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 where we're gonna be writing generalised statements about specific number properties.

We're gonna be looking at describing a situation using algebraic symbols and starting to move towards proving some algebraic results.

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

So we're gonna start by looking at how to set up a proof.

We get into the really good bit now where we've got those skills to prove some exciting results.

So proof requires all relevant cases to be considered and accounted for.

One of the easiest ways to do this is to write a generalised statement about the relevant cases and then use all our algebraic manipulation skills.

So how do we write the generalised form of an odd number? For example, what do you reckon? Right, the easiest way is 2n + 1 or 2n - 1 where n is an integer.

How about this one? Is 6n + 5 an odd number when n is an integer? Yes it is.

The easiest way to explain this is to partly factorise.

So 6n + 5 can be written as 2 lots of 3n + 2, add 1, which shows that it's one more than an even number, so an odd number, or you might have said that 6n has to be even, five more than an even number will be an odd number, even add odd is odd.

Right, why would 6n + 5 not be a generalised form for any odd number? Why did we go with 2n + 1? Why would 6n + 5 not be good? Right, this is an important point.

It is odd and it is always odd, but not all odd numbers can be written in that form.

That is very specifically odd numbers which are five more than a multiple of six.

So 11 would work as 6n + 5, but I couldn't write the number 3 in the form 6n + 5, or the number 7 in the form 6n + 5.

So I need a way that would generalise any odd number.

And that's why we use 2n + 1 or 2n - 1.

Sam and Andeep want an expression for any positive odd number.

Sam says, "I am going to write 2n + 1." Andeep says, "I'm gonna write 2n - 1." What is wrong with both their generalisations? They have not defined what n can be.

There you go, they remembered now where n is an integer.

Right, but Sam only wants positive odd numbers.

What's the smallest n can be in this case if they only want positive odd numbers? Right, 0, 2 x 0 + 1 is 1, which is the smallest positive odd number.

So Sam can rewrite their statement as I want all numbers of the form 2n + 1 where n is an integer and n is greater than or equal to zero.

What does Andeep need to add to his generalisation so it only includes positive odd numbers? What do you reckon? Right, well, we can't have zero this time, 'cause 2 x 0 - 1 is -1 , that's not a positive odd integer.

So he needs a way of excluding zero, but having all values greater than zero.

So he could have gone with n is greater than or equal to one or n is greater than zero.

Because we've specified these are integers, then these two inequalities represent the same values, being greater than zero is the same as being greater than or equal to one if we're only referring to integers.

Sam says, "This time I want any two consecutive multiples of five.

How can I express that?" Well, the general form for any multiple of 5 is 5n.

How could we write the next multiple of five? So the next multiple of 5 would be 5n + 5.

So any two consecutive multiples of 5 could be written as 5n and 5n + 5 where n is an integer.

Andeep says, "What about two consecutive negative odd numbers?" This is tricky.

So if you work out how we can write this.

Right, you can have 2n - 1 and 2n - 3 where n is an integer less than or equal to zero.

That's gonna be the easiest way.

You could always try out some numbers and check that it works.

So if n was 0, you'd have -1 and -3, n has to be less than or equal to 0.

So n could be -1 and then you'd have -3 and -5 and so on.

So that seems to work.

All right, quick check, the expression 2n where n is greater than zero represents all positive even numbers.

What do you think? Well, then if you said false, which is the correct justification for your answer? Right, this doesn't include zero.

We've said n is greater than zero.

So that's not a problem.

The problem is this includes any decimal multiplied by two, which wouldn't be an even number.

We need to say that n is an integer and n is greater than zero.

Which of these shows the general form for multiples of 3 greater than 10? What do you reckon? Right, there's lots to think about here.

B or D would work.

For A, this doesn't include 12, 'cause n has to be greater than 4, it can't be 4.

And 3 times 4 is 12 and 12 will be fine.

We'd like to include 12.

For C, these will be 10 more than multiples of 3, not multiples of 3 themselves.

We need the number itself to be a multiple of three.

So that leaves B, which would work nicely, that includes 12 and then all multiples of 3 greater than 12.

Or D, and again, n can't be zero.

So the first integer would be one which would make our first number 12, which is our first multiple of 3 greater than 10.

When proving something algebraically, writing a generalisation is often the first step.

Let's think about how we could prove that the sum of any odd number and any even number is always an odd number.

Now it's a proof, we wanted to cover all relevant cases, so we need to generalise an odd number and an even number.

