Lesson video

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

My name is Ms. Davies 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 logical arguments.

By the end of this lesson, you'll be able to use algebra and algebraic manipulation to construct logical arguments.

You might have been doing some work on proof already.

We're gonna look at a slightly different type of proof today.

It might be a little bit different to some of the things you've looked at already.

We're gonna talk in today's lesson about rational and irrational numbers.

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

So we're gonna start then by something called setting up a contradiction.

You're gonna see why this is useful as we progress through the lesson.

So sometimes, a statement can be proven true by showing that it cannot be false.

So this is a different way of proving things are true.

To do this, we need to be able to write opposing statements.

For example, say we want to prove that all frogs are amphibians.

What we could do is we can assume instead that this is false.

That means that there exists a frog which is not an amphibian.

And if we can show that that is logically unsound and therefore impossible, we can prove that all frogs have to be amphibians.

So what is the opposing statement to the claim that all rich people are happy? Izzy says, all rich people are sad.

Sofia says, some people who are not rich are also happy.

What do you think? If you're not sure about what an opposing statement needs to look like at the moment, have a discussion about the difference between what those two have said and how that relates to the statement that all rich people are happy.

So Izzy's statement is actually too strong.

In order to show that this statement is false, you only need one rich person to not be happy.

You might have also had a discussion as to whether being not happy and being sad are the same thing or is there a middle ground? Can you be not happy but also not be sad? Sofia's statement is incorrect.

It's an opposing statement to the idea that all happy people are rich, but that's different to all rich people are happy, and that's a useful thing to be aware of as well.

So the correct opposing statement would be, there is a rich person who is not happy.

And that would show that all rich people are happy is false.

Okay, let's see if you can get this one.

What would be the opposing statement to the claim that all birds can fly? What do you think? Well done if you said D.

The opposing statement to all birds can fly, is that there is a bird that cannot fly.

Try this one.

What is the opposing statement to the claim that all countries in the northern hemisphere are cold? What do you think? Well done if you said A, there is a country in the northern hemisphere which is not cold.

Again, some things to think about is the opposite of being cold, being warm, or is it possible for it to be not cold but also not be warm? It's easier to use the phrasing in the statement and use a negative and say not cold rather than worrying about what is and isn't an opposite.

Okay, how about this one? It always rains on at least one day in August.

What's the opposing statement to this? This one's a bit trickier.

Well done if you picked B.

If the statement is it always rains on at least one day in August, we need there to be in August in which it didn't rain on one day, which means it didn't rain at all.

Now that's different to being sunny.

It could not rain but also not be sunny, can't it? Equally, it's possible for a day to be both sunny and rainy.

So saying there's sunshine every day doesn't also mean that it never rained.

Right, let's try writing some opposing statements to mathematical conjectures now.

So for example, any integer squared is positive.

If this is false, what would have to be true? Right, there'll be an integer which when squared is not positive.

We have to be careful here as positive and negative are not exact opposites.

If something is not positive, it doesn't have to be negative, it could be zero.

Think about that when you are writing those opposing statements.

Use the actual wording in the question and then use a negation.

So say not positive rather than the word negative.

Jun says this is a bit like when we were finding counterexamples.

I guess it is thinking about what a counterexample would look like if it existed.

And this is quite a nice way of thinking about it.

So think about if there was a counterexample, what would it look like? So consider the statement, all multiples of four are even.

Jun says, if a number is not even, it is odd.

Jacob says, I don't think that's true.

There are numbers which are neither odd nor even.

All right, what do you think to those two ideas? Jacob is correct.

If we're considering all numbers, even and odd are not opposites.

Non-integer values are neither odd nor even.

2.

5 isn't an odd number, but it's not an even number either.

So being not even is not always the same as being odd.

However, if we were talking about integers, odd and even are opposites, and all multiples of four by definition are integers.

So if you said all multiples of four are even, it would be okay for an opposing statement to say there is a multiple of four that is odd.

Sometimes using algebra can make our statements clearer and help us see the first step in the proof.

So for example, the sum of two consecutive integers is odd.

If this is false, what would have to be true? Can you think of an opposing statement to this one? Right, there would be two consecutive integers, which when added together give an even result.

You could say give a result which is not odd.

But because we're talking about integers and adding two integers will give us an integer, we could say even instead of not odd.

Right, to make this easier, we could use algebra.

We could write the conjecture as if a and b are consecutive integers, then a plus b is odd.

What would our opposing statement look like now? Could we write it using algebra? We could say there are consecutive integers, a and b, such that a plus b is even.

