Operations with Decimals

Learning goals

Add and subtract two decimals that end by lining up the points, padding with trailing zeros
Explain why lining up the points is what stacks like place values
For two decimals that end, multiply as whole numbers, then count the decimal places of both factors
Divide by a decimal that ends by sliding both points until the divisor is whole
Slide the point when multiplying or dividing by ten, a hundred, a thousand

Adding and subtracting: line up the points

When you add whole numbers in columns, you line up the ones with the ones. You line up the tens with the tens, and the hundreds with the hundreds. The reason is simple: you may only add digits that count the same thing. Three hundreds and four hundreds combine into seven hundreds. But three hundreds and four tens do not collapse into a single column, because they are different sizes.

Decimals work the very same way, and the decimal point is what keeps the columns honest.

Before stacking two decimals that end, give both numbers the same number of decimal places by padding with trailing zeros. You proved in the last lesson that a trailing zero does not change a decimal’s value ( $3.4 = 3.40$ ). Then every column is filled, and there is no empty slot to misjudge.

Stack $3.40$ over $1.25$ with their points in one column and add:

\begin{array}{r} 3.40 \\ +\ 1.25 \\ \hline 4.65 \end{array}

Aligning the points puts the $4$ tenths over the $2$ tenths, and the $0$ hundredths over the $5$ hundredths. So every column once again holds digits of a single place value. That is the whole rule: line up the decimal points, then add or subtract column by column exactly as you do with whole numbers. The point in the answer drops straight down beneath the points above it.

Why lining up the points is the same as lining up place values#

Write each decimal in expanded form, the way you learned to split a number into its places. For example $5.3$ is $5 + \tfrac{3}{10}$ and $2.16$ is $2 + \tfrac{1}{10} + \tfrac{6}{100}$ . Adding them means collecting like place values together, because only quantities of the same size combine:

\left(5 + \tfrac{3}{10}\right) + \left(2 + \tfrac{1}{10} + \tfrac{6}{100}\right) = (5 + 2) + \left(\tfrac{3}{10} + \tfrac{1}{10}\right) + \tfrac{6}{100}.

The ones gather with the ones, the tenths with the tenths, and the six hundredths stay put. That regrouping is allowed because addition lets you add in any order. The regrouping is also exactly what stacking the numbers with their points aligned does on paper. Each column of the stack already holds a single place value. So the column sums are exactly these grouped sums. A tenth and a hundredth can certainly be added ( $\tfrac{1}{10} + \tfrac{1}{100} = \tfrac{11}{100}$ , or $0.11$ ); the expanded form just shows they need their own separate places to do it, the same way three hundreds and four tens stay as $340$ instead of collapsing into a single digit. Now suppose you misalign the points. That writes a tenth and a hundredth into the very same column, as if they were the same size, and adds their digits together as though they matched. The expanded form shows that column sum has no correct place value to belong to, so it is not a real digit of the answer. Lining up the points is just the visual shortcut for keeping every column to the one place value it is allowed to hold.

Lining up the decimal points stacks tenths over tenths and hundredths over hundredths, so every column holds digits of a single place value. The point in the answer drops straight down.

Worked example 1 Add $12.6 + 3.45$

The two numbers have a different count of decimal places, so first pad the shorter with a trailing zero: $12.6$ becomes $12.60$ . That trailing zero does not change its value, but it fills the hundredths column so nothing is left to guess.

Stack them with the points aligned and add column by column from the right, carrying just as with whole numbers. The hundredths give $0 + 5 = 5$ , the tenths give $6 + 4 = 10$ , so write $0$ and carry a $1$ into the ones, and the rest follows:

\begin{array}{r} 12.60 \\ +\ 3.45 \\ \hline 16.05 \end{array}

Drop the point straight down into the answer, and the sum is $16.05$ .

Worked example 2 Subtract $8.3 - 5.74$

Pad $8.3$ to two decimal places so both numbers have a hundredths digit: $8.3$ becomes $8.30$ . Without that placeholder zero there is nothing in the top hundredths column to subtract from.

Now subtract column by column from the right, borrowing when a top digit is too small. The hundredths need a borrow first, since $0 - 4$ cannot be done, so take one tenth (ten hundredths) from the tenths column. That makes the hundredths $10 - 4 = 6$ , and it drops the tenths from $3$ down to $2$ .

