Lesson video

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

My name is Miss 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 looking at general algebraic forms for specific number properties.

This is the first part of Proof, and by the end of today's lesson, you'll be able to fluently write general algebraic forms for different number properties.

We're gonna look at multiples and integers today.

So if you need a reminder of what those are, pause the video now.

So we're gonna start by looking at divisibility tests.

Aisha, Izzy and Jacob each have a pile of chocolates.

Aisha can tip all of her chocolates into one box.

Izzy splits hers equally between two boxes, and Jacob splits his equally between three boxes.

What I want to know is who must have an even number of chocolates and who could have an even number of chocolates? Pause the video and have a think.

Well done, if you said that Izzy must have an even number of chocolates.

She can split hers equally between two boxes and only even numbers are divisible by two.

It is possible for Aisha to have an even number.

They're all in one box.

Her chocolates must be divisible by one, but all integers are divisible by one.

We know Jacob must have a multiple of three, but that could be odd or even.

Multiples of three can be odd or even.

Izzy says, "I could write the number of items I have as 2b." If she does this, what would b represent in her expression? Right, b would be the number of items. So in this case, chocolates, in one box.

Again, we are assuming that all chocolates are left whole, so b has to be a positive integer in this case.

So if b is a positive integer, Izzy has 2b and that must be a multiple of two.

Jacob says, "I could then write the number of items I have as 3b where b is a positive integer." Now can you spot a potential problem with Jacob expressing his chocolates this way? Well done, if you spotted this.

Izzy has already used the variable b to represent the number of items in one of her boxes.

So if Jacob also uses b, that might suggest that the number of items in one of his boxes is the same as the number of items in one of Izzy's, and that doesn't have to be the case for this scenario.

So in order for Jacob to make it clear that the number of items in one of his boxes could be different to the number of items in one of Izzy's boxes, he needs to use a different letter.

So if a, b, and c are positive integers, a and 3c could be even, but 2b has to be even.

Okay, let's change the scenario a little bit.

This time they each have a pile of counters, they keep one counter and then Aisha tipped all of hers into one box.

Izzy is able to split hers equally between two boxes, and Jacob splits the rest of his equally between three boxes.

Who had an even number of counters to start with this time? Don't forget, they each kept one counter before putting the rest in the boxes.

Right, so Aisha could have an even number or an odd number.

Jacob could have an even number.

So if he had three in each box for example, or then one leftover, that would be 10, which would be an even number or he could have an odd number.

If he had two in each box for example, that would be six in the boxes and one leftover.

So seven in total.

Now it's impossible for Izzy to have an even number this time.

Her boxes must contain an even number, like they did with the chocolates, but this time she has one extra.

So her total number of counters must be one more than an even number.

One more than an even number is an odd number.

We could write their expressions this time as, a + 1, 2b + 1, and 3c + 1.

Right, we're gonna formalise that now.

So if n is an integer, then 2n is always an even integer.

Negative numbers can count as even integers.

That's absolutely fine, -2 would be even.

We could call this the general form for an even number, 'cause any even number could be expressed this way and any number expressed this way has to be even when n is an integer.

There are other expressions though that could be even.

If n is an in integer, which of these are going to be even? What do you think? Well done, if you said, 2n + 2 and 2n + 4.

Because 2n is even, 2n + 1 is odd.

So 2n + 2 is even.

And then odds and even numbers alternate when you add one.

We could also demonstrate this by factorising.

2n + 2 can be written as two lots of n + 1.

n + 1 is an integer and then multiplying by 2 will make it an even integer.

We can do the same for 2n + 4 to show that that is even.

So the general form for an odd integer is 2n + 1 or 2n - 1, where n is an integer.

However, just like with even numbers, there are other expressions that could be odd.

Izzy says, "3b + 2 is an odd integer 'cause if b = 3, then 3b + 2 is 11, which is odd." Do you agree? Right, well, 3b + 2 could be odd.

She's shown that it's odd when b = 3.

