Lesson video

Well done for making the 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 to make sure you've got everything you need and let's get started.

So, this lesson we are proving or disproving a statement.

By the end of the lesson you'll be able to appreciate what constitutes a proof of a statement and what is required to disprove.

A conjecture then is a mathematical statement that is thought to be true but not been proven yet.

We're gonna talk about generalising.

So to generalise is to formulate a statement or rule that applies correctly to all relevant cases.

So let's get started with proving a statement.

Sophia, Jun, and Alex are trying to prove that the sum of two odd numbers is always even.

Sophia does this.

Jun does that.

And Alex does that.

Take some time to have a look through what they've done.

Whose method do you find most convincing and why? And remember they're trying to prove that the sum of two odd numbers is always even.

What do you reckon? All right, well let's think about a proof, then we'll come back to your ideas.

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

If there's a limited number of cases or types of cases, it can be possible to prove them by testing them all.

I'll give you an example.

If I wanted prove that all positive integers less than 10 have fewer than 5 factors.

Well there's only 9 positive integers less than 10, so I could test all 9, and then I'd know that's definitely true.

Most of the time there'll be too many cases or an infinite number of cases, so testing cannot prove a conjecture true.

What we need to do instead is to generalise each case, algebraically or pictorially and then construct a mathematical argument.

What we can do is we can use rules and structures that we know are true in mathematics to build up these arguments for other facts.

So let's look at Sophia.

Now, Sophia has shown three examples that work.

3 add 5.

1 add 3.

7 add 7.

However, this does not prove it works for any two odd numbers.

The pictorial representation is good 'cause hopefully it's helping her see a bit of a structure as to why that works.

She's not generalised each place, she's just shown three examples.

This is a very good demonstration.

You might be convinced with what she's done.

It's not quite a proof.

Now Jun's shown lots of examples that do work and that can be quite convincing.

If someone shows you loads and loads of examples you might be thinking, "Yeah, it must be true, then.

It works for all these numbers." However, it is not enough to show it's true for all cases.

It's enough maybe to make a conjecture, but we've not shown it's true for all cases.

What he needs to do is find a way of generalising two different odd numbers so that he can show it's true for any two odd numbers.

Now Alex's should be the most convincing.

It might not have been what you said, but see if I can convince you.

It looks a bit like a single example.

So it looks a bit like one of Sophia's examples.

But what he's done that Sophia didn't is he is generalised odd numbers.

You'll see that there's no number associated with the pink rod.

It's just a pink rod.

So the rods have no set value.

But what he is been able to say is that an odd number can be written as double some value add 1.

So if you look at that top diagram with the two pink rods and the one white rod, that's representing an odd number.

And an odd number can be written as two lots of a number plus 1.

Then the green is second odd number.

Again two lots of some number add 1, and it doesn't have a value.

It could be the same size as the pink one.

Then when he adds the two representations, he's shown that the extra 1 match up to give a total which is a multiple of 2.

So it's similar to what Sophia did, but there's no set value given to his rods.

Therefore he's generalised for any odd number.

Now his argument could be improved by explaining what the rods represent.

He might need some algebra, he might need a conclusion to go with this, but he is got a really good start.

Do you think there is a better way to prove this than Alex's way? You might wanna pause a video, just have a discussion with a partner.

What do you think is a good way to prove that the sum of two odd numbers is always even? A formal proof can be algebraic or it can use pictorial representations, or graphs, or logical arguments.

These are all examples you may see if you continue developing your proof skills.

Now a concept or conjecture can be demonstrated convincingly, you could be really convinced that this is true, but it not be a proof.

The important thing is that all defined cases are accounted for and any assumptions are mathematically sound.

So for example, what we just looked at, Alex used the fact that all odd numbers can be represented as one more than an even number.

Now it's okay to assume that as a fact.

We're gonna need to assume some things are facts in order to build proofs.

Let's have a look at Pythagoras.