The definitions of certain types of numbers will help us when writing opposing statements.

Let's look at this one.

Root two is an irrational number.

Well, what is an irrational number? Pause the video, can you remember? It was in our definitions to start with.

Right, an irrational number is one that cannot be written in the form a over b, where a and b are integers and b is not equal to zero.

Divided by zero is undefined.

So if root two was not an irrational number, it would be possible to write it as a fraction a over b, where a and b are integers and b is not equal to zero.

Now that's quite a tricky concept and that does lead into a very famous proof that we'll look at in the second part of the lesson.

But just getting that idea of what an opposing statement looks like at the moment.

So what would be a good opposing statement to the claim all numbers greater than 10 are positive? What do you think? Well, we need to use the language in the statement.

So there is a number greater than 10 which is not positive.

Remember, say not positive is better than negative 'cause something that's not positive could be negative or zero.

How about this one? If a-squared is odd then a has to be odd.

Where a is an integer, what would the opposing statement be? Right, so his is a little bit trickier, so well done if you got this.

So there is a value for a where a-squared is odd, but a is not odd.

We can say even here 'cause it says a is an integer.

So the one you're looking for is the last one.

There's a value for a where a is even and a-squared is odd.

What would be the opposing statement to the claim that if a is an even integer and b is an odd integer, a plus b is odd? So it's that second one.

There are integer values for a where a is even and b, where b is odd such that a plus b is even.

Again, you could also say not odd, because a and b are integers, those are opposites.

Okay, your turn.

I'd like you to fill in the gaps in the opposing statements to each conjecture.

Give it a go for A, B and C.

You got a little bit more freedom this time.

I'd like you to write an opposing statement to each of these.

Think about the wording in the question, use that to help you write your statement.

And for three, we've got the sum of a rational number and an irrational number is an irrational number.

So I'd like you to write an opposing statement.

For b, Jun decides to go even further and use the features of an irrational number, which is fantastic, but there's some mistakes.

So can you spot the mistakes that he has made? Give those two a go.

Well done.

This is quite an advanced mathematical skill, so well done if you got these.

We've got, there is an integer value for a, where a-cubed is odd, but a is not odd.

Even would be fine as well 'cause we're talking about a being an integer.

Right, if a and b are integers, there is a value for a and a value for b where 8a plus 12b does equal one.

For c, there is a largest positive even in integer, which can be written as a = 2n where n is an integer.

I've brought in some of our generalisation skills there as well to help me set up that proof.

There is a value for a, where a is even, but a-squared is not even or odd.

A has to be an integer because it is even.

Saying something is even implies it is an in integer.

Therefore a-squared is an integer.

So saying not even and saying odd are the same thing here.

For b, there is an integer value for a where a-squared minus 4a is odd, but a is not odd.

Again, even is fine because a has to be an integer.

For c, there are integer values for a and b where 10a plus 30b equals five.

For three, simplest opposing statement would be to say, there's a rational number a and an irrational number b, where a plus b equals c and c is a rational number.

There's other ways you could have written that.

I think that was the simplest.

This was really quite tricky.

One of the problems is it's used the same variables, m and n, for both numbers.

Well that's suggesting that those fractions have to be the same, and if you are adding a rational number to an irrational number, we're not saying that you get back to the number you started with, we're just saying you get to a different rational number.

He also needs to define m and n, 'cause we need to say where m and n are integers and n cannot be zero 'cause dividing by zero is undefined.

Apart from that, he set that up really nicely.

Right now, I'm gonna have a go at using this.

We're gonna have a go at proving with contradictions.

Now once we've written an opposing statement, we then need to prove this cannot be the case.

This can be a really useful tool in mathematics as there are rules and structures, which means that some things cannot be true.

Let's try this one.

The sum of two consecutive integers is odd.

So let's assume that there are consecutive integers, a and b, where a plus b is even.

So that's our opposing statement.

Just like other forms of algebraic proof that you might have done before.

We can generalise specific number properties.

So how could we write consecutive numbers? What do you think? We've got two options.

We could say where a is n and b is n plus one.

Or we could say where b is a plus one.

I've gone the other way.

I've said assume that there are consecutive integers, a and b, where a plus b is even.

If a equals n, then b equals n plus one.

Their sum can then be written as n plus n plus one.

Assuming a plus b is even, means that a plus b equals 2m, where m is an integer.

I'm just using another variable that I haven't used already.