But it also could be even, if b = 2 and 3b would be 6, add 2 would be 8.

So it does not represent an integer, which must be odd.

Izzy says, "I think 6b + 1 is always odd." Do you agree this time? Yeah, this time she's correct, 6b is even, because it's two lots of 3b, so it's got a factor of 2.

An even integer add 1 has to be an odd integer.

So Izzy is correct this time.

Time for you to have a go.

If n is an integer, which of these must be even? Give this a go.

Well done, if you said A and D, We can factorise A to show that it is a multiple of 2.

B could be odd or even.

It's definitely a multiple of five, but multiples of five, alternate between odd and even.

C actually has to be odd, because if it's a multiple of 6 it's even, and this is one more than a multiple of 6, so odd.

And D has to be even.

10 lots of a number is always even, and then if we subtract 2, it will still be even.

If we expand it and refactorized, we could show that's 2 lots of 5n + 19.

So definitely even.

Okay, which of these could represent 2 different even numbers, m and n are both integers.

What do you think? For A, 2n and 2n have to be the same even number, because n will represent the same number.

2n and 2m definitely could be different even numbers, but they could be the same if n equaled m.

So there is that option to remember.

2n and 4n will definitely be different, except for if n is zero.

The other thing to realise with this is the second one does have to be double the first one there is that extra element to those expressions.

Again, these could be different even numbers and most of the time they will be, they will actually be the same number if n = m + 1.

So just be aware of that as well.

So we can use similar methods to see if an expression is a multiple of other numbers.

So which of these must be multiples of 5? What do you think? Yeah, of course we need to factorise out a 5 to see if they are gonna be multiples of 5.

And we can do that for the first one and the last one.

The second one actually is 3 more than a multiple of 5.

For this third one, it could be a multiple of 5 for some values of n, but not necessarily.

And this one is 2 less than a multiple of 5.

So it cannot be a multiple of 5.

Time for you to have a practise.

I'd like you to sort the expressions into the table.

For all of these, n is an integer, so which are always odd, which are always even, and which could be either, depending on the value of n? Give this a go and then come back for the next bit.

So for question two, we're gonna assume n is a positive integer.

And then I want to know which of these expressions must be a multiple of 6, which could be a multiple of 6 depending on n, and which cannot be a multiple of 6, whatever value of n you use? When you've sorted those expressions, see if you can add your own expression to each part of the table.

Try to be a bit different to the ones you've seen before.

Give this a go and then come back for the last question.

And finally, Aisha says, "I'm going to express two different odd numbers as 2a + 1 and 2b + 1." What extra information does Aisha need to add to make sure these are definitely two different odd numbers? For B, Jacob says, "I'm gonna write two multiples of 3 as 3c and 3c + 3." Will his expressions show two different multiples of three? And what else is gonna be true about Jacob's numbers? Give those three a go.

Let's have a look then.

Pause the video and check that you've got the right ones in the right places.

Just want to draw your attention to a couple of different ones.

So H, you've got two lots of n + 3 and then we're adding 2.

That's still gonna be even.

So if you add 2 to an even number, you get the next even number.

The one below it, K, is interesting as well.

Even though multiples of 3 can be odd or even, 4n + 2 is even.

And if you multiply an even number by an odd number, you're gonna get an even answer.

So just be aware of that for K.

Let's look at question 2.

So again, pause the video and check you've got the right ones in the right places.

Again, a couple to draw your attention to.

If we look at F in that first column, the expression in the bracket is a multiple of 3.

That means the whole expression is a multiple of 2 and a multiple of 3, therefore it must be a multiple of 6.

You could factorised out a further factor of 3 and have 6 lots of n - 3.

Let's also look at G, in that final column.

Two lots of n + 3 + 1 is an odd number, and of course multiples of 6 are always even.

If we look at K, 5(2n - 1).

Where, 5 is odd and 2n - 1 has to be odd, and an odd multiplied by an odd number is another odd number.

So again, that cannot be a multiple of 6, because it cannot be even.

So a few interesting results there.