Sophia says, "My teacher gave us lots of right angle triangles with squares drawn.

We measured the squares and recorded our results." Jacob said, "We did something similar except we just did one example with square toy bricks, and we showed that the bricks in the smaller squares could be moved to make the bigger square." Now I want you to think about how you discovered Pythagoras.

Can you remember how you did it? Think about how Sophia's teacher and Jacob's teacher showed them.

What do you think to those methods? So, the methods are very similar.

Sophia used more examples.

So you might say that she's better placed to make a conjecture.

Jacob just did one example.

Sophia did loads.

However, testing some examples is not gonna constitute a proof, not unless we account for all possible cases, and that's not gonna be possible here.

Now Jacob's method is more practical and you might have liked the idea that he can move the toy bricks and that makes it feel like a better proof.

However, it's still just one example.

It only worked for the specific sized triangle that Jacob's teacher gave him and the number of bricks that Jacob's teacher gave him.

It's not showing it's true for all cases, it is not a generalisation.

Right, Jun said, "My teacher drew this shape with a smaller square inside a larger square.

They said the area of the whole shape can be worked out two ways." Right, if we look at the larger square, we could do A plus B, because that's one side length, multiplied by A plus B, which is the other side length.

Or A plus B all squared.

The other way we could do it is we can calculate the area of the four triangles and add the area of the smaller square.

So that would be A times B over 2 for each triangle.

So four lots of A times B over 2.

Add C squared.

All right, well let's manipulate those.

So A plus B all squared can be written as A squared plus 2 AB plus B squared is the product of two binomials.

And the other way of writing it, we have four lots of AB over 2 plus C squared.

If we rearrange that, we can write that as 2 AB plus C squared.

Now those are two different ways of representing the same area, so they must be the same thing.

So you can put them equal to each other.

Now we can solve them like an equation.

So subtract 2 AB from both sides and we've got A squared plus B squared equals C squared.

That is Pythagoras' theorem.

What do you think to that method? How did it compare to Sophia's and Jacob's methods from earlier? See if you can explain it to your partner or explain it to me.

All right, Alex said, "My teacher drew this diagram and showed that if you cut the smaller squares along the lines, the pieces can be arranged to fit into the larger square." So you'll see we have a right angle triangle, we've got the squares of the two shorter sides, and I've drawn lines in a certain place 'cause something is gonna happen.

Let's have a look.

We put A there and rearrange the pieces we get the larger square.

We've shown that the two smaller squares add to the larger square, which is Pythagoras's theorem.

What do you think to that method? How does it compare to the other three? Let's have a look at these three different methods.

Now, having examples can help you understand how the theorem works.

So Sophia and Jacob's methods of showing some examples can be a really good way for you to learn how something works.

And it can be quite convincing if it works for lots of different numbers.

However, it doesn't prove the theorem unless all possible cases are accounted for.

Now the third example is a nice demonstration.

You might have liked it, but it is not a proof.

Again, we haven't shown it's true for all triangles, just the one that I have drawn.

Now, that second method that Jun did is a proof.

We've used variables to represent the sides.

This would work with any right angle triangles.

And we've got this generalisation, so we could definitely turn that idea into a proof with a little bit more structure and a conclusion.

Alex wants to prove that angles in a parallelogram sum to 360 degrees.

Alex has drawn three parallelograms and measured the angles.

They always sum to 360 degrees.

Has Alex proved this? Explain your answer.

No, of course not.

He's only demonstrated it works for three cases, not all possible parallelograms. All right, Jun's playing around with some shapes as well.

He wants to prove the angles in any isosceles triangle sum to 180 degrees.

Jun says, "I've drawn a square, then cut it in half along a diagonal, I get an isosceles triangle with angles 90, 45, and 45 and these sum to 180.

All right, what pieces of mathematical information did Jun use to set up his proof? Right, you might have said the fact that squares have four right angles.

Or the fact that the diagonal of a square bisects the vertices.

