then add c". The right hand side (RHS) is like saying "do b plus c first, then add a". Due to the associative law, these two are the same

With these laws, be very careful with negative numbers. A classic mistake is to do this

\[a - b = b - a \text{ <- FALSE!!}\]

This is not true!! The LHS is saying "a add minus b", so we need to bring the minus sign with us

\[a - b = -b + a\]

which is correct.

Multiplication

Moving onto multiplication now, it also has commutative and associative laws.

\[ab = ba\]

\[(ab)c = a(bc)\]

You notice that these rules are analogous to addition rules. Also notice here I dropped the \(\times\) sign. This is very common when working with only variables. You will also often see a dot like this \(a \cdot b = b \cdot a\).

There is another multiplication rule too, called distribution

\[a (b + c) = ab + ac\]

Workshop: Solving Equations

The goal of algebra is to solve for unknown values. If we can set two expressions equal to one another, if there is only one unknown, we can often find that value. For example

\[x + 3 = 2\]

Can we find the value of \(x\)? Yes, and the most important idea in algebra needs to be understood in order to do that. So long as we do exactly the same thing to both sides of an equation, we can freely manipulate it however we like. So consider taking away 3 from both sides

\[x + 3 - 3 = 2 - 3\]

\[x = -1\]

For all of these workshop problems, find the value of \(x\)

(a) \(3x = 6\)

(b) \(5(x - 1) = 10\)

(c) \(2x + 3 = 4x - 5\)

For all of these workshop problems, isolate \(x\) onto the LHS of the equal sign (i.e. make them of the form \(x = ...\))

(d) \(2x + u = 3(x - u) + 4\)

(e) \(4(5x) + 6z = 6z - u\)

Solutions

(a) Diving both sides by 3 will isolate x and give the solution

\[3x \div 3 = 6 \div 3\]

\[x = 2\]

(b) We need to use the distributive rules

\[5x - 5 = 10\]

\[5x - 5 + 5 = 10 + 5\]

\[5x = 15\]

\[5x \div 5 = 15 \div 5\]

\[x = 3\]

(c) We need to get all the \(x\) and all the constants on opposite sides first

\[2x + 3 + 5 - 2x = 4x - 5 + 5 - 2x\]

\[8 = 2x\]

\[2x \div 2 = 8 \div 2\]

\[x = 4\]

(d) It's common to see "skipping" steps in algebra. Compare your solution against this one and see which parts were skipped.

\[2x + u = 3(x - u) + 4\]

\[2x + u = 3x - 3u + 4\]

\[x = -4u + 4\]

Skipping basic algebra steps like this is so common from now on I will do it. You can carry on writing it out verbosely for as long as you need, but eventually you'll find that you won't need to.

(e) Notice how we can just "cancel" out the \(z\) terms?

\[20x = -u\]

\[x = -u \div 20\]

Fractions

So far we've only really explicitly considered whole numbers, namely integers. We can also have bits of numbers, which we call the rational numbers. Rational numbers have the following form

\[\frac{m}{n}\]

where \(m\) can be any integer and \(n\) can be any integer except 0.

An important fraction property is the following

\[\frac{a}{a} = 1\]

Fractions have their own rules too, the most important one is cancellation

\[\frac{am}{an} = \frac{m}{n}\]

The most common mistake here is the following, so don't fall into it!

\[\frac{a + m}{an} = \frac{m}{n} \text{<- FALSE!!}\]

You can only cancel when the part you want to cancel is multiplying everything on the top and the bottom.

One more convenience with fractions comes when dealing with negative numbers. We can put the minus sign in different places

\[\frac{-m}{n} = \frac{m}{-n} = - \frac{m}{n}\]

Multiplying fractions is easy

\[\frac{m}{n} \frac{a}{b} = \frac{am}{bn}\]

Powers with fractions is also very easy

\[\left( \frac{m}{n} \right)^a = \frac{m^a}{n^a}\]

Adding fractions is not so easy however, and they can only be added if the denominator is the same

\[\frac{m}{a} + \frac{n}{a} = \frac{n + m}{a}\]

What do we do if the bottoms are not the same? A very common technique is to make the bottoms the same by multiplying the fractions by 1 in a very unusual way.

\[\frac{m}{n} + \frac{r}{s} = \frac{s}{s} \frac{m}{n} + \frac{n}{n} \frac{r}{s} = \frac{sm}{ns} + \frac{nr}{ns} = \frac{sm + nr}{ns}\]

Remember that if the top and bottom of a fraction are the same the value is 1, and multiplying by 1 makes no difference. We are not changing the values of the fractions, so we're free to do this. It does help us get the fractions into an easier form to deal with. Hold onto this idea of getting things into an easier form - it's a really core idea for a lot of mathematics.

Exponents

When multiplying a thing by itself many times, it might be helpful to have some kind of short hand.

\[aa = a^2\]

\[aaa = a^3\]

\[aaaa = a^4\]

This superscript is called an exponent or power. Powers have their own rules too, like this one

\[(a^m)^n = a^{mn}\]

There is also additions of exponents

\[a^{x+y} = a^x a^y\]

And a distributive property

\[(ab)^x = a^x b^x\]

Negative powers also have a special property

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

Until now, we've worked only with numbers that are rational. That is, they can be written down fully as a fraction or a decimal. However, not all numbers are like this. Many useful numbers can not be ever fully represented as a fraction or a decimal. Some examples include

\[\sqrt{2}, \sqrt{5}, \pi\]

Looking more at those square root terms, they have some interesting algebraic properties. The most important of these to remember is a root can also be written as an exponent

\[\sqrt{a} = a^{\frac{1}{2}}\]

In general

\[\sqrt[n]{a} = a^{\frac{1}{n}}\]

Workshop: More Advanced Algebra

Now we have all the basic elements of algebra, we can start exploring some of the more complex equations.

In these problems solve for \(x\)

(a) \(x^2 - 14 = 0\)

(b) \(x^5 = 9x^3\)

(c) \((x -2)^2 - 36 = 0\)

(d) \(\frac{5x + 1}{3(x + 2)} = 6\)

Solutions

(a) Remember the laws of square roots! Did you get both answers?

\[x^2 = 14\]

\[x = 7, -7\]

(b)

\[\frac{x^5}{x^3} = x^2 = 9\]

\[x = 3, -3\]

(c)

\[(x-2) = \pm 6\]

\[x = 8, -4\]

(d)

\[\frac{5x + 1}{3x + 6} = 6\]

\[5x + 1 = 6(3x + 6) = 18x + 36\]

\[13x = -35\]

\[x = - \frac{35}{13}\]

Further questions

Some questions with answers but no worked solutions for you to mull over. Find \(x\)

  1. \(\frac{x}{15} + \frac{2x}{3} = 7\)
  2. \(\frac{7}{2x} - \frac{6x}{3} = \sqrt{2}\)
  3. \(\left( \frac{x}{4} \right)^2 + 6(x^2 - 2) = 0\)
  4. \(\frac{2}{\sqrt{x}} + \frac{\sqrt{x}}{7} = \frac{2}{7 \sqrt{x}}\)
Solutions
  1. \(x = \frac{105}{11}\)
  2. \(x = \frac{7}{2(\sqrt{2} + 2)}\)
  3. \(x = \pm \sqrt{\frac{192}{97}}\)
  4. \(x = -12\)

Footnotes

1. Don't worry about what "exists" means in this context