The tenths column now reads $2 - 7$ , which again cannot be done. So borrow a second time, this time taking one whole (ten tenths) from the ones column. The $8$ ones becomes $7$ , the tenths becomes $12$ , and $12 - 7 = 5$ . Finally the ones give $7 - 5 = 2$ :

\begin{array}{r} 8.30 \\ -\ 5.74 \\ \hline 2.56 \end{array}

The point drops straight down, giving $8.30 - 5.74 = 2.56$ . A quick sanity check: $5.74 + 2.56 = 8.30$ , so the subtraction undoes correctly.

Check your understanding

What is $6.7 + 0.48$ ?

7.18

11.5

0.115

6.118

Multiplying decimals that end: multiply, then count the places

Multiplication breaks the lining-up habit, and that trips many students. To multiply two decimals you do not line up the points.

For example $0.3 \times 0.2$ : ignore the points and multiply $3 \times 2 = 6$ . Each factor has one decimal place, so the product needs $1 + 1 = 2$ decimal places, which makes the answer $0.06$ , not $0.6$ . That matches common sense: multiplying two positive numbers that are each smaller than one gives something smaller still.

Every product of two decimals that end works the same way. You ignore the points entirely and multiply the two numbers as if they were whole numbers. Only at the very end do you put a point back in the answer. The single question is where the point goes, and a count of decimal places settles it:

For two decimals that end, count the total number of decimal places in the two factors, and give the product that many decimal places. That count tells you where to put the point. If the result then ends in a zero, you can drop it afterward exactly as you always could, since $0.60$ and $0.6$ name the same number.

Why the product's decimal places add up#

Every terminating decimal is a fraction. Its bottom number is ten, or a hundred, or a thousand, and so on. One decimal place means a denominator of ten. Two places mean a hundred, and three places mean a thousand. So a factor with one decimal place is some whole number over ten. In the same way, a factor with two decimal places is some whole number over a hundred.

Take $2.1 \times 0.05$ as the model. Rewrite each factor as a whole number over a power of ten:

2.1 \times 0.05 = \frac{21}{10} \times \frac{5}{100}.

Multiplying fractions multiplies the tops and multiplies the bottoms, a rule you proved in the fractions chapter:

\frac{21}{10} \times \frac{5}{100} = \frac{21 \times 5}{10 \times 100} = \frac{105}{1000}.

The tops multiplied to give the whole-number product $21 \times 5 = 105$ , and the bottoms multiplied to give a thousand. A denominator of a thousand means three decimal places, so $\tfrac{105}{1000} = 0.105$ . Here is the general pattern. One factor’s bottom contributes its zeros, and the other factor’s bottom contributes its zeros. Multiplying the two bottoms simply adds those two collections of zeros together. The count of zeros in the bottom is the count of decimal places. So the product’s decimal places are counted by the zeros in its bottom. That count is the places of the first factor plus the places of the second. The point lands by counting, not by lining up.

A special, very common case is multiplying a decimal by $10$ , by $100$ , or by $1000$ . Since $10$ has no decimal places, the digits of the answer are just the digits of the original times ten. So the only effect is that every digit moves up one place. On paper that looks like the decimal point sliding to the right: one place for $10$ , two places for $100$ , three for $1000$ .

4.72 \times 10 = 47.2, \qquad 4.72 \times 100 = 472, \qquad 0.6 \times 1000 = 600.

This is just place value at work: multiplying by ten makes each digit worth ten times as much. So the tenths digit becomes a ones digit, the ones digit becomes a tens digit, and so on up the line.

Dividing by $10$ , $100$ , or $1000$ slides the point the same number of places, but the other way, to the left, because dividing by ten makes each digit worth ten times less:

4.72 \div 10 = 0.472, \qquad 4.72 \div 100 = 0.0472, \qquad 6 \div 1000 = 0.006.

Watch what happens in that last one. Sliding the point three places left runs the number out of digits, so you fill the empty places with zeros: $6$ becomes $006$ , and the point lands in front, giving $0.006$ . Skipping a placeholder zero is the single most common mistake in this slide.

Multiplying by ten slides the decimal point one place to the right, because every digit becomes worth ten times as much. Each extra zero in the multiplier slides it one more place.

Check your understanding

What is $6.3 \div 100$ ?

