University of Tasmania, Australia

Learning Activity 1
Place Value Numeral Expander

View the following video:

Dr Paul Swan - Number Expander

What did you learn from this video about place value?

Write your thought down on a piece of paper first, then move on to Understanding 1 and 2.

Understanding 1
Each digit in a number has a place value depending on its position

For example,  look at this number and identify each place value:

604

 

 

Check your answer here

Understanding 2
Numbers can be partitioned and renamed in terms of the position of each digit.

Here is another example:

The number 25 604.378 is partitioned or broken up to show the value of the digits.

The 2 is in the ten thousands place, so the value of the 2 is 20 000 (2 x 10 000)

The 5 is in the thousands place, so the value of the 5 is 5 000 (5 x 1 000)

The 6 is in the hundreds place, so the value of the 6 is 600 (6 x 100)

The 0 acts as a place holder in the tens place, indicating that there are no tens in the tens place

The 4 is in the ones place, so its value is 4 (4 x 1)

The 3 is in the tenths place, so the value of the 3 is 3 tenths or 0.3 (3 x 110 )

The 7 is in the hundredths place, so it represents 7 hundredths or 0.07 (7 x 1100 )

The 8 is in the thousandths place, so it represents 8 thousandths or 0.008 (8 x 11000 )

So, the number can be renamed or expressed in a different way, in terms of its place value parts.

25 604.378 =2 ten thousands + 5 thousands + 6 hundreds + 4 ones +3 tenths+ 7 hundredths + 8 thousandths

or

25 604.378 = (2 x 10 000) + (5 x 1000) + (6 x 100) + (4 x 1) + (3 x 0.1) + (7 x 0.01) + (8 x 0.001)

or

25 604.378 = 20 000 + 5 000 + 600 + 4 + 0.3 + 0.07 + 0.008

The above are examples of expanded notation.

Understanding 3
Numbers can be partitioned and regrouped into different base-ten values to assist with computation.

The number 6 512 could be expressed in various ways, for example:

 

What we have done here is partition the whole number, and them regroup it into different base-ten values (such as exchanging thousands and hundreds for tens)

 

This understanding will make solving the following everyday problem easier:

 

$2,750 is withdrawn in cash, the request being for $100 notes, the largest possible denominations.

How many hundred dollar notes will be given?

 

 

Have you noticed: a comma is used in the above example ($2,750) but not in other numbers?  The reason for this is that when we discuss money, we use a comma to separate the hundreds and thousandths place as well as the hundred thousandths place and millions place.  For example we write twelve million dollars as $12,000,000.

Check your answer here

Understanding 4

The value of place value positions increase in powers of 10 from right to left and decrease in powers of 10 from left to right. The decimal point separates the whole number values from the decimal values.

We know that each digit in a number has a place value depending on its position.

The chart below shows the place value of positions in columns, either side of the decimal point. The decimal point separates the whole numbers from the decimals.

Place Value Chart

 

The digits to the right of the decimal point are decimals, not whole numbers.

 

Our place value system uses the digits 0 – 9 to represent numbers, with ten as a base. The value of place value positions increase in powers of 10 from right to left and decrease in powers of 10 from left to right.

This can be demonstrated in the table below which uses ones to thousands to show the increase by a power of ten each time the position moves to the left, or increase by a power of ten each time the position moves to the right.

Thousands	Hundreds	Tens	Ones
1000 = 10x10x10=103	100 = 10x10=102	10 =101	1 =100

   
Look at the number 2 385 on the place value chart below. The position on the chart shows the place value of each digit. The digit with the largest value is read first, so we read the number from left to right.

 

Thousands	Hundreds	Tens	Ones
2	3	8	5

We read this as number as: two thousand, three hundred and eighty five

Using expanded notation, 2 385 can be expressed in a number of ways, such as:

2 thousands, 3 hundreds, 8 tens and 5 ones

Using expanded notation, the numeral can also be represented as:

(2 x 1 000) + (3 x 100) + (8 x 10) + (5 x 1)

Using expanded notation with powers of ten, the numeral can be succinctly represented as:

(2 x 10³) + (3 x 10²) + (8 x 10¹) + 5

On the place value chart below, look at the number 67.51

 

The position on the chart shows the place value of each digit, with the decimal point separating the whole numbers from the decimals. The digit with the largest value is read first, so we read the number from left to right.

