Decimals, Percents, and Real Numbers

7.2 Operations on Decimals

Adding and Subtracting Decimals

You can add decimals by place value, by fraction conversion, or by standard algorithm with decimal alignment.

All three methods are really the same idea in different forms. The whole point is to keep matching place values together so that tenths are added to tenths and hundredths to hundredths.

Example: Using fraction conversion

Compute $2.16 + 1.76$ by converting the numbers to fractions.

Show Solution

Convert to hundredths:

$$2.16=\frac{216}{100},\quad 1.76=\frac{176}{100}.$$

Then

$$\frac{216}{100}+\frac{176}{100}=\frac{392}{100}=3.92.$$

This also matches the standard stacked algorithm if decimal points are aligned.

The decimal point is the anchor. Once it is lined up, the rest of the digits fall into the right place naturally.

Example: Lining up the decimal

Compute $14.23 + 8.913$ using the standard addition algorithm.

Show Solution

Start By lining up the decimals. Add in trailing 0's to make sure everything lines up. These do not change the vaules and gives us numbers for our algorithm to work with.

op tens ones . tenths hundredths thousandths
$$1$$ $$4$$ . $$2$$ $$3$$ $$0$$
+ $$8$$ . $$9$$ $$1$$ $$3$$
.

Now we can treat it like addition for whole numbers, just keeping that decimal point in place.

op tens ones . tenths hundredths thousandths
$$1^{+1}$$ $$4^{+1}$$ . $$2$$ $$3$$ $$0$$
+ $$8$$ . $$9$$ $$1$$ $$3$$
$$2$$ $$3$$ . $$1$$ $$4$$ $$3$$

And we've completed our addition!

MyOpenMath: Decimal addition

Example: Subtracting with the standard algorithm

Compute $123.45 - 9.87$ using the standard addition algorithm.

Show Solution

Start by lining up the decimals.

op hundreds tens ones . tenths hundredths
$$1$$ $$2$$ $$3$$ . $$4$$ $$5$$
- $$9$$ . $$8$$ $$7$$
.

We cannot do $5-7$ in the hundredths place, so borrow $1$ tenth from the tenths place.

op hundreds tens ones . tenths hundredths
$$1$$ $$2$$ $$3$$ . $$4^{-1}$$ $$5^{+10}$$
- $$9$$ . $$8$$ $$7$$
. $$8$$

Now the tenths place is $3-8$, so borrow $1$ one from the ones place.

op hundreds tens ones . tenths hundredths
$$1$$ $$2$$ $$3^{-1}$$ . $$3^{+10}$$ $$15$$
- $$9$$ . $$8$$ $$7$$
. $$5$$ $$8$$

Continue left to right:

  • Ones: $2-9$ needs borrowing from the tens place, so it becomes $12-9=3$.
  • Tens: after borrowing, $1-0=1$.
  • Hundreds: $1-0=1$.

So the final difference is

$$123.45-9.87=113.58.$$

MyOpenMath: Decimal subtraction

Multiplying Decimals

Because mutliplication is commutative, there are some really easy ways to multiply decimal numbers. For example, take $9.81\times 43.5$. We can rewrite both of those numbers as a whole number times a power of ten.

$$9.81=981(10^{-2})$$ $$43.5=435(10^{-1})$$

The benefit to doing this is that we can use the commutative property and some of the properties of exponents to make this problem easier.

\[ \begin{align*} 9.81\times 43.5 &= 981(10^{-2})\times 435(10^{-1}) \\ &= (981\times 435)(10^{-2}\times 10^{-1}) \\ &= 426735(10^{-3}) \\ &= 426.735 \end{align*} \]

Multiply as whole numbers first, then place the decimal so total decimal places match the sum of decimal places in factors.

This rule comes straight from rewriting each decimal as a fraction with a power of 10 in the denominator. The decimal point does not disappear; it just gets tracked through the computation.

Using our standard algorithm for multiplication it goes something like this:

First, line up the factors by place value.

op $$10^1$$ $$10^0$$ . $$10^{-1}$$ $$10^{-2}$$
$$4$$ $$3$$ . $$1$$ $$5$$
$\times$ $$9$$ . $$8$$ $$1$$

Now ignore the decimal points and multiply $4315\times 981$ using the standard algorithm. We will place the decimal back at the end.