Now we need to manipulate this until we get either a statement which is mathematically unsound and cannot be true, or a statement which is a contradiction to one of the assumptions we have stated.

So let's see what we can do now.

So that means n plus n plus one equals 2m.

We must be able to add a and b together and get something which is even.

So 2n plus one equals 2m.

We can see this is gonna be a problem.

If I rearrange that, I've got that 1 = 2m - 2n.

If I factorised, that's 2(m-n).

So this states that one is an even number.

Well, that is mathematically unsound.

So sum of two consecutive integers cannot be even.

Now, I need my conclusion.

Therefore the sum of two consecutive integers is odd.

So let's just look at the key features of that proof again.

So we've started with an opposing statement, then we've made sure we've generalised all the things that we've said were gonna be true.

So we need to make sure that a and b are consecutive integers, we need to make sure that a plus b is even.

So we've generalised both of those things.

Then we've manipulated this into a form that is mathematically wrong.

With our conclusion, we've said that this is mathematically unsound, so the sum of two consecutive integers cannot be even.

Therefore, we have then proven that the sum of two consecutive integers has to be odd.

Pause video if you want to read through that again.

Now there are other ways to prove this statement.

You might have been thinking about this.

You might have proven this one algebraically before.

There's the algebraic proof.

Jacob says, I think this is much easier than assuming the opposite and trying to reach a contradiction.

Jacob may well be right here, and you might be thinking the same.

The reason is there are some things that are really difficult to prove algebraically.

And actually there are much easier to prove by finding a contradiction.

We'll look at a famous example in a moment.

It's also just nice to be aware that there are different ways of going about things.

Right, let's try this one.

Prove that there are no integers a and b for which 10a+30b=5.

You are gonna do a similar one in a moment, so watch my steps.

So we're gonna start with an opposing statement.

Assume that there are integers, a and b, where 10a plus 30b is five.

Well, 10a plus 30b, they have a common factor so I can divide them both by 10 and I can do the same on the right hand side.

So that would mean that a + 3b = 1/2.

If b is an integer, then 3b is an integer.

Remember we started by saying a and b are integers.

If I add two integers I get an integer.

So a plus 3b must be an integer, therefore it cannot be a half.

That means there are no integers a and b for which 10a + 30b = 5.

Read through that one again if you need to.

And I'd like you to then prove that there are no integers, a and b, for which 8a+12b=1.

See how you go.

Right, so let's say that there are some integers, a and b, where 8a plus 12b is one.

Let's divide them all through by four.

So 2a plus 3b would equal a quarter.

If a is an integer, then 2a is an integer and if b is an integer, then 3b is an integer.

We do that for both this time.

We know an integer plus an integer is an integer, therefore the sum of two integers cannot be a quarter.

There are no integers a and b, for which 8a+12b=1.

Well done.

That is a really tricky skill.

Really proud of you for giving that a go.

Let's try another one.

Use the contradiction to prove that for any integer a, if a-squared is odd then a must be odd.

Well, let's start with an opposing statement.

Assume there is an integer value for a, where a squared is odd but a is even.

So that means a can be written as 2m, where m is an integer, because it's even.

And a-squared can be written as 2n + 1, where n is an integer.

So the a-squared is all squared.

So I can write a-squared as 2m all squared, and I can write it as 2n plus one because it's odd.

So I've got 2m or squared equals 2n plus one, which gives me 4m squared equals 2n plus one.

We can see there's gonna be a problem here.

The easiest way to show it is to do 4m squared minus 28 because I can show that that's even.

If I factorised that I get two lots of 2m squared minus n equals one.

And we can see here that that's gonna be a problem 'cause we're saying that one is a multiple of two, which is just not correct.

We'll come on to our conclusion in a minute.

Try getting to a similar point with your one.

Right, assume there's an integer value for a where a-squared is even and a is odd.

A would be 2m plus one where m is an integer but a-squared would be 2n.

So a all squared would be 2m plus one all squared, and that would have to be equal to 2n for some values of m and n.

While you square that binomial and simplifying gives me 4m-squared plus 4m plus one equals 2n.

We can rearrange that.

One equals 2n minus 4m-squared minus 4m.

So 1=2(n-2m squared-2m) and that's a very similar point to what I got on the left hand side.

Let's look at our conclusions then.

So this states that one is an even number which is mathematically unsound.

So it is not possible for a-squared to be odd and a to be even.

Therefore for any integer a, if a-squared is odd, a must be odd.

See if you can write a conclusion for your one.

This states that one is an even number which is mathematically unsound.

