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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, and lower bounds, and lots of key vocabulary in between.
You may know some of these keywords, you may not, but I'll make them clear during the lesson.
Great to be learning with you.
Let's make a start.
Hi everyone and welcome to this lesson on upper and lower bounds in additive calculations and it's under the unit rounding, estimation, and bounds.
By the end of the lesson, you'll be able to calculate upper and lower bounds for calculations involving rounded numbers.
Our keywords today will include the 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, 6 has been rounded to the nearest integer.
6.
5 is the upper bound.
6.
3 has been rounded to one decimal place means that 6.
35 is the upper bound.
And the lower bound for a rounded number is the smallest value that the number could have taken prior to being rounded.
Look at this in our lesson today.
We'll also be looking at the word error interval and the error interval for a number x shows the range of possible values for x and it's written as an inequality where a is less than equal to x which is less than b.
Now today's lesson will be broken into two parts.
We'll be using bounds in additive calculations.
And then, we'll be using bounds in context.
So let's make a start using bounds in additive calculations.
Now we use error intervals to help us see the upper and lower bounds.
Remember the error interval, a is less than or equal to x which is less than b shows a to be the lower bound and b is represented as the upper bound of the number x according to a given degree of accuracy.
However, it's really important to know which bounds from our error intervals to use in order to calculate the largest and smallest values for a given calculation.
So let's have a look at an example.
If we're given two measurements of string: p is equal to 3.
4 centimetres measured to one decimal place, and q is given as 8 centimetres measured to the nearest integer.
Can you work out the error intervals of length p and q? Have a little think and press pause if you need more time.
Well, hopefully you've spotted the error interval of p would be 3.
35 centimetres less than or equal to p which is less than 3.
45 centimetres.
And the error interval for q would be 7.
5 centimetres less than or equal to q which is less than 8.
5 centimetres.
Now using these error intervals, can we identify the bounds of p and q? Well, we know the lower bound of p is 3.
35 centimetres and the upper bound of p is 3.
45 centimetres.
The lower bound of q is 7.
5 centimetres and the upper bound of q is 8.
5 centimetres.
I just want to stress the notation that I'm using in this lesson.
This is an example notation to show the upper bound and the lower bound of p and q.
You can see the subscripts of the letters I've used.
The reason for this is because we are dealing with lots of variables and lots of letters today so it's a really concise, efficient way to show what bounds we are looking at, but you might see different notation in different books.
So now we know the error interval for p and q are given here.
If p and q are added together to make length l, how would you work out the error interval for l, in other words, the error interval for the sum of the lengths? See if you can have another think.
Well, to work out the error interval, we know it has to be in this form: something is less than equal to l which is less than something.
But how do we find these bounds? Well, to find the lower bound of l, it would be the lower bound of p, add the lower bound of q.
And to find the upper bound of l, it would be the upper bound of p, add the upper bound of q.
Substituting our values in, we have 3.
35 add 7.
5 and we have 3.
45 add our 8.
5.
This means we've identified our error interval for our length l.
10.
85 is less than or equal to l which is less than 11.
95.
Notice how the lower bound is included in the inequality as the lower bounds of p and q were also included.
You can see that by the less than or equal sign.
But I also want to stress how the upper bound is not included in the inequality and this is because the upper bounds of p and q were not included.
That's really important when forming your error interval.
So in summary, the upper bound for the sum of two numbers, a and b, is found by summing the upper bound of a with the upper bound of b.
And the lower bound of the sum of the two numbers, a and b, is found by summing the lower bound of a with the lower bound of b.
If you want to jot this down and press pause, please do.
Well done, so let's have a look at a quick check.
Here, we're given a to be 4.
6 centimetres rounded to one decimal place and we're given b to be 3.
45 centimetres rounded to two decimal places.
We're also given c which is 12 centimetres rounded to the nearest integer.
And the question wants you to work out the bounds for a and b and the bounds for b and c.
Notice how we have an error interval written for you.
All you need to do is fill in those gaps and it'll help you work out those bounds.
So you can give it a go.
Press pause if you need more time.
Great work, let's see how you got on.
The most important step when identifying bounds is knowing your error intervals.
So let's work out our error interval for a.
We have 4.
55 centimetres less than or equal to a, which is less than 4.
65 centimetres.
Working out the error interval for b, 3.
445 centimetres less than or equal to b, which is less than 3.
455 centimetres.
And for c, it's 11.
5 centimetres, which is less than or equal to c, which is less than 12.
5 centimetres.
Using these error intervals allows us to quickly and efficiently identify the sum of any two numbers.