Sam says we can write the odd number as 2n + 1 and the even number as 2n, when n is an integer.

It's a good idea from Sam, but there is a problem.

Can you spot it? Right, this doesn't generalise any two numbers where one is odd and one is even.

If we use those two values, that's only showing it's true when they're consecutive numbers.

How can we generalise any odd number and any even number? What do you reckon? This is what I went for.

Assume there is an odd number 2n + 1 and an even number 2m.

So I've used different variables where n and m are both integers.

Doesn't matter if they're the same integer, that's not a problem, they just need to be integers.

And then 2n + 1 is odd and 2m is even.

Let's try this one.

The product of any two odd numbers is odd.

Why would we not use 2n + 1 and 2n + 1 where n is an integer as our odd numbers.

What do you think? Yeah, this would only cover cases where the odd numbers are the same number.

We want to show it's true for any two odd numbers.

What about 2n + 1 and 2n - 1? Why would that not be a good idea? Well, that would only show it was true for cases when the odd numbers are consecutive so they'll have a difference of two.

We wanna show it for any two odd numbers.

How are we gonna write this? Can you think of an idea? I went for this one.

Assume there are two odd numbers, 2n + 1 and 2m + 1, where n and m are both integers.

Notice that I'm writing this first step, it has a complete sentence.

So it's really clear for whoever's reading it to see exactly how I am setting up my proof.

Right, which of these show consecutive odd numbers? What do you reckon? B or D show consecutive odd numbers.

A is consecutive integers not consecutive odd integers.

C, they're two different odd integers, but they're not necessarily consecutive.

Right, time for you to put this into practise.

You're gonna have a go at writing the first step of some different proofs.

So for question one, you're gonna need to read the statement that you're trying to proof and tell me why that would be a bad generalisation.

And then for C, can you fill in the steps of this proof with a better generalisation? Take note of how I write those first steps, 'cause you're gonna be writing your own in a minute.

Question two is you've got a different proof this time, so make sure you read that carefully.

Why are the numbers I picked for A and then for B bad generalisations? And then can you write a good generalisation below? All right, time for you to write your own first steps.

Think about the structure of mine when you're writing your own.

I'd like you to write the first step of each proof by creating a generalisation for each case.

Give those a go.

Right, you've got three more to try here.

Again, think about how you're structuring those with a full sentence and make sure you're defining any variables.

Let's have a look then.

The problem with 2n and 2n is it only covers the cases where the two even numbers are the same.

We want any two even numbers.

For B, that only covers cases where one even number is double the other.

So again, we're adding an extra constraint, we want any two even numbers.

The best way then is to use different variables.

So assume there are two even numbers, 2m and 2n, where m and n are integers.

So for question two, we want the difference between two consecutive square numbers.

Well, n + 1 and n - 1 are not consecutive numbers.

So n + 1 squared and n - 1 squared are not consecutive square numbers.

Well, I dunno if you spotted that.

For B, n squared is a square number, but n squared plus one isn't a square number, it's one more than a square number.

Well done then if you went with assume there are two consecutive square numbers n squared and n + 1 squared where n is an integer.

For question three, check that you have full sentences.

Assume there are two odd numbers, 2m + 1 and 2n + 1, where m and n are integers.

It's okay if m and n are the same, 'cause we're just proving the sum of any two odd numbers is always even.

So for B, I've written this in two different ways.

So you could have your first step starting with take the square of an even number to n squared where n is an integer, or you might not want to go that far with your first step.

You might want to just start with assume there's an even number 2n where n is an integer and then do the squaring in step two.

That's fine as well.

For C, again, I've done it two different ways.

Firstly, you could say take the square of an odd number 2n + 1 all squared where n is an integer.

Or you could have said, assume there is an odd number 2n + 1 where n is an integer and then do the squaring part in the next step.

We do need to have an generalisation of an odd number as 2n + 1 or 2n - 1.

That's absolutely fine, but it would not work just to have n squared where n is an odd integer.

We need that property of odd integers to be part of our proof.

Then we've got assume there is an odd number 2n + 1 and an even number 2m where n and m are both integers.

For E, assume there are four consecutive odd numbers, 2n + 1, 2n + 3, 2n + 5 and 2n + 7 where n is an integer.

And finally assume there are two odd numbers, 2m + 1 and 2n + 1, where m and n are integers.

You might have also added they have squares, 2m + 1 squared and 2n + 1 squared.

Right, we've done the hard work really now setting up our proofs.