So it's not possible for a-squared to be even and a odd.

Therefore for any integer a, if a-squared is even a must be even.

Well done.

We're now gonna have a look at one of the most common things to prove using this method and that's the fact that the number root two is irrational.

This is not a proof you need to memorise at the moment.

I'm showing you this because showing you why this method of proof is really useful in some cases.

If you find this a little bit tricky to follow, do not panic.

We are pushing that higher level math skills here.

So if root two was not a irrational number, it would be possible to write it in the form a over b where a and b are integers and b is not equal to zero.

So that's where we're gonna start.

Assume that root two is rational.

So we can write root two as a over b where a and b are integers and b is not equal to zero.

Now this is the really important step that's gonna be useful later on.

Assume any common factors have been cancelled.

So a over b is in its simplest form, therefore the highest common factor of a and b is one.

If they're not in their simplest form, you could divide them by a value and get it into a simplest form.

So assuming that's been done and a over b is in its simplest form and now we do some manipulation.

So I've decided to square both sides and that would give me two equals a-squared over b-squared.

I'm doing that because I don't want a square root to be dealing with.

So if I square both sides, I've got an integer now.

Two equals a-squared over b-squared.

What could we do next? Pause the video.

What ideas have you got? So what I'm gonna do is I'm gonna multiply both sides by b-squared.

It gets rid of the fraction, makes things easier to deal with.

So I've got two b-squared equals a-squared, because a and b are integers a-squared must be even.

It's a mathematical fact if a-squared is even a is even, and that's something we just proved previously.

So if a is even then a equals 2n for some integer n, so that means that 2b-squared would be able to be written as 2n all squared.

So 2b-squared will be equal to 4n-squared.

So b-squared would be able to be written as 2n-squared.

What does that tell you then about b-squared and therefore b? Right, they're both even.

B-squared is even, therefore b is even.

Well, hang on a minute, let's look at what we've just said.

We've just said that a has to be even, but if a is even then b is also even.

But that's a problem 'cause they now have a common factor of two and we started off by saying that a and b have been simplified so they only have a common factor of one.

And this is a contradiction to one of our original assumptions.

So the assumption that root two could be rational leads to this contradiction.

And even if you then simplified a and b and did this all over again, you'd get exactly the same contradiction.

So that means that root two has to be irrational.

It's not possible for it to be rational, therefore it has to be irrational.

And that is one of the most common ways of using this idea of proving, starting with an opposing statement and trying to reach a contradiction.

Do not panic if you couldn't follow absolutely every step there or you weren't sure why we chose to do certain things.

This is completely new, which means you can build up to it slowly.

Time for you to have a go then.

I'd like you to fill in the gaps for this proof that for any integer a, if a-squared minus 4a is odd, then a is odd.

And I'm proving this by starting with an opposing statement.

All you do is fill in the gaps and then write a conclusion.

Off you go.

Well done.

Question two, I've started off a proof that for any integer a, if a-cubed is odd, then a is odd.

I want you to finish it off.

You can use as many steps as you need to make this really clear.

Don't forget your conclusion.

Right, Jun is trying to prove that for any integer a, if a-squared is a multiple of three, then a must be a multiple of three.

I'd like you to read through his proof and then have a think about these questions.

He has written integers, which are not multiples of three as 3n plus one.

There's a slight problem with this.

His proof is really good as a start, but there is a problem with him writing things that are not multiples of three as 3n plus one.

Can you spot it? Then for b, can you build on June's work to show that a contradiction is reached for all values of a, which are not multiples of three? He's actually made a good start and that can be the first part of the proof, but you're gonna need to do something else to prove what he wants you to prove.

And then do you think there would've been an easier way to prove this? And your final challenge, if you choose to accept it, is to prove that root three is irrational.

You're gonna do it by assuming it's rational and showing this leads to a contradiction.

Now this is very similar to the proof for root two.

So you could go back in the video or back in your notes and have a look at that.

And it's gonna use the fact that for any integer a, if a-squared is a multiple of three, then a must be a multiple of three, which is what Jun was trying to show in the previous question, so that is gonna be useful to you so remember that.

Give this one a go and we'll see how far you got when you're done.

Fantastic.

It's absolutely fine if you found some of that a bit tricky.

That's what happens when you look at something that's completely new to you.

It takes time and it takes practise for you to wrap your head around what you are doing.

So well done for persevering with that.

So I'd like you to pause the video and check that you filled in the gaps correctly.

