Lesson video

Hi everyone.

My name is Ms. Ku, and I'm really happy that you're joining me today.

Today we're going to be looking at bounds, upper bounds, lower bounds, degrees of accuracy, error intervals, and more importantly, the real life application of these.

I hope you enjoy the lesson.

I know I will.

Let's make a start.

Hi everyone, and welcome to this lesson on upper and lower bounds in multiplicative calculations.

And it's under the unit rounding, estimation and bounds.

And by the end of the lesson you'll be able to calculate upper and lower bounds for calculations involving rounded numbers.

We'll be looking at these key words, upper bound and the upper bound for a rounded number is the smallest value that would round up to the next rounded value.

For example, six has been rounded to the nearest integer, means 6.

5 is the upper bound.

6.

3 has been rounded to one decimal place, means 6.

35 is the upper bound.

The lower bound for a rounded number is the smallest value that the number could have taken prior to being rounded.

We'll also be looking at the keywords error interval.

And the error interval for a number x shows the range of possible values of x and it's written in the form a is less on equal to x, which is less than b.

Today's lesson will be broken into two parts.

We'll be looking at using bounds in multiplicative calculations in the first part, and then we'll be using bounds with a combination of operations.

So let's make a start using bounds in multiplicative calculations.

Well remember we use error intervals to help us see the upper and lower bounds.

And the error interval is where a is less than equal to x, which is less than b, and it shows a to be the lower bound and b represents the upper bound of the number x according to a given degree of accuracy.

But remember, it's so important to know which bounds from our error intervals to use in order to calculate the largest and smallest values from a given calculation.

So let's have a look in example.

An Oak teacher has tiles to cover 24 metres squared exactly, and he measures this wall to be seven metres by three metres to the nearest metre, but it doesn't understand why the tiles do not fully cover the wall.

Why do you think the tiles did not fully cover the wall? Have a little think.

Well, it's because the Oak teacher did not consider bounds as a result of rounding.

So let's identify those error intervals for the length and the width of the wall that he measured.

Well, the error interval for the length would be 6.

5 less than, equal to our length, which is less than 7.

5.

Remember it was rounded to the nearest metre.

The width would be 2.

5 metres less than, equal to w, which is less than 3.

5 metres.

Same again, we rounded to the nearest metre.

So from these error intervals, what do you think the upper and lower bounds for the length and width of the wall are? Well, the lower bound of our length would be 6.

5 metres, and the upper bound of our length would be 7.

5 metres.

The lower bound of our width would be 2.

5 metres and the upper bound of our width would be 3.

5 metres.

I just want to stress how this is an example of notation showing the lower and upper bound of length and width.

You might see lots of different ways in different textbooks showing the upper and lower bound of certain variables.

For me, we're using subscripts as we're dealing with lots of letters and it's just a more efficient way.

So now let's see if we can identify the error interval for the area.

How do you think we can work out the error interval for our area? Well, to do this, you need to multiply the lower bound of the length by the lower bound of the width.

Remember the formula to work out the area of a rectangle is a length by width.

And to work out the upper bound of the area, it be the upper bound of the length multiplied by the upper bound of the width.

Notice how my error interval has a less than or equal to, and this is because the less than or equal to is included in the lower bounds of our inequalities.

So that means it's included in the lower bound for our area.

Now for the upper bound of our area, notice how the upper bound is not included in the inequality and this is because the upper bounds of l and w were not included.

Now substituting in, we simply have 6.

5 multiplied by 2.

5 and 7.

5 multiplied by 3.

5, thus giving me the error interval for our area to be 16.

25 metre squared, less than equal to a, which is less than 26.

25 metre squared.

So that means, we now know the lower bound of our area of the wall to be 16.

25 metre squared, and the upper bound of our area of a wall is 26.

25 metres squared.

So therefore the Oak teacher didn't have enough tiles to cover the wall because the upper bound of the area is greater than that 24 squared.

So in summary, the upper bound of the product of two numbers a and b, is found by the upper bound of a multiplied by the upper bound of b.

And the lower bound of the product, the two numbers a and b is found by the lower bound of a multiplied by the lower bound of b.

