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 multiple approaches to logical arguments.

We're gonna look at different ways that you can prove things.

This might be different to some of the proof things you might have looked at already.

By the end of the lesson, you're gonna be really good at constructing logical arguments.

We're gonna have a look at some geometric proofs today.

So I'm gonna use the term apex.

So the apex is the point or the vertex which is the greatest perpendicular distance from the base.

And I'm gonna talk about it in respect to triangle.

So where you have the base of a triangle, the vertex, which is the greatest perpendicular from the base, is called the apex.

We're gonna also talk about congruent shapes.

So if one shape can fit exactly on top of another using rotation, reflection, or translation, then the shapes are congruent.

And I'm also gonna refer to the hypotenuse, 'cause we're gonna have a look at some Pythagoras.

So the hypotenuse is the side of the right-angle triangle, which is opposite the right tangle.

Let's have a look then at some geometric demonstrations.

So sometimes we can demonstrate or prove a result geometrically using known properties of shapes.

For example, we're gonna prove that alternate angles in parallel lines are equal.

The first thing we're gonna wanna do is draw a diagram.

So I need a pair of parallel lines, and in order to draw alternate angles, I need a transversal as well.

So I've drawn that in.

And then it's always useful to label everything so we can refer to them later.

So I'm using L1 and L2 for our two parallel lines.

And then I've labelled the transversal as T.

Right, a and b are alternate angles.

So they are what we're trying to show are equal.

So again, I'm gonna label them a and b.

And we're gonna use the known fact that corresponding angles are equal to help us.

So I'm gonna draw in a pair of corresponding angles, and again, I'm gonna label them so I can refer to the correct angle.

Anything we write on our diagram can form part of the proof.

So it's absolutely okay to label things on the diagram, use notation on the diagram as part of what you're proving.

We can start by stating any assumptions that will help us set up our proof.

So my first statement's gonna be, assume L1 and L2 are parallel, and T is a straight line intersecting both.

So explaining how I'm setting this up, it goes with my diagram.

Angle c and angle d are corresponding angles, so are equal.

And now I can think about how I'm gonna show that a and b are equal.

So 180 minus angle c equals angle a, as adjacent angles on a straight line sum to 180 degrees.

And I'm explaining what mathematical facts I'm using.

Equally, 180 subtract angle d, equals angle b, as adjacent angles on a straight line sum to 180 degrees.

Because angle c and angle d are equal, 180 minus angle c is equal to 180 minus angle d.

Because that's how we defined a and b, that means that angle a must equal angle b, therefore, alternate angles in parallel lines are equal.

Have a read through that.

Are you convinced by this demonstration? Do you think it's a proof? Right.

Well, we use variables for the angles.

So what we've done is we've shown this works for angles of any size.

I can move that transversal and change the size of the angles and the maths is exactly the same.

We started by stating all assumptions we were gonna use.

We did a really good job at stating those.

And each step followed logically from the step before.

You don't need to have lots of math skills to be able to follow what we wrote.

We used words to help us explain things as well as the mathematics.

And we finished with a conclusion.

I think that was a pretty good proof.

What you might have said is we had to use the fact that adjacent angles on a straight line sum to 180 degrees, so you have to already be happy that that's true.

We also use the fact that corresponding angles in parallel lines were equal.

So you have to be happy that that is already true.

You might have been a little unsure as to whether you wanted to use that fact.

But if we're happy assuming those facts, then that was a complete proof.

I'm gonna try a go.

Given that the interior angles of a triangle sum to 180 degrees, show that the interior angles in a hexagon sum to 720.

We're gonna do this this way.

So take a point C within the hexagon.

A hexagon has six sides.

So six triangles can be drawn with an apex at C.

And I've demonstrated that, so C is the apex of every triangle.

The base of all the triangles are the sides of that hexagon.

Given that the interior angles in a triangle summed to 180 degrees, the sum of the interior angles in all six triangles is 180 times six, which is 1,080.

Now, the interior angles of a hexagon does not include the angles around the point C.

So we've got the interior angles in all six triangles.

We don't want to include those angles around C 'cause they are not interior angles of the hexagon.

Angles around a point sum to 360 degrees.

Therefore, the interior angles of the hexagon is 1,080 subtract 360, which is 720.

