Lesson video

Hello, Mr. Robson here.

Great choice again to join me for maths.

I'm particularly excited, because what we're about to learn is just beautiful.

So a learning outcome for today, is that we'll be able to use knowledge of expressions and equations to solve problems. Solving problems is just a wonderful thing in mathematics.

What makes a good artist? What makes a good historian? What makes a good mathematician? Problem solvers, that's what we are, isn't it? We like to figure out, what's going on in a problem.

We quite like to prove why certain things are true.

So, let's have a look at some of that today.

We're gonna need some keywords, throughout this lesson.

Expression.

We're gonna use a lot of expressions.

An expression contains one or more terms, where each term is separated by an operator.

For example, 3ab plus 7b minus four.

That's an expression.

3ab is a term.

7b is a term.

Negative four is a term.

Variable.

We're gonna use the word variable.

A variable is a quantity, that can take on a range of values.

For example, the A and the B in the expression above, are variables.

Two parts to the lesson.

We're gonna use algebra to explain number tricks.

And we're going to use algebraic expressions in proof.

Let's get started with some number tricks.

Pick a single digit number.

I've got mine.

What are we gonna do with this single digit number? Double it.

I've done mine.

Add 10.

Yep, got mine.

Halve that.

Add eight.

And then subtract your original number.

Okay, what do you notice? If you are doing this alone, you might not have noticed anything significant.

If you're doing this in a room full of people, you might have together noticed something quite interesting.

If we haven't noticed anything yet, I'd like you to pick another single digit number and try again.

And then answer the question, what happens? I'll ask you to pause this video, and take a few moments to conclude what is happening here.

So, you should have got the same result as me.

Regardless of what number you started with.

I had eight in my head and I doubled it, and I added 10.

And I halved that, and I added eight, and I subtracted my original number.

And I ended at 13 And you ended at 13.

All of you.

Whatever single digit number you started with, you ended up with 13.

So, that's interesting.

But is that enough evidence to say this is a really interesting number trick.

Are we gonna test some more things? We like testing things in mathematics.

So let's test double digit numbers.

Same number trick.

We're gonna double your number.

Add 10, half it, add eight.

Subtract your original number.

But we're gonna do it for double digit numbers.

Let's see if it works for those too.

Try it.

And then tell me, what you notice.

Pause this video, give it a go.

So, whilst we like to test a lot of things in mathematics, we like to test them as easily as possible.

So my double digit number was 10.

I'll just start with the easiest double digit number I how to work with.

I doubled it, I add 10, I halve that.

I add eight, and I subtract my original number, and there we go again.

13.

Interesting.

So, it works for single digit numbers, and it works for double digit numbers.

And whatever double digit number you picked, I know you ended up with 13.

So does it work for any number? So does it work for three digit numbers? Four digit numbers? Five digit numbers? Don't be afraid to use a calculator if you're gonna test that.

Does it work for negative numbers? Decimals, fractions? Will it work for any number? I'm going to ask you to pause this video, and substitute in any number.

Not a single digit number, not a double digit number.

Let's get some variety to this.

Pause this video and substitute in any number.

Did anything change? No, there was no escaping the outcome of 13.

I tested negative 0.

5.

Because I can kill two birds with one stone here.

I can test a decimal and negative, at the same time.

And I started with negative 0.

5 deliberately.

Because when I double it, I get negative one.

And the whole thing's about to become easier.

Add 10 to that, halve that, add eight.

And then subtract my original number.

12.

5 minus negative nor 0.

5.

When I take negatives out my life, things get more positive.

That gives me 13.

There was no escaping 13.

I'm excited as the variety of numbers like you might have substituted in there.

Did anybody go crazy and try a six digit number? It still works.

There was no escaping the end result of 13.

So, we've demonstrated pretty trustfully that it works.

We always end up at 13.

But we like to go further and understand why things work.

It's what mathematicians do.

Why does that work? So, we'll do that by writing a generalisation to explain what is happening here.

Same number trick.

But to represent any number.

And I'm sure in the experiment we've done so far, you'll appreciate the variety of numbers, that we could have begun with.

We can express that variety of numbers, by using a variable.

We could say let's start with X.

X being a variable.

It can take on any one of the infinite values available.

Double it.

Okay, we started with X.

Double X is two x.

Add ten.

Two x plus 10.

Yes, they are algebra tiles.

Algebra tiles are really useful for modelling mathematics in all sorts of ways.

So, I'm running them concurrently with the algebraic expressions that I'm writing.

Because this next step is really easy to see.

You can see with my algebra tiles, how I turned my starting X into two lots of that starting X.

And I added 10 to it.

The next step catches a lot of people out.

We want to half two x plus 10.

Two x plus five.

X plus 10.