Press pause to copy this down as this is a nice useful summary.

Well done.

So let's have a look at a quick check.

Here we have a, which is three rounded to one significant figure and we have b, which is 12 rounded to two significant figures and it wants you to work out the error interval of , the error interval of b, and then the lower bound of the product of a and b, and the upper bound of the product a and b.

The second part of the question gives you a to be 5.

6 rounded to one decimal place, and b to be 8.

74 rounded to two decimal places.

Same again, you are asked to find the error interval of a, the error interval of b and identify the bounds of the product a and b.

So you can give it a go.

Press pause if you need more time.

Well done.

Let's see how you got on.

Well you should have these error intervals for a and b.

So the lower bound would be, the product of the lower bound of a and b, giving us 28.

75.

The upper bound is the product of the upper bounds of a and b, giving us 43.

75.

For the second part of the question, we should have these error intervals, so that means the lower bound would be the product of the lower bound of a and b giving us 48.

47925, and the upper bound would be the product of the upper bounds of a and b giving us 49.

40925.

Great work if you got this.

So when finding the upper and lower bounds using division, we can explore this by changing the divisors and the dividend.

For example, if we're given a divided by b, let's have a look at this using bars.

A divided by b simply means, how many b's fit into a.

So this is just an example, if we were to increase the dividend a, do you think more b's would fit into a? Let's have a look.

I'm going to increase the size of a and increasing the size of a, you might notice I can fit more b's into a, so therefore, yes, we know when using the maximum dividend it will increase the overall answer.

So now let's have a look at reducing the size of b.

Do more b's fit into a? well in reducing the size of b you can see yes.

Therefore we know when using the minimum divisor we'll increase the overall answer.

So therefore, when given two numbers a and b, where a is the dividend, the biggest possible quotient comes from the largest possible value of a divided by the smallest possible value of b.

In other words, the upper bound of a divided by the lower bound of b.

So how do you think we can find the smallest possible quotient when we're given two numbers a and b where a is the dividend, have a little think.

Well it can be found by the smallest possible value of a divided by the biggest possible value of b.

In other words, the lower bound of a divided by the upper bound of b.

Press pause if you want to copy this summary down as it is really important.

Great work everybody.

So let's have a look at a check.

Here we're given p, and p is found by x divided by y.

We know x is 3.

3 correct to one decimal place and y is 5.

8 correct to one decimal place.

And the question wants us to identify the upper and lower bounds of p giving our answer to three significant figures.

Firstly, always identify the error intervals of what we have.

So showing the error intervals of x, we have 3.

25 less than equal to x, which is less than 3.

35 and we have the error intel level y, which is 5.

75 less than equal to y, which is less than 5.

85.

Once we have this, we know to work out the upper bound of p, it must be the upper bound of x divided by the lower bound of y.

Substituting this in, we should have 3.

35 divided by 5.

75 giving me 0.

5826, so on and so forth.

Rounding this, the three significant figures gives us nor 0.

583.

Next let's find that lower bound of p.

Well to find the lower bound of p, it's the lower bound of x divided by the upper bound of y.

Substituting this in, we have 3.

25 over 5.

85, giving us knot 0.

55, so on and so forth.

Rounded to three significant figures gives a 0.

556.

Notice how finding the error intervals allows us to structure our working out easily and concisely so to work out those bounds.

Now what I'd like you to do is do a question.

You have to work out the upper and lower bound of the density where the formula is density is equal to mass over volume.

You're given the mass to be 3.

42 kilogrammes, correct to two decimal places, and the volume to be five centimetres cubed correct to one significant figure.

I want you to work out these bounds giving your answer to three significant figures.

So you can give it a go.

Press pause if we need more time.

Great work.

Let's see how you got on.

Well firstly, always identify those error intervals as they really do help.

So the error interval over the masses here, and the error interval of volume is here.

Then to work out the upper bound of our density, we know it's the upper bound of the mass divided by the lower bound of the volume.

This gives us our answer to be 0.

7611, so on and so forth.

Rounding it to three significant figures gives me 0.

761 to work at the lower bound of the density while it's the lower bound of the mass divided by the upper bound of our volume.