Of course for M, N and O, you could have gone with all sorts of expressions.

I've gone with 6 lots of something, to show that it's clearly always a multiple of 6.

For n, I've chosen another multiple.

So I've gone with 4n.

And some things that are divisible by 4 are also divisible by 6, but not all of them.

And then the easiest ones cannot be a multiple of 6 is 6n + 1.

Because it's one more than a multiple of 6.

And for 4, Aisha needs to add that a and b are integers.

Otherwise those don't have to be odd.

And she also needs to say that a is not equal to b, otherwise those would be the same odd number and she wanted them to be different.

So for B, yes, they will always be two different multiples of 3.

There's no values of c, where they'll be the same, and they're both gonna be multiples of 3.

We can factorise 3c + 3 into three lots of c + 1 one to show this.

The other thing that's gonna be true about Jacob's number is the second number is always 3 greater than the first or it's gonna be the next multiple of 3.

So if he's fine with that, that's good, but if he wants two different multiples of 3, where they're not necessarily consecutive and we'll look at that, what that means in a moment, then you'd need to write them in a different way.

Let's have a look now then at general algebraic forms. So if n is an integer, we can write the general form of a consecutive integer as n + 1.

That would be the next integer.

Or n - 1, that would be the previous integer.

Izzy says, "I want to write 3 consecutive integers.

Can I write them as n - 1, n + 1 and n + 2?" What do you think? No, these are not gonna be consecutive.

Because if you take 1 away from a number and you add 1 to a number, those two numbers are now gonna have a difference of 2.

So they're not consecutive integers.

Yeah, she would need to have n, n + 1 and n + 2.

Or n - 1, n, n + 1.

For consecutive integers, we need to make sure they have a difference of 1.

However, there's lots of different ways we can write consecutive integers, but sometimes we want a way of writing every possible case.

So every set of 3 consecutive integers can be expressed below.

That's what we said previously.

What do you think to 5n? 5n + 1 and 5n + 2? Yes, they're consecutive integers.

However, be aware that if you write them like this, you're also adding an extra number property.

You're dictating that the smallest number is a multiple of 5.

So if you wanted 5, 6 and 7, or 10, 11, 12, this would express that.

However, if you wanted 1, 2 and 3, you could not write them this way because 1 is not a multiple of 5.

So just be aware of that extra element.

Okay, if the general form for an even integer is 2n, what are the next two even integers? 2n + 2 and 2n + 4.

Remember every other integer is even.

So if you want consecutive, even integers, you need to have a difference of 2.

So the easiest way to write consecutive even integers would be 2n, 2n + 2, 2n + 4, 2n + 6, 2n + 8 and so on.

If we wanted to, we could show that they were even by removing a factor of 2.

So I factorised those expressions.

What do you notice? You might have noticed that the values in the brackets are consecutive integers.

We've got n, n + 1, n + 2 and n + 3.

And that makes sense.

If you multiply consecutive integers by 2, you're gonna get consecutive, even integers.

So Aisha, Izzy, and Jacob are trying to write expressions for 3 consecutive odd integers this time.

We're gonna assume a is an integer.

Read those expressions.

What is the problem with each of their suggestions? Can you spot it? So for Aisha, a does not have to be an odd integer.

We need it to be an odd integer.

Also, if a was odd, then a + 1 would be even.

So they wouldn't be consecutive odd integers.

Izzy's only works when a is 1.

If a was 2, we'd have 2, 6, 10, and they wouldn't be consecutive odd integers.

Jacob is possibly the closest.

2a + 1 is the general form for an odd integer.

So he's got a good start.

But then 2a + 1 would be even, if you add 1 to an odd integer, you get an even integer.

So let's build on what Jacob's got.

What would be the next odd integer after 2a + 1? So there's a difference of 2 between odd integers, so you'd get 2a + 3.

So you could write consecutive odd integers as 2a + 1, 2a + 3, 2a + 5.

What would the previous odd integer be before 2a + 1? Yeah, it'd be 2a - 1.