Your conclusion might say something like, therefore we reach a contradiction, if a-squared minus 4a is odd and a is even.

So a-squared minus 4a is odd, a must be odd.

Read that through and then we'll look at the next one.

For question two, I started it for you and now I need some algebraic manipulation.

2m all cubed is 2m times 2m times 2m or 8m-cubed, where I'm remembering your indices skills.

Then we've got 8m-cubed minus 2n is one.

We can factor out a two to show this as a multiple of two.

And again we're stating that one is an even number which is mathematically unsound.

A conclusion then, if a-cubed is odd and a is even, we get a contradiction.

So for any integer a, if a-cubed is odd, then a has to be odd.

The problem with what Jun has written is integers which are not multiples of three could also have the form 3n plus two.

So we know that 3n is a multiple of three, 3n plus one isn't, 3n plus two isn't, but 3n plus three would then be a multiple of three again.

So he has got a contradiction for when he uses numbers of the form 3n plus one, we also need to show there's a problem if we're using numbers of the form 3n plus two.

You might have also said he's not included a final statement, so his proof is incomplete.

So if we build on Jun's work, then we need to do exactly the same, but for numbers of the form 3n plus two.

So he can take his first seven steps where he's set everything up nicely and say.

Okay, let's assume a-squared is a multiple of three so a-squared is 3m and a is 3n plus two where m and n are integers.

You might have chosen to use different integers so they're not the same as the ones you've used before.

That's absolutely fine as well.

So that means 3n plus two all squared would have to be equal 3m.

So you've got 9n-squared plus 12n plus four equals 3m.

Well done if you've got that bracket squared correct.

If we subtract all the multiples of three, so we've got four equals 3m minus 9n-squared minus 12n and take out a factor of three, that shows us the four has to be equal three lots of m minus 3n-squared minus 4n.

Well, this says that four is a multiple of three, which is mathematically unsound.

Therefore, if a-squared is a multiple of three, then a cannot be an integer of the form 3n plus one or 3n plus two.

So for any integer a, if a-squared is a multiple of three, then a must be a multiple of three.

That's a long proof.

Well done if you followed that through to the end.

Even if you didn't get it right yourself, being able to follow it through when I have done it and correct your own work is a really important skill.

I wonder whether you thought about whether there was an easier way to prove this.

It'd be straightforward to prove it the other way around.

It would be easier to prove that if a is a multiple three, then a-squared is a multiple of three 'cause you just take a equals 3n square and factorise to show it's also a multiple of three.

But that's the wrong way around.

We're trying to show it the other way around that if a-squared is a multiple of three, then a has to be a multiple of three and that's the different thing.

Now, if we want to do that, just like with Jun before, we'd have to categorise all integers as having the form either 3n, 3n plus one or 3n plus two.

And showing that 3n-squared is a multiple of three, but 3n plus one squared and 3n plus two squared are not.

Well that's kind of what we did with the contradiction before.

So it's a personal preference as to whether you prefer to start with a contradiction and work backwards, or whether you want to start with the forms of the integers and work up to show why 3n squared is a multiple of three, but 3n plus one squared isn't and 3n plus two squared isn't.

As I say, there's real high level skills there, so well done for stretching your brain and having a think about that.

And then we've got root three is irrational.

Assume that root three is rational.

So root three can be written as a over b, where a and b are integers and b is not equal to zero.

Assume any common factors have been cancelled, so a over b is in its simplest form.

We're gonna square both sides, so that gives us three equals a-squared over b-squared and multiply both sides by b-squared, so 3b-squared equals a-squared.

And because a and b are integers, a-squared must be a multiple of three, so a is a multiple of three, and we've just proven that with Jun.

So if a is a multiple of three, then a can be written as 3n for some integer n.

So 3b-squared can be written as 3n all squared or 9n-squared.

But that means that b-squared can be written as 3n-squared if we divide both sides by three.

And because b and n are integers, b-squared must be a multiple of three now so b is a multiple of three.

But if a and b are both multiples of three, they have a common factor of three.

And that's a contradiction to the fact that a over b was in its simplest form.

This is a contradiction to one of our original assumptions, so the assumption that root three could be rational leads to a contradiction, therefore root three is irrational.

If you follow that all the way through to the end, you should be incredibly proud of yourself.

That is a really tricky proof.

Thank you for joining us today.

We've looked at something really interesting, but maybe you found a little bit tricky.

That's absolutely fine.

Really important to remember that there are different ways to prove things, some you might find easier than others.

Thank you for joining me today.

I really look forward to seeing you again.