Lesson video

Well done for making a decision to learn using this video today.

My name is Ms. Davis and I'm gonna be helping you as you work your way through this lesson on proof.

We've got lots of exciting things coming up so make sure you've got everything you need and let's get started.

Welcome to this lesson on writing a proof.

By the end of this lesson, you'll be able to manipulate and transform algebraic expressions in order to prove or show a result.

We're gonna look at some really interesting proofs today.

A couple of key words that we're going to use.

If you need a reminder of what a conjecture is or what generalising means, pause the video and read through those now.

(no audio) So let's start by writing a formal proof.

There are many different types of proof.

Using algebra and key number facts can be a useful way to prove a range of results and that's what we're gonna be looking at today.

For example, we're gonna prove that when n is an integer, n plus 8 squared, minus n plus 2 squared is a multiple of 12.

Now, from the way this statement is worded, we can see that algebraic manipulation is going to help us prove this.

We've already got an algebraic expression, we're gonna want to manipulate that to show it's a multiple of 12.

So with any algebra proofs, we need to start by defining the variables we are using.

So my first step would be take all numbers of the form, n plus 8 all squared, minus n plus 2 all squared where n is an integer.

Now that information was in the title, we are just rewording it to say that that is gonna be our start point to our proof.

Now I need to manipulate this.

So these numbers can also be written as n squared plus 8n, plus 8n, plus 64.

Subtract n squared, plus 2n, plus 2n, plus 4.

And I've tried not to skip too many steps.

So I've squared that bracket and I've given all four partial products for that first bracket and then I've shown that I'm subtracting the whole of that second bracket squared.

So I've left that product in a bracket.

Now in the next step I can do some simplifying, so I'm collecting like terms, got n squared, plus 69, plus 64, subtract all of n squared, plus 4n, plus 4.

I've decided now to expand that second bracket.

So I've got n squared, plus 60n, plus 64, subtract n squared, subtract 4n, subtract four, and now I can simplify.

So if you look at the steps in my proof, some of them were just algebraic manipulation, that's fine.

Some of them can include words, some of them can just be entirely written in words.

That's absolutely fine.

Right, how could we now show that this is a multiple of 12? Pause the video.

What would you do? (no audio) Well done if you said factorise.

The easiest way to show this is a multiple of 12 is to remove a factor of 12 and show that this can be written as 12 lots of n plus 5.

Well, n is an integer, so n plus 5 is an integer, 12 lots of an integer will be a multiple of 12.

So I've done the factorising and then I've added in words, this is an integer multiplied by 12, so must be a multiple of 12.

And then I just need a conclusion to finish.

Therefore, all numbers of the form n plus 8 squared, minus n plus 2 squared where n is an in integer are multiples of 12.

And that's essentially me rewriting what we started with.

If you want to read through that proof in full, pause the video and look through those seven steps now.

(no audio) sometimes the statement we want to prove has to be written as a generalised algebraic statement first.

For example, prove that for any three consecutive integers, the difference between the squares of the first and the last number is four times the middle number.

There's a lot going on in this proof, so let's pick it apart.

So you want to generalise three consecutive integers, Pause the video.

How could we do this? (no audio) There are different ways you can write three consecutive integers, but if I think about what I'm going to do in a minute, I'm gonna need to square the first and the last number.

So we're gonna want to use the simplest forms possible.

I think with that in mind, you can write it as n, n plus 1, n plus 2, where n is an integer or n minus 1 and n plus 1 where n is an integer.

There's positives and negatives of choosing either 1 of those, but either is gonna give us a nice result.

I have chosen any three consecutive integers can be written as n, n plus 1 and n plus 2 where n is an integer.

Right, now I need to return to my proof.

So I'm trying to show that the difference between the squares of the first and the last number, we'll come on to the rest in a minute.

So let's start with that bit.

So the difference between the squares of the first and the last number can be written as n plus 2 all squared minus n squared.

And then I'm just gonna have a look.

I'm trying to show that that is four times the middle number, so that's what I'm gonna work towards.

Right, what do you think we should do in our next step? (no audio) This is similar to the last one.

We're gonna want to expand our brackets and then we're gonna want to collect like terms. N plus 2 all squared is n squared, plus 2n, plus 2n, plus 4, minus n squared we had as well.

If you prefer to write n plus 2 all squared as n plus 2, times n plus 2 first to remind you that it's the product of two binomials, hen that's absolutely fine.

Now I'm gonna collect like terms. N squared, subtract n squared is zero n squared, 2n, plus 2n is 4n, plus 4.