 

Tens	Ones	.	Tenths	Hundredths
6	7	.	5	1

  
We read this as number as: sixty seven point five one

67.51 can be expressed in a number of ways using expanded notation, such as:

 

6 tens, 7 ones, 5 tenths and 1 hundredth

or

(6 x 10) + (7 x 1) + (5 x 0.1) + (1 x 0.01)

 

Using expanded notation with powers of ten, the numeral can be represented as:

 

(6 x 10¹) + (7 x 10⁰) + (5 x 10^-1) + (1 x 10^-2)

 

What do 10^-1 and 10^-2 represent?
 

Practice Task 1  (testing understandings 1 & 4)

Write the following numbers in numerals.

a) Fifty eight thousand and five

b) Six million, four hundred and twelve thousand, two hundred and twenty 

c) Seventy five point one two five 

d) Four hundred and eight point zero two 

e) One hundred and sixty two and fifteen thousandths 

f) Two hundred and seven million and twenty
 

Check your answer here

Practice Task 2 (testing Understandings 1, 2 & 4)

 

Example: Express 6 548 using expanded notation

6 548 = 6000 + 500 + 40 + 8
6 548 = (6 x 1000) + (5 x 100) + (4 x 10) + 8

 

Now complete the following practice tasks.

Express the following numbers in at least two other ways using expanded notation.

a) 25 408

b) 6 357 250

c) 92.5

d) 6.25

 

Check your answers here

 

Practice Task 3 (testing Understanding 3)

Example: How many tens are there in 43 560?

Remember that this is asking how many tens are in the entire number, not in the tens place.  The answer is 4 356. We can check this by multiplying 4 356 by 10 (43 560).

Now complete the following practice tasks.

a) How many hundreds are in 120 800?

b) How many ones are in 609?

c) How many tens are in 2 530?

d) How many thousands are in 9 689 000?

Check your answers here (PDF 164.7 KB)

Practice Task 4 (testing Understandings 1,2, 3 & 4)

 

Example: Partition and rename the number 4 506 using regrouping

4 506 = 4 000 + 500 + 6
4 506 = 45 hundreds + 6
4 506 = 450 tens + 6  

 

Now complete the following practice tasks.

Partition and rename the following numbers using regrouping.

How many different ways can you do this?

a) 3 651

b) 9 005

c) 15 508

d) 650 432

e) 98.4

f) 9.15

 

Check your answers here

For your interest...

The table below shows numbers in powers of ten up to a million (10⁶).

Try to imagine the enormous magnitude of a number such as 10²⁴.  It seems impossible to comprehend!

On the other hand, 10^-3 is  (a tenth of a tenth of a tenth), or 0.1 x 0.1 x 0.1, which is 0.001 or one thousandth.  Now try to comprehend the size of 10^-16. 

When would numbers of such a small or large magnitude be used?

The following fascinating video helps us to understand the magnitude of these numbers.

 

A million	1 000 000	10 x 10 x 10 x 10 x 10 x 10	106 (ten to the power six)
A hundred thousand	100 000	10 x 10 x 10 x 10 x 10	105 (ten to the power five)
Ten thousand	10 000	10 x 10 x 10 x 10	104 (ten to the power four)
A thousand	1 000	10 x 10 x 10	103 (ten to the power three)
A hundred	100	10 x 10	102 (ten to the power two)
Ten	10	1 x 10	101 (ten to the power one)
One	1	x 10	100 (ten to the power zero)
A tenth	0.1	x 10	10-1 (ten to the power negative 1)
A hundredth	0.01	x 10	10-2 (ten to the power negative 2)
A thousandth	0.001	x 10	10-3 (ten to the power negative 3)

     
   

Check your Understanding of Big Idea 1

Big Idea 1 has aimed to demonstrate the following four understandings:
1. Each digit in a number has a place value depending on its position.
2. Numbers can be partitioned and renamed in terms of the position of each digit.
3. Numbers can be partitioned and regrouped into different base-ten values to assist with computation.
4. The value of place value positions increase in powers of 10 from right to left and decrease in powers of 10 from left to right. The decimal point separates the whole number values from the decimal values.

Does this make sense to you now?

