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Hi, my name's Mr. Peters.
Thanks for choosing to learn with me today.
In this lesson, we're gonna take our understanding of decimal numbers a little bit further and we're gonna think about using a range of representations to be able to round decimal numbers with tenths to the nearest whole number.
You ready to get started? Let's get going.
So our learning outcome for this lesson is, that by the end of this lesson, you should be confident enough to say that I can explain that decimal numbers with hundredths can be partitioned in lots of different ways.
Throughout this lesson, we've got a couple of keywords here that we're gonna keep referring back to.
I'm gonna have a go at saying them first and then you can afterwards.
My turn, partitioning.
Your turn.
My turn, decimal number.
Your turn.
So partitioning is when we think about taking an object or a value and splitting that object into smaller parts.
Breaking it down into smaller parts.
A decimal number is a number that has a decimal point.
And we'll find after the decimal point some digits which refer to the fractional part of a number.
So this lesson today is gonna be broken down into three cycles.
The first part we're thinking about partitioning decimal numbers and writing equations additively.
The second part, we'll be thinking about partitioning decimal numbers and writing those equations multiplicatively.
And the third part, we'll think about taking those decimal numbers and partitioning them in lots more different ways.
Let's get started with the first cycle.
Throughout this lesson, you are gonna meet Jun and Aisha as well.
And they've got lots of great thinking to share with us as we go throughout the lesson.
So let's get started here.
First of all, we have a number.
Our number is 2.
43.
And Jun is saying that we can partition this number into ones, tenths and hundredths.
We refer to this as standard partitioning.
Partitioning it by its place values.
Let's have a look.
The number is 2.
43.
And we can see here that we can place the two ones into one of our parts.
We can place the four tenths into another one of our parts, and we can place the three hundredths into the other part.
So we've now changed our counters to represent the numbers themselves.
So we can see that 2.
43 can be broken down into two, 0.
4 and 0.
03.
We can record this as an addition equation as well.
And you can see that below.
2 + 0.
4 + 0.
03 = 2.
43.
Have a look at this example now.
As we partition this number, ask yourself what do you notice this time? There we go.
What did you notice? What was different this time? That's right.
Well done if you spotted it.
Actually, we placed some of these parts into the different parts, didn't we? We placed the tenths into the part on the right-hand side, and we placed the hundredths into the middle part.
Does it matter? No, it doesn't matter does it? Because here we can position the addends in any order we like.
The sum will still remain the same.
So we could write this as 2 + 0.
03 + 0.
4 = 2.
43.
And again, as I said before, the position of the addends has changed, isn't it? We can see that 0.
03 has come before 0.
4, but it doesn't matter because when we recombine all of those addends together, it still makes the same sum of 2.
43.
So I think we can write this equation in lots of different ways.
Let's have a look at how we could represent this.
We could start with the whole, couldn't we? On the left-hand side of our equal sign.
And then on the right-hand side of the equal sign we can represent our addends.
And at the moment, we started with the ones, then we added the tenths, then we added the hundredths.
We could also represent it, couldn't we, by swapping some of these addends around.
So in the previous example, we had the ones, then the tenths and then the hundredths, whereas now we've got the ones plus the hundredths plus the tenths.
And again, we could take that another step further.
We could start with the hundredths this time, plus the ones, plus the tenths.
Or we could start with the hundredths plus the tenths, then plus the ones.
Or again, we could start with the tenths this time, then adding the ones and then adding the hundredths.
Or finally, starting with the tenths again, but then adding the hundredths and then adding the ones.
Can you see how I work through that systematically? I started with the ones being the first addends and then exhausted all the possibilities of how that could be done.
I then started with the hundredths being the first addend and exhausted all the possibilities of how that could be represented.
And then finally, I started with tenths, and again, exhausted all the different ways in which I could write that equation with the tenths being the first addend.
So we've got eight equations here.
We can then take this a step further by swapping the position of the expressions and the numbers on either side of the equations.