Now I'm looking to show that this is four times the middle number.

Important to go back to your proof and look at what you're trying to show.

How are we gonna show this? (no audio) Yeah, we're looking at factorising again, aren't we? So I've got four lots of n plus 1 and I'm making sure I'm factorising it this in a useful way, not just fully factorising it without thinking about what I'm trying to do.

So I'm trying to show that it's four times the middle number, so I'm trying to write it as four lots of n plus 1, which I can do.

So then my sentence is, this is four times the number n plus 1, which is the middle number, and don't forget your conclusion.

Therefore for any three consecutive integers, the difference between the squares of the first and the last number is four times the middle number.

If you haven't come across it before, the three dots arranged in the triangle at the beginning of step six means, therefore.

Right now we are gonna have a go.

So I'm gonna show you this one on the left.

Then you are gonna go at a similar proof on the right.

So proof that the sum of any multiple of three and a multiple of six is divisible by three.

So I'm gonna start by generalising, take a multiple of three: 3n and a multiple of six: 6m where n and m are integers.

Their sum is written as 3n plus 6m.

I'm trying to show that this is divisible by three.

I'm gonna take out a factor of three.

Three lots of n plus 2m.

This is a multiple of three, so it's divisible by three.

Therefore, the sum of any multiple of three and multiple of six is divisible by three.

Right, your turn.

You're gonna have a go at proving that the sum of any multiple of 12 and multiple of eight is a multiple of four.

You don't have to write the numbers by the steps, but it's really important that you are writing each step on a new line, working down the page, give it a go.

(no audio) So take a multiple of 12, 12n and a multiple eight, 8m where n and m are integers.

Their sum is written as 12n plus 8m.

Want to show it's a multiple of four, so in fact try that, four, so four lots of 3n plus 2m.

Then your conclusion, this is a multiple of four, therefore the sum of any multiple of 12 and multiple of eight is a multiple of four.

And you could have put those two bits in two separate sentences.

That's fine as well.

Right, Andeep wants to prove the divisibility rule for positive two digit multiples of three.

Let's see what he means.

Andeep says, "I know for any multiple of three, the sum of the digits is also a multiple of three, but how would I generalise this?" This might have been a result that you've seen before.

So divisibility rule to help you see if something is a multiple of three, if you add the digits of the number and that's a multiple of three, then your number will be a multiple of three.

So how can we show that this always works? Well, this rule relies on summing the digits.

So we need to express all positive two digit numbers in terms of their digits.

Right, if the ones digit is p and the 10s digit is q, how can we write an expression for a two digit number? What do you think? (no audio) This was a bit tricky, so well done if you got this.

We can write it as p plus 10q.

If we had 29 for example, the ones digit would be nine and the 10s digit would be two.

We can write that as nine plus 10 times 2, nine plus 20, 29.

So the start of our proof would be, take any number of the form, p plus 10q, where p is an integer and p has to be between zero and nine inclusive because it has to be a single digit number.

Q is an integer, but q has to be between one and nine inclusive.

If q was zero, it wouldn't be a two digit number, it would only be a one digit number.

So q needs to be between one nine inclusive, p can be zero, that would be fine.

Andeep say, "Now I need to show that if the digit sum to a multiple of three, then the number itself is a multiple of three." How could he set up this next step? Pause the video, what would you do next? (no audio) Right, he needs to generalise the cases where the digit sum to a multiple of three.

So now we just wanna look at what happens when p plus q is a multiple of three.

So I can say, assume the digits sum to a multiple of three.

So p plus q equals 3n where n is a positive integer.

Now we need a way to substitute 3n into the original expression.

This is where this proof gets a little bit tricky to see what we need to do.

So what we need to do is we need to isolate p plus q 'cause we want to rewrite that as 3n to use that piece of information we've just written.

So if we look at p plus 10q, if we pull out p plus q, we could rewrite this as p plus question, plus 9 q, and then we can substitute 3n for that p plus q.

So we're then using that information that the digits sum to a multiple of three.

So I've got 3n plus 9 q.

Now I want to show us a multiple of three, so I just need to factorise.

And then my conclusion.

Therefore for any positive two digit number P plus 10q, if it's digits p and q sum to a multiple of three, then the number is a multiple of three.

If you want to pause and look through all the steps of that proof, feel free to do that.

Now you are gonna do a similar one in the task with three digit multiples of three.

So you might wanna check the key features of that proof.

Quick check then.