This wouldn't work for a rectangle, for example.

That wouldn't give us an isosceles triangle.

So it's that fact that that diagonal bisects the vertices the side led to the squares are the same, which gives us that isosceles triangle, and the square has four right angles.

Well done if you said any of those ideas.

All right, has he proven then that angles in any isosceles triangle sum to 180? You might wanna discuss this with a partner, or you might want to just write yourself a sentence, or tell me what do you think.

All right, he's only shown it for right-angled isosceles triangles.

Not other types of is triangles.

The other thing to be aware of is it does rely on us knowing all those properties of a square first.

Time for a practise.

Here is one way to demonstrate that the interior angles of a triangle sum to 180 degrees.

You could do this as you read through it, or you could just read through the explanation.

Once you're happy with that demonstration, can you explain how this demonstrates the interior angles in a triangle sum to 180? You need to add some explanation.

And then what mathematical rule do you need to already know in order for this to work? When you're happy with your answers, we'll move on to the next one.

All right, here's another demonstration for the interior angles of a triangle.

I'd like you to tell me why the green angles, which are the ones labelled B on the left-hand side, why do they have to be the same size? The same for the blue angles, those ones labelled A on the right-hand side.

Why do they have to be the same size? What mathematical rule do we need to use? And then how does that show that interior angles in a triangle sum to 180 degrees? Have a go at those, and then we'll look at the next bit.

Okay, let's put those methods together then.

So here are three methods for demonstrating the interior angles of a triangle sum to 180.

You've got Sophia using that ripping the corners off method.

You've got Alex using that diagram that we've just looked at.

And Jun says, "I drew 10 different triangles and measured all the angles." I would like you to explain in words which method could become a proof and why.

See if you can try and mention all three methods.

What are the positives or the negatives of some of these different methods? What do you think is going to be the best proof? Give this one a go.

Well done.

That was lots of thinking about triangles.

We're now gonna have a look at perpendicular lines.

Jacob is trying to remember the rule about the gradients of perpendicular lines.

He has tried three examples.

I'd like you to write a conjecture for him.

And then tell me why what he's drawn so far is not a proof.

Give it a go.

But this time Jacob has drawn the gradient triangles for 2/3 and -3/2.

And he's shown that if you rotate one of those 90 degrees you get the other.

So he says if the gradients have a product of -1, they are perpendicular.

So is this argument more convincing than what he did on the previous example? Explain your answer.

Right, so Jacob copied down the proof from his revision notes.

So this is the proof that if two lines are perpendicular, then the gradients have a product of -1.

I would like you to give one positive of this proof, and then why is it not a complete proof? And then thirdly, I'd like you to fill in the gaps in this conclusion for Jacob's proof.

Come back when you're ready for the answers.

All right, this is a nice way to demonstrate interior angles of a triangle sum to 180.

I wonder if it's something you've done before.

So what you can do is you can rip the angles off, rearrange them, and show that they form a straight line.

We need to already know though that adjacent angles on a straight line sum to 180 degrees.

For this method, the black lines marked are parallel.

Therefore the green angles, the ones labelled B, are alternate angles in parallel lines.

So that means they must be the same.

The two angles labelled A are also alternate angles in parallel lines, so they must be the same.

So let's look at why this shows the angles in a triangle sum to 180 degrees.

Well, A, B, and C lie on a straight line at the top, so that means they sum to 180.

If we use those parallel lines rules we just talked about we know that the base of the triangle must also be A and B.

Well that means A and B and C are the interior angles in the triangle, and we already know that A plus B plus C is 180 because they lie on a straight line.

So the mathematical rule that we have to already know to use this is that alternative angles in parallel lines are the same, and that angles on a straight line, if adjacent, sum to 180.

So let's talk about these methods then.

So, Jun's method is nice and easy.

It's a good way of convincing you that this might work, but it is not a proof.

It does not show its true for all triangles.