Loop 1: multiply by the ones digit ($1$ in $981$).

op $$10^3$$ $$10^2$$ $$10^1$$ $$10^0$$
$$4$$ $$3$$ $$1$$ $$5$$
$\times$ $$1$$
$$4$$ $$3$$ $$1$$ $$5$$

So the first partial product is $4315$.

Loop 2: multiply by the tens digit ($8$ in $981$), so shift left one column.

op $$10^4$$ $$10^3$$ $$10^2$$ $$10^1$$ $$10^0$$
$$4$$ $$3$$ $$1$$ $$5$$
$\times$ $$8$$
$$3$$ $$4$$ $$5$$ $$2$$ $$0$$

So the second partial product is $34520$.

Loop 3: multiply by the hundreds digit ($9$ in $981$), so shift left two columns.

op $$10^5$$ $$10^4$$ $$10^3$$ $$10^2$$ $$10^1$$ $$10^0$$
$$4$$ $$3$$ $$1$$ $$5$$
$\times$ $$9$$
$$3$$ $$8$$ $$8$$ $$3$$ $$5$$ $$0$$

So the third partial product is $3883500$.

Add the three partial products.

op $$10^6$$ $$10^5$$ $$10^4$$ $$10^3$$ $$10^2$$ $$10^1$$ $$10^0$$
$$3$$ $$8$$ $$8$$ $$3$$ $$5$$ $$0$$
+ $$3$$ $$4$$ $$5$$ $$2$$ $$0$$
+ $$4$$ $$3$$ $$1$$ $$5$$
$$4$$ $$2$$ $$3$$ $$3$$ $$0$$ $$1$$ $$5$$

So $4315\times 981=4233015$.

Finally, put the decimal back. The original factors had $2+2=4$ decimal places total, so the product has 4 decimal places:

$$43.15\times9.81=423.3015.$$

Example: Decimal multiplication

Compute 6.2 times 1.43.

Show Solution

Ignore decimal points initially:

$$62\times143=8866.$$

Total decimal places: 1 from 6.2 and 2 from 1.43, so 3 in product:

$$6.2\times1.43=8.866.$$

MyOpenMath: Decimal multiplication

Dividing Decimals

For a physical model for division, imagine that the unit is divided into smaller pieces. For example, consider $0.47$ divided by $0.2$. This is the same as asking how many $0.2$'s can fit into $0.47$. If the whole grid below represents the number $1$, then $0.2$ is equal to $2$ columns. We can see $4$ full columns, and then $7$ smaller squares left over. It takes $20$ of the smaller squares to make one $0.2$, so we have $2$ copies of $0.2$ with a remainder of $\frac{7}{20}$. All together,

$$0.47\div0.2=2+\frac{7}{20}=2.35.$$

Area model for $0.47\div0.2$, showing two full groups of $0.2$ and $7$ hundredths left over.

Physical models are not always practical, so we can also use the standard long division algorithm. The issue is that the decimal point is not always obvious at first glance. A reliable strategy is to move the decimal point in both numbers the same amount until the divisor is a whole number. Dividing $0.47$ by $0.2$ is the same as dividing $47$ by $20$. As long as the decimal in the quotient stays lined up with the decimal in the dividend, the decimal point will land in the right place.

\[ \require{enclose} \begin{array}{rll} 2.35 && \hbox{(Quotient)} \\[-3pt] 20 \enclose{longdiv}{47.00}\kern-.2ex \\[-3pt] \underline{40\phantom{.00}} && \hbox{} \\[-3pt] \phantom{00}70 && \hbox{} \\[-3pt] \phantom{00}\underline{60} && \hbox{} \\[-3pt] \phantom{000}100 && \hbox{} \\[-3pt] \phantom{000}\underline{100} && \hbox{} \\[-3pt] \phantom{0000}0 && \hbox{} \\ \end{array} \]

Shift the decimal point in the dividend and divisor by the same amount until the divisor is a whole number, then divide.