If n is an integer, select the statements that are true for 6n plus 15.

What do you think? (no audio) well done if you said B and D.

Let's look at them all separately.

If I factorise out a two, I can write that as two lots of 3n plus seven, but that only gets me 6n plus 14.

So I've got plus 1 on the end.

This is one more than an even number, so definitely odd.

I can factorise out three.

So three lots of 2n plus 5.

So it is a multiple of three.

It's always one less than a multiple of four.

Or we could try one for this one and show why it doesn't have to be.

Six times one, for example, will be six plus 15, which is 21, and that's not one less than a multiple of four.

And it's always three more than a multiple of six.

Well that would be six lots of n plus 2, plus 3, and that's our way of showing that it is 3 more than a multiple of six.

Time for you to have go at some proofs then.

So you're gonna prove that the sum of two consecutive integers is odd to start with.

Then in question two, want you to prove that if n is an integer, the numbers of that form are a multiple of three.

And then for B, using the same manipulation, can you prove that numbers of this form are also one less than a multiple of four? (no audio) Right, question three, here are the steps to three different proofs in a random order.

I'd like you to rearrange them to form three distinct algebraic proofs.

So you're gonna have to think about what is the generalisation for each of the proofs, what's the conclusion for each of the proofs, and then that should help you fill in the steps of algebraic manipulation.

Give those a go.

(no audio) And question four, Andeep wants to prove the divisibility rule for positive three digit multiples of three.

I'd like you to fill in the blanks for the first two steps and then complete this proof for three digit numbers.

This is similar to the one that we did in the explanation.

Andeep wants to show that if the digit's sum to a multiple of three, then the number itself is a multiple of three.

Give that one a go.

(no audio) For question five, Aisha has found a number trick.

I'd like you to test three different integers and see if you can work out what this trick seems to do.

So what conjecture could you make? Then I'd like you to prove that this works for all integers.

Give that one a go.

(no audio) And finally, I'd like you to prove that the Product of two consecutive positive integers add the larger integer is always a square number.

So what I mean by that is if you have two consecutive positive integers, so for example five and six, and you have the product of those, so 30 and then you add the larger one, so six, that will be a square number.

We can see it is in that case 'cause it's 36.

I'd like you to prove that that is always the case.

Give that one a go.

(no audio) Well done.

Lots of things to be thinking about in that activity.

So for question one, we want to take two consecutive integers n and n plus 1.

Their sum is n plus n plus 1, which can be written as 2n plus 1.

This is one more than an even number, so an odd number.

Or you could say this is the general form of an odd number if you prefer.

Therefore, the sum of any two consecutive integers is odd.

For question two is we want to take any number of the form 2n plus 6 all squared, minus 2n plus 3 all squared where n is an integer.

Then we want to manipulate this.

It's up to you how you manipulated this, what sort of steps you did, as long as all your working is really clear and then I can simplify.

So that gives me 12n plus 27.

And to start with, I wanted to show that as a multiple of three.

So take out a factor of three.

You've got three lots of 4n plus 9.

Therefore numbers of the form 2n plus 6 all squared subtract 2n plus 3 all squared where n is an integer are multiples of three.

Then I wanted to do the same thing but show that there are one less than a multiple of four.

So I'm gonna pick up at step five.

So instead of factorising out a three, I'm gonna partially factorise out a four.

So I'd have four lots of 3n plus seven, but that gives me 12n plus 28.

So I then need to take away one and then I've shown that numbers of the form, 2n plus 6 squared minus 2n plus 3 squared where n is an integer are always one less than a multiple of four.

So for question three, here are your three distinct proofs.

The first one is showing that the sum of three consecutive integers is always divisible by three.

The middle one, the sum of three consecutive even integers is a multiple of six.

And the last one, the sum of three consecutive odd integers is three more than a multiple of six.

I'd like you to pause the video and check that you've got the steps to those three proofs in the right order.

(no audio) So for question four, we want to take any number of the form p plus 10q plus a hundred r where p, q and R are integers.

P and q can be any single digit integer.

So between zero and nine inclusive, but r cannot be zero.

So r can be between one and nine inclusive.

The digit sum to a multiple of three.

Then p plus q plus r equals 3n.

Where n is a positive integer.

You could use a different variable instead of n as long as it's not p, q or r.

P plus 10q plus a hundred r equals p plus q, plus r plus 9 q, plus 99 r.

I'm trying to isolate p plus q plus r, so I can substitute in 3n, then I'm substituting 3n, so 3n plus 9 q plus 99 r.