Therefore, interior angles in a hexagon sum to 720.

Pause the video if you want to go over any of that again.

Right, your turn now.

I'd like you to use the fact that interior angles in the triangle sum to 180, to show that interior angles in an octagon sum to 1,080.

Think about what we just did with the hexagon.

Off you go.

See if you can finish off my proof.

So we've got eight sides of an octagon, so eight triangles are drawn with an apex at C, and the base of each triangle are the sides of the octagon.

Given that the interior angles in a triangle sum to 180 degrees, we're stating the fact that we're using, the sum of the interior angles in all eight triangles is 180 times eight, which is 1,440.

And I've written my calculation to show where the 1,440 has come from.

The interior angles of the octagon do not include the angles around the point C.

As angles around a point sum to 360, it's important that we state that rule that we're using, the interior angles of the octagon sum to 1,440 subtract 360, just 1,080.

Therefore, the interior angles in an octagon sum to 1,080.

You might have thought that was really wordy, and that actually it's a really obvious thing to prove, however, the point of proof is that it's really clear to follow from one step to the next.

It's important that we've shown our calculations, and we've explained which rules we're relying on for our proof.

Okay, you may have seen some geometric demonstrations before.

This is a proof for Pythagoras' theorem.

Sam says, "Now we've learned how to write proofs, I think we can improve this." I agree with you Sam.

What could Sam and Izzy do now to improve this proof? What do you reckon? Right, well, we need to explain the generalisation and explain the diagram.

Where is it coming from? Why are we using it? What are our a, b, and c representing.

We need to explain how the algebraic expressions were formed and any assumptions.

Where's the a squared plus 2ab plus b squared come from? We need to write each step of the algebraic manipulation, we don't wanna skip steps, and we need a conclusion to explain what we've shown.

So let's give it a go.

Take four congruent right-angled triangles with side lengths a, b, and c, where c is the hypotenuse.

Arrange the triangles so the hypotenuse form as square as shown in the diagram.

The area of the larger square can be calculated by multiplying the length by the height, so you've got a plus b multiplied by a plus b.

Alternatively, the area can be calculated by adding the area of the smaller square to the area of the four triangles.

So you've got four lots of ab over 2 plus c squared.

What could we do now? Pause the video.

What would your next step be? Right, well, these expressions represent the same area, so they must be equal.

So you can put them equal to each other.

So there's the two expressions equal to each other.

We can expand all the brackets, and then subtract 2ab from both sides.

And then we've got Pythagoras theorem, for any right-angled triangle with side lengths, a, b, and c, where c is the hypotenuse, a squared plus b squared equals c squared.

Pause the video if you want to read through all those steps.

We can use our proof skills to prove theorems involving circles, which you can then use to solve problems. We are gonna prove that the angle in a semicircle is a right angle.

So this is a circle theorem that you may have come across before.

The great thing about proving a circle theorem is that you know you can always use it.

So setting up the diagram is gonna be really important here.

So we need a circle.

Let's label the centre as O.

And the angle in the semicircle is a right angle.

So we need a semicircle with an angle.

So essentially we've got a triangle drawn where the base is the diameter and the apex is the point on the circumference.

So we can refer to all the relevant angles and sides.

We can label the vertices of the triangle.

We're gonna label them A, B, and C.

If you label the angles instead, sometimes things can get a little bit confusing if you then divide the angles into smaller angles.

So just by labelling the vertices, it's easier to refer to the right angles.

Okay, so we're gonna take a triangle ABC with vertices on the circumference of a circle, where the base AC is the diameter.

The radius OB can be drawn to split the triangle into two isosceles triangles.

I'm gonna do that on my diagram so we can see what we're referring to, and that can form part of my proof.

As OB and OA are both radii, the triangle OAB is isosceles.

And the angle OAB equals the angle OBA.

You can see now why I labelled the vertices so that I can explain what angles I'm referring to.

Now I'm gonna mark that on my diagram, that's gonna be really helpful to me.

As OB and OC are both radii, the angle OBC equals the angle OCB, and I'm gonna label that on my diagram to make that easier.

We have labelled these as x and y so we can now refer to these variables in our proof.

It's easier to do the algebraic manipulation now.

What could we do next? Pause the video.