That is the entire trick. Multiplying both numbers by the same power of $10$ does not change the quotient, but it does make the division much easier to carry out.

Example: Decimal division

Compute 13.39 divided by 0.13.

Show Solution

Multiply both numbers by 100:

$$13.39\div0.13=1339\div13.$$

Now divide:

$$1339\div13=103.$$

So the quotient is 103.

The decimal point in the quotient should line up with the decimal point in the dividend. Keeping those points aligned prevents a lot of avoidable mistakes.

MyOpenMath: Decimal division

Scientific Notation

Scientific notation is a compact way to write very large or very small numbers without losing track of the meaningful digits. It takes advantage of the properties of multiplication and exponents like we have already.

For example, the distance from the Earth to the Sun is about $$93,000,000$$ miles and the distance from the Earth to Proxima Centauri, the next nearest star other than the Sun, is about $$25,000,000,000,000$$ miles. If I were to ask you to how many trips from the Earth and Sun would it take to reach Proxima Centauri, you're likely to immediately panic when thinking about calculating something with those numbers.

Scientfic notation solves this by getting rid of all of those extra zeros. We first just write down the non zero part of the number, sometimes called the mantissa. For earth, the only numbers we really care about are the $93$. We write this down with a decimal place after the first digit.

$$93 \rightarrow 9.3$$

Then we need to count up how many times we'd multiply the number by 10 go get back to original number. We had $6$ zeros originally, and we also need the decimal to move one more time to get past the $3$, so all together we have

$$9.3\times 10^7$$

This way we're only keeping track of the mantissa and how many times we move the decimal. Similarly, that distance to Proxima Centauri is $$2.5\times 10^{13}$$

Our division problem becomes

\[ \begin{align*} \frac{\text{distance to P. Centauri}}{\text{distance to Sun}}&=\frac{2.5\times 10^{13}}{9.3\times 10^7} \\ &=\frac{2.5}{9.3}\times 10^{6} \\ &\approx 0.372 \times 10^{6} \\ &= 3.72 \times 10^{5} \end{align*} \]

Expanding that number back out, it would take about $$372000$$ trips from the Earth to the Sun to cover the same distance from earth to Proxima Centauri.

Definition: Scientific Notation

A number is in scientific notation when written as

$$a\times10^n$$

with $1\le |a|<10$ and integer $n$.

Example: Scientific notation conversion

Write 413,682,000 and 0.00000231 in scientific notation.

Show Solution
  • 413,682,000 = 4.13682 x 10^8
  • 0.00000231 = 2.31 x 10^-6

Move decimal to make leading factor between 1 and 10, and count places moved.

When the decimal moves left, the exponent is positive. When it moves right, the exponent is negative.

MyOpenMath: Decimal multiplication and scientific notation

Rounding Decimals and Significant Digits

Rounding keeps useful precision while reducing computational burden.

Sometimes we do not need every digit a number gives us. And other times, the it the level of precision is dictacted by the instrument used to measure. For example, if we were dividing up miles given to us as $\frac{4}{7}$ miles, getting 10 decimal places here doesn't makes sense. At best, we might get one decimal place and even that is beyond our real level of precision. Rounding lets us keep the level of detail that matters and discard the rest.

Fact: Rounding Workflow

To round to a target place:

  • locate target place,
  • inspect next digit,
  • keep target digit same if next digit is 0 through 4,
  • increase target digit by 1 if next digit is 5 through 9,
  • drop remaining trailing digits.

Example: Rounding practice

Round 7.456 to nearest hundredth, tenth, and unit.

Show Solution
  • Nearest hundredth: 7.46 (next digit is 6)
  • Nearest tenth: 7.5 (next digit is 5)
  • Nearest unit: 7 (next digit is 4)

The same pattern works for whole numbers too. You just choose a different target place.

Round-Off Errors

When a number is rounded, some information disappears. That is usually fine, but it is still worth knowing which digits are actually carrying meaning.

  • Non-zero digits are always significant.
  • Zeros between non-zero digits are always significant.
  • Leading zeros are never significant.
  • Trailing zeros are significant if there is a decimal point.
MyOpenMath: Rounding decimals