Then I can factorise three lots of n plus 3 q plus 33 r.

Three lots of any integer is a multiple of three.

Therefore for any positive three digit number p plus 10q plus a hundred r, if it's digits, p, q and r sum to multiple three, then the number is a multiple of three.

And then let's look at Aisha's number trick.

So you could have tried any three different integers and hopefully you saw that whatever number you started with you'd get the same number to finish.

So your conjecture might have been, I conjecture that for any integer if you double it, add six, double again, subtract four, divide by four and subtract two you'll get the integer you started with.

If you didn't want to write all those words out, you might have used algebra.

I conjecture that for any in integer n, two lots of 2n plus 6, subtract four all over four, subtract two is equivalent to n.

And then we're gonna prove it works.

So take any in integer n, apply the operations detailed to give the number of the form, two lots of 2n plus 6 minus 4 all over four, subtract two and then we're gonna manipulate that, which is n, which is the value we started with.

Therefore, for any positive integer n doubling, adding six, doubling again, subtracting four, divided by four, then subtracting two will produce the integer n.

In fact, this works for all numbers, not just integers.

And finally, the product of two consecutive positive integers at the larger integer is always a square number.

So let's take two consecutive integers and we want positive integers.

So n has gotta be greater than zero.

The product of the integers at the larger integer can be written as n lots of n plus 1, plus n, plus 1, which is n squared, plus n, plus n, plus 1 or n squared, plus 2n, plus 1.

If we factorise that we're trying to show it's a square number, well that factorise into n plus 1, times n plus 1 or n plus 1 all squared.

This is written as an integer squared, so must be a square number or zero.

But since n is greater than zero, n plus 1 cannot be zero.

So n plus 1 squared is a square number, therefore the product of two consecutive positive integers at the larger integer is always a square number.

Right, now we've worked really hard writing our own proofs.

We're gonna have a look at analysing other people's proofs.

Lucas is trying to prove that the difference between the squares of any two consecutive odd numbers is always a multiple of eight.

There's his proof.

He says, "That didn't work.

Four n plus 8 is not always a multiple of eight." I'd like you to read through his proof.

Why has it not worked? (no audio) Right, well, he is not generalised consecutive odd numbers properly.

This proof is supposed to be about the squares of any two consecutive odd numbers.

Right, he thinks he's corrected it now.

What else does he need to do to improve his proof? (no audio) Right, that works much better now, He's defined consecutive odd numbers as 2n plus 1 and 2n plus 3.

However, he hasn't actually told us what he's doing.

He hasn't defined what n is, n needs to be an integer, doesn't it? In order for those to be odd.

And he hasn't explained why 8n plus 8 is always a multiple of eight and he hasn't written a conclusion.

Algebraic proofs should start by defining any variables used and if a generalisation is made, this should be explained.

So he needs to write, take two consecutive odd integers, 2n plus 1 and 2n plus 3 where n is an integer.

Now he can do the algebraic manipulation just as he did before.

Luca says, "I know this has to be a multiple of eight," but how does he show this? Well of course we need to factorise, don't we? So if you factorise out an eight, we get eight lots of n plus 1 and proof should always end with a conclusion that explains what has been proven.

Therefore, the difference between the squares of any two consecutive odd numbers is always a multiple of eight.

Right, read through this proof.

What is wrong for the proof that the difference between two different odd numbers is always even, give it a go.

(no audio) Right, so this proof looks like it was working and we've proved that something is even, we've got two lots of something at the end.

However, the algebraic manipulation has gone wrong.

So it's just a coincidence that we've ended up with something that has to be even.

The difference between two odd numbers should be 2m plus 1, subtract all of 2n plus 1.

We need brackets round the 2n plus 1, then we get 2m minus 2n.

So yeah, that's true, Jun, in this case it's possible for the proof to look correct, but from incorrect working, we haven't proven what we set out to prove.

A proof is a mathematical argument.

Someone should be able to follow your working logically from your initial assumption to your conclusion.

So take a quadratic equation of the form axe squared plus bx plus C equals zero where A, B, and C have any numerical values (no audio) Right, this is a proof of the quadratic formula, but what is wrong with it? Pause the video, give yourself time to read through it.

(no audio) Right, it skips too many steps.

I don't know if you felt this as well as you're trying to read through it.

It's not easy to follow all the algebraic manipulation.

So let's look at this in more detail.

So for this first step we can see what we're doing there.

We're dividing every term by A, we might not be sure why.

It's because we're gonna complete the square.

