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How to construct a ….

When I was teaching at the local FE college, there was one part of the GCSE where I sort of cheated,,,  I let YouTube do the work for me!

This was the set of skills called ‘Constructions’ – which can be hard to recreate on a whiteboard, especially when was one my ‘Motor Skills’…

The premise of these skills is that the student needs to

  • Construct a perpendicular bisector of a line
  • Construct a perpendicular to a line from any given point
  • Construct a line that bisects an angle

The extra catch is these have to be done using a pair of compasses and a straight edge only

There is something a little ‘old fashioned’ about these skills, but they remain on the GCSE Maths syllabus. I’ll also add that I haven’t seen them as much on recent exams papers – Maybe once at most across all three appears for any given season…

For all that they are rather fun to do and understand.

They are hard to do in front of a class; and on a blog too!   So let me give you some links to show how to do them.


Angling for some Fun – Part two

In the last post I started to describe how patterns in angles can help us find the size of missing angles without doing any measuring. In this post I am continuing with that theme. First I will show what we mean by Corresponding and Alternate angles.

Then I will show how we can put all this together to answer a question.

For both of these situation, we need one line crossing two others which are parallel.  If we don’t know the two lines crossed are parallel, we can’t use these rules, got it?

And you know how to tell if two lines are parallel? The two lines are never going to meet, however far we extend them.

So, the two angles marked in red – they are going to be equal.  So if we are told the lower one is 50˚  then the higher one is also 50˚.

These is what are called ‘Corresponding angles’

This is like the rule in the last post where we know two angles must be the same if they are ‘opposite’. Angles are also the same if the are ‘corresponding’

Now look at this diagram.

The two red angles are ‘corresponding’ and the green angle is ‘opposite’ one of the red angles.

So all three angles will be the same.

We say the bottom Red angle and the green angle are ‘Alternate’.  ‘Alternate’ is like a combination of the ‘Corresponding’ rule and the ‘Opposite’ rule.

Some people think of this as the Z rule – because the angles in a Z are the same.

I’ve been looking for an exam question that uses all of these but they are hard to find, and I only want one for this post, which would be too long otherwise…  so…  let’s have a look at this


A few points about the wording and notation here
1. I across two lines shows those lines are of equal length : AB = AC
2. > on two lines shows they are parallel. AP is parallel to CB
3. ‘Diagram Not Drawn to scale’ means take the information given as true. Don’t check them with your ruler or protractor.

ABC is an isosceles triangle, so <ACB = <CAB = 70˚.  < ABC = 180 – 2 x 70 = 40˚ – because of angles in Triangle ABC add to 180˚

<BAP = <ABC because they are Alternate – (to see the Z shape turn it round a bit)

So x = 40.

To Infinity and Beyond

I’ve been posting a lot recently on answers to GCSE questions – mainly because that is what I have been doing with students –  and I do have ten more days of that.

For today something different a look at something quite unusual…  As the title suggests – INFINITY.

Today’s post is based on some of the work of German mathematician Georg Cantor, a man not often understand by his contemporaries – In fact one said he was ‘aha ed of his time by one hundred years.

What Cantor was trying to do was explore what infinity means, and his revolutionary idea was that there is more than one infinity!  In fact…  there are an infinite number of infinities!

I’m not planning to show where they all come from in this post – I’ll just show that there is more than one.



If you go out with six friends this week, how will you know there are six of them? Well, OK, that seems like a simple question – because you can count, right?. Yes, but understanding how we can count is the starting point of understanding where all the extra infinities come in. You know you have six friends with you because you have given each a number, starting with 1..  and you got up to six! OK so may not have done that explicitly, but that is what you did.

And that is what we are going to use in a moment to demonstrate infinities.

Two quantities are the same if you can match up one item from one ‘list’ with one ‘item’ of the other… and have none left over. So the numbers 1 to 6 can be matched with your 6 friends with no numbers left over.

Now, this is where we come to some surprising ‘facts’…..  Ask,  are there more whole numbers than ‘even’ whole numbers.  You think, yes? Well look at this…..





Even numbers can be paired up with whole numbers with no numbers from each group left out.  This is one of the surprising things we learn about infinity. The number of even numbers is the number all numbers.

Now imagine you are on a ‘chessboard’ floor that stretches for infinity in all directions.

How many squares are there. Infinity – or more than infinity. This, after all ‘infinity squared 




Let’s try counting the squares in the same way that we counted your friends.  Start with one square – it could be the one you are standing on. Thats 1.  Now count the 8 squares around you – 2,3,4… 8

Then the squares around that.

 You can keep going like that for ever  – but that’s OK, we can keep counting for ever.

So even on this infinite board, the number of squares is the same as the number of numbers – Infinity.