Now we're gonna have a go at constructing some simple proofs.

So algebraic proofs start with a generalisation.

We're really good at this now.

We need to make sure all variables are defined and then we can use our algebraic manipulation skills.

We're gonna try to prove that the sum of any odd number and any even number is always an odd number.

Now we know how to set this one up, we've done it already.

So assume there's an odd number 2n + 1 and an even number 2m where n and m are both integers.

What do you think our next step is going to be? Right, we're looking for the sum, so we're gonna add them together.

Now notice when I write proofs, I write them down the page.

I use full sentences when necessary, and I'm really clear with how one step moves onto the next.

So step two, the sum of these numbers can be written as 2n + 1 + 2m.

So now I'm gonna manipulate that expression, so I can write it as 2n + 2m + 1.

And then if I partially factorise, I've got 2 lots of n + m + 1.

Now the sum can be written as one more than an even number, so it must be an odd number.

And I need to write that so it's really clear why I'm saying this must be odd.

Therefore the sum of any odd number and any even number is always an odd number.

And that's it, we've proved it.

It might be something that you took for granted, but now we've shown why it must work.

The point of a proof is that someone should be able to read your proof and follow your working and be convinced by your arguments.

We don't want to skip any of our algebraic manipulation.

Even if you think it's really easy or really obvious, why one thing is equivalent to another thing, you still need to show it for someone else to follow.

And we always need to end with a conclusion.

In the example we had before, we said therefore the sum of any odd number and any even number is always an odd number.

It's a little bit like repeating the question we started with, we're just showing that we've proved what we wanted to prove.

Little bit of notation, the three dots arranged in a triangle shape is the therefore sign.

So we can use that instead of the word therefore if we want to be a little bit more concise, if we're short on space.

Let's have a go then.

I'm gonna do one and then you're gonna try one.

Prove that the product of any two odd numbers is odd.

So I'm gonna start with my generalisation.

Assume there are two odd numbers, 2n + 1 and 2m + 1, where n and m are both integers.

Their product is written as 2n + 1, 2m + 1.

Now I can manipulate that into the form 4mn + 2n + 2m + 1.

And I'm trying to show that this is odd.

So if I partially factorise, I can show that this is 2 lots of 2mn + n + m + 1.

I need to explain what I've done here.

So this is one more than a multiple of two, so odd.

And finally my conclusion, therefore the product of any two odd numbers is odd.

Right, so if you use a similar structure to show that an odd number squared is always one more than a multiple of four.

Give it a go.

Well done, hopefully yours looks really nice and neat like mine that somebody could follow it and be convinced by your argument.

So assume there's an odd number 2n + 1 where n is an integer.

You can of course use any variable you like.

Its square can be written as 2n + 1 all squared or 2n + 1, 2n + 1 one in brackets.

Now we can expand those.

We've got 4n squared + 2n + 2n + 1, which is 4n squared + 4n + 1.

You can write those in two steps or you can write those in one line.

I was a little bit short on space, which is why I wrote them on one line.

Now I wanna show this one more than a multiple of four.

So I need to try and take out factor of four, so I can write 4 lots of n squared plus n plus 1.

This is one more than a multiple of four.

And my conclusion, therefore any odd number squared is one more than a multiple of four.

Hopefully, yours followed the same structure as mine.

Yours might have been even better.

If you had more space, you could lay that out going down the page a little bit more.

Let's have a look at this one then.

Here is Sam's attempt to prove the conjecture, the sum of any two even numbers is even.

Read through their three steps.

What have they done well and what could they do to improve? What do you reckon? Right, there's lots of good things about this proof.

They've started with a generalisation, and they've correctly generalised two different even numbers, 2m and 2n, where m and n are integers, is a really good start.

And they have written a conclusion.

They've said that this must be a multiple of two and then they've referred back to the original proof.

Therefore, the sum of any two even numbers is even.

The problem a little bit is in the middle.

They have jumped to the conclusion from line two.

Line one and line two work nicely, but they haven't shown why 2m + 2n is even.

So it'd be better if they factorised and then shown that it has to be a multiple of two.

There you go, that looks better.

We've got that extra step showing that this is an integer multiplied by two and that's why it has to be even.

And it says that is really convincing proof.

Now Sam, I can follow your steps and I agree you must be correct, and that's what we're asking for from a good proof.

Thanks Andeep, it is great to now prove things that we used to take for granted, and that's what you're gonna do in the task in a moment, is see if you can prove some things that you might just have been taking for granted all this time.