Let's have a look at that.
So as you just saw, I was able to rotate the expression on the right to the left-hand side and the number from the left to the right-hand side.
And these represent exactly the same thing as we have just seen in the previous eight equations.
So altogether now we have 16 different ways of representing this part-part-whole model so far using addition equations.
We could also represent this part-part-part-whole model as a subtraction equation as well.
For a subtraction equation, we need to have one of the parts on the left-hand side and then we need on the right-hand side the whole minus the other two remaining parts.
So to start off with, I've chosen two to be the part on the left-hand side, and then on the right-hand side we're gonna have the whole, which is 2.
43, and then we're gonna minus the two other remaining parts.
So I could minus 0.
4 and I can also minus 0.
03.
And if we think about that a little bit more, having 2.
43 and then minusing the tenths of 0.
4 and then minusing the hundredths of 0.
03, that would leave us with the ones, wouldn't it? And the ones is two, which is placed on the left-hand side.
So that is why 2 = 2.
43 - 0.
4 - 0.
03.
We could also express it like this by swapping around the order of the parts in which we're subtracting.
So in the first example, we subtracted 0.
4 first and then 0.
03 second.
Whereas in the second example, we are gonna subtract 0.
03 first and then subtract the 0.
4 after that.
We could also have a different part remaining, couldn't we? So on the left hand side now, I've got 0.
4 as the part that will be remaining.
And in order to achieve that, we would need to have the whole, 2.
43, and we'd need to minus the other two parts.
So first of all, we'll minus 0.
03 and then we'll minus the two.
And we could also do it this way around as well.
What do you think the last two equations would be? Take a moment to have a think.
That's right.
The remaining part that would need to be left for itself would be 0.
03 and we could place that on the left-hand side of our equation.
And then on the right-hand side with our expression, we could have 2.
43 minusing the other two parts again.
We can minus the two and then we could minus the 0.
4, couldn't we? We could also swap that around.
So we've got 2.
43 - 0.
4 - 2.
So again, we've got another eight equations here.
This time we've got eight subtraction equations and I think we could extend that again to 16 equations, couldn't we? By again rotating the expressions and the numbers to either side of the equal sign.
So let's have a look at that again.
If I do this, we can now see that the expression is on the left-hand side and the remaining part of the number is on the right-hand side.
And we can do this for each example once again.
So altogether, we've now got 16 different ways of writing addition equations and 16 different ways of writing subtraction equations.
Well done, if you've managed to see how we can relate to this part-part-whole model to all of those equations.
We can also take decimal numbers and represent them as mixed numbers and partition these mixed numbers as well.
This should be learning that we're already familiar with.
Let's have a look at how we could partition this mixed number.
Our number is two wholes and 43 hundredths.
We can create this using place value counters and we could partition this number into its parts.
There we go.
So now you can see we've partitioned it into two wholes, three hundredths and four tenths.
And again, we could record this as an equation.
2 43/100 = 2 + 4/10 + 3/100.
You may notice that the equation is in a different order to the way that we split the parts.
And that doesn't matter again, does it? The order of the addends in addition equation do not matter because no matter what order they're in, they will always sum to the same amount.
Right, time for us to check our understanding.
I'm gonna have a go first and then you are gonna have a go with the part-whole model on the right-hand side.
So, what we're going to do is we're gonna partition this number and then we're gonna write an addition and a subtraction equation that would represent the part-whole model.
So it says 6.
52, I could partition that into six, 0.
5 and 0.
02.
6 ones, five tenths and two hundredths.
And then I can record it as an addition equation.
6.
52 = 6 + 0.
5 + 0.
02.
Or I could record it as a subtraction equation where the part that will be remaining will be the six and then we're gonna be equal to the whole 6.
52 minus the two remaining parts.
0.
5 and 0.
02.
Right, your turn.
Have a go.
That's right.
We could have partitioned 7.
14 into seven, 0.
1 and 0.
04.
And two possible equations that you could have written were 7.