So which of these show consecutive multiples of 3? And again, a is an integer, Right, we've got C and D.

So for A, they're increasing by 3s, but a doesn't have to be a multiple of 3.

If a was 1, we'd have 1, 4, 7, 11, and they wouldn't be multiples of 3.

B, the first one, 3a is a multiple of 3, but 3a + 1 isn't.

So we need to have 3a, 3a + 3, 3a + 6.

Or if we factorised, that'd be 3a.

3 lots of a + 1 and 3 lots of a + two.

Okay, which of these show consecutive multiples of 4? What do you think? It is B.

If you've got caught out with a, remember that a - 1 and a + 1 are not consecutive integers.

It should be 4 lots of a - 1.

4a, 4 lots of a + 1 and so on.

You can expand the brackets to show this as well.

For B, they are consecutive multiples of 4.

Important to note that we're starting on 4 lots of an odd number.

So that's that extra specification we've got there.

For C, that only works if a is 1.

If a is 2, we'd have 8, 16, 24 and they're not consecutive multiples of 4.

And for D, they have a difference of 4, but 11 less than a multiple of 4 is not a multiple of 4.

So 4a - 11, is not a multiple of 4.

So there are other number properties that can be useful to express.

Aisha says, "I could write a square number as n squared.

What would the next square number be?" Well, then if you said n + 1 squared, if we add 1 to the integer and then square it, we'll get in the next square number.

If you wanted to, you could expand the brackets and write it as n squared + 2n + 1.

I agree with Aisha, though.

That first form is gonna be easier to use.

All right, what would the next square number be? Well done if you said n + 2 squared.

Again, we could expand that.

The n + 2 squared is gonna be an easier form to use most of the time.

So which of these must be a square number? What do you reckon? Well done, if you said B and D.

So A is the sum of 2 square numbers, but that doesn't have to be a square number.

So for example, 4 + 9 = 13 and that's not a square number.

It does work for some cases, 9 + 16 = 25, and that is a square number.

But that's not a good way to write a square number.

B is definitely square, because it's the sum of two integers then squared.

C actually can't be a square number, because n squared is a square number and adding 2 to a square number won't give us another square number.

And D must be a square number, 'cause it's an integer squared.

Right, which of these must be a square number? What do you think? So for the top one, we could factorise that as n + 3 or squared.

The second one is (n + 2)(n + 6).

That doesn't have to be a square number.

So C, n - 2 or squared is the other square number.

Izzy wants to write the general form for a positive 2-digit number.

She says, "I can write any 2-digit number as ab, where a is the digit in the tens column and B is the digit in the ones column." Let's have a look at why this doesn't work.

ab means a x b.

So if we wanted the number 32 with a = 3 and b = 2 then ab would actually mean 3 x 2 or 6 not 32.

Okay, Izzy's had another thought.

"How about a + b, where a is the digit in the tens column and b is the digit in the ones column?" How could we show Izzy that this will also not work? What do you reckon? Right, if we use the same example, so 32 and a = 3 and b = 2, and 3 + 2 = 5, not 32, which is what we're looking for.

Jacob says, "I think we need to use what you said about the place value columns, Izzy.

The important thing is 3 needs to be in the tens column, so it's actually 30." If we think about 32, we can write that as 3 lots of 10 + 2.

So to write the general form for a two digit number based on its digits, we can write that as 10a + b, where a is the digit in the tens column and b is the digit in the ones column.

So again, if we think about 32 when a = 3 but 10a is that's 30 + b, so 32.

Time for you to have a go.

If n represents any integer, how could you show each of these? What do you reckon? And for question two.

In each question, I'm writing consecutive numbers with a certain property.

Which expression is incorrect in each one? Give this one a go.

For 3, Izzy wants to write an expression for any 3 consecutive even integers.

What is the problem with her choice? See if you can write your answer in a full sentence.

Jacob wants to write an expression for any 3 consecutive odd integers.

See if you can think about the problem with his choice.

When you're happy with your explanations, come back.