Sophia's is a convincing demonstration as well, because she could have drawn any triangle she liked and shown that this works.

However, this is still not a proof.

Not showing it's true for all possible cases.

Alex's method does prove this is true, or will do if he added a bit more explanation.

As long as we already know the rules for angles in parallel lines, we can use this to show the interior angles of a triangle add to 180.

So your conjecture may be something like, "The product of the gradients of two perpendicular lines is -1." It's not a proof, we've not shown this works for every possible case.

Now you might have been more convinced by this.

That's absolutely fine.

He's using a representation, and he is demonstrating this underlying structure with perpendicular lines and how they relate to each other.

However, this is not a proof.

It's a nice demonstration, but not proving it works for all pairs of perpendicular lines.

Let's have a look at Jacob's proof then.

So some of the positives.

It states a clear start point.

We're assuming that two lines L1 and L2 are perpendicular.

In line two we have a generalisation that can apply to any two linear graphs.

It also uses logical steps most of the time.

And we have shown that M1 multiplied by M2 is -1.

Now, it's not a complete proof because it skips steps.

For example, Q and R are used, but we're not told what Q and R are standing for.

There's also quite a lot of algebraic manipulation skipped in the last three lines.

It's hard to see how to get from one line to the next.

And it's missing a conclusion.

So let's see if we can fill in Jacob's conclusion.

So something like, "If two perpendicular lines have gradients M1 and M2, then the product of their gradients is -1." And if Jacob had done this, included all the steps, and a conclusion, it would be a good proof.

It does rely on Pythagoras's theorem already being proved.

If you want to see the complete proof, here it is now.

You'll notice there's a diagram on the right hand-side as well.

Diagrams can form parts of proofs, they can be really useful tool.

Right, now for the easier bit we're gonna have a look at counterexamples.

Izzy and Laura are investigating prime numbers.

"3 subtract 2 is 1.

Laura, what are the next pair of consecutive prime numbers with an odd difference." Laura's tried some, they're consecutive prime numbers and she's looking at their difference.

She's got 2, 2, 4, 2.

"I've tried several and I cannot find any," says Laura.

What do you think? Can you help Laura out here? Pause the video.

What do you reckon? Right, I was a little vague with my instructions because I was hoping you're gonna notice this.

There are no other consecutive primes with an odd difference.

Why not? Can you provide a convincing argument? Okay, well you might have said something like, "The difference between two odd numbers is always even.

Because 2 is the only even prime number, then using 2 is gonna be the only way of getting a difference between prime numbers that's odd.

So the only two consecutive primes with an odd difference, are 2 and 3.

All other consecutive primes will both be odd, therefore their differences will be even." Proving a statement can be quite difficult.

We looked at some quite complicated examples in that first part of the lesson, and how much detail you have to go in to actually prove something, not just show it.

Knowing certain number properties can be helpful when constructing proofs.

But also they can be helpful when disproving a statement.

Disproving a statement can be a lot easier, 'cause all we need is one example that does not work.

And we call this a counterexample.

It can be useful to work systematically to find counterexamples.

But also it can be useful to know certain number properties, like the fact that 2 is the only even prime number.

All numbers of the form 3 to the power N plus 2, where N is a positive integer will be prime.

Can you find a counterexample for that statement? Give it a go.

There are many examples that do work.

There are some there.

The quickest way, though, to find a counterexample will be to start at N equals 1 and work systematically.

It's not always the case, but it will be here.

So let's do 3 to the 1, add 2.

That is prime.

3 to the 2, add 2.

That's prime.

3 to the 3, add 2.

That's prime.

3 to the 4, add 2.

That's prime.

3 to the 5, add 2.

That's a multiple of 5, not prime.

So it took us a few goes, but working systematically brought us to a counterexample.

Let's try another one.

If M is an integer and N is an integer, M squared plus N squared will be a multiple of 5.

Can you find a counterexample to this statement? Give it a go.

Again, we can work systematically.

So, 1 squared and 2 squared.

1 squared and 3 squared.

1 squared and 4 squared doesn't work.

That's not a multiple of 5.

We have found a counterexample.

You may have found a different counterexample.

We now need to write a concluding statement.

"The conjecture, 'If M is an integer and N is an integer and M squared plus N squared will be a multiple of 5,' does not hold when M is 1 and N is 4.

Therefore this statement cannot be true." Laura says, "I think it would've been quicker to think of single digit numbers which do not add to a multiples of 5 and work from there.

6 plus 5 is 11, so any square numbers ending in 6 and 5 will be a counterexample." So Laura's trying to think a little bit about the properties of number to help her.

There's an example.

4 squared plus 5 squared equals 41.

4 squared ends in a 6, 5 squared ends in a 5.

Something to be aware of.

In this case it did not take us long to find a counterexample working systematically.

It was a really good choice.

Laura's may be quicker, but no square numbers end in 2, 3, 7 or 8.

So by trying to work out some features, she may have spent time looking for examples that don't exist.

Laura is correct, though, that sometimes using number facts can be quicker.

If you can't see an easy way to find a counterexample, working systematically is a good way to start.

Jun says, "I conjecture that for any three-digit multiple of 11, the first and last digits sum to the second digit." We're gonna look at this one together.

The first 9 three-digit multiples of 11 work.

I'll give you some examples.

10 times 11 is 110, and 1 add 0 is 1.

11 times 11 is 121, and 1 add 1 is 2.

12 times 11 is 132, and 1 add 2 is 3.

Working systematically you'd find a counterexample on the 10th try.

19 times 11 gives you 209, and 2 add 9 is not 0.

Now this worked for many multiples of 11 because 11 lots of a number is 10 lots, add 1 lots.

So it's sort of a structure about the multiples of 11.

I'll show you 13 times 11, 14 times 11, and 18 times 11.

You can see why the outside digits are gonna add to that middle digit.

This stopped working when the tens columns summed to the value of 10 or more.

So if you had 19 and 190, 1 lot and 10 lots, 9 add 1 is then 10, it takes you over a multiple of 10 and this pattern stops working.

The point I'm trying to make is that identifying this structure could make finding a counterexample easier.

It's important to be aware of which cases are included in a conjecture.

Izzy says, "I conjecture that 10 N will be greater than N for any value of N." Jun's trying some.

"If N is 1 10 N is 10, so 10 N is bigger." He tries N is 5, and then tries lots more.

He says, "I think multiplying a number by 10 always makes it bigger." What do you think? Right, if you only try values of N greater than 0, you will not find a counterexample.

But Izzy said any value of N.

So this includes negative numbers.

If N was -1, 10 N would be -10.

So 10 N is smaller.

So just think about negative numbers, decimals, things like that when you're looking for a counterexample Jacob says, "I conjecture all multiples of 10 have an even number of factors." Test the first two multiples of 10.

Does this conjecture hold? So, 10 has 4 factors, and 20 has 6 factors.

It does hold the first two multiples of 10.

I'd like you to find a counterexample and write a conclusion.

This does not hold for the 10th multiple of 10 because 100 has 9 factors, therefore this statement is false.

What knowledge helps us find this counterexample quickly? What do you think? Right, it's the fact that the square numbers have an odd number of factors.

So all you need to do is pick a square number that's a multiple of 10 and then you'll have your counterexample.

So knowing that fact can speed things up.

Time for you to have a go.

For each conjecture, I'd like you to find a counterexample and write a conclusion.

Off you go.

For question two, you've got a conjecture.

Laura tries two values.

Why might these values not be a sensible starting point? And then see if you can find a counterexample.

Don't forget to write your conclusion.

So three, Izzy playing around with this idea.

Take any injured greater than 12.

Reverse its digits.

So if I went with 36, I'd have 36.

And if I reverse the digits, I get 63.

Find the difference between those numbers.

So 63, subtract 36, 27.