Can you think about how we could then manipulate this? Because interior angles in a triangle sum to 180 degrees, the angle AOB can be written as 180 minus 2x, but BOC can be written as 180 minus 2y.

I haven't really got space to write that on my diagram.

If you have, you might wanna put that on your diagram.

Because AOB and BOC are adjacent angles on a straight line, they're the two angles at O, that means they sum to 180.

So I can write 180 minus 2x, plus 180 minus 2y, equals 180.

Then I can rearrange, that gives me 360 minus 2x minus 2y equals 180, or 180 equals 2x plus 2y.

That means that 90 equals x plus y.

Well the reason why I did that is 'cause we're trying to show that the angle in a semicircle is a right angle, so 90 degrees.

Therefore, for any triangle with vertices on the circumference of a circle and base as the diameter, the angle at the apex must be 90 degrees.

That's the long way of saying the angle in a semicircle is a right angle, and that's the proof for that circle theorem.

So, true or false? Annotations on a diagram can form part of a proof.

What do you think and what's your justification? Yeah, absolutely.

Diagrams, labels, and notation can help the reader follow stages of the proof.

It's absolutely fine to annotate your diagram and refer to that in your proof.

Right, time to give this a go.

I'd like you to arrange the steps of this proof into the correct order to show that the sum of the interior angles in any polygon is 180 times n minus 2, where n is the number of sides.

So this is similar to the one that we just did with the hexagon and the octagon, but now I'm showing it for any polygon with n number of sides.

Can you put those in the right order? Okay, using the diagram or otherwise, I'd like you to prove that the interior angles in a triangle sum to 180 degrees.

Give this one a go.

Okay, this diagram shows a rectangle ABCD with lines connecting to adjacent vertices to a point O.

I'd like you to show that the angle x plus the angle y equals the angle z.

Give this one a go.

Now I'd like you to complete this proof for the circle theorem, that the angle at the centre of the circle, AOC, is twice the angle at any point on the circumference, ABC.

This is similar to the proof that we just did, that the angle in the semicircle is 90, so you might be able to use some of the same ideas.

Give that one a go.

And finally, I'd like you to read through this proof for, "The tangents to a circle from external point are equal in length," and tell me how it can be improved.

As many ideas as you can, what would make this a better proof? We need to start by taking any polygon with n sides, and then we need a point C inside the polygon.

So it's important those two steps are that way around.

The point C can be joined to each vertex to form n triangles with the size of the polygon as the bases and C as the apex of each.

Given the interior angles of a triangle sum to 180 degrees, the sum of the angles for n triangles can be written as 180n.

So that's really similar to what we did with the hexagon and the octagon.

But the interior angles of the polygon do not include the angles around the point C, so these must be subtracted.

As angles around a point sum to 360, the interior angles of the polygon sum to 180n minus 360.

Then we can factorise, that gives us 180 lots of n minus 2.

And our conclusion, therefore, the sum of the interior angles for a polygon with n sides is 180 multiplied by n minus 2.

Well done if you've got those in the correct order.

Now we know we can use that fact when we're looking at angles in a polygon in the future.

Question 2, labelling the lines and the angles is gonna help us here.

You could have used different variables, you just need to compare what you've written to what I've done.

So lines L1 and L2 are parallel.

So angle a and angle x are equal, as they are alternate angles in parallel lines.

Equally, angle c and angle y are equal, as they are alternate angles in parallel lines.

Adjacent angles on a straight line sum to 180 degrees, so angle a plus angle b plus angle c is 180.

'Cause angle x and angle y are equal to angle a and angle c respectively.

angle x plus angle b plus angle y equals 180 degrees, therefore, interior angles in a triangle sum to 180.

Well done if you got that one.

So for 3, I've labelled the other two angles in the triangle as angle p and angle q.

ABCD is a rectangle, so angle DAB and angle CDA are 90 degrees.

Therefore, p equals 90 subtract x, and q equals 90 subtract y.

Because interior angles in a triangle sum to 180, p plus q plus z, equals 180.

So 90 minus x plus 90 minus y plus z, must equal 180.

So 180 minus x minus y plus z, equals 180.

Bit of rearranging and we have z equals x plus y.

Therefore, if a rectangle ABCD has adjacent vertices A and D connected to a point inside the rectangle O, then angle OAB plus angle ODC, equals angle AOD, or from our diagram angle x plus angle y equals angle z.