14 = 7 + 0.
1 + 0.
04.
Or 7 = 7.
14 - 0.
1 - 0.
04.
Well done, if you managed to write the equations in a slightly different way to really extend yourself in your thinking.
Okay, so onto our first task for today then.
What I'd like you to do here is to complete the part-part-whole models and then think about writing two addition and two subtraction equations that could be represented using these part-part-whole models.
Good luck, and I'll see you shortly.
Okay, let's see how you got on.
Well, the parts on the left-hand side were five, 0.
01 and 0.
3 and the whole was 5.
31.
The equations we could have written, could have looked a little bit like these.
Check your equations against these ones to see if you managed to get these or whether yours are slightly different.
Okay, the part-part-whole model on the right-hand side could have been partition to eight, 0.
2 and 0.
09.
And again, here are some of the equations that you may have chosen to write.
Check against these and see if you wrote the same or if you wrote something similar, that still works.
Right.
Let's move on to phase two of our lesson now.
We're gonna think about partitioning decimal numbers and we're gonna be expressing this multiplicatively.
Okay, let's start here and think about how a decimal number can be written as a multiplication equation where 0.
01 is actually one of the factors.
Have a look.
On the left-hand side, you can see my 0.
24.
My 0.
24 is made up of two lots of 10 hundredths and an additional four hundredths.
On the right-hand side, you can see that I've broken it down into 24 lots of 0.
01, 24 hundredths.
You can see how they're all individually separate from each other and we can record that as 24 multiplied by 0.
01.
Where 24 is a factor and 0.
01 is a factor, and then on the left-hand side of the equal sign, 0.
24 would be the product.
Here's another example.
And this time, we've used place value counters.
On the left-hand side, you can see I've got 0.
24 again, but what'd you notice this time on the right-hand side? That's right.
I've separated it into its tenths and into its hundredths.
And we could say that there are two tenths and there are four hundredths.
Can you see how I've recorded this? I've recorded the two tenths as 0.
2 and I've recorded the four hundredths as four multiplied by 0.
01, where the four represents the four green counters and the 0.
01 represents the value of each one of those counters.
So we can say that 0.
24 is equal to 0.
2, the two tenths, plus four lots of 0.
01.
Taking this even further now, we could make both the tenths and the hundredths a factor in our equation.
Well, did you notice this time? That's right.
In the last example, we had 0.
24 = 0.
2 + 4 x 0.
01.
But here we've now breaking down our tenths to have a factor of 0.
1.
So we could say that we've got two lots of 0.
1, which is the yellow counters and we've got four lots of 0.
01.
So we can say that 0.
24 = 2 x 0.
1, two lots of 0.
1 or one tenth plus four multiplied by 0.
01.
Four lots of a hundredth or four lots of 0.
01.
Now look how we've taken this even further this time.
In the previous examples we had 0.
24, but now we've got 3.
24.
What did you notice about how I've recorded these equations this time? Let's look at the first example.
I've got 3.
24 and that's equal to three multiplied by one, three lots of one, those are the red counters, plus two lots of 0.
1, two lots of the yellow counters, plus four lots of 0.
01 or four lots of the green counters.
That's an interesting way of writing these equations, isn't it? Making us think a little bit more about where these numbers are represented within are part-part-whole models.
I could also write this in a bit of a mixture of way of doing it.
So 3.
24 is equal to three, we know that 'cause three multiplied by one is three.
Plus 0.
2 and we know that 0.
2 comes from the two lots of 0.
1, plus four multiplied by 0.
01.
So again, lots of different ways we could express this part-part-whole model as an equation.
Time for you to check your understanding again.
Have a look at this number here created with the place value counters.
Which one of these equations could represent the number shown? That's right, it's B, isn't it? B shows four lots of one, four multiplied by one, that would be the red counters, plus two multiplied by 0.
1, which would be the yellow counters, plus five multiplied by 0.
01, which again would represent the green counters in this case.
All of that would be equal to 4.