Now we have to half both terms. We have to halve the two X term, and we have to halve the 10.

And we get x plus five.

You can see that in the algebra tiles.

If I said there's two x plus 10 in algebra tiles, remove half those tiles, you would remove half those tiles, and leave me with X and five ones.

So we've halved two X plus 10, we're at x plus five.

We're gonna add eight to that.

That's quite a simple move.

Add the constant eight to the five, to get to X plus 13, and then subtract your original number.

Your original number was x.

I'm gonna remove that from my algebra tile pile, 13.

No avoiding it.

You can put any value into X at the beginning, and you are going to end up at 13.

We've used algebra, algebraic expressions to demonstrate why you will always end up at 13.

Let's check, you've got that.

A different trick this time.

I wonder if this will be 13.

Are all tricks 13? I don't know.

But something we can explore.

This number trick, pick any number.

Subtract eight, double that result.

Add 16.

Halve that result.

And then subtract your original number.

Allow you to test this number trick, with a wide variety of numbers.

Single digit, double digit, negatives, decimals.

What do you notice? And then the highest level skill.

Can you write a generalisation to explain what is going on? Pause this video, and take a couple of minutes to do that.

I know what happened.

So I'm gonna go straight to the generalisation, to demonstrate to you, why what happened, happened.

Now you might have written the same generalisation.

You may have used a variable of X.

I'm gonna start with N.

N representing any number.

N subtract eight.

That's N subtract eight.

I'd like to double that.

Two lots of the expression, N minus eight.

And then I can expand those brackets.

Are you starting to see some of the algebraic skills you've been developing so far? Coming into play here.

Multiplying an expression by constant.

Multiplying out the brackets.

Two lots of bracket, and minus eight is the equivalent to having two N minus 16.

We're gonna add 16 to that.

That's handy.

Two N minus 16.

When I add 16, leaves me just with two N.

Half of two N is N.

And then subtract your original number.

Well that means I'm going to do N minus N, and be left with nothing.

No matter what you started with, in your experiments, you came out of that number trick with nothing.

And these algebraic expressions can demonstrate why.

Practise time now.

This is a number pyramid.

What does a number pyramid do? We're gonna put the digits one to five in the bottom row.

You can put them in any order you like.

When you come to do this task, you can put one, two, three, four, five, in that bottom row in any order you like.

I'm gonna show you one example.

I'm gonna put them in this order.

One, five, four, two, three.

When you experiment with this, you can use any order you like for those digits.

As long as there's one, two, three, four, and a five.

A block is equal to the sum of the two blocks upon which it sits.

Okay.

So one plus five makes six.

Five plus four, nine.

The sum of four and two.

The sum of two and three.

And our algebra pyramid just keeps.

Our number pyramid just keeps adding up and up and up.

So in the next row, six and nine will make 15.

Nine and six will make 15.

Six and five will make 11.

There we go.

And then the next row, I'm gonna have 15 plus 15.

And 15 plus 11.

And then the top block's going to be 30 plus 26.

56.

That's how your number pyramid works.

What I'd like you to do for this task, is insert the numbers one, two, three, four, five, into your bottom row.

Add them and then answer this.

What's the largest total you can make in the top box? Who knows? Is it 56? Is the smallest total I can make in the top box, 56? Because why would it change? We're just using one, two, three, four, five.

Or does it change? Should we experiment a little? Like we did with the number tricks.

Should we test a few things? Try an order to begin and then mix up that order and see if anything changes.

What we do as mathematicians.

We experiment, we explore.

Once you've established if there is a larger total than 56 to be made, what strategies did you employ in order to try and explore this? What did you change when you were doing multiple repetitions of this task? Part C, later write a generalisation to explain what is happening here.

Pause this video and give that a go.

What's the largest total you could make in the top box? The answer is 61.

Provided you started with one, two, three, four, five, in the bottom boxes.

Your order would've looked something like this.

Now you might have 61, and have done it slightly differently.

Your one and two might be mirrored.

You might have two on the very left hand side, one on the very right hand side.

Your three and four might be mirrored in positions.

But crucially, if you've made 61 as a total at the top, you have five in the middle.

As we sum up, we hit four, eight, nine, six, 12, 17, 15, 29, 32 making 61.

You may have made 61 in a slightly different route to that.

But guaranteed you had five in the middle of the bomb.

One and two on the very outside of the row.

And you had 61 in that top block.

So things changed, and why did they change? What strategies did you employ to make the largest total? You would've noticed that, the totals were bigger.

As we had the bigger numbers in the central block, of our bottom row.

When we had five in the central block of the bottom row, we started to make larger numbers.

So, you might have said, I tried putting the largest number in the middle.

Since there's more blocks on top of that central block.

There are more blocks on top of that central block.