It's easier to do that when the coefficient of x squared is one.

If you found this next step hard to follow, it's possibly 'cause you didn't realise we need to complete the square.

So it might be better to write the words by completing the square and then put this step in.

And then we skip a lot of steps to get to where we were for step four.

So let's see if we can add in some steps.

Right, if we expand the brackets for that second part, the B over two A all squared, we get B squared over four A squared.

Then we want to rearrange, so it's on the other side of the equation.

Then we need a common denominator so that we can then make it a single fraction and then it's gonna look like step four.

And here is the complete proof.

We can see now that each step is really clear to follow, we could have added some words in to explain what we were doing if we wanted to make it even clearer, but we needed that many steps to really show all that algebraic manipulation.

Pause video now if you want to look at that in more detail.

(no audio) Right, Your turn, what is missing from this proof for the product of any two odd numbers is odd? Read through it.

What do you think? (no audio) Well done if you said the variables haven't been defined.

M and n are used but we don't know what they represent.

The rest of the proof was pretty good.

In fact, 2n plus 1 only has to be an odd number where n is an integer.

So that sentence is so important.

Okay, your go, I'd like you to read through these two proofs and identify the error in each one.

(no audio) Okay, you've got two more to have a look at, Where is the error in each of these? (no audio) For question two, I'd like you to read through this proof that the product of any two odd numbers is odd and tell me how it could be improved.

Off you go.

(no audio) Can you do the same now for this proof that any odd number at any even number is always an odd? (no audio) And the same for C.

This time we're trying to prove that the sum of the squares of two odd integers is even.

(no audio) And finally, how could we improve this proof that if n is an integer 4n plus 3 squared minus 12n times n plus 1 is a square number.

(no audio) Well done.

Let's see if we can spot the mistakes.

There's actual algebraic mistakes in this one.

Two n plus 1 all squared is not 4n squared plus 1.

It's the product of two binomials.

We should have four partial products.

The correct conclusion was reached but from incorrect working.

So we've not proved what we wanted to prove.

For B, n squared plus 1 is not a square number.

So it's our start point, our generalisation, which is incorrect.

What we should have is n plus 1 all squared.

For C, we've got a couple of issues here.

First one is we need brackets to show that we were subtracting the whole of the second bracket squared.

Because we didn't have the brackets, we only ended up subtracting the 9n squared, not the 6n and the one that we needed to also subtract.

And then that caused us problems, that's probably why the next mistake happened, 'cause we didn't know what to do.

18n plus 5 is two more than a multiple of three, so it is not a multiple of three.

So that should have told us that we made a mistake somewhere, we couldn't factorise out three.

And for D, it's the incorrect factorisation and there's two issues with this.

To start with, that's not how we factorise.

So that's an incorrect factorisation.

If we want to show this as a multiple of three, it's three we need to remove as a factor.

So we're writing this as three lots of 2n plus 3.

Then we've shown it's a multiple of three.

Right, and how could we have improved these? I wonder if you had any of these points in your answer.

We need to define the m and nr integers, otherwise those two numbers don't have to be odd.

And we need to write a conclusion to explain what has been proven.

So we haven't actually shown that we're adding these two things together that clearly.

So you might have wanted to say the sum of the 2 numbers is 2n plus 1 plus 2m, and then rearranged.

More importantly after step two, we need to show why that is an odd number.

So you might want to partially factorise and say two lots of n plus m plus 1.

We need to show why that is an odd number.

It is one more than an even number, so an odd number.

For C, we need to explain the generalisation.

We have defined the variables, but we need to write something like two odd integers can be written as 2m plus 1 and 2n plus 1.

We're trying to explain why we've chosen to write it in this form.

You might have said this could be further improved by adding more steps.

There's a little bit of a jump from the expression in step one to the expression in step two.

We could add a few more steps to help us see that.

And finally, this is not the way to show an expression.

It is a square number, so it's that last bit that is the problem.

When we got to 4n squared plus 12n plus 9, we need to show that it could be written as an expression all squared.

So if you factorise it into double brackets, you get 2m plus 3, 2m plus 3 or 2m plus 3 all squared.

This is either a square number or zero, but because n is an integer, 2m plus 3 cannot be zero.

Well done, we worked really, really hard with those proofs today and I'm hoping you can see some of those things that make a proof really good so that you can avoid some of those mistakes when you're writing your own proofs.

Thank you for joining me today.

I've had a lot of fun.

I hope you have as well.

And I look forward to seeing you back for another video soon.

(no audio).