Now, at this point, if I didn’t know what was coming next, I might be starting to doubt my earlier remark – that there is more than one sort of infinity.

Lets consider now, decimal numbers, by which I mean numbers of the form


I have 20 digits after the pint there. Lets say we mean numbers that DO end, but we will not specify how many digits they have before they do.

So lets start with the number 1.5

Where do we go to next to count; 1.6  or 1.55? Or 1.5000000001.

There are actually an infinite number of next steps, and thats before we get off our first ‘square’.

And THAT is where the next order of Infinity comes in. Its not possible to ‘count’ the decimals  with the numbers 1, 2, 3 …….  even they go on for ‘infinity’.

There are more than infinity decimals!

Crazy Curves

One last post before I move on from ‘continuous’ curves….

I was going to include this in the last post, but that was already too long.

We have looked at curves which are continuous everywhere, and some which are not – but are continuous for most of the way.

Is it possible for a curve to be discontinuous everywhere?  In theory yes, though we need to consider rational and irrational numbers.

A rational number is any that can be written down, accurately, with numbers.  This includes numbers there are ‘recurring’ like 0.3333333 because this can be written as 1/3, and be accurate

Pi is an example of an irrational number

So if we say that y = f(x) where f(x) = 1 when x is rational and f(x) = 1/x where x is irrational…  then that would define a curve, but one that is so chopped up is would be continuous in only very small sections between rational numbers

Let’s have a look at curves

I’ve been looking at exam questions a lot recently in this diary – well it is coming up to exam time!

In June, the focus is going to be on famous Mathematicians, Over the summer I’m looking to write some entries on mathematical puzzles – If you know some good ones please send them in!

Today though, I’m looking at curves – the lines we can draw to represent certain mathematical equations.

In particular, today and tomorrow I am writing about curves that are ‘continuous’ and curves which are not, which we call ‘discontinuous’. Also, what that means for how we can use these curves to answer questions – That bit is for tomorrow!

What do I mean by continuous? Basically this means that an ant can follow the curve and get to all parts of the curve, without flying off the page or leaving the line. There is a Mathematically precise way of defining ‘continuous’ but I think my ant gives you the idea.

Lets look at some examples. (For this I am going to borrow screenprints from my favourite websites, the link on the Links page)

y = x2

You can imagine the any being able to walk around this line. In fact any line of the form
Axa + Bxb + Cxc+ ….   will be continuous so long as a, b, c etc are positive integers.





y = sin (x)


And again the ant can walk up and down these curves.


Since the line for y = cos x looks similar, we can say that is continuous too.

Lets have a look at some Non-Continuous curves now, The easiest to show is  y = 1/x

This curve is in two parts. Our ant isn’t going to get from one part of the curve to the other without a jump.






Going back to our trigonometry, Tan (x)  – unlike Sine and cosine – is discontinuous. There are many and regular breaks in the line.



In tomorrow’s post I will look at some more eccentric examples and show why its important to understand if a curve is continuous or not.




Completing the powers.

Yesterday’s post built up some of the things we need to answer a question where the power is negative and a fraction.

One key point is that for all numbers n

n0 = 1  and n1 = n.

With negative powers we still need to maintain the rule of adding powers.

2x x 2-x  = 2x-x  (and since x – x x = 0)  = 20 = 1

Re-arrange this and we get  2-x = 1/2x

And there the first new rule – a negative power is 1/the positive power

32 = 9 so 3-2 = 1/9

To get the second rule we need to consider how powers can be combined.

(n2)3 = n6  – When you raise a power to a power – multiply the powers

[Consider  n x  n     x     n x n     x    n x n]


Now look at

(n2)1/2  =  n1  = n

So what does raise to power of half mean if this involves getting from n2 to n?   Taking the square root!  We could replace 2 and 1/2 with 3 and 1/3 – so see n1/3 means take the cube root – and so on.

n1/x means take the xth root of n.

Just before we get back to the question given, lets just complete the patterns by considering what x3/2 means. I have seen some exam questions that do pose questions like this.

I suggest you split the 3/2  into  1/2 x 3  or,  n3/2 = (n1/2)3

so 43/2  =  (sqrt(4)3)  = 23 = 8. It is generally easier to take the ‘root part first. In a non calculator paper you’ll only be asked about roots you know.

Let’s get back, at last, to the original question.

64-1/2–   Take the 1/2 part first, that means square root, and the square root of 64 is 8.  The – part means take the reciprocal  1/8

What Carol did was take 1/2 of 64, not the square root of 64, so your answer should include a sentence.  ‘Carol didn’t know that a power of 1/2 means square root, not multiply by 1/2’ – then give the correct answer of 1/8.