Please proceed to Big Idea 2

Learning Activity 1

Counting forwards and backwards in base-ten increments

An important understanding to assist with counting and adding multiples of powers of ten is to be able to identify how many of that multiple is in the entire number.  This is important to enable regrouping (see Place Value Big Idea 1).

 

Example 1:

This example demonstrates counting by tens. This is quite simple:

 158   168   178   188   198
but what happens when we add another ten?
158   168    178    188    198   208

Most people would recognise that 198 is made up of one hundred, nine tens and eight ones.  A better level of understanding is to also recognise that 198 can also be expressed as 19 tens and eight ones.  The one hundred has been regrouped into ten tens, which added to the nine tens make a total of 19 tens.
So one hundred, nine tens and six ones can be renamed as nineteen tens and eight ones.
By knowing this we can easily see what happens when we add another ten:

nineteen tens and eight ones, plus ten = twenty tens and eight ones
or 
198 + 10 = 208

 

Example 2:

This example demonstrates counting by hundreds. 

4 672   4 772    4 872    4 972    5 072

Most people would recognise that 4 972 is four thousands, nine hundreds, seven tens and two ones.  A better level of understanding is to recognise that 4 972 can also be expressed as 49 hundreds, 7 tens and two ones. 
The four thousands have been regrouped into forty hundreds, which added to nine hundreds makes a total of 49 hundreds.

To make it easier to calculate 4 972 + 100, four thousands, nine hundreds, seven tens and two ones can be renamed as forty nine hundreds and seventy two.

By knowing this it is easy to add another hundred:

forty nine hundreds and seventy two, plus one hundred = fifty hundreds and seventy two
or
4 972 + 100 = 5 072

Exercise for counting forwards and backwards in base-ten increments.

Go to the following link from ICT games:

Flip Counter

This is the type of website you would embed in your future teaching.

Complete the following flip-counter activities:

1. Counting forwards and backwards by tens

This example reinforces the importance of regrouping and partitioning in different ways to assist with computation.

2 000 + 500 + 90 + 8 can be renamed as 259 tens + 8

259 tens + one more ten is 260 tens

2 598 + 10 = 2 608

2. Counting backwards and forwards by hundreds

Make the four digit number 5 302 on the flip-counter.

This example reinforces the importance of being able to partition in different ways to assist with computation

5 002 = 5000 + 2

5 002 = 50 hundreds + 2

The second equation assists in recognising that 50 hundreds subtract one hundred is 49 hundreds. 5 002 - 100 = 4 900 + 2, which is 4 902.

 YOU CAN USE A CALCULATOR TO SEE HOW THE DIGITS CHANGE WHEN COUNTING FORWARDS AND BACKWARDS IN PLACE VALUE PARTS

Understanding 1

Ordering and Comparing Numbers

The place value chart (see PV Big Idea 1) below can be used to revise the value of each position in which the digit is placed.
The number in the chart is twenty five thousand, seven hundred and eighty three.

 

Ten Thousands	Thousands	Hundreds	Tens	Ones
2	5	7	8	3

 

 But what if we put these same digits in different positions or places?

 

Ten Thousands	Thousands	Hundreds	Tens	Ones
7	2	3	8	5

The number of the chart below now reads seventy two thousand, three hundred and eighty five.

There is a considerable difference between the two numbers, especially if for example each was a prize in a lottery.

25 783 is smaller than 72 385. Written in mathematical notation this is 25 783 < 72 385.

72 385 is larger than 25 783. Written in mathematical notation this is 72 385 > 25 783.

Using the same digits 2, 5, 7, 8 and 3, think about how you could make the largest number possible.  Which number would you place in the ten thousands column? Which number would you place in the ones column?

The largest possible number using the digits 2, 5, 7, 8 and 3 is 87 532, shown on the chart below.
It is read as eighty seven thousand, five hundred and thirty two.

Ten Thousands	Thousands	Hundreds	Tens	Ones
8	7	5	3	2

 

Just as the place value positions decrease by a power of ten from left to right(see PV Big Idea 1), the digits are placed in order from largest to smallest from left to right.

The greatest place value on the above chart is ten thousands, so to make the largest number possible with our digits the highest number (8) must be placed in the ten thousands column.

The next column of highest value on the chart is the thousands place, so the next highest number (7) is placed in the highest place.