So the first thing you've gotta prove is that the product of any two even numbers is always even.

I've started it off usually to fill in the gaps.

Then I'd like you to construct your own proof that the square of any even number is a multiple of four.

That might be something that you've never thought of before.

Can you prove it? Well done, so we've got the sum of two odd numbers is even.

I've started it off for you.

Can you finish that proof? And then can you write your own proof for the sum of any four consecutive odd numbers is a multiple of eight? Think about consecutive odd numbers.

So you could have one, three, five, and seven.

When you add those, you get a multiple of eight.

Now can you prove it's true for any four consecutive odd numbers? Try it out.

Question five, we've got the sum of the squares of two odd integers is even.

I've started it off for you, can you finish it? And then this is a nice result and it's a good thing to remember.

So can you prove that the difference between two consecutive square numbers is odd? Give that one a go.

Well done, let's have a look then.

So there are two even numbers 2m and 2n where m and n are integers.

Their product is 2m times 2n, which can be written as 4mn.

This can be written as 2 lots of 2mn.

So it is a multiple of two and therefore even.

The product of any two even numbers is always even.

Then you're having to go on your own.

So something similar to this.

Any even number can be written as 2n where n is an integer.

But you could have said assume there is an even number 2n where n is an integer.

It's square is then 2n all squared, which is equivalent to 4n squared.

This is four lots of an integer, so a multiple of four.

The square of any even number is a multiple of four.

These are simple proofs in that there's not too many steps to the algebraic manipulation, so it's important that you really explain what it is you're showing.

Let's have a look at three.

There are two odd numbers 2m + 1 and 2n + 1 where m and n are integers.

Their sum can be written as 2m + 1 + 2n + 1 or 2m + 2n + 2.

Then we can factorise, that's 2 lots of m + n + 1.

This is two lots of an integer, so it must be even.

Therefore the sum of two odd numbers is even.

Question four is probably my favourite one so far.

So we've got four consecutive odd numbers.

There are other ways that you could write this.

I think this is possibly the easiest, because it avoids using negative values.

2n + 1, 2n + 3, 2n + 5 and 2n + 7 where n is an integer.

Don't forget that, particularly if you're doing lots of proofs in a row, it's easy to miss out that important bit.

Their sum then is 2n + 1 + 2n + 3 + 2n + 5 + 2n + 7.

Of course, I'm gonna want to simplify that to 8n + 16.

That is the same as 8 lots of n + 2.

Therefore the sum of four consecutive odd numbers is a multiple of eight.

Now it can be really tempting to skip steps if you're feeling a little bit lazy and don't want to write out all those numbers added together.

However with proof, it's one part of mathematics where it is not appropriate to skip steps and try and do things more efficiently, 'cause we need somebody else to follow what we are doing.

Needs to be mathematically sound how we've got from one step to the next.

Question five, there are two odd numbers, 2m + 1 and 2n + 1, where m and n are integers.

The sum of their squares can be written as 2m + 1 squared + 2n + 1 squared.

Now we're gonna need some algebraic manipulation.

We've got 4m squared + 2m + 2m + 1 + 4n squared + 2n + 2n + 1.

Well done, if you put that step in.

I'm gonna simplify.

We've got 4m squared + 4m + 4n squared + 4n + 2.

Now I can factorise, I've got 2 lots of 2m squared + 2m + 2n squared + 2n + 1.

And my conclusion, this is two lots of an integer, so even.

The sum of the squares of two odd integers is even.

And then probably my favourite one of all is this result.

The difference between two consecutive square numbers is odd.

So assume there are two consecutive square numbers, n squared and n + 1 squared, where n is an integer.

We want their difference.

This can be written as n + 1 squared - n squared.

N + 1 squared is n squared + n + n + 1, and then subtract n squared.

If I simplify, I get 2n + 1.

And you've done it.

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

Therefore, the difference between two consecutive square numbers is odd.

It might have sounded like it was gonna be quite a complicated proof that, but we got to the generalisation of an odd number really quickly.

I hope you had as much fun as I did with that task.

There's lots of interesting results that we've now proved, and you can now use this basic structure to prove even more interesting mathematical results.

Well done for working so hard and for putting that effort into laying out your work really clearly down the page.

You don't have to write the numbers in the circles.

That's not necessary at all.

But we do need to see clear steps working down your page.

Thank you so much for joining me.

We've had a really, really good lesson today.

And I hope to see you again for some more proof lessons maybe in the future.