0.063

630

0.63

63

Worked example 3 Multiply $1.4 \times 0.6$

Ignore the decimal points and multiply the numbers as whole numbers:

14 \times 6 = 84.

Now count the decimal places in the two factors. The factor $1.4$ has one place and $0.6$ has one place, so the product needs $1 + 1 = 2$ decimal places. Put the point two places from the right of $84$ , which requires a placeholder zero in front:

1.4 \times 0.6 = 0.84.

Check the size: $1.4$ is a little more than one and $0.6$ is a little more than half, so a product near $0.8$ is reasonable.

Worked example 4 Multiply $0.25 \times 0.3$

Strip the points and multiply as whole numbers:

25 \times 3 = 75.

Count the places: $0.25$ has two decimal places and $0.3$ has one, so the product needs $2 + 1 = 3$ decimal places. Placing the point three from the right of $75$ needs a leading placeholder zero:

0.25 \times 0.3 = 0.075.

The leading zero is essential. Without it, $0.75$ would have only two decimal places and would be ten times too big.

Check your understanding

How many decimal places does the product $0.12 \times 0.4$ have, and what is it?

3

places,

0.048

2

places,

0.48

1

place,

4.8

3

places,

0.480

Dividing decimals: make the divisor a whole number

Division is the operation where a decimal point in the wrong place does the most damage. So there is a clean rule that removes the danger. To compute $4.5 \div 0.9$ , slide both points one place right, so the divisor becomes the whole number $9$ and the dividend becomes $45$ :

4.5 \div 0.9 = 45 \div 9 = 5.

That slide multiplied $4.5$ by ten and $0.9$ by ten. You are allowed to do this because a division is a ratio. Multiplying the top and bottom of that ratio by the same number leaves its value unchanged. Ten over ten is one, so $45 \div 9$ and $4.5 \div 0.9$ have the same answer. What matters is that the dividend slides the same number of places as the divisor.

Here is the rule. For a divisor that ends, slide the decimal point in the divisor (the nonzero number you are dividing by) to the right until it becomes a whole number. Next, slide the point in the dividend (the number being divided) the same number of places, then divide and bring the point straight up. Dividing by a whole number is the long division you already know.

Why scaling both numbers by ten leaves the quotient unchanged#

The division $1.2 \div 0.5$ is the fraction $\tfrac{1.2}{0.5}$ , so multiply its top and its bottom by ten:

\frac{1.2}{0.5} = \frac{1.2 \times 10}{0.5 \times 10} = \frac{12}{5}.

Multiplying by $\tfrac{10}{10}$ is multiplying by one, so the value has not moved: both $1.2 \div 0.5$ and $12 \div 5$ come out as $2.4$ . The bottom $0.5$ has become the whole number $5$ , which is the whole point of the slide. Nothing there depended on the two particular numbers $1.2$ and $0.5$ : multiplying top and bottom by the same nonzero number is always allowed, because you are multiplying by a disguised form of one. So slide as many places as you need to clear the point out of the divisor, always moving the dividend the same number of places. If the dividend runs out of digits, append zeros on the right, which a trailing zero never changes.

Dividing a decimal by a nonzero whole number is the simplest case of all. There is no sliding to do, because the divisor is already whole. You just do long division and place the point in the quotient directly above the point in the dividend.

Worked example 5 Divide $7.2 \div 0.8$

The divisor $0.8$ has one decimal place, so slide both points one place to the right. The divisor $0.8$ becomes the whole number $8$ , and the dividend $7.2$ becomes $72$ :

7.2 \div 0.8 = 72 \div 8.

Now it is whole-number division:

72 \div 8 = 9.

So $7.2 \div 0.8 = 9$ . Check it by multiplying back: $0.8 \times 9 = 7.2$ , which matches.

Worked example 6 Divide $9.6 \div 0.04$

The divisor $0.04$ has two decimal places, so both points must slide two places to the right. The dividend $9.6$ has only one digit after its point, so first write it with two: $9.6 = 9.60$ . That trailing zero does not change its value, but it gives the second slide a digit to land on.

Now slide both points two places right. The divisor $0.04$ becomes the whole number $4$ , and the dividend $9.60$ becomes $960$ :

9.60 \div 0.04 = 960 \div 4.

Now divide:

960 \div 4 = 240.

So $9.6 \div 0.04 = 240$ . It makes sense that dividing by a very small number gives a large answer: $0.04$ fits into $9.6$ many times over.