So rather than thinking of this block being the sum of those two blocks, we could think of this block being added into both of those blocks.

And both of those blocks.

So, having large numbers in the middle meant they had a greater impact on that total at the top.

Part C, was to write a generalisation to explain what's happening here.

The fact that we had to put the five in the middle, the largest number in the middle to make the largest total at the top, An algebraic generalisation will explain why that was so.

If we start with variables in the bottom of our pyramid, rather than constants, one, two, three, four, five.

Let's start with variables.

A, B, C, D, and E.

What are we gonna write in the row above it? Well, we need to have the sum of the two blocks on which it sits.

So that first block on the left, will be A plus B.

The next block will be B plus C.

And so on.

Our number pyramid is just adding up as we did with number.

We're just leaving it as algebraic expressions for now.

In the third row up A plus B summed with B plus C.

So I'm gonna write, in the third row, left block A plus B plus B plus C.

Yeah? No.

You are shouting at the screen, Mr. Robson, you can simplify that.

B plus B, well, that's 2B.

So you are gonna write A plus two B plus C.

Absolutely.

We'll keep those expressions as simple as possible, in the next row.

So in the middle block there, B plus C plus C plus D.

Well that's B and two Cs and D, et cetera.

The next row above it.

Those expressions are gonna get a little bit long for me to fit in the blocks on my diagram.

But we would have A plus two B, plus C plus B plus two C plus D.

I'm gonna simplify the B terms and the C terms. So, on the left hand side I've got A plus three B, plus three C, plus D.

B plus three C, plus three D plus, E.

Even in these expressions now, we can see the power of the numbers that were in the middle at the bottom.

We had B, C, and D in the middle at the bottom.

They're the ones with the coefficient of three.

They're the ones being used multiple times by the time we get up to this row.

By the time we get to the top row, it's really clear to see why the middle number was key.

Because we have in total, A plus four lots of B, plus six lots of C, plus four lots of D plus E, A and E on the outside only appear in the total at the top once.

Whereas the number in the middle, the five which we had earlier, was used six times by the time it reached the top.

This algebraic expression demonstrates the fact that the number in the middle of our pyramid was multiplied or summed a greater number of times, by the time it reached the top of the pyramid.

So, part one was about demonstrating why certain things work using algebra.

We're gonna go to a really high level skill now, and I'm excited.

Proof, this is huge in mathematics.

Huge in algebra.

Can we prove why certain things are true? Absolutely.

Let's have a look.

Demonstrate that the sum of three consecutive integers, is a multiple of three.

Three consecutive integers.

Okay, let me start simple.

One, two, three, integers, consecutive.

That means they are one after the other.

One, two, three.

Demonstrate that the sum of three consecutive integers is a multiple of three.

Okay, I'll sum them together.

One plus two plus three.

They sum to six.

Is six a multiple of three.

Oh yes, it's the second multiple of three.

So, six is three times two.

What could I try next? Demonstrate the sum of three consecutive integers is a multiple of three.

Any other three consecutive integers? Let's five, six, seven.

Do they sum to 18? Is 18 a multiple of three? It's the sixth, multiple of three.

Okay.

Should we skip to the little larger? Single digit numbers seems to be working.

10, 11, 12.

Let's step into the double digits.

They sum to 33.

Is 33 a multiple of three? It's the 11th, multiple of three.

Okay, let's go crazy.

Let's skip a few.

Let's go to 357, 358, 359.

What do we think? That gonna be a multiple of three? Well they sum to 1074.

Is 1,074, a multiple of three? I absolutely at this stage, I'm using my calculator.

I think I'm onto something, but I don't want to spend ages having to add those three numbers using a column method and then get the bus stop method out, to divide 1,074 by three.

I'll use my calculator at this stage, because I want to be as efficient as possible.

1,074 divided by three gave me 358.

So 1,074 is a multiple of three.

It's the 358th, multiple of three.

Okay, I'm pretty sure we've demonstrated that the sum of three consecutive integers are a multiple of three.

Did you notice anything? Look more closely.

It's not just any multiple of three.

It's three lots of the middle number.

That's interesting.

So will this always be true? Can we test it for all different groups of three consecutive numbers? Because I can see it works for single digit numbers.

I can see it works for double digit numbers.

I can see it works for treble digit numbers.

Or that group of treble digit numbers.

Do I want to test all the treble digit numbers? And then all the four digit numbers? All the five digit numbers? Do I want to test 1,000,001 1,000,002, 1,000,003? Not really.

So this is where proof becomes incredibly useful.

We think we've spotted something.

Take any three consecutive integers, and not only will it be a multiple of three.

It'll be three lots of the middle number.

But rather than trying to exhaust all the numbers that exist in order to test this, which would be exhausting, we'll use proof.