Substituting in our values gives us 0.

6209, so on and so forth.

Rounding it to three significant figures gives us 0.

621.

Really well done if you've got this one right.

Generally, in the question the units are given.

So don't worry if you didn't write these units down.

Great work everybody.

Now it's time for your task.

I want you to work out the upper and lower bounds of the following, giving your answers to three decimal places.

So you can give it a go, press pause if you need more time.

Great work.

Let's move on to question two.

Question two gives you another formula where the area is equal to a half times a times b times sin c.

You have to work out the upper bound only of the area of the triangle correct to three significant figures.

So you can give it a go.

Press pause if you need more time.

Well done.

Let's move on to question three.

A jug holds 160 millilitres of juice correct in the nearest 20 millilitres and a cup holds 40 millilitres of juice correct in the nearest five millilitres.

Lucas says the lowest number of full cups this jug could fill is three cups.

Is he correct? Explain.

You can give it a go.

Press pause if you need more time.

Great work everybody.

Let's have a look at question four.

Multiple choice answer given x is equal to a over b, which of the following is the correct notation for the error interval of x? And explain why the other notations are incorrect.

So if you can give it a go, press pause if you need more time.

Well done.

Let's move on to these answers.

Well for question one, remember show those error intervals as they really help you working out.

From here, here's our upper and lower bound of the product of a and b, and our upper and lower bound of a divide by b.

Press pause if you need more time to mark.

Well done.

For question two, working out our error intervals again as they always help.

From here we can work out the upper bound of the area of the triangle.

Great work, press pause if you need.

And for question two, working out our error intervals enables us to identify that Lucas is correct as we truncate 3.

52 does give us three full cups.

Lastly for question four the answer was C.

And these are the reasons why the other notations are incorrect.

Well done if you got this.

Great work everybody.

So now it's time for the second part of our lesson where we are using bounds with a combination of operations.

Now it's important to know how the biggest and smallest possible answers can be achieved from certain operations.

So I want you to fill in the table below to summarise how these bounds can be found.

The first one I've done for you, copy it down, fill in what you can, press pause as you'll need more time.

Well done.

Well we know that the lower bound of the product of a and b is the lower bound of a multiplied by the lower bound of b.

And the upper bound of the product of a and b is the upper bound of a multiplied by the upper bound of b.

And if you were to sum a and b, well, there'd be the sum of the lower bounds and the sum of the upper bounds.

But if you'd asked to find the lower bound through subtraction, it has to be the lower bound of a, subtract the upper bound of b.

The upper bound would be found by the upper bound of a, subtract the lower bound of b, and to work out the lower bound of a divided by b, it's the lower bound of a divided by the upper bound of b.

And to find the upper bound, it would be upper bound of a divided by the lower bound of b.

Now this helps us when answering questions using a combination of operations.

For example, a is equal to v subtract u all over t and we're given v to be 34.

4 correct to three significant figures, u to be 24 correct to two significant figures, and t to be 12.

5 correct to one decimal place.

And the question wants us to work up the upper bound of a, giving our answer correct to three significant figures.

Well just like before, always identify your error intervals.

We have these error intervals for v, u, and t.

Now notice how the question wants us to find the upper bound of a.

In this question you can spot we're using division, but we're also using subtraction as well.

So we have a combination of operations.

So in order to get the upper bound of a, we are using division.

Clearly, to find the upper bound, it would be the upper bound divided by the lower bound.

So notice by notation here, it has to be the upper bound of the subtraction of v takeaway u over the lower bound of t.

So how do we find the upper bound of v subtract u? Well to find the upper bound of v subtract u, remember the upper bound is calculated by the upper bound of v, subtract the lower bound of u.

So notice how I've written the calculation now using our bounds.

To find the upper bound of v subtract u, it's the upper bound of v subtract the lower bound of u, all divided by the lower bound of t.

Then we can substitute our values from the error intervals.

So we're substituting these values to give me the upper bound of a to be 0.

880 to three significant figures.

The structure of the working out on the screen shows you the correct substitution of each bound in order to identify the upper bound of a.

Great work.

So let's have a look at a check.

I'll do the first one and then like you do the second one.