Well done for that when you constructed that proof by yourself with very little structure.

So question 4, we want to split our shape into two isosceles triangles with a radius.

As OB and OA are both radii, the triangle OAB is isosceles, and OAB equals OBA.

As OB and OC are radii, angle OBC equals angle OCB.

Because interior angles in a triangle sums 180, I can write AOB is 180 minus 2x, and BOC is 180 minus 2y.

Because angles around a point sum to 360, AOC equals 360 minus 180 minus 2x, minus 180 minus 2y.

Be careful with your rearranging here.

So we've got 360 minus 180, subtract negative 2x, so that's plus 2x, subtract 180, subtract negative 2y, which is plus 2y.

So just check your rearrangement there.

We get AOC equals 2x plus 2y.

Well, that means AOC is two lots of x plus y, and that's the angle at the circumference.

So for any circle, the angle at the centre of the circle, AOC, is twice the angle at any point on the circumference, ABC.

Well done for proving that circle theorem.

Right.

Couple of things that are wrong with this one.

We didn't start with the generalisation of the diagram.

We didn't explain what diagram was showing us.

So to improve it, we could say, "The diagram shows a circle was centre O.

The tangent AB intersects the circle at point A.

The tangent BC intersects the circle at point C." So for point 3, we need to explain why those two triangles are congruent.

It might help to actually draw in the line OB to show the two triangles we're talking about on the diagram.

And then because both triangles share the same hypotenuse, they both have another side which is the same length, OA and OC are the same because they're radii, and both triangles have a right angle because of the rule that a tangent meets the radius at a right angle, that means that those two triangles are congruent.

Now there's a law for that.

There's a right angle hypotenuse side law for congruency, which says if you know two triangles have a right angle, the hypotenuse of both triangles are the same, and one of the other sides of both triangles are the same, that means the triangles are congruent.

I made a conclusion, something like, "Therefore, for any circle, the tangents to the circle from external point are equal in length." All right, now we're gonna have a look at some graphical demonstrations.

So sometimes we can demonstrate a mathematical statement or disprove a mathematical statement using a graph.

Sofia says, "I had this conjecture, 'for any value of x, x squared is always greater than x.

' But we disproved it with a counterexample." Andeep says, "I can't remember the counterexample, and every value I try works.

Squaring a value always makes it bigger, right?" Well, plotting this on a graph is gonna help Andeep see where the counterexample is.

So let's plot the curve with equation y equals x squared, and align with equation y equals x, 'cause we're comparing x squared to x.

Where can we see values of x squared which are not greater than x? Most values for x squared are greater than x, where can we see the counterexamples? Well done if you said, "When x is between zero and one." So we're looking for where the graph of y equals x squared is below y equals x, so it has smaller y values.

All right, let's try it.

When x is zero, zero squared is zero.

Well, zero is not bigger than zero, so x squared is not larger than x.

Equally, when x is 1, 1 squared is one, 1 isn't bigger than 1, so x squared is not larger than x.

In fact, x squared equals x, when x is zero and x is 1.

That's why the two graphs intersect.

All values between those are counterexamples as well, 'cause that's when the curve is below the line.

Andeep says, "I remember now, we use the counterexample x equals 0.

5, 'cause 0.

5 squared equals 0.

25, which is smaller." Let's try another one.

Show for all values of x, that 2 to the power of x is greater than x.

"Testing lots of examples could convince us this is true, but it's not a proof." Well done, Jun, for remembering that.

We need to show it's true for all values of x.

Let's use a graph then.

So y equals 2 to the x.

So quick reminder as to what this looks like.

If I start in the middle of the table, 2 to the power of zero is one, 2 to the power of 1 is 2, 2 to the power of 2 is 4, 2 to the power of 3 is 8.

If I then move to the left of my table, 2 to the power of negative 1 is 1 over 2, 2 to the power of negative 2 is 1 over 2 squared, and 2 to the power of negative 3 is 1 over 2 cubed or 1 over 8.

So now I can draw that, and if you've got graphing software, you can type in the equation, it'll draw it for you.

And there's y equals x.

So we can see from the graph that the line and the curve will never intersect.

The values of 2 to the power of x will always be bigger than x.