So we change that language from demonstrate that, to prove that.

Prove that the sum of any three consecutive integers is a multiple of three.

I want to prove it.

Not show some examples that it's true.

Prove it.

This is where algebra becomes really powerful.

So, let's start with any integer.

Now you've already seen me use N to represent any integer in this lesson.

Let's start with N.

That's gonna be our first number.

So, it's the sum of any three consecutive integers.

So, what's the next number that comes after N? It's consecutive integers.

So it's gonna be one more than the first number.

How do we say one more than N? That's right.

N plus one.

What's the next number gonna be? What's gonna be one more than that? How do you say one more than N plus one? N plus two.

That's how we can use algebra to represent three consecutive integers.

N, N plus one, N plus two.

I'm proving something about the sum of these three integers.

So let's sum them together.

N plus N, plus one, plus N plus two.

Well done.

You are screaming at the screen.

You can simplify that.

Absolutely.

Let's gather those N terms together.

Let's gather the constant terms together.

And we should have three N plus three.

And then have a variety of mathematical skills comes into play again.

We've got the expression, three N plus three, and you know you can factorise that into three lots of bracket N plus one.

You've taken out the factor of three, and the factor of N plus one.

In fact, it doesn't matter what number N is.

Sorry, It doesn't matter what integer N is.

This will always be true.

We'll always get three lots of N plus three as the sum of those three consecutive integers.

Which we can express as three lots of bracket N plus one.

Did you notice therefore, it's always going to be multiple of three.

And it's always going to be that middle number.

N plus one, was the expression for the middle number.

Hence we always got three lots of our middle number.

Joyous, hey.

Next, I wanna check you've got that really high level skill list.

So we might not do it perfectly, first time.

But we're certainly gonna give it a go.

And if we make mistakes, we'll learn from those mistakes.

Prove that the sum of any five consecutive integers, is a multiple of five.

That's it.

That's the task.

That's what I'm asking you to do.

It's tricky, but it's doable.

We want to give it a go.

We wanna be brave.

If you try to have a go now, and you need a hint, you might want to just go back in this video a moment and look at that again.

There's the proof, that the sum of any three consecutive integers and multiple of three.

If you get a little bit stuck proving that the sum of any five consecutive images and multiple of five, you might just want to rewind and pause, and have a look at how we did the proof for three.

Anyway, pause.

Be brave, give it a go.

So, the start looked very familiar.

Let N be any integer.

Therefore, five consecutive integers will be N, N plus one, N plus two, N plus three, N plus four.

That would represent five consecutive integers.

When we sum those together, we get that.

Let's simplify.

Five lots of N plus 10.

And it factorises.

And it factorises to five lots of bracket N plus two.

And there is the proof.

It will always be a multiple of five.

But once you've got that expression, not only do we know it's a multiple of five, it's a multiple of five lots of N plus two.

Oh, N plus two is in the middle again.

So, we take any five consecutive integers, and we always get five lots of the middle one, as their sum.

Practise time now.

Question one.

I'd like to prove that the sum of any seven consecutive integers, is a multiple of seven.

Again, you might want to rewind, and have a look at the proof for three and for five.

If that helps you get started with that task.

Part B is slightly different.

Prove that the sum of any four consecutive integers is a multiple of two.

I won't say any more than that one slightly different, because it might give the game away.

But, use algebra to prove that the sum of any four consecutive integers is a multiple of two.

Pause this video, and give that a go.

Okay, question one.

Prove that sum of any seven consecutive integer is a multiple of seven.

Let's N be any integer.

There are seven consecutive integers.

There is the sum, of seven consecutive integers.

And I certainly want to simplify.

Because that's a long expression.

Seven lots of N plus 21.

That is the sum of seven consecutive integers.

And wonderfully it factorises.

We can take out the factor of seven, and be left with the expression N plus three in our brackets.

So we can see that the sum of seven consecutive integers will always be a multiple of seven.

And not just any multiple of seven.

It will be seven lots of N plus three.

N plus three, is right in the middle of our list again.

The sum will always be equivalent to seven multiplied by the middle number.

Part B.

Prove that the sum of any four consecutive integers is a multiple of two.

Let N be any integer.

There's four consecutive integers.

N, N plus one, N plus two, N plus three.

Let's sum them together.

There's a long sum.

Let's simplify.

Four lots of N plus six.

Now notice we can't take our a factor of four here.

So it won't be a multiple of four, the sum.

But we can take out a factor of two.

Right, that's the end of the lesson now.

I thoroughly enjoyed it.

I love algebra.

I love proof.

I hope you did too.

We showed that algebraic terms can be used to explore patterns in number problems. And found that algebraic expressions, can be used to prove why certain outcomes are inevitable.