25, which is a number represented at the top.
Shall we see if we can explain the issues in option A and option C? Well, in option A, it says that 4.
52 is equal to four plus five lots of 0.
1.
But look at our representation.
Do we have five lots of 0.
1? No, we don't.
We've got five lots of 0.
01, haven't we? So actually, we know that that should represent five multiplied by 0.
01 plus 0.
2 and that 0.
2 would represent the two tenths, wouldn't it? At the bottom, we've got four plus seven multiplied by 0.
1.
Let's have a look.
Oh, I can see what they've done here.
Instead of separating it into tenths and to hundredths, they've felt that the hundredths were the same as the tenths.
So there's seven counters there if you combine the yellow and the green ones together.
And they've said that all of them had a value of 0.
1, which we know is not true because two of them have a value of 0.
1 and five of them have a value of 0.
01.
Well done if you managed to get that.
Here's another check.
Which picture represents the following expression? Take a moment to have a think.
That's right.
It's C, isn't it? We can see that there are five lots of one, or one multiplied by five.
We know that the commutative law, it doesn't matter which order the factors are placed, therefore, we could write it as five multiplied by one or one multiplied by five.
Plus two lots of 0.
01.
We can see those two green counters, can't we? Plus four lots of 0.
1.
And again we can see those four yellow counters.
Does it matter that I added the hundredths first before adding the tenths? It doesn't matter, does it? We know that in addition, it doesn't matter the order of the addends 'cause it will still remain the same sum.
Okay, onto your tasks now for our second phase.
First things first, I'd like to look carefully and filling the missing numbers here for these equations.
And then for task two, I'd like you to write as many addition or multiplication equation as you possibly can that could be expressed by this number here below.
If you wanna take that step further, then I'd like you to start thinking about how you could write it as both an addition and multiplication equation.
Good luck, and I'll see you shortly.
Okay, let's go through these ones here then.
0.
7 = 7 x 0.
1.
0.
47 = 47 x 0.
01, 'cause we know the 0.
01 is the hundredths and we have 47 of those hundredths.
0.
47 = 4 x 0.
1 + 7 x 0.
01.
That means we've got four tenths and seven hundredths.
We know that four tenths and seven hundredths is the same as 47 hundredths.
And at the bottom, 3.
47 is equal to 347 hundredths, because we know that 100 hundredths is equal to one whole.
So 300 hundredths would be equal to three wholes.
We know that four tenths is equal to 40 hundredths and seven hundredths obviously remains seven hundredths.
And you combine all that together, that creates 347 hundredths.
And then, we can also express 3.
47 as three lots of one, three multiplied by one, plus four lots of 0.
14 multiplied by 0.
1, plus seven lots of 0.
01, seven multiplied by 0.
01.
Well done if you've got all of those.
Aisha wrote down some examples that she came up with.
Here, you can see that she's written a mixture of addition and multiplication equations here.
The first one is an addition equation.
2 + 0.
3 + 0.
05 = 2.
35.
Then, she's written it as, 2 + 3 x 0.
1 + 5 x 0.
01 which is again equal to 2.
35.
And then at the bottom, she's written as 2.
35 = 2.
3 + 0.
01 x 5.
So that's quite interesting.
That's a different way of doing it, Aisha.
Well done for thinking like that.
You've combined both the ones and the tenths together into the same number and then you've added on the additional a hundredths using 0.
01 as a factor, haven't you? Well done.
Really nice way of representing that.
I wonder if you came up with something similar to Aisha? Well done if you did.
And for any of your own examples that you came up with as well.
Okay, now onto the final part of our lesson.
Decimals can be partitioned in different ways.
So we can start thinking about partitioning decimal numbers in a standard way to start off with.
That is again, when we partition a number into its place values.
So let's have a look at this number here.
We've got three tenths and we've got five hundredths.
This is 0.
35 and we've partitioned it into its tenths and to its hundredths.
This can be written as an equation.
0.
35 = 0.