I’ve made this into 2 blogs posts with lots of background but if you can remember the rules given in these posts, this question need not take you long in an exam,


Taking it to the power…

Not because I’ve been watching election results today, but because I was discussing ‘powers’ with a student this week – Today’s subject is ‘powers of numbers’…  Its the first of a two-parter.




Before starting on the matters of powers, let’s consider the form of this question.

It’s another of those ones that suggests someone has done the question, and you have to give comment on their answer.  In this case the question does say that Carol has made an error, but this won’t always be the case. I have done a question recently where the ‘answer’ given  was correct, and the marks were to be gained by saying so.

With these question types, I think you just have to do the question yourself. You may be able to spot an error in something you haven’t tried yourself, but I’ll be honest, as a Maths tutor I can’t always do so myself.

Let’s get back to the question in hand and consider what a power means if is a)  negative and b) a fraction – since in this case the power is negative and a fraction.  I could tell you both answers, but I like to show how these answers fit into the wider picture.

You will recall that if were want to multiply two ‘powers’ together you can do this by adding the powers  – so long as the ‘base’ is the same. The base is the number ‘raised to’ the power. In this case the base is 2.

22 x  23  = 25.

This makes sense when you think of 22 as 2 x 2 and 23 as 2 x 2 x 2.  Put the big multiply together as

2 x 2       x      2 x 2 x 2 and you can see this can be written as 25

That rule of powers works well when the powers are positive whole numbers. What we want is for it to work with other numbers.

First we need to consider two special cases.

What does 21 mean?  And what does 20 mean?

21 means 2 multiplied together one time – and that is just 2!  Remember this holds for all base numbers and you can’t go wrong.

41  = 4    361  = 36   40001 = 4000

With 20 we want to keep to the ‘adding indexes’ rule so that

23 x 20 = 23+0 = 23

8 x 20 = 8  – We can soon see that for this to work, 20 must have the value 1

Again, that works for all numbers

40  = 1    360  = 1   40000 = 1

Now we know what powers of 0 and 1 are, can you see what consequence this has for negative and fraction powers?  This is what I’ll be looking at in the next post.



Find the distance from a Speed time graph

Today’s post os about an exam question I was prepping for my students yesterday. I was surprised to see this question because it cover’s two points I associate more with A-Level work.

a)  Area under the curve as the distance
b) Using the ‘ trapezium rule’

Before I say more, here is the question








The question asks for the distance the car travels, but the graph is of speed over time. This is where the first of my ‘surprise facts’ comes in – You can find the distance travelled by measuring the ‘area under the curve’. I won’t here show why that is the case – that is for A-Level students (OK I might post on that later). For now though, I will just apply the rule.

The area ‘under the curve’  is the area of the shape made by the curve at the top and the line ‘speed = 0’ at the bottom, bound to the left and right by the times we are interested in : In this case t=0 and t=20.

We can only estimate the area of this shape, given how irregular it is – the top side anyway – and this is where the ‘Trapezium rule’ comes in. To apply this we have to divide the area into strips, each one a trapezium (OK, the first one is a triangle, but we can apply a similar rule. To be fair, the question does give a clue to what to do by saying we should use ‘4 strips of equal width’

We divide up like this.

The first area is a triange with area

1/2 x 5 x 22*

*The value of the speed at t=5


The other areas are trapeziums with area 5 x 1/2(Height at start + height at end).

The ‘height’ in each case is the speed values at t= 10, 15 and 20.

If we look at the calculation closely we can save time by noting the middle times are included twice each, 1/2 each time.

so the area of the trapeziums are 5 x (22 + 28 + 32 + 35 x 1/2)

This is a quicker way that fiddling about with all the 1/2s but The 35 – value at t=20 – still needs the 1/2 because it is a measurement only on one trapezium.

We get the value 5 x 99.5.   This is where I took into account that we are only estimating, so using a value of 100 instead of 99.5 is justifiable.    This gives us an estimate of 500m

There is a part(b) for this question on the next page.

“Is this an over-estimate or an under-estimate for the distance. Give a reason for you answer.’


The answer in this case is that it is an underestimate, because all those trapeziums still left a small space under the curve. I don’t think our rounding up to 500m stops this being the answer.

A simple algebraic proof

As we get close to the exams, my students mostly want to do previous exam questions with me, so today I am sharing one which I found rather interesting – Perhaps because it foreshadows the kind of Maths one does at A-Level and degree level but its aimed at GCSE students.

The skills required are not that different to solving equations, but the level of thinking required is perhaps one level further on.

The question is…

Prove that the square of any odd number is one more than a multiple of 4.

The first thing to notice are those words ‘prove that’..  for students who are used to Maths being about doing a sum and finding an answer, this could seem quite new.  We are told what to expect. We just need to show the statement is true.