For question 5, I'd like you to write a general form for each of these.

You need to make sure you define your variables.

So if you are using n, you need to tell me if n has to be an integer, maybe n has to be positive, whatever it is to make that the correct general form.

For 6, how could we write a positive 3 digit number with hundreds digit a, tens digit b and one's digit c? So similar to what Izzy was trying to work out with a 2 digit number, but I'd like you to do it with a 3 digit number.

When you're happy with your answers, come back and we'll have a look together.

Let's have a look then.

If n is an integer, the next integer would be n + 1.

A different integer, consecutive to n has to be n - 1.

Any even integer, 2n.

Other forms are possible, but if you want to gather any possible even integer, then 2n is the best.

A different even integer.

You could have come out with all sorts here.

You might have gone with 2n + 2.

Because that'd be the next one.

You might have gone with 4n or 8n.

You might have gone with 2m and used a different variable.

If we want it to be different, we need to make sure we say that m and n are different.

An odd integer, the easiest way is 2n + 1 or 2n - 1.

A square number, again, the easiest is n squared.

A different square number.

You could have gone with m + 1 squared, that's the next square number.

Or again, if you went with m squared, you need to make sure you say that m is an integer and m and n are not the same.

For question 2, 2 lots of n + 3 is incorrect.

If you write those all in the same forms, if you expand all of those, you'll have 2n, 2n + 1, 2n + 2, 2n + 3.

So the next one should be 2n + 4.

Or 2 bracket n + 2.

For B, again, if you expand them, it'll be easier to see.

The second one is incorrect.

It's actually the same as the first.

So we've got 4n + 4, and then we should have 4n + 6, 4n + 8, 4n + 10.

For C, it consecutive odd integers, 2n is the problem.

That one's even, isn't it not odd.

The next odd number after 2n - 1 is 2n + 1, which I've actually written as 2n + 1 - 1.

So actually the first and the third are consecutive odd integers.

We don't need one in the middle.

And for D, it's that second one that's incorrect.

That should be n - 1 or squared or n squared - 2n + 1.

If you factorise that first one, you'd see that it was n - 2 or squared.

So the next one's n - 1 or squared.

Then n squared, then n + 1 or squared.

So with Izzy, the problem is they're all even numbers.

However, they're only consecutive when n is 1.

So if n was 5 for example, we'd have 10, 20, 30 and they're not consecutive even integers.

For Jacob, they are all consecutive odd numbers.

The problem is this only covers cases starting on a number, one more than a multiple of 10.

So we wanted 11, 13, 15, or 21, 23, 25, that would be fine, but this would not work if you wanted 3 consecutive odd numbers that didn't start one more than a multiple of 10.

And for question 5, there's other answers that are possible.

I've gone with 5n, 5n + 5 and 5n + 10, where n is an integer.

For B, I've gone with n squared and m squared.

How you define your integers here may depend on whether you consider zero as a square number.

If you want to include zero, then n and m just have to be integers.

If you don't want to include zero as a square number, then you could have n and m as positive integers, and that will avoid that problem.

They need to be different square numbers, so m is not equal to n.

For C, I've got 2n where n is an integer and n is less than zero.

So n could be -1 and then 2 x -1 = -2, which is a negative even integer.

And for 6, we could write this as 100a.

So 100 x the hundreds digit plus 10b, which is 10 times the tens digit plus C.

And then a, b, and c, have to be positive integers less than 10 and a has to be bigger than zero.

Otherwise, this wouldn't be a 3 digit number.

If a was zero, b could be zero, and c could be zero, but a couldn't be, if we want a 3 digit number.

If you didn't add that extra bit to the end, that's fine for the moment, but just have a think about that.

You might wanna add that into your notes.

Well done today.

We made a really good start towards Proof.

and these skills, if you go on to do further Proof, will be really, really useful for you.

Have a read through of what we've learned today, and then I really hope that you join us again, especially come back for some exciting lessons where we go on to prove some results using these.

I look forward to seeing you then.