Sliding the points only clears the divisor. It does not promise a whole-number quotient, since the dividend can still fail to divide evenly once the divisor is whole.

Worked example 7 Divide $5.4 \div 1.2$

The divisor $1.2$ has one decimal place, so slide both points one place to the right. The divisor $1.2$ becomes $12$ , and the dividend $5.4$ becomes $54$ :

5.4 \div 1.2 = 54 \div 12.

The divisor is now whole, so finish exactly like the case above: divide and keep the point in the quotient directly above the point in the dividend. Twelve goes into $54$ four times ( $4 \times 12 = 48$ ) with $6$ left over, which is $60$ tenths. Append a zero to the dividend ( $54 = 54.0$ ) and continue: twelve goes into $60$ tenths five times exactly.

5.4 \div 1.2 = 4.5.

Clearing the divisor turned a decimal-by-decimal division into a decimal-by-whole-number division. It did not turn it into a whole-number answer, and it does not need to.

Once the divisor is a whole number, long division works exactly as it does for whole numbers, and the quotient's decimal point sits directly above the dividend's. Subtracting 48 from 54 leaves 6; bringing down the 0 makes 60; and 12 goes into 60 exactly 5 times, so the remainder ends at 0.

Worked example 8 Divide $3.5 \div 4$

The divisor is already the whole number $4$ , so there is nothing to slide. Do long division and keep the point in the quotient directly above the point in the dividend. Since $4$ does not go into $3$ , the quotient starts with a $0$ in the ones place; bring the point up and continue. Four goes into $35$ tenths eight times ( $8 \times 4 = 32$ ) with $3$ tenths left over, which is $30$ hundredths:

3.5 \div 4 = 0.875.

When the remainder does not clear, append zeros after the decimal point in the dividend ( $3.5 = 3.500$ ) and keep dividing. Here $30$ hundredths give $7$ with $2$ left, and $20$ thousandths give a clean $5$ , so the remainder reaches $0$ and the division terminates at $0.875$ . Not every division ends this way: some remainders never reach $0$ , and the digits repeat forever instead of stopping. That case is covered in the next lesson.

Check your understanding

To compute $6.3 \div 0.7$ by making the divisor a whole number, what division do you actually carry out, and what is the answer?

63 \div 7 = 9

6.3 \div 7 = 0.9

63 \div 70 = 0.9

630 \div 7 = 90

Common mistakes

Lining up the points to multiply. Aligning decimal points is for addition and subtraction only. For two decimals that end, multiply by ignoring the points, multiplying as whole numbers, then counting the total decimal places.
Forgetting to count places in a product. $0.3 \times 0.2$ is $0.06$ , not $0.6$ . The two factors have one place each, so the product has two. Multiplying two positive numbers below one gives something smaller, not larger. (This place-counting shortcut is for two decimals that end.)
Sliding the point the wrong way when multiplying by ten. Multiplying by $10$ , $100$ , or $1000$ moves the point to the right (the number grows); dividing by them moves it left. Count the zeros to know how many places.
Sliding only one number when dividing. When you move the point in the divisor, you must move it in the dividend by the same number of places. Moving just one of them changes the quotient.
Dropping placeholder zeros in the answer. A product like $0.075$ or a sum like $16.05$ needs its internal and leading zeros. They hold the other digits in the right places; erasing one changes the value tenfold.

Key takeaways

For two decimals that end, add and subtract by lining up the decimal points (pad with trailing zeros so every column is filled). Then work column by column as with whole numbers, and the point drops straight down.
Lining up the points works because it stacks like place values, and you may only combine digits of the same place value.
For two decimals that end, multiply by ignoring the points, multiplying as whole numbers, then giving the product as many decimal places as the two factors have together.
Multiplying by $10$ , $100$ , or $1000$ slides the point right by one, two, or three places; dividing by them slides it left.
For a divisor that ends, divide by sliding its point right until it is a whole number. Slide the dividend the same number of places, then divide and bring the point straight up.
Sliding both points the same amount is multiplying top and bottom by the same number (ten, a hundred, or a thousand). So sliding both points the same amount never changes the quotient.

Operations with Decimals

Learning goals

Adding and subtracting: line up the points

Multiplying decimals that end: multiply, then count the places

Dividing decimals: make the divisor a whole number

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice