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A useful exercise in your department or with a planning buddy or when helping NQTs or non-specialists is to think about:

Examples and boundary (nearly) examples you might use

Which examples might show the greatest range of the concept

The precise language (a mini script) one might use to describe the positive and negative versions of the concept to really lock down concise explanation.

The different ways the same concept could be tested (different command words or wording of questions testing the same skill)

The choice and sequence of questions you ask to test the concept on a formative assessment level.

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STEM Learning Ltd offers the largest collection of learning resources for teachers and educators of science, technology, engineering, computing and mathematics in the UK.

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Chapter 21, in Engelmann's Magnus Opus, Theory of Instruction has changed the way I sequence worked examples to communicate a concept. This change took place greatly whilst I was at Michaela Community School when I started experimenting how to teach a UK Maths Challenge After school club. This is written about in more detail, here.

What I noticed, which may sound obvious, is that if you teach using examples which demonstrate the explicit features of a concept first and then progressively make those features implicit then the probability of success for each child in the learning process is greater. Engelmann refers to this as Covertization, " instruction that involves prompt shifts progressively from highly-prompted examples to unprompted examples."

One example which will be detailed below, would look like this:

So a sequence of examples with explicit features transitioning to implicit features provides a process where pupils take overt steps to go from the first line of working to the last line of working. This means that the first example pupils encounter is structured within the simplest context.

Covertization results in examples that communicate a concept, or part of a concept, to be communicated in its most explicit and misconception-proof form.

The best way to appreciate covertization is to look at some examples which aren't explicit.

Here are examples of fractions:

Here pupils can develop the following misconceptions:

all fractions have a one at the top

all fractions increase by '1' in the bottom value

fractions get bigger when the bottom value is greater

Another example:

Here are examples of expressions commonly known as a 'difference of two squares'

Here pupils can develop the following misconceptions:

The first term is always positive and a

The second term is always a constant

The second term is always an integer

The second term is always negative

What are the examples that I would use instead? To communicate an expression as a 'difference of two squares' I would start with the following, by stating that an expression must meet the following conditions:

1. The coefficient of the variable is a square number

2. The power of the variable must be even

3. The power of the variable is never an odd number

4. One term will be negative AND the other term will be positive

5. The constant must be a value that can be square rooted.

Here are the following examples that I would use to reiterate the conditions stated above:

There are other examples that I could include here, however, what I've done is limit the examples to only meet the conditions I've stated. But look at the complexity of what pupils can identify as an expression known as a difference of two squares by selecting the examples that I have.

More importantly, I've started with an example which is explicit with the features that make an expression a difference of two squares. With the initial set of examples, you can see that the coefficient of 1 for isn't obvious because pupils can't see it, and '1' as a constant isn't an obvious example of a square number. Explicit features allow pupils to build a clear and misconception proof understanding of a concept, which is what makes an example powerful.

My final point, we can see that the process of covertization has highly prompted examples allowing pupils to respond successfully to writing the product of two expressions because the coefficients and powers provide them that structure. Starting the other way around with implicit features now looks unstructured and poorly guided for a pupil to know how to rewrite the expression as a product of two expressions.

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Engelmann: Communicating through Covertization - Conception of the good

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Chapter 21, in Engelmann's Magnus Opus, Theory of Instruction has changed the way I sequence worked examples to communicate a concept. This change took place greatly whilst I was at Michaela Community School when I started experimenting how to teach a UK Maths Challenge After school club.

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The Count On Races game, played by an adult and a child, helps children learn to count on mentally. A coin or a dice with 1 or 2 on each face is used. Children say the next number when counting on. For example, when starting from 7, for a jump of 2 we say '8,9'.

The Cube Model (4-6) and Cube Model Advanced (4-6) tasks, children recreate the models. This task develops visual-spatial skills and helps children to recognise two parts within the whole.

The Reasoning in Addition Puzzles get children thinking deeply about the value of different shapes. This Mathsbot link allows you to generate new reasoning in addition puzzles.

Visual Representations

This article explains the importance of children's finger discrimination skills and this article gives different ways in which fingers can be used. Here are some Finger Discrimination tasks by Youcubed.

These Base-5 and Base-10 Number Tracks are coloured to help children to see quantities within a number track.

This Online Base-5 Abacus by Math Rack represents quantities 10-100 using a base-5 colouring.