So that means to work out the bounds of a and b, it's the lower bound of a add the lower bound of b and the upper bound of a add the upper bound of b.
Thus, skimming goes the error interval of 7.
995 is less than or equal to a + b, which is less than 8.
105.
So that means we have our lower bound and our upper bound.
Really well done if you got this.
For the error interval of b and c, we're summing the lower bounds of b and c and we're summing the upper bounds of b and c, giving us the lower bound to be 14.
945 and the upper bound to be 15.
955.
Really well done if you got this one right.
So now we know the error intervals of p and q are given here.
Now let's look at the difference between p and q and I've lined them up so you can see we're going to identify the difference as d.
How would you work out the error interval for d? Have a good think about this and press pause if you need more time.
Well, to work up the error interval of d, let's look at the lower bound of d.
The lowest possible value of the difference between q and p is the smallest possible length of q subtract the biggest possible length of p.
This would give us the smallest difference between q and p and the upper bound would be the biggest possible length of length q, subtract the smallest possible length of length p.
Substituting in our values, we can therefore identify what our lower and upper bound of d would work out as.
4.
05 is less than d which is less than 5.
15.
Now you might ask, why are we using less than signs in our error interval? Well, it's because we use the lower bound of q and the upper bound of p.
Notice how the upper bound of p does not include the 3.
45.
So therefore, our inequality must be less than.
Also notice how we identify the upper bound of d by doing the upper bound of q, subtract the lower bound of p.
The upper bound of q is not included in our error interval for q.
So therefore, we can only use the less than sign in our error interval for d.
So in summary, when a is greater than b, the upper bound of the difference between a and b is found by the upper bound of a, subtract the lower bound of b.
And same again when a is greater than b, the lower bound of the difference between a and b is found by the lower bound of a, subtract the upper bound of b.
This is a really important summary.
So if you want to press pause and write it down, please do.
Great work, so let's move on to a check.
Very similar to before, we have a, which is 4.
6 rounded to one decimal place; b, which is 3.
45 rounded to two decimal places; and c, which is 12 centimetres rounded to the nearest integer.
And we're asked to work out the bounds of a - b and we're asked to work out the bounds of c - b.
See if you can give it a go.
Press pause for more time.
Great work, let's see how you got on.
Just like before, identifying the error intervals of a, b, and c is so important as it helps us to work more efficiently.
Now in order for us to work out the error interval of a - b, we need to recognise which bounds we're going to be using.
So it must be the lower bound of a, subtract the upper bound of b, would be our lower bound of a - b.
And the upper bound of a subtract the lower bound of b would be the upper bound of a - b.
Substituting in our values, we now have our lower and upper bound of a - b, which is 1.
095 and 1.
205.
Well done of you got this.
Now let's have a look at c - b.
Same as before, you need to recognise how do we find these bounds.
Well, the lower bound is found by the lower bound of c, subtract the upper bound of b.
And the upper bound is found by the upper bound of c, subtract the lower bound of b.
Substituting in our values gives us the lower bound to be 8.
045 and the upper bound to be 9.
055.
Massive well done if you got this.
Laura brings up a really good point, she says, "If the error interval says 4.
5 centimetres is less than equal to a, which is less than 5.
5 centimetres, then why does she use 5.
5 in her calculations?" Because the inequality says we don't include 5.
5.
Shouldn't it be 4.
5 centimetres less than or equal to a, which is less than or equal to 5.
49 recurring centimetres? This is a really good point, so let's show algebraically the reason why 5.
49 recurring is equal to 5.
5 and this will show you why we use 5.
5 in our calculations.
Well, first of all, let's state x as the recurring decimal.
We know x is equal to 5.
49 recurring, which we know is 5.
4999 going on forever.
Now we multiply this value of x by 10.
This is because there is only one digit recurring.
This gives us 10x to be 54.
9 recurring.
Rating it as a decimal will be 54.
999 going on forever.
Now we're going to subtract our x from our 10x, thus giving us 9x.
And subtracting these values means we just focus on our 54.
9 takeaway our 5.
4, which gives us our 49.
5 to be 9x.
Solving for x gives 49.
5/9 but let's form a simplified fraction now, which is 495/90, which then simplifies to 5.
5.
So we've just shown algebraically the reason why 5.
49 recurring is equal to 5.
5.
So that's why we use 5.
5 in our calculations with our bounds.
Great work, everybody.
So now let's move on to your task.
Question one wants you to work out the bounds for the following calculations.
See if you can give it a go.
Press pause if you need more time.
Great work, let's move on to question two.
Izzy thinks of three different positive numbers where a is less than b which is less than c.
And then, she thinks of a fourth number d.
I want you to insert the letters a, b, or c so to make the largest possible solution.
Give it some thought.
Press pause if we need more time.
Great work, let's move on to question three.
Question three is a lovely little cross number puzzle.
See if you can give it a go.
Press pause, you definitely need more time here.
Great work, let's go through these answers.
Well, for question one, always identify those error intervals.
From here, you're able to work out your calculations more efficiently and concisely.
Here's all the working out.
Please press pause in order to mark.
Great work, now let's move on to question two.
Here are all our answers.
Same again, press pause if you need more time to mark.
Great work, for question three, here's all our answers to our cross number and some working out too.
Really well done.
Press pause if you need more time to mark.
Great work, everybody.
So let's move on to using bounds in context.
Laura says, "Why are bounds so important? Why can't you just use the number given?" And Sofia says, "Well, it's important to know the possible maximum and minimum values of a number before they are rounded.
These are called limits of accuracy and are so important in real life.
For example in medicine, engineering, et cetera." Laura asks, "Can you give me an example?" So let's have a look at an example.
In medicine, there are different sizes of syringes.
What do you notice about the dosage each syringe can hold in this image? See if you could look at this picture and have a little think.
Well, hopefully you can spot the large syringe holds more, but the error intervals are one millilitre.
This means that the dosage is rounded to the nearest millilitre.
Now the middle syringe does hold less than the big syringe.
And if you look, the intervals are 0.
2 millilitres.
In other words, the dosage is rounded to the nearest 0.
2 millilitres.
And lastly, the small syringe.
The small syringe holds less but the intervals are 0.
1 millilitres.
In other words, the dosage is rounded to the nearest 0.
1 millilitre.
So using the large syringe rounded to the nearest millilitre, what do you think the error interval would be if it contained three millilitres of medicine? Have a little think.
Well, the bounds are found using 3 plus or minus our 0.
5.
The 0.
5 is found because it's the degree of accuracy, the nearest millilitre divided by 2.
So that means we have our error interval for our large syringe.
It's 2.
5 millilitres less than or equal to the large syringe capacity, less than 3.
5 millilitres.
Now looking at the middle syringe, we know it's correct in the nearest 0.
2 millilitres.
So what would the error interval be if it contains 3.
0 millilitres of medicine? Have a little think.
Well, we know that the bounds are found by 3.
0 plus or minus 0.
1.
Remember, it's the degree of accuracy because it's 0.
2, divided by 2 gives us plus or minus our 0.
1.
That means the error interval for our medium syringe would be 2.
9 millilitres less than equal to the capacity of our medium syringe, less than 3.
1 millilitres.
Now let's have a look at the smallest syringe.
Now the smallest syringe is correct in the nearest 0.
1 millilitre.
So what would the error interval be if it contained 3.
0 millilitres of medicine? Well, same again, the bounds are found by 3.
0 plus or minus our 0.
05.
This is because the degree of accuracy which is 0.
1 is divided by 2.
So that means the error interval for our small syringe is 2.
95 millilitres less than or equal to s, less than 3.
05 millilitres.
Now putting all these error intervals together, Sofia says, "So if a dosage had to be as accurate as possible, which syringe would you use and why?" Have a little think.
Well done, well, Laura spots, "It's gotta be the smallest syringe because the difference in the bounds is the smallest." So whether it be engineering, medicine, architecture, et cetera, it's so important to know the impact rounding can have on an overall calculation.
Equally, it's important to know how to improve accuracy by improving the limits of accuracy.
Let's think about the instruments that we have.
If you had to be more accurate in medicine and had to give a dosage, then the smallest syringe with that smallest degree of accuracy would be the best choice.
Let's have a look at a quick check question.
Sofia and Sam are asked to measure the lengths and widths of a tennis court with a trundle wheel and the length they measured is 37 metres to the nearest metre; and the width, they measured it to be 11 metres to the nearest metre.
Part a says, work out the upper and lower bounds of the perimeter of the tennis court.
And part b says, how can Sofia and Sam improve their accuracy? See if you can give it a go.
Press pause if you need more time.
Well done, let's have a look at working out the upper and lower bounds of the perimeter.
Well, to work out the lower bound of the perimeter, it would be two lots of the lower bound of the length add the lower bounds of the width, thus giving us the lower bound of the perimeter to be 94 metres.
To work out the upper bound of the perimeter, it'd be two lots of the upper bound of the length add the upper bound of the width.
Working this out gives us the upper bound of the perimeter to be 98 metres.
So now let's have a look at accuracy.
How can Sofia and Sam improve their accuracy? Well, they can improve their accuracy by changing how they measure the lengths and the widths.
So instead of using a trundle wheel, maybe a tape measure will be more accurate as tape measures measure using metres, centimetres, and millimetres.
Well done if you got something like that.
Let's have a look at another check question.
Here we have a bookcase and the bookcase has a maximum capacity of 100 kilogrammes measured to the nearest integer.
Now it currently holds 95 kilogrammes to the nearest integer, and Aisha has three more books to put in the bookcase.
The wonders of Mathematics is 1.
8 kilogrammes to one decimal place; Mathemagic surprise, which is 1.
5 kilogrammes to one decimal place; and Hardback Hard Maths, which is 2 kilogrammes to one significant figure.
Aisha says it's not going to exceed the capacity if we consider bounds.
Explain if Aisha is correct.
See if you can give it a go.
Press pause if you need more time.
Well done, let's see how you got on.
Just like before, it's so important to identify the error intervals.
So let's work out the error intervals of our bookcase.
Well, you know the maximum capacity is 100 kilogrammes.
So that means we have this error interval for the capacity of our bookcase.
Now let's have a look at the error intervals of what it's currently holding.
We know it's 95 kilogrammes for the nearest integer.
So this is the error interval for it's?? currently holding.
Now let's work out the error interval for each of our books.
We know the Wonder of Maths is 1.
8 kilogrammes to??for one decimal place.
So this is our error interval.
We know Mathemagic surprise is 1.
5 kilogramme to one decimal place.
So this is our error interval.
And we know Hardback Hard Maths is 2 kilogrammes to one significant figure.
So here is our error interval.
Identifying those error intervals is really good practise as it'll help you concisely and efficiently do the calculations later.
So considering bounds, let's identify the lower bound of the total capacity.
Well, it would have to be the lowest possible capacity of our bookcase, subtracts the highest possible masses of the book so far.
Subtract the upper bound of our first book, the upper bound of our second book, and the upper bound of our third book.
Substituting this in means we have a -1.
9.
Now if we were looking at the upper bound of the total capacity, well, that means we're looking at the absolute maximum total capacity of our bookcase.
Subtracting the lower bound of all the masses of the books, subtract the lower bound of our three books.
This gives us 1.
3 kilogrammes.
So considering bounds, there is a chance that the bookcase will not exceed capacity with potential room of 1.
3 kilogrammes more.
Great work, everybody.
So now let's have a look at your task.
For question one, read the questions carefully.
Press pause for more time.
Well done, let's move on to question two.
Question two is a great question.
Same again, read the question carefully.
Press pause if you need more time.
Great work, everybody.
Let's go through these answers.
Always identify those error intervals as they help you out.
From here, you can work out the upper and lower bounds of the perimeter.
And then considering bounds, hopefully you've identified there is no chance he does not have enough concrete as the upper bound of the perimeter is less than the lower bound of the concrete.
Well done if you got this one.
For question two, same again, always identify those error intervals as they help you out.
And for part b, is it possible there's, okay- And for part b, is it possible there's too much milkshake for the glass? If yes, how can the manufacturers ensure the machine does not overfill the glass? Yes, it is possible for the machine to overfill the glass.
This is because the upper bound of the strawberry syrup add the upper bound of the milk is greater than the lower bound of the glass.
So how can the manufacturers ensure that the machine does not own fulfil the glass? Well, firstly, they could improve the degree of accuracy of the amount of strawberry syrup given and the amount of milk given.
For example, if they improve the degree of accuracy to the nearest 5 millilitres, you can see from our calculations that the upper bound of the strawberry add the upper bound of the milk will be less than the lower bound of the glass.
So therefore, the glass will not overfill.
So if the manufacturers simply improve the degree of accuracy, that means they can ensure the glass will not overfill.
Alternatively, they can just provide a bigger glass.
Great work if you've got this one.
Great work, everybody.
So in summary, we use error intervals to help us see the upper and lower bounds.
And the error interval of a less than or equal to x less than b shows a to be the lower bound and b to be the upper bound of the number x according to a given degree of accuracy.
Now remember when adding numbers, the total lower bound is the sum of the lower bounds of each number.
And the total upper bound is the sum of the upper bounds of each number.
But when subtracting, the lower bound of the difference is the subtraction of the upper bound from the lower bound of the numbers and the upper bound difference is found by the subtraction of the lower bound from the upper bound of the numbers.
Great work, everybody.
It was wonderful learning with you.