One of my students said ‘does this mean I have to look at every odd number’.  That would be quite difficult because there is a never -ending list of them. But its not a bad way to start, to convince yourself it might be true.

52 = 25  – and that is 1 more than 24 which is 4 time 6.

112 = 121 and that is 1 more than 120 which is 4 times 30

I could keep going like that – and you should do maybe a couple of your own – but I will never prove its true for ALL numbers like that.

Its easier in Maths – and in life – to prove something isn’t true. We only need to find ONE odd number where the square is not 1 more than a multiple of 4 to prove that statement is not true*. A person only needs to show he was in Birmingham on Tuesday evening to show he didn’t commit the crime in Paris.

(*You could try but don’t waste your time looking,  it is true!)

Proving things are true is harder.  But we can still do it.

We need to think of numbers more generally.  What is the ‘general’ odd number.  Take a number, any number. Will it be odd? Well there is a 50-50 chance.  But if we double the number we know it will be even (131 doubled is 262), then add 1. We then know the answer we get will be odd.

And in fact, all odd numbers can be found using  2n + 1 – That is a general odd number!!

So take our general odd number, and square it.  (2n+1)(2n + 1).  I could have written this with a 2 but I find its easier to multiply brackets if you can see both.

We get 4n2 + 4n + 1

Factorise just part of this you get 4(n2 + n) + 1

And that is actually what we are looking for! Let me explain. In 2n + 1, our ‘general’ odd number, n is just any whole number, and so n2 + n is also just a  whole number.

So 4(n2 + n) must be a multiple of 4 [ 4 times (n2 + n)] and 4(n2 + n) + 1 is 1 more!

Finding a turning point

For today’s post I’m going to look at an exam question I saw posted on Facebook.

I confess this is the sort of stuff I get nerdy about! I love the way that equations can describe pictures, and vice versa.




Before we start, lets make sure we understand the terms used. What exactly is a ‘turning point’?   Well, it is the point where the line stops going down and starts going up (see diagram below).   That point at the bottom of the smile.

There are three different ways to find that but in all cases, we need to start by finding the equation – finding out the values of ‘a’  and ‘b’.

‘b’ is the easier. take the case where x=0 and the two terms with x can both be disregarded and so b = -5 = the point where the curve crosses the y-axis (also known as the line ‘X = 0’.  That is similar to the reason why the term ‘c’ is often called ‘the y intercept’ for straight lines, equation y = mx+c

The other point we know is (5,0)  so we can create the equation

25 + 5a – 5 = 0 (By substituting the value of 5 in for x)

We can solve this for a giving a=-4


The full equation is  y = x2 – 4x – 5

I usually check my work at this stage  52 – 4 x 5 – 5 = 0 – as required

Now, I said there were 3 ways to find the turning point. I will give all three here bu, be warned, the third does require some A-Level maths. This diary is more aimed at GCSE students but method 3 is actually the way I would usually find the turning point, so I will give a brief description here.

Method 1 – The ‘parabola’ is symmetrical.

A Parabola is the name of the shape formed by an x2 formula


On this version of the graph

I’ve marked the turning point with an X and the line of symmetry in green. 

This tells us the value of x on the turning point lies halfway between the two places where y=0 (These are solutions, or roots, of x2 – 4x – 5 = 0

We know one of these is is x=5. We can get the other by factorising  to give (x-5)(x+1) = 0. So x = -1 is the other solution. Halfway between x = -1 and x = 5 is x = 2.

when x = 2, y = 22 – 4 x 2 – 5 = -9.  So the turning point is (2,-9).

Method 2  Complete the Square

If we ‘complete the square’ on this equation we get

x2 – 4x + 4 – 4 – 5  –  I’ve added in 4 and taken 4 away which looks like an eccentric thing to do, but this means we can factorise the first part into a square

(x – 2)2 – 9 (also combined the -4 and – 5 to make -9)

This where we use our knowledge that a ‘square’ is never less than 0, but it can be 0. So its minimum value is when that square = 0 – so x – 2 = , so x = 2. And we can also see the value of the whole thing there is 0 – 9 = -9

Method 3 – By differentiation

This is A-Level stuff really, so I’ll only give an overview. This is the way I would usually do this, but then I have studied Maths through A-level (and beyond!)

Differentiation is one branch of Calculus, the mathematics of measuring change.

By a rule you will learn if and when you first study calculus, the equation of how much x2 – 4x – 5 is changing is given by 2x – 4.

The turning point is where the line isn’t changing, so 2x – 4 = 0 (Zero change) so 2x = 4 and x = 2. y = 9 can be as above.



I’ve given three methods here. Fortunately they all give the same answer! (They wouldn’t be good alternatives otherwise!)

Choose which of the first two you feel most comfortable with.  The third method is the easiest to extend to where we have equations of than an a x2