The Coloured 100-Squares help to identify patterns in the 100-square and can support children as they count in 2s and 5s.

The Teens & Tens Cards help children to discern between teen and tens numbers. The Place Value Sorting Task show different representations of teen numbers and challenge children to spot misconceptions.

Progression in Pattern

The Progression in Pattern guidance shows how children's understanding of repeating patterns can be built. The Grids for Repeating Patterns can be used as children look to create circular patterns with a fixed number of spaces.
Number Foundations Assessment Fluency in Additive Reasoning, 1-20 Fluency in Calculation Within 20 Spatial Reasoning in Early Childhood, Early Childhood Maths Group

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The Count On Races game, played by an adult and a child, helps children learn to count on mentally. A coin or a dice with 1 or 2 on each face is used. Children say the next number when counting on. For example, when starting from 7, for a jump of 2 we say '8,9'.

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A collection of links to radio and podcast episodes (in English) with a mathematics bent - to inspire teachers to inspire, to feed the minds of our young mathematicians in the making, or, simply, to entertain. Programmes have been (loosely) categorised as shown below, and arranged alphabetically within each category by programme title.
Themes

For a list of links to the programmes from which episodes have been included in this selection, click here.

For a list of links to a selection of other, solely maths-focused podcasts,

Teaching of Mathematics & the Mathematical Mind
(Click here to go back to the menu)

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A collection of links to radio and podcast episodes (in English) with a mathematics bent - to inspire teachers to inspire, to feed the minds of our young mathematicians in the making, or, simply, to entertain. Programmes have been (loosely) categorised as shown below, and arranged alphabetically within each category by programme title.

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GCSE Questions

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Maths ROTD: A selection of GCSE exam questions on algebraic fractions, complete with answers #TesMathsROTD https://t.co/DzPMcLnX7k

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Maths ROTD: A selection of GCSE exam questions on algebraic fractions, complete with answers #TesMathsROTD https://t.co/DzPMcLnX7k

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Maths ROTD: A selection of GCSE exam questions on algebraic fractions, complete with answers #TesMathsROTD https://t.co/DzPMcLnX7k
— Tes_Maths (@Tes_Maths) December 27, 2018

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Maths ROTD: A selection of GCSE exam questions on algebraic fractions, complete with answers #TesMathsROTD https://t.co/DzPMcLnX7k

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Maths ROTD: A selection of GCSE exam questions on algebraic fractions, complete with answers #TesMathsROTD https://t.co/DzPMcLnX7k

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Maths ROTD: A selection of GCSE exam questions on algebraic fractions, complete with answers #TesMathsROTD https://t.co/DzPMcLnX7k

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Two functions tasks written during non-contact this afternoon.

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Two functions tasks written during non-contact this afternoon. https://t.co/o2hGCmdfMn

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Two functions tasks written during non-contact this afternoon. pic.twitter.com/o2hGCmdfMn
— Chris McGrane (@ChrisMcGrane84) September 13, 2018

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Two functions tasks written during non-contact this afternoon. https://t.co/o2hGCmdfMn

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Two functions tasks written during non-contact this afternoon. https://t.co/o2hGCmdfMn

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Two functions tasks written during non-contact this afternoon.

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How to solve an adfected quadratic @mathsjem.pptx - Google Drive

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How to solve an adfected quadratic @mathsjem.pptx

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Can anyone send me some weird, challenging ratio questions?

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Can anyone send me some weird, challenging ratio questions?

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Can anyone send me some weird, challenging ratio questions?
— Ben Gordon (@mathsmrgordon) October 6, 2018

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Can anyone send me some weird, challenging ratio questions?

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Can anyone send me some weird, challenging ratio questions?

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Can anyone send me some weird, challenging ratio questions?

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A slightly more systematic approach to factorising quadratics (starting with prime factorisation of c).  

From ‘The Essentials of School Algebra Parts II - III’, first published 1938.

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The Essentials of School Algebra Parts II - III

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A slightly more systematic approach to factorising quadratics (starting with prime factorisation of c).  

From ‘The Essentials of School Algebra Parts II - III’, first published 1938. https://t.co/ScAcGQtJr1

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A slightly more systematic approach to factorising quadratics (starting with prime factorisation of c).

From ‘The Essentials of School Algebra Parts II - III’, first published 1938. pic.twitter.com/ScAcGQtJr1
— Jo Morgan (@mathsjem) April 14, 2018

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A slightly more systematic approach to factorising quadratics (starting with prime factorisation of c).  

From ‘The Essentials of School Algebra Parts II - III’, first published 1938. https://t.co/ScAcGQtJr1

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The Essentials of School Algebra Parts II - III

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A slightly more systematic approach to factorising quadratics (starting with prime factorisation of c).  

From ‘The Essentials of School Algebra Parts II - III’, first published 1938. https://t.co/ScAcGQtJr1

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A slightly more systematic approach to factorising quadratics (starting with prime factorisation of c).  

From ‘The Essentials of School Algebra Parts II - III’, first published 1938.

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The Essentials of School Algebra Parts II - III

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Links to above

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Not sure if this qualifies as ‘weird’ enough but it might be potentially challenging:

Find ratios which are half way between 1:2 and 1:3 then 1:3 and 1:4, generalise for

1:n and 1:(n+1)

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Those of you who follow this blog will know I have a thing for explaining with colours. This isn't just a gimmick for younger students, it also works for 16-18 year olds.

In the picture below we were looking at proving a statement involving reciprocal trigonometric functions and fractions. A common source of misconception with this kind of question is that students split the question into working with the numerator and denominator separately, then make mistakes when they put them back together. They can't see the big picture.
Image credit: Mathssandpit
When I discussed this on the board I used separate colours for the expressions in the numerator and denominator. The class could follow the logic so easily. It's probably my most successful introduction to this topic. I saw that some students used highlighter on their notes after I'd gone through it, so they could track the solution.

The second type of question we looked at was solving a trigonometric equation. The straight forward expansion was all in one colour, but the roots of the quadratic were highlighted in different colours. The reasoning behind this was that students often solve half the quadratic and neglect the other impossible solution. Our exam board likes to see students consider the other solution and formally reject it. It makes the solution complete. By using a colour, the impossible solution stands out and reminds students to provide a whole solution.
Image credit: Mathssandpit
So when you are planning for misconceptions at A-level, remember that coloured pens aren't just for younger students.

This entry was posted in A-Level, Algebra, General, Problem Solving and tagged A-level, Fraction, reciprocal, trigonometry on by .

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Those of you who follow this blog will know I have a thing for explaining with colours. This isn't just a gimmick for younger students, it also works for 16-18 year olds. In the picture below we were looking at proving a statement involving reciprocal trigonometric functions and fractions.

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#Maths ROTD: Angles Chase II is a great problem solving activity for angles in polygons & parallel lines #TESMathsROTD https://t.co/Cy8Ccj711E

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#Maths ROTD: Angles Chase II is a great problem solving activity for angles in polygons & parallel lines #TESMathsROTD https://t.co/Cy8Ccj711E

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#Maths ROTD: Angles Chase II is a great problem solving activity for angles in polygons & parallel lines #TESMathsROTD https://t.co/Cy8Ccj711E
— Craig Barton (@mrbartonmaths) September 7, 2018

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#Maths ROTD: Angles Chase II is a great problem solving activity for angles in polygons & parallel lines #TESMathsROTD https://t.co/Cy8Ccj711E

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#Maths ROTD: Angles Chase II is a great problem solving activity for angles in polygons & parallel lines #TESMathsROTD https://t.co/Cy8Ccj711E

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#Maths ROTD: Angles Chase II is a great problem solving activity for angles in polygons & parallel lines #TESMathsROTD https://t.co/Cy8Ccj711E

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Teaching hints

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This is my biggest set of lesson stuff yet.

The PowerPoint introduces the topic. It sets up some example-problem pairs.

It then leads onto some vary and twist questions, which are available here as Word file.

The PowerPoint then goes onto talking about solving when rearranged and includes a Victorian textbook exercise and some questions where you have to find x, given the area of some shapes (this took ages to do. I posted it to Twitter twice. Both times the questions were written wrong. Doh!)

I've also included an open middle problem solving thing and a timed learning check, that graduates in difficulty.

Talking about learning checks...I've also written some Timed Questions that increase in complexity every 3 questions. I now have over 6000 questions in my Timed Questions database!

As always, comment and critique.

There's probably 2 lessons of stuff here, easily.

EDIT: One of the area problems was still wrong! Thanks to Professor Smudge for catching that 🙂

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New lesson: Finding roots of quadratics by factorising

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This is my biggest set of lesson stuff yet. The PowerPoint introduces the topic. It sets up some example-problem pairs. It then leads onto some vary and twist questions, which are available here as Word file.

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Shiny Pebbles and other stuff: A minus times a minus is a plus

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The reason why, we really should discuss.

So, why is the result of multiplying two negatives numbers together a positive number? There are various answers to that.
First, let's think a little about why we care.

We start off with the non-negative integers \[ \mathbb{N} = \{0,1,2,3,\ldots\} \] which we all learn to manipulate from a very young age. We know how to add and multiply them, and (at least sometimes) to subtract, at least as long as the number we want to subtract is no bigger than the one we want to subtract it from.
Then somebody comes along and says that when you subtract \(2\) from \(1\), the answer is \(-1\): a bit like \(2-1\), but with a minus sign in front of it. This is a new kind of number called a negative number, and they're useful for situations where you have to subtract a bigger number from a smaller one, and in certain unpleasant circumstances to describe the amount of money you have in the bank.
But now we have to ask: when we do arithmetic with them, how do they work?
One solution is just to present some rules, including the rule that the product of two negative numbers is the same as the product of their positive counterparts. In other words:

Ooh, not at all satisfactory. The rule seems completely arbitrary. But did we really have any choice? Are there any constraints that determine this? What if we'd chosen the rule that the product of two negative numbers was negative?
Let's hold this thought for the moment and think about how \(\mathbb{N}\) behaves.
So, let's take a careful look at the non-negative integers, and what we know about them. In fact, there is a bunch of stuff that we know. I'm too lazy to write in an explicit multiplication symbol, so I'll use \(mn\) to represent the product of \(m\) and \(n\).
In the list below, \(m,n\) and \(p\) are arbitrary elements of \(\mathbb{N}\). Then

1. \(m+n = n+m\) (addition is commutative)

2. \(m+(n+p) = (m+n)+p\) (addition is associative)

3. \(mn = nm\) (multiplication is commutative)

4. \(m(np)= (mn)p\) (multiplication is associative)

5. \((m+n)p = mp+np\) (multiplication distributes over addition)

6. \(0+n = n\) (\(0\) is the additive identity)

7. \(1n = n\) (\(1\) is the multiplicative identity)

8. If \(m+n = p+n\) then \(m=p\) (cancellation law for addition)

9. If \(mn=pn\) and \(n \neq 0\) then \(m=p\) (cancellation law for multiplication)

There are lots of other rules which we could write down, but these are all familiar, and the others follow from them, so we'll take them as a useful working collection.
In particular, there's one very familiar one which is conspicuous by its absence. It is the rule \(0n = 0\). I left it out on purpose, because we can show that it follows from the others, by the following argument:
We know that \(0+0=0\), because \(0\) is the additive identity. But that means that \((0+0)n=0n\), no matter what \(n\) is. By the distributive law, then, \(0n+0n=0n\), so that \(0n+0n=0n+0\), and then by the cancellation law, \(0n=0\).
There is an important lesson hiding in here. It is that our familiar rules aren't independent: once we have agreed on some of them, others are forced upon us. (So it's a good idea to choose a useful collection of rules to start with!)
This should raise an ugly possibility. What if my collection of rules isn't even consistent to begin with? How can I check that?
That's actually quite hard. Let's settle for the moment by accepting that there really is a set of non-negative integers with operations of addition and multiplication which satisfies the above rules. So there are lots of consequences from them that I haven't written above, but at least it all makes sense.
All this doesn't really help yet, but I can make use of it. Whatever the negative integers are, I want arithmetic to keep obeying the same rules as before. I want \(0\) and \(1\) to have their familiar properties, and I want the usual laws of algebra to continue to hold. So for every positive integer \(n\) I introduce a negative integer \(-n\) with the property that \(-n+n=0\). The set of all integers, positive, zero, and negative, is denoted \(\mathbb{Z}\). Now we can investigate what happens if we insist that all the rules above continue to hold.
Then \(\ldots\)
Once we have all the stuff above, it then follows by a fairly short argument that \((-1)(-1)=1\), if we insist that all the above rules continue to hold when we allow \(m,n\) and \(p\) to be negative.
First, we all agree that whatever we mean by \(-1\), it has the property that \(-1+1=0\). Then we know that \[ (-1+1)(-1) = 0(-1) = 0 \] and so that \[ \begin{split} (-1+1)(-1)&=(-1)(-1)+1(-1)\\ &=(-1)(-1)+(-1)\\ &=0 \end{split} \] But now adding \(1\) to each side gives \[ (-1)(-1)+(-1)+1=0+1 \] so that \[ (-1)(-1)+ 0 = (-1)(-1)= 1 \] So if we insist that all the algebraic rules for non-negative numbers continue to hold, we have no option. A minus times a minus is a plus.
Well, now we can see that if we decide that a minus times a minus is minus, then the ordinary rules of algebra must break somewhere. So this isn't really an arbitrary rule, it's forced upon us by the requirement that the algebraic properties of \(\mathbb{Z}\) are the familiar algebraic properties of \(\mathbb{N}\).
We can see various other things, too: one very important one is that \((-1)n = -n\). The argument is fairly similar: \[ \begin{split} (-1)n & =(-1)n + 0\\ &= (-1)n + n + (-n)\\ &= (-1)n+1n + (-n)\\ &=(-1+1)n + (-n)\\ &=0n + (-n) \\ &= -n \end{split} \]
So, this tells us that if we extend the non-negative integers to include negative ones, and insist that algebra works as before, there are certain logical consequences, including the familiar a minus times a minus is a plus. If. But how do we know that these rules describe anything at all? What are these negative numbers that we have introduced, and how do we know that it is even possible to introduce negative numbers while preserving the algebraic rules?
You might be tempted to think I'm about to kick up a lot of dust and then complain about how hard it is to see, but when the dust settles the view is wonderful.
So here's the awkward question \(\ldots\)

Negative numbers are there to be the solutions to equations of the form \[ m+x = n \] when \(m>n\). In particular, \(-n\) is the solution to \(n+x=0\), if this makes sense.
So, let's make sense of it.
We're going to do this a bit indirectly, thinking about pairs of integers \((m,n)\). Secretly we want to think of this pair as \(m-n\), but we don't yet have a meaning for this when \(m \lt n\).
So we're going to lump together certain pairs.
We say that the pairs \((m,n)\) and \((p,q)\) are equivalent if \(m+q=n+p\). A collection of pairs which are all equivalent to each other is called an equivalence class.
This is actually quite a familiar idea: it's analogous to thinking of \(3/4\) and \(6/8\) as different ways of representing the same fraction. There are infinitely many pairs of integers \((m,n)\) which we can use to do it, namely all the ones satisfying \(3n=4m\). We don't (often) talk explicitly about a fraction being an equivalence class of pairs of integers, but that's what's really going on.
We're going to do the same trick, but with pairs where it's the difference than matters, not the ratio.
We need to do three things now:

1. Define addition on these equivalence classes of pairs.

2. Define multiplication on these equivalence classes of pairs.

3. Show how we can think of these equivalence classes as the familiar integers.

Addition

This is the easy one.
We define addition on pairs by \[ (m,n)+(p,q)=(m+p,n+q) \]
But we're not done yet, because the collection that the answer lies in should only depend on the collections of the original pairs, not on the specific pairs. We can check this.
So suppose that \((m,n)\) and \((M,N)\) are equivalent to each other, as are \((p,q)\) and \((P,Q)\). Then we need to check that \((m+p,n+q)\) is equivalent to \((M+P,N+Q)\).
First, we know that \[ m+N=n+M \] and \[ p+Q=q+P \] so adding these together we have \[ m+N+p+Q=n+M+q+P \] i.e. \[ (m+p)+(N+Q)=(n+q)+(M+P) \] which is exactly what we need.
So this operation of addition works in the sense that the equivalence class of the sum depends only on the equivalence classes of the pairs we add.
Multiplication is a bit tougher.

Multiplication

This is not so obvious, so let's cheat a little. We are thinking of \((m,n)\) and \(m-n\), and \((p,q)\) as \(p-q\). Then the product ought to be \[ \begin{split} (m-n)(p-q) &= mp-mq-np+nq\\ & = (mp+nq)-(mq+np) \end{split} \] So we use the definition \[ (m,n)(p,q)= (mp+nq,mq+np) \] We can quickly check that \[ \begin{split} (p,q)(m,n) &= (pm+qn,qm+pn)\\ &=(mp+nq,mq+np)\\ &= (m,n)(p,q) \end{split} \] so that this is commutative, but now we have to check that the result is also independent of the choice of pairs.
So we consider \((m,n)\), \((p,q)\) and \((P,Q)\) where \((p,q)\) and \((P,Q)\) are equivalent. Then \[ (m,n)(p,q) = (mp+nq,mq+np) \] and \[ (m,n)(P,Q) = (mP+nQ,mQ+nP) \] and we see that \[ \begin{split} mp+nq+mQ+nP &= m(p+Q)+n(q+P)\\ &=m(P+q)+n(Q+p)\\ &=mP+nQ+mq+np \end{split} \] so that the two answers are equivalent.
Nearly there.
In just the same way, if \((m,n)\) is equivalent to \((M,N)\), then \((m,n)(p,q)\) is equivalent to \((M,N)(p,q)\). Putting this together, \((m,n)(p,q)\) is equivalent to \((M,N)(p,q)\), which is equivalent to \((M,N)(P,Q)\), so the equivalence class of the product only depends on the equivalence classes of the pairs we are multiplying.
Now we know that we have a well-defined arithmetic on our (equivalence classes of) pairs. But we still need to see how we can think of this as supplementing the non-negative integers with the negative integers.
We make use of the fact that we can choose any representative we like for each equivalence class. So we use the representative of the form \((n,0)\) for all pairs of the form \((p+n,q)\) where \(n \geq 0\), and the representative of the form \((0,n)\) for the pairs of the form \((p,q+n)\) where \(n \gt 0\).
Using this, we can now see that \((m,0)+(n,0)=(m+n,0)\) and \((m,0)(n,0)=(mn,0)\), so the non-negative integers live inside this strange new object.
We also see that \((n,0)+(0,n)=(n,n)\), which is equivalent to \((0,0)\). If we think of \((n,0)\) representing \(n\), then \((0,n)\) represents \(-n\).
It is then tedious but straightforward to check that all the rules of algebra for the non-negative integers still work. (This is mathematics code for 'I can't be bothered to write it all out, but I don't see any reason why you shouldn't suffer.')
So, now we know that is is, indeed, possible to add negative integers into the mix while preserving the well-known algebraic structure of the non-negative integers, and that the argument up above for why \((-1)(-1)=1\) does actually apply.
Of course, now that we have defined multiplication of our pairs of positive numbers, we can also do this explicitly and see the same answer from an entirely different perspective. Given than \(-1\) is represented by \((0,1)\), we have \[ (0,1)(0,1)=(0+1,0+0)=(1,0) \] and \((1,0)\) represents \(1\) in our model.

After all this, you might wonder what the point of all that peculiar stuff about equivalence classes was. It just made a complicated picture of the positive and negative numbers, which everybody was happy with anyway. The snag is that you can't always just add a new kind of number into the mix and preserve the algebra that you like, but now we know that it can be done in this particular case of interest.
Maybe the most famous example of this is Hamilton's search for a three-dimensional version of complex numbers. You can start off with the real numbers, then include a new quantity \(i\) which satisfies \(i^2=-1\), and just do algebra with combinations of this \(i\) and the real numbers, and it all works: complex numbers make sense, and we can think of their algebra as a way of adding and multiplying two-dimensional vectors.
Hamilton spent some time trying to add in another quantity like \(i\) so get a three-dimensional version of algebra. In fact, that isn't possible. He eventually realised that it could almost be done in four dimensions, but you had to give up on the multiplication being commutative.
So there are problems with adding the new kind of numbers in and just assuming that the old rules of algebra will continue to hold. But now, after quite a lot of work, we can say with confidence that it is possible to introduce negative integers in a way which is compatible with the algebra of the non-negative ones. This is a Good Thing.

Food for thought

1. I wrote down a multiplication rule which was motivated by me knowing how the answer should behave. Could I have chosen another rule with difference consequences?

2. I wrote down a bunch of rules above, which are followed by the non-negative integers. But are there other sets of quantities which satisfy those same rules? What happens if I try this trick on these other sets?

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The reason why, we really should discuss. So, why is the result of multiplying two negatives numbers together a positive number? There are various answers to that. First, let's think a little about why we care. We start off with the non-negative integers \[ \mathbb{N} = \{0,1,2,3,\ldots\} \] which we all learn to manipulate from a very young age.

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A Padlet with links to loads of resources

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A Padlet with links to loads of resources

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Part 1

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Part 2

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Teaching hints

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Ideas for formulae to use when teaching substitution

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What are your favourite formulae to use in key stage 3 to practice substitution into formulae? Go! #teammaths #mathschat #mathscpdchat

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What are your favourite formulae to use in key stage 3 to practice substitution into formulae? Go! #teammaths #mathschat #mathscpdchat

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What are your favourite formulae to use in key stage 3 to practice substitution into formulae? Go! #teammaths #mathschat #mathscpdchat
— Ben Gordon (@mathsmrgordon) August 15, 2018

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What are your favourite formulae to use in key stage 3 to practice substitution into formulae? Go! #teammaths #mathschat #mathscpdchat

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What are your favourite formulae to use in key stage 3 to practice substitution into formulae? Go! #teammaths #mathschat #mathscpdchat

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What are your favourite formulae to use in key stage 3 to practice substitution into formulae? Go! #teammaths #mathschat #mathscpdchat

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Teaching hints

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Yesterday, I shared two posts on how I have taught simplifying surds in the past, including the plethora of mistakes I have made. You can find Part 1 at https://t.co/EHbufPVBkM and Part 2 at https://t.co/CyvNnc4kEs.

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Yesterday, I shared two posts on how I have taught simplifying surds in the past, including the plethora of mistakes I have made. You can find Part 1 at https://t.co/EHbufPVBkM and Part 2 at https://t.co/CyvNnc4kEs.

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Yesterday, I shared two posts on how I have taught simplifying surds in the past, including the plethora of mistakes I have made. You can find Part 1 at https://t.co/EHbufPVBkM and Part 2 at https://t.co/CyvNnc4kEs.
— Jess Prior (@FortyNineCubed) June 29, 2018

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Yesterday, I shared two posts on how I have taught simplifying surds in the past, including the plethora of mistakes I have made. You can find Part 1 at https://t.co/EHbufPVBkM and Part 2 at https://t.co/CyvNnc4kEs.

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Yesterday, I shared two posts on how I have taught simplifying surds in the past, including the plethora of mistakes I have made. You can find Part 1 at https://t.co/EHbufPVBkM and Part 2 at https://t.co/CyvNnc4kEs.

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Yesterday, I shared two posts on how I have taught simplifying surds in the past, including the plethora of mistakes I have made. You can find Part 1 at https://t.co/EHbufPVBkM and Part 2 at https://t.co/CyvNnc4kEs.

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45 Free Maths Lesson Plans for KS3 and KS4 by Colin Foster

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45 Free Maths Lesson Plans for KS3 and KS4 by Colin Foster

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Ordering surds

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Basic Angle properties https://t.co/DuU1xPfi9m pic.twitter.com/SFNPpUXWLU
— @mhorley (@mhorley) May 13, 2018

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A set of questions to practise working backwards from a volume to find a missing length in cuboids, cylinders, pyramids, cones and spheres. The volume is fixed each time at 1 litre, allowing pupils to focus on the process of reversing the operations.</...

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Write down a positive integer.

Make a list of all its factors.

Now write down the number of factors each of those factors has.

The square of the sum of those numbers will be the sum of their cubes.

Extension ideas:

By trying numbers with different numbers of factors, try to spot a pattern. e.g. a, ab, abc; a, aa, aab etc.

Prove it

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Fractions in this worksheet are represented as circles divided into sectors. The task is to work out what fraction of each circle has been shaded.
answers

Each fraction on this worksheet is shown using an arrow above a point on a straight line.
answers

The task here is to work out what fraction of each oblong has been shaded.
answers

29 circles need to be shaded in order to represent given fractions.
answers**

This worksheet shows 20 partially shaded polygons. The task is to work out what fraction of each polygon has been shaded.
answers

This can be used as a fast paced warm up game. Fractions are represented by 35 partially shaded circles and are arranged on a coordinates grid. Percentage or decimal related questions might also be asked.

24 illustrations (pies and slices) provide a visual representation of improper fractions. As an extra activity, learners can also give their answers as mixed fractions (also known as mixed numbers).
answers

Mixed fractions are illustrated here as points on a line. Answers may also be given as improper fractions.
answers

This worksheet gives learners practice in finding fractions of whole number quantities represented by blocks of squares. This may also provide an opportunity to explore and discuss equivalent fractions.
answers**

The task here is to find fractions of small numbers represented by groups of shapes. Learners can then simplify the answers as far as possible.
answers

On this worksheet, learners shade in circles or blocks to visually illustrate equivalent fractions.
answers**

This worksheet is designed to help learners reinforce their concept of equivalent fractions by dividing sectors of partially shaded circles.
answers

This worksheet is like the one above, but this time the task is to divide rectangles.
answers

This worksheet allows learners to demonstrate their understanding of equivalent fractions by shading in a range of shapes.
answers**

The following 3 worksheets use lines to demonstrate the idea of fractions of whole number quantities:

12 linear scales aimed at those familiar with the 2, 5 and 10 multiplication tables
answers

12 linear scales aimed at those familiar with the 3, 4 and 6 multiplication tables
answers

12 linear scales aimed at those familiar with the 7, 8 and 9 multiplication tables
answers

These worksheets use dials to show fractions of whole number quantities:

20 dials, each with 8 intervals
answers

20 dials, each with 10 intervals
answers

20 dials, each with 12 intervals
answers

30 questions e.g. 1/2 x 10.
answers

30 questions e.g. 4/5 x 55.
answers

Cut up along the DOTTED lines only.
You should now have 32 rectangles (16 from each sheet).
The rectangles are given out so each player has at least one each.
Encourage the players to look at the number on the left of their rectangles and consider what questions might give their number.
Any player can start by reading aloud the 'question part' on the right of their rectangle.
If another player has the answer on their card they call it out, and then ask their question.
This continues until the loop ends when the player who asked the first question gives the answer to the very last question.
Tips:
It's worth printing out two extra sheets just so you can keep track!
For smaller groups some players may be given 2 or more rectangles, or if you prefer the loop itself may be shortened with a little cutting and sticking.

30 fractions (e.g. 5/15) need to be written in their simplest form
answers

30 more fractions (e.g. 81/90) need to be written in their simplest form
answers

20 questions e.g. 4/5 + 3/5
answers

The following 2 worksheets allow learners to practice adding fractions with different denominators:

20 questions e.g. 1/8 + 1/4
answers

20 questions e.g. 1/3 + 1/12
answers

Here are 20 questions involving mixed fractions with different denominators e.g. 2 11/12 + 2/3
answers

These worksheets give learners practice subtracting fractions with different denominators:

20 questions e.g. 3/4 - 1/2
answers

20 questions e.g. 5/12 - 1/6
answers

20 questions e.g. 1/2 x 5/6.
answers

Hopefully this mnemonic will come in handy when demonstrating the algorithm...

20 questions e.g. 1/4 &divide 1/10.
answers

This worksheet has 2 sections. The first tests learners' ability to change fractions into decimals. The second section allows learners to show that they can change decimals into fractions.
answers

**Note: There are many different ways to correctly shade in the shapes.
These are not the only "answers".

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Fractions in this worksheet are represented as circles divided into sectors. The task is to work out what fraction of each circle has been shaded. answers Each fraction on this worksheet is shown using an arrow above a point on a straight line.

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Uses congruency

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Equilateral triangles

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Labelling sides, find the relevant ratio etc

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Labelling sides, find the relevant ratio etc

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Labelling sides, find the relevant ratio etc

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Labelling sides, find the relevant ratio etc

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Just discovered some cool calculator tricks - checking prime factorisation, converting decimal time to mins, verifying algebra! Mind blown! https://t.co/n6d23gTmxf

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Just discovered some cool calculator tricks - checking prime factorisation, converting decimal time to mins, verifying algebra! Mind blown! https://t.co/n6d23gTmxf

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Just discovered some cool calculator tricks - checking prime factorisation, converting decimal time to mins, verifying algebra! Mind blown! https://t.co/n6d23gTmxf
— Caroline Cheetham (@MathsCaroline) May 11, 2018

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Just discovered some cool calculator tricks - checking prime factorisation, converting decimal time to mins, verifying algebra! Mind blown! https://t.co/n6d23gTmxf

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Just discovered some cool calculator tricks - checking prime factorisation, converting decimal time to mins, verifying algebra! Mind blown! https://t.co/n6d23gTmxf

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Just discovered some cool calculator tricks - checking prime factorisation, converting decimal time to mins, verifying algebra! Mind blown! https://t.co/n6d23gTmxf