Given that x is equal to three over y add k, and we know y is 6.

8 correct to one decimal place and k is 0.

7 correct to one significant figure.

The question wants us to work out the lower bound of x giving our answer correct your three significant figures.

Just like before, always identify the error intervals as they really do help.

So here my error intervals of y and k.

Now, given the fact that we want to find the lower bound of x to do this, we need the lower bound of three over y add the lower bound of k.

Because we have addition so we're trying to add the two lower bounds.

Now in order for us to work out the lower bound of three over y, it's found by three divided by the upper bound of y as this will give us the smallest possible value of three over y.

Same again, we're still adding the lower bound of k.

From here, I can substitute in three over 6.

85.

And by 0.

065, gives me the final answer of 0.

503 correct to three significant figures.

Same again, you can see the structure of my working out identifying which correct bound to substitute in, in order to find the lower bound of x.

Great work.

Now it's time for your check.

Here, we're given A is equal to x over w add p.

Where x is 4.

8 correct to one decimal place, w is three correct to one significant figure, and p is 3.

2 correct to two decimal places.

You're asked to work out the upper bound of a giving our answer correct to three significant figures.

See if you can give it a go.

Press pause if you need more time.

Well done.

Let's see how you got on.

Well, in order to identify the upper bound of a, let's identify the error intervals first as they always do help.

Here are our error intervals for x, w and p.

So to work out the upper bound of a, look at the operations that we have in the question, we have division.

So therefore it should be the upper bound of x divided by the lower bound of w and p.

Now in order to work out the lower bound of w and p, it's the sum of the lower bound of w and the lower bound of p.

So substituting our values in, we should have 4.

85 all over 2.

5 at 3.

15 giving me 0.

85, eight to three significant figures.

Really well done if you got this one.

Great work everybody.

So now it's time for your task, given that x is equal to a plus b all over c, and we're given the values of a, b, and c to a degree of accuracy, you have to calculate the lower bound of x, gimme your answer correct to three significant figures.

So you can give this a go.

Press pause if you need more time.

Well done.

Let's have a look at question two.

Question two says m is equal to x square over y plus z.

Now we're give an x, y, and z to a certain degree of accuracy and you're asked to calculate the upper bound of m, giving your answer correct to three significant figures.

So you can give it a go.

Press pause if you need more time.

Well done.

Let's have a look at question three.

Question three has quite a few operations here.

We have D is equal to the square root of three a over b squared takeaway c squared where we're given a, b, and c to a degree of accuracy and we're asked to work out the upper bound of d giving your answer correct to three significant figures.

So you can give it a go.

Press pause if we need more time.

Well done.

Let's have a look at question four.

Izzy has come up with a rule to find the upper bound of any calculation where x is greater than zero and y is greater than zero.

She says, "I just need to use the biggest possible value each time." Put a tick or a cross in the box for the calculations where it would work or where it would not work.

So you can give it a go.

Press pause if you need more time.

Well done.

Let's see how you got on with these answers.

Or remember, always identify your error intervals as they're really helpful.

To work out the lower bound of x, it would be the lower bound of a and the lower bound of b, all divided by the upper bound of c.

Substituting this in, you should have got 1.

79.

For question two, identifying those error intervals, is always a good starting point.

Then to work out the upper bound of m, it would be the upper bound of x squared, all divided by the lower bound of y and to the lower bound of z giving us 42.

9 to three significant figures.

Well done.

For question three, this was a tough one, but always start by identifying those error intervals.

Now from here we can substitute in to find the upper bound of d, which gives us 1.

29.

Press pause if you need more time to have a look at that working out.

For question four, when does she have to use the biggest possible value each time in order to find the upper bound? Well, it have to be for the addition, not the subtraction, not the division, the product of x word and y.

Not the subtraction, and not the subtraction and the division.

Well done if you got this? Great work everybody.

So in summary, this table nicely summarises how to find the upper lower bounds of calculations given certain operations.

However, sometimes there are questions which involve a combination of operations, so it's important to know which bounds to substitute in order to achieve the biggest or smallest possible value.

Great work, everybody.

It was tough lesson today, but it was wonderful learning with you.