Jun says, "I conjecture that for any value of x, 2 to the x is greater than 2x." What do you think to Jun's conjecture? Let's draw y equals 2x to help us.

Well done, Jun.

I can see from the graph that this is not true, when x is between 1 and 2 inclusive.

Let's show for all positive values of x, that 10x is greater than 5x.

Andeep says, "I have drawn the lines y equal 5x and y equals 10x." What more does Andeep need to do to show this fact? Right, it's the fact that we need all positive values of x, so he could restrict the domain so that x is greater than zero.

If you are using graphing software, you can type that in in curly brackets.

And now we can see that 10x is greater than 5x for all positive values of x.

We could show that they intersect when x is zero, but zero isn't a positive number.

So anything bigger than zero is absolutely fine.

Right, your turn.

Using the graphs shown, which of these statements are true? What do you think? Right, let's have a look at each of them in turn.

So for all values of x, 2x is greater than x minus 3.

Well, that's not true.

You can just about see them intersecting at negative 3, negative 6, which means at that point they're exactly the same.

And then for x values smaller than negative 3, x minus 3 is actually gonna be bigger.

For B, for all values of x, x squared is greater than x minus 3, that is gonna be true.

You can see the curve x squared is always above the line x minus 3.

For all positive values of x, x squared is greater than 2x.

We're looking at all positive values now, so we don't need the negative values of x.

But that's still not true.

Between zero and 2, 2x is bigger than x squared.

So the last one that's true is for all positive values of x, 2x is greater than x minus 3.

In fact it's any value where x is bigger than negative 3, isn't it, 'cause that's where they intersect.

But all positive values, 2x is definitely bigger than x minus 3.

Right, your turn.

I'd like you to use the graph to explain why that conjecture does not hold.

And then give a numerical counterexample to that conjecture.

Give it a go.

Right, for question 2, I'd like you to plot your own lines and explain why Alex's conjecture is false.

Off you go.

For 3, I'd like you to show that for any value of x, x squared plus 1 is greater than x.

You can use the graph to help you.

For question 4, I'd like you to plot the graphs with equations, y equals negative x, and y equals x squared, and then tell me which of those statements are true.

So if you gonna explain why, your graphs are gonna help you here.

Let's have a look then.

So use the graph to explain why it doesn't hold, or we can see for values of x between zero and 1, the graph of y equals x cubed is below the graph of y equal x squared.

You can also see the same is true for negative x values.

Give a numerical counterexample.

Well, we can have when x is 0.

5, 0.

5 squared is 0.

25, but 0.

5 cubed is 0.

125.

So x cubed is less than x squared.

Also, because the two graphs intersect when x is zero and when x is one, they'll be counterexamples to the statement that x cubed is greater than x squared, so those two points, x cubed equals x squared.

Equally, you could go for any negative value as well.

So there are our lines.

Well done for plotting those correctly.

So we can see from the graph that Alex's conjecture is only true when x is greater than 1.

5.

If you wanted to, you could have used a counterexample.

We can see from the graph that the line y equals x, and the curve y was x squared plus 1, do not intersect.

The curve y was x squared plus 1 always has larger y values than the line for any value of x.

Therefore that's true.

So which of these statements were true? So hopefully you've drawn your curve and your line.

So for any value of x, x squared is greater than negative x.

Well that is false.

We can see that the curve is below the line when x is between negative 1 and zero.

For any positive value of x, x squared is greater than negative x.

That's true.

Zero is not a positive value, and the curve is above the line when x is greater than zero.

For any integer x, x squared is greater than negative x.

That's false.

Zero and negative 1 are integers, and the line in the curve intersect at those two points.

So that means x squared equals negative x, x squared is not bigger than negative x 'cause they're equal.

And there you go.

So we've shown now how we can use graphs to prove or disprove certain conjectures as well.

So we've used our geometry skills, and we've used our graphing skills today.

Well done for all your hard work again today.

I hope you enjoyed having a look at some of those geometric proofs.

Maybe it reminded you of some of the circle theorems that you might have done before, Pythagoras that you might have done before.

And the great thing about proving things is that we know we can use them now and we can trust that they're always going to hold.

Rather than just having to trust your teacher when they tell you that these things are true, now you've actually proven it for yourself.

I hope you're feeling really proud of your hard work today, and I look forward to seeing you again.