The digits are placed in order from highest to smallest, resulting in the smallest (2) in the ones place.

To make the smallest possible number with the same digits, the digits would be placed in reverse order, as seen on the chart below. This number is read as twenty three thousand, five hundred and seventy eight.
 

Ten Thousands	Thousands	Hundreds	Tens	Ones
2	3	5	7	8

Here are a few different numbers that can be made with the same five digits.

 35 782  

  72 583  

72 538

  53 872  

53 278

38 572

These numbers can be sequenced or ordered from smallest to largest.

The smallest number is 35 782, as all of the other numbers have more than 35 thousands.

The next number in the sequence is 38 572, as once again all of the other remaining numbers have more than 38 thousands.

The next number would be 53 278. Even though 53 872 also has 53 thousands, the next column of highest value is the hundreds, and 53 278 has the smallest number of hundreds.

53 872 would therefore be the next number in the sequence.

The two remaining numbers both have 72 thousands and five hundreds. If the thousands are regrouped into hundreds, we can say that both numbers have 725 hundreds.

The next number must be 72 538 as it has five tens, and the last or largest number in the sequence is 72 583 as it has eight tens.

So the order of numbers from smallest to largest is:

 35 782  

 38 572  

 53 278  

   53 872   

72 538

72 583 

Understanding 2

Ordering Decimals

The chart below shows place values from tens to thousandths, with the decimal point separating the whole numbers from the decimals.

As we know from Place Value Module 1, the value of place value positions increase in powers of 10 from right to left and decrease in powers of 10 from left to right.

The number shown on the chart below is expressed as twelve point five seven nine. In mathematical notation this is written as 12.579

 

Tens	Ones	.	Tenths	Hundredths	Thousandths
1	2	.	5	7	9

The highest of the five digits used is nine, but this has the smallest value of thousandths.

Each thousandth is ten times smaller than a hundredth, which is ten times smaller than a tenth, which is ten times smaller than one, and so on.

This understanding is very important to be able to recognise the value of numbers and compare and order them.

The number shown on the chart below is expressed twelve point zero eight nine. In mathematical notation this is written as 12.089

 

Tens	Ones	.	Tenths	Hundredths	Thousandths
1	2	.	0	8	9

The zero cannot be ignored. It shows that there are no tenths in the tenths place.

Comparing the two numbers in the charts above, we can say that:

Both numbers have 12 wholes. Even though the number 12.089 has more hundredths than 12.579, it has no tenths. The number 12.579 has the greater value as it has 5 tenths. Tenths are ten times the value of hundredths.

 

The two numbers on the chart below are fifty point one two five (50.125) and fifty point three one (50.31).

  

Tens	Ones	.	Tenths	Hundredths	Thousandths
5	0	.	1	2	5
5	0	.	3	1

Comparing the two numbers in the charts above, we can say that:

Both numbers have 51 wholes, so the decimals are compared next, starting with the tenths place which has the highest value. The number 50.31 is the greater of the two as it has three tenths, while the other number has only one tenth.

A common misconception is that 50.125 would be the larger of the two numbers, as it has 125 after the decimal point, which is greater than 31. If this is the incorrect thinking, it shows that place value understanding may not extend to decimals and that the decimal numbers are being incorrectly viewed as whole numbers.

Placing a zero on the right-hand-side of a decimal can sometimes help with the interpretation of the values. For example, when comparing the two numbers in the chart above, they can be written as:

50.125
and
50.310

This helps us recognise that the total amount of hundredths after the decimal point is 125 and 310. It is clear to see that 50.310 is the bigger of the two numbers.

 

IMPORTANT NOTE:

The implications for this misunderstanding place value, especially the value of decimals, can be dire, such as when administering medication. Unfortunately, overdosing of infants is quite common.
Instructions are now given to pharmacists to avoid using decimal points followed by zero. For example, 55 mg written as 55.0mg may be incorrectly seen as the whole number 550 if the decimal is ignored.
Another precaution which has been adopted in recent years in the pharmaceutical industry is to always have a zero preceding the decimal point for numbers less than one. For example, .55mg should be written as 0.55mg. This dose is just over half a milligram, and if mistaken as 55mg the administered dose would be nearly a hundred times too much.

Learning Activity 2

Ordering and Comparing Decimal Numbers

Use your understandings from Place Value Big Idea 1 and Place Value Big Idea 2 to complete the following activity.