And reverse those digits.

So 27 and 72.

Add those two numbers together and you get 99.

Izzy says, "I conjecture you'll always get the answer 99." Right, you might wanna read that through again.

See if you can try a few numbers first and then can you find a counterexample? Why am I starting at 13 and trying every integer not be the most efficient way? Play around with that idea and then we'll look at the last bit.

Right, I'd like you to find a counterexample to Alex's conjecture and write a concluding statement.

Then could you rewrite his conjecture so that it is true? Off you go.

And now you need to do the same for Sophia's conjecture.

Read through it and find a counterexample.

Then see if you can change it to make it true.

Let's have a look, then.

There are many counterexamples that could be used.

I'm just gonna give you an example.

The conjecture that any odd number and any prime number is an even number doesn't hold if you use the prime number 2.

So for example, 9 plus 2 is 11, that's a counterexample.

Therefore this statement is false.

For B, the ones digit of any multiple of 3 is a multiple of 3 itself.

Well that's not gonna work for the fourth multiple of 3 which is 12 because, 12 is a multiple 3, but the ones digit, 2, is not a multiple of 3.

Therefore this statement is false.

With kites this conjecture does not hold.

To be a kite two of the opposite angles have to be the same, and less than 180.

And all four have to add to 360.

So there's a possible kite with angles of 110, and 110, 80, and 60.

Now only two of those are obtuse, so they don't have to have three obtuse angles.

Drawing sketches of these may have helped you with your explanation.

Starting with N is 17 and N is 41 was not the best idea 'cause we had to square our values for N.

It would've been a lot easier to pick smaller values of N which were easier to square.

Equally, after all that hard work, those two values did work.

It'll be easier to find a counterexample if we work systematically.

Start with the first prime number, which is 2, then 3, then 5, and 7, and go until we find a number that's not prime.

You get a counterexample when N is 7.

7's a prime number, but 7 squared plus 7 plus 1 is 57, and that's not prime.

A counterexample has been found when N is 7.

Therefore this conjecture does not hold for all prime numbers.

Right, I hope you had fun playing around with this one.

This works a lot of the time.

It doesn't work with any two digit number where the digits are the same.

So, 22.

Also doesn't work for three digit numbers.

So, why am I starting at 13 and trying every integer not be the most efficient way? Well, because there's a structure to the ones that don't work.

So it would be easier to spot that pattern rather than going through every example.

Alex's conjecture.

Right, this does not hold when X is 1, for example,.

1 plus 3 is 4, 1 times 3 is 3, and 4 is greater than 3, not less than 3.

In fact, you'll find any counterexample when X is less than or equal to 1.

5.

So Alex could change conjecture to say that for any value of X greater than 1.

5, X plus 3 is less than 3X.

He's just getting rid of those counterexamples.

Let's look at Sophia's.

This conjecture does not hold when X is 0.

5, or 1/2, for example.

There's 0.

5 squared is 0.

25, and that's less than 0.

5.

So squaring the number has actually made it smaller.

In fact, this is gonna be the case for values between 0 and 1, and 0 squared itself is equal to 0.

1 squared is equal to 1, so not greater than.

So any value from 0 to 1 inclusive will be a counterexample.

So she could say that any value of X where X is greater than 1 or less than 0, X squared is greater than X.

So she's got rid of that section that doesn't work.

Or I conjecture that for any value of X except where X is between 0 and 1 inclusive, X squared is greater than X.

Right, there was loads of ideas today.

I'm hoping that you had fun playing around with those demonstrations of Pythagoras.

Might have reminded you of things that you've done before.

We had a look, as well, at the properties of perpendicular lines.

Does that remind you of how you learned that the first time? And then we've had a go at playing around with some number patterns and spotting counterexamples.

There was lots of hard work today and I know some of those tasks required you to do a lot of thinking.

So, well done.

You've been absolutely fantastic.

Thank you for joining me, and I really look forward to seeing you again.