3 + 0.
05.
We can also then extend this to think about partitioning in a non-standard way.
Look carefully.
This was a standard partitioning.
Then what do you notice changes here? That's right.
We've actually moved one of our tenths over to sit with the hundredths.
We've redistributed our tenth to sit into the hundredths part and we can record that as an equation.
0.
35 = 0.
2 now + 0.
15.
What do you notice this time? How do you think we'd record that as an equation? Again, we have redistributed one of our tenths, haven't we? And we could record this now as 0.
35 = 0.
1, which is the left-hand side addend, plus 0.
25 which is the right-hand side addend.
So did you notice anything about these three equations? What was happening each time? That's right.
Each time we redistributed one counter to the other side, didn't we? One of our tenths counters to the other part.
And so, in this example, what we can say is that when an addend decreases by one tenth and then the other addend increases by one tenth, the sum remains the same.
So far we've been redistributing the tenths.
I wonder if we can redistribute the hundredths as well? Let's have a look.
Here is our standard partitioning of 0.
35 = 0.
3 + 0.
05.
And let's take one of the hundredths over to the other side.
Yep, there we go.
We can represent this now as 0.
35 = 0.
31 + 0.
04.
We can continue this.
0.
35 = 0.
32 + 0.
03.
And 0.
35 = 0.
33 + 0.
02.
Finally, 0.
35 = 0.
34 + 0.
01.
Can you see how we've worked systematically through this and found lots of different ways that we can represent this part-part-whole model through this partitioning and redistribution of the hundredths? In the previous one, we said when an addend decreased by one tenth and the other addend increased by one tenth, the sum remain the same.
Does that still apply here for hundredths? That's right.
When an addend decreases by one hundredth but the other addend increases by one hundredth, then the sum remains the same again.
So now I'm starting to think, does this always work? If one of the addends increases and the other addend decreases by the same amount, the sum remains the same at all times.
I wonder if you could test that out on some different numbers.
We can say that all of these examples that we've just written are non-standard partitioning.
So we can see that they're not partitioned by place value this time.
The place values have been combined when we've been partitioning them.
Now let's look at this example.
What you notice has changed this time? This time, we have a one, don't we, in our number? The previous number was 0.
35.
Now we've got 1.
35.
Can you see, actually, I haven't partitioned this in a standard way to start off with either.
I've gone for a non-standard way because we have three place values here.
We have the ones, the tenths and the hundredths.
But can you see in the left-hand part, I've combined both the ones and the tenths and in the right-hand part I've just left the hundredths.
So as Jun said, "We can continue to be even more flexible when we think about how we can partition these numbers." At the moment, we've got 1.
3 and 0.
05, but I could just swap the one to the other side.
And now I've got 1.
35 = 0.
3 + 1.
05.
Again, another way of expressing this part-part-whole model, by increasing and flexibility.
Now you can see that I've moved a tenth over, so now I've got 0.
2 + 1.
15.
And then, oh, look, we did this time.
We moved a hundredth back to the other side.
We're being really flexible here, aren't we? We've now got 1.
35 = 0.
21 + 1.
14.
Lots of different ways of representing this number.
Look, carefully at this one, what do you notice this time? That's right.
We moved two counters across this time.
All the other examples so far, when we've been redistributing have redistributed one counter.
Whereas now, we've redistributed more than one counter, haven't we? So actually we could say 0.
23 + 1.
12 = 1.
35.
And that was a change in two hundredths from one of the parts to the other parts.
We can now start thinking about solving missing number problems through using our understanding of partitioning.
Let's have a look here.
We've got 3.
72 = 3.
02 plus something.
Hmm.
So we're missing a part here.
You can see here we've got 3.
72.
And we know that 3.
72 is made up of three ones, seven tenths and two hundredths.
Let's have a look at one of the parts.
One of the parts has three ones and two hundredths.
Therefore, we must be missing the tenths, aren't we? And therefore, the missing part would be seven tenths because we don't have the tenths recorded within our parts yet.
3.
72 = 3.
02 + 0.
7.
Okay, time to check your understanding.
True or false? 6.
83 = 6 + 0.
3 + 0.
08.
That's right.
It's false, isn't it? And which justification helps us out here? A or B? That's right.
It's A, isn't it? We've got an issue here about what the numbers represent.
The six represents six ones, the eight represents eight tenths and the three represents three hundredths.
Well, actually in our equation, on the right-hand side, we've got six plus three tenths plus eight hundredths, which is not correct, is it? We've got eight tenths and three hundredths.
Well done if you spotted that.
Okay.
And now have a look at these equations here.
Which ones of these equations could represent 7.
24 being partitioned? Take a moment to have a think for yourself.
That's right, it's C and D.
7.
24 = 7.
2 + 0.
04.
And 7.
24 = 7 + 0.
24.
Why are A and B incorrect then as well? Well, A, 7.
24 = 7.
2 + 4.
Well, that's not right, is it? Because 7.
2 + 4, well, that's seven ones and four ones, which is 11 ones, so that would be 11.
2.
And then the second one is 7.
4 + 0.
2.
And again, that represents seven ones and four tenths plus another two tenths, so that would be 7.
6.
And again, we know that's not correct 'cause we're looking for 7.
24.
Okay, onto our last task for today then.
Task number one is asking you to fill in these missing numbers for me.
Task two is having a look at these part-part-whole models here and asking you if you can complete those part-part-whole models.
And if you manage that, then maybe Jun is saying have a go at making your own part-part-whole model for someone else to solve.
And then finally, the last task here.
How many different ways can you partition 6.
23.
And record that as an equation.
It'd be really interesting to see what you can come up with.
Good luck, and I'll see you shortly.
Okay, here are the answers to Task A.
Did you manage to get all of those? Well done if you did.
It helped me to maybe think about drawing a part-part-whole model next to this to help me think about where the wholes in the parts were and therefore what the missing whole and part would be each time.
It's Task two.
4.
1 and 0.
05 would be equal to 4.
15.
0.
92 is the whole and then one of our parts we've got 0.
4.
Hmm, that means I've got four tenths and in my whole I've got nine tenths so I need another five tenths and the 2 hundredths, so that would be 0.
52.
And then the last part, whole model.
I've got three, 0.
4 and 0.
12.
Well, that would be 3.
52.
And that's because I've got three ones, four tenths.
And then on the right-hand side, I've got another one tenth plus two hundredths.
And then finally, how many different ways could you have partitioned 6.
23? Aisha's got some examples here.
She partitioned it into six ones and then combined the tenths and the hundredths together to make 0.
23.
She partitioned it by combining the ones and the tenths together and then adding on the hundredths.
She then decided to take the whole of 6.
23 and subtracted the wholes and the tenths together the 6.
2, leaving her with just the hundredths.
And then look what Aisha's done here.
She's combined both addition and multiplication here.
She's got 6.
23 is equal to 6.
2.
She's put the tenths and the ones together and then she's added on, but she's written the hundredths as a factor of 0.
01.
So there are three lots of 0.
01, weren't there, altogether.
So she's written 6.
23 = 6.
2 plus three lots of 0.
01.
Nice thinking, Aisha.
Well done if you came up with something similar to Aisha or anything of your own.
Ah, well done today.
That's the end of our lesson.
Hopefully again, you are starting to feel a lot more confident now thinking about decimal numbers and how we can partition these in lots of different ways.
So to summarise our lesson today into the key points, we know that decimal numbers can be partitioned into two or more parts.
We know that partition decimal numbers can be written as either addition, subtraction, or a combination of addition and multiplication equations.
And finally, partition decimal numbers can be partitioned in a standard way, which is by their place values and a non-standard way, which is where their place values become more mixed up.
Thank you for learning again with me today.
I've really enjoyed myself.
Hopefully you have too.
Take care and I'll see you again soon.
