3.01 Decimals and place value

Up until now, the smallest value we have looked at on the place value table was ones (units). In this chapter we are going to consider what happens when we want to show numbers smaller than $1$1 on a place value table.
We saw in the previous chapter how we can write numbers smaller than $1$1 using fractions. But how can we write them using a place value table?

 

Just like whole numbers, where $10$10 ones make up $10$10, and $10$10 tens make up $100$100, we can continue this pattern the other way. $1$1 one is made up of $10$10 tenths, and $1$1 tenth is made up of $1$1 hundredth, and so on. As with whole number place values, the further to the left a column is, the larger its place value. 

 

Partitioning Numbers

Partitioning numbers means breaking numbers down in to two or more parts (or splitting them into smaller units). To do this, we must understand the value of each digit in a number. 

There are different ways of partitioning a decimal number. For example, $1.23$1.23 is equal to

• $1$1 one, $2$2 tenths and $3$3 hundredths ($1+0.2+0.03$1+0.2+0.03 or $1+\frac{2}{10}+\frac{3}{100}$1+210​+3100​)
• $1$1 one and $23$23 hundredths ($1+0.23$1+0.23 or $1+\frac{23}{100}$1+23100​) 
• $12$12 tenths and $3$3 hundredths ($1.2+0.03$1.2+0.03 or $\frac{12}{10}+\frac{3}{100}$1210​+3100​)

By breaking up numbers into smaller components, we can more easily compare which numbers are larger. Understanding of this will also make addition and subtraction of decimals easier.

 

Comparing decimal numbers

Using place values, we can compare the relative sizes of decimal numbers in a similar way to how we compare whole numbers - the difference being that we can't necessarily just look to see which number has a digit in the higher place value. While this method works for numbers greater than $1$1, it won't work for numbers less than $1$1 as there will always be a digit in the tenths column.

Exploration

Consider the numbers $12.34$12.34 and $2.34$2.34.  We can see that $12.34$12.34 is larger than $2.34$2.34 as it has a number in the tens column, and $2.34$2.34 doesn't. 

Now consider $0.123$0.123 and $0.19$0.19, we can see they both have a $1$1 in the tenths column, so can we just compare $123$123 to $19$19? If so, $123$123 is definitely a larger number than $19$19, so $0.123$0.123 is clearly larger! But is it really?

Let's stop and think about what these numbers mean by representing them both in a place value table.

First let's write $0.123$0.123 in a place value table.

Ones	$\cdot$·	Tenths	Hundredths	Thousandths
$0$0	$\cdot$·	$1$1	$2$2	$3$3

We can see that $0.123=\frac{1}{10}+\frac{2}{100}+\frac{3}{1000}$0.123=110​+2100​+31000​. We could also express this as $\frac{12}{100}+\frac{3}{1000}=\frac{120}{1000}+\frac{3}{1000}$12100​+31000​=1201000​+31000​ or simply $\frac{123}{1000}$1231000​.

Now let's write $0.19$0.19 in a place value table.

Ones	$\cdot$·	Tenths	Hundredths	Thousandths
$0$0	$\cdot$·	$1$1	$9$9	$0$0

Notice that we included a $0$0 in the thousandths column to show that there were no thousandths in the number, and also to make sure the two numbers we are comparing are the same length.

We can now see that $0.19=\frac{1}{10}+\frac{9}{100}$0.19=110​+9100​, which we could express as $\frac{10}{100}+\frac{9}{100}=\frac{100}{1000}+\frac{90}{1000}$10100​+9100​=1001000​+901000​ or simply $\frac{190}{1000}$1901000​.

This means we are actually comparing $123$123 thousandths to $190$190 thousandths, so in fact the number $0.19$0.19 is greater than the decimal number $0.123$0.123.

If we put both numbers in the same place value table it becomes obvious when we compare columns from left to right:

Ones	$\cdot$·	Tenths	Hundredths	Thousandths
$0$0	$\cdot$·	$1$1	$2$2	$3$3
$0$0	$\cdot$·	$1$1	$9$9	$0$0

The second number has more hundredths, so it is the larger value.

Practice question

Question 1

Multiplying and dividing by powers of 10

Multiplying or dividing by powers of $10$10 ($10$10, $100$100, $1000$1000, and so on) is much easier than multiplying by other numbers, because our number system is base $10$10. Let's have a look at what this means.

Consider the number $253$253. We can write this as:

$253=2\times100+5\times10+3\times1$253=2×100+5×10+3×1

When we multiply $253$253 by $100$100, we can write it like this:

$\left(2\times100+5\times10+3\times1\right)\times100$(2×100+5×10+3×1)×100	$=$=	$2\times10000+5\times1000+3\times100$2×10000+5×1000+3×100
	$=$=	$25300$25300

Notice that when we multiplied by $100$100 the place value of the "$5$5" changed from tens to thousands, since $10\times100=1000$10×100=1000. In fact, all of the digits moved up by two place values.

Worked example

Example 1

Evaluate $5.62\times1000$5.62×1000.

Think: Since we are multiplying, the digits will all move up in place value. Since there are three zeros in $1000$1000, each digit will move up by three place values.

Do: We can evaluate the multiplication by moving each digit up by three place values.

So $5.62\times1000=5620$5.62×1000=5620.

Reflect: Notice that we added a zero in the Ones place, to properly represent the new places that our values ended up in. We also removed the decimal point, since we didn't need it after the multiplication.

Example 2

Evaluate $7.43\div100$7.43÷​100.

Think: Since we are dividing, the digits will all move down in place value. Since there are two zeros in $100$100, each digit will move down by two place values.

Do: We can evaluate the division by moving each digit down by two place values.

So $7.43\div100=0.0743$7.43÷​100=0.0743.

Reflect: We moved the digits up in place value for multiplication and down for division. The number of zeroes in the conversion unit told us how many place values to move by. We ended up adding in some extra zeros to keep track of the places for each of our values.

Practice questions

Question 2

Question 3