Click on the link below to practise ordering decimal numbers.

Top Marks - Decimals 
 

NOTE: You can change the order of the decimals once you have dragged them to the hooks. Click on ‘done’ when you are sure they are in the correct order.

Practice Task 1

Use a calculator or the flip counter to:

a)  Count forwards by tens from 842 to 1 142

b)  Count backwards by tens from 3 053 to 2 863

c)  Count forwards by hundreds from 16 739 to 20 239

d)  Count backwards by hundreds from 120 378 to 118 378

e)  Count forwards by tenths (0.1) from 2.5 to 4.2

f)  Count backwards by tenths (0.1) from 1.6 to 0.3

 

The flip counter is located here

Practice Task 2

Which is the greater of the two numbers?

a)  25 409 603  or 25 099 750

b)  345.99 or 355.02

c)  3.075 or 3.41

d)  65.650 or 65.69

e)  0.034 or 0.008

f)  0.01 or 0.003

 

Check your answers here

Practice Task 3

Order these numbers from smallest to largest.

a)    241 702  

214 875  

214 400  

400 001  

241 075 

 

b)    1.01   

0.707    

0.85     

1.1     

1.9          

1.008

 

 

Check your answers here

Extended thinking

 
1) Without using a calculator as an aid, orally count:

Backwards by tens from 2512 to 1812

Forwards by one hundreds from 7901 to 12,901

Count forwards by tenths (0.1) from 0.5 to 2.1

 

2) Solve the problem.

Today's interest rate is 6.65%. The loan is due to be paid back after twelve days.

The interest rate will increased by 0.05% each day for the next 12 days.  What will the interest rate be when the loan is due to be paid back?

 

Check your answers here

1. Counting forwards and backwards in base-ten increments

2. Ordering numbers

3. Comparing numbers

Learning Activity 1

The standard algorithm for addition using regrouping

The following video uses money to demonstrate exchanging base 10 values when using the standard algorithm for addition. 

This is also referred to as regrouping.

For example:

• 10 ones can be exchanged for a ten
• 10 tens can be exchanged for a hundred
• 10 hundreds can be exchanged for a thousand
• 10 thousands can be exchanged for ten thousand, and so on.

To watch the video, click here (Prof Timothy Peil, Minnesota State University, Moorhead http://web.mnstate.edu/peil/MDEV102/U1/S4/StandardAddAlg4/StandardAddAlg4.htm)  

Understanding 1

The base-10 system of numeration allows for the existence of fast and efficient algorithms.

The base-10 system of numeration allows for the existence of fast and efficient algorithms. Standard algorithms are tried and true quick methods, especially when dealing with larger numbers. However, if place value understanding is unclear these methods are prone to error.

Incorrect lining up of the numbers to be added can lead to error. If the standard algorithm has been relied on and without an understanding of place value, the error may go unrecognised.

Here is another typical error which indicates a lack of place value understanding:

 

Incorrect

  5  8
+2  4
 7 1 2

The sum of the ones is 12. The two digits have been entered into the ones column.

 

Correct

  5¹ 8
+2  4
  8  2 

The sum of the ones is 12, which can be partitioned into one ten and two ones. The 2 (ones) is placed in the ones column and the one ten is added to the tens column.

  
Our students need to be able to estimate to check reasonableness of answers.

Now we will look at how this skill can be developed.
  
  

Learning Activity 2

Multiplication and division strategy using expanded notation and partial products/quotients

The following video models appropriate mathematics language, including the terms product, partial product, expanded notation and sum.

In the video 76 x 45 is expanded to (70 + 6) x (40 + 5)

The partial products are added. The sum of the partial products is the final product.

     

Example 1:

852 x 7= 5946

  852
X    7
                    5 600  (7 x 800)   
                 350   (7 x 50)
                 14   (7 x 2)
5 964

7 x 852 = (7 x 800) + (7 x 50) + (7 x 2)
As in the example on the video, the partial products in this multiplication example are added to find the answer, that is, the product of 7 and 852.

 

Example 2:

7 254 ÷ 6 = 1209 

7 254
                  -  6 000 (1 000 x 6)
                      1 254 (remaining)
                - 1 200 (200 x 6)
                          54 (remaining)
                 -54 (9 x 6)
          0        

This division method uses partial quotients.  7 254 is partitioned into numbers which are easily divided by 6.   
7 254 = 6 000 + 1 200 + 54 
The partial quotients of 1 000, 200 and 9 are added to find the answer.

 

Learning Activity 3

Computation strategies resembling mental computation

Anyone is capable of developing sound mental computation skills.

There is a common myth that people who are quick with being able to calculate ‘in their head’ are mathematically talented.  Most often, they’ve just be taught, or learned, a range of strategies.   

Using a non-standard method to calculate, which resembles mental computation, helps us to be more skilled with estimation.  Estimation is an everyday life skill, which enables us to recognise the reasonableness of answers.

Examples of non-standard or informal methods which resemble mental computation:

Example 1:

491 + 307 

          400 + 300 = 700
       90 + 0 = 90
       7 + 1 = 8
Total of partial sums = 798

The two numbers are easily partitioned into hundreds, tens and ones:
(400 + 90 + 1) + (300 + 7)
When adding mentally, most people start with the highest value, in this case, hundreds.
Hence, the sum of 400 and 300 is 700. The next highest value is tens, therefore 90. The sum of the ones is 8. This is shown as 7 + 1 rather than 1 + 7 as it is easier to commence with the larger amount when adding on. 

 

Example 2:

4 350 + 1 398

4 000 + 1 000  = 5 000
Double 300 = 600
98 + 50 = 148 (150 - 2)
Total of partial sums = 5 748

The two numbers to be added are once again partitioned, but in this case with the exception of the tens and ones.
(4 000 + 300 + 50) + (1 000 + 300 + 98)
A similar strategy to the one in the previous example has been used for the thousands and hundreds. 
However, for the tens and ones the compensation strategy (explained below) is used to find the sum.  This is appropriate as 98 is close to 100. 100 + 50 is easily solved (150), but as 100 is two more than 98, the extra two that was added to 98 must be subtracted from 150.

Example 3:

500 - 237

500 - 200 = 300
300 - 30 = 270
270 - 7 = 263

In this subtraction example the number to be subtracted is partitioned into 200 + 30 + 7.  These three numbers, starting with the hundreds, are subtracted from the 500 one at a time to find the difference between the two numbers.

 

Practice Task 1

Solve these addition examples using the standard algorithm. Be sure to identify the place values of all the digits.

 

   7 983
 +2 899

 

  5 209
+7 891
 

 

Check your answer here
  

Practice Task 2

 

Identify the possible errors made in the following calculations.  These are typical classroom errors.
 

           _{1 1}     
     479
+  567
     536 
        

     208
+  799
           9 917          

 

 

Check your answer here

Practice Task 3

Solve these examples using partitioning and partial products or quotients:
 

a) 35 x 27

b) 621 x 8

c) 1 576 ÷ 8

 

 

 

Check your answers here

Practice Task 4

Solve this division problem using partitioning and partial quotients:

 

A syndicate of seven people won a lottery prize of $48,898.50.  If the prize is shared equally among the seven, what will each person’s share be?

 

Any remainder will need to be expressed in monetary terms.

 

 

Check your answer here

Practice Task 5

Solve the following examples using strategies resembling mental computation or non-standard written methods: 
 

a)  90 - 35

b)  48 + 52

c)  31 x 5

d)  556 - 105

e)  755 + 246

f)  4 321 + 5 226

g)  19 000 – 8 010

h)  803 – 110

i)  120 x 98

j)  5 600 ÷ 50

 

Check your answers here

Ask friends and family members to solve some of the above examples and ask them to tell you about their method. Identify the range of personal strategies. Note strategies that give the right answer, but appear inefficient.

  

Check your Understanding of Big Idea 3

The purpose of this Big Idea is to recognise how place value understanding and computation skills develop together. This recognition should come about through use and further extension of the skills and understandings from Place Value Big Idea 1 by:

1. Working with algorithms, both formal (standard) and informal

2. Using partitioning and expanded notation for computation.

3. Using a number of strategies for computation.

You are now ready to take the quiz for the Place Value and Appropriate Computational Skills.

Good Luck!

Click here to take the quiz.

Please click here to complete a survey about the Place Value & Appropriate Computational Skills Module on the Mathematics Pathway to Education