# Decimal fractions

## What are decimal fractions?

Fractions whose denominator is a power of ten (10, 100, 1000, etc) are *decimal fractions*.

$$\frac{30}{10},\frac{3}{10},\frac{23}{100},\frac{12345}{1000}$$

Any fraction whose denominator is a product of a power of 2 and a power of 5, can be expressed as a decimal fraction. And vice versa: reducing a decimal fraction results in a fraction whose denominator is a product of a power of 2 and a power of 5. This is because powers of 10 (10, 100, 1000, etc.) have only twos and fives as prime factors.

$$\begin{array}{l}\frac{1}{2}=\frac{1}{{2}^{1}\times {5}^{0}}\\ \phantom{\frac{1}{2}}=\frac{5}{2\times 5}=\frac{5}{10}\end{array}$$

$$\begin{array}{l}\frac{1}{5}=\frac{1}{{2}^{0}\times {5}^{1}}\\ \phantom{\frac{1}{5}}=\frac{2}{2\times 5}=\frac{2}{10}\end{array}$$

$$\begin{array}{l}\frac{125}{1000}=\frac{{5}^{3}}{{2}^{3}\times {5}^{3}}=\frac{1}{{2}^{3}\times {5}^{0}}\\ \phantom{\frac{125}{1000}}=\frac{1}{{2}^{3}}=\frac{1}{8}\end{array}$$

## Decimal notation

Any decimal fraction can also be expressed in decimal notation, and vice versa:

$$3=\frac{30}{10}$$

$$123.45=\left(1\times {10}^{2}\right)+\left(2\times {10}^{1}\right)+\left(3\times {10}^{0}\right)+\left(4\times {10}^{-1}\right)+\left(5\times {10}^{-2}\right)$$

$$\begin{array}{l}23.45=\left(2\times 10\right)+\left(3\times 1\right)+\left(4\times \frac{1}{10}\right)+\left(5\times \frac{1}{100}\right)\\ \phantom{23.45}=\frac{2\times 1000}{100}+\frac{3\times 100}{100}+\frac{4\times 10}{100}+\frac{5}{100}\\ \phantom{23.45}=\frac{2345}{100}\end{array}$$

$$\begin{array}{l}\frac{2345}{100}=\frac{2\times 1000}{100}+\frac{3\times 100}{100}+\frac{4\times 10}{100}+\frac{5}{100}\\ \phantom{\frac{2345}{100}}=2\times 10+3\times 1+4\times {10}^{-1}+5\times {10}^{-2}\\ \phantom{\frac{2345}{100}}=23.45\end{array}$$

...

${10}^{2}=100$, the hundreds position,

${10}^{1}=10$, the tens position,

${10}^{0}=1$, the ones position,

${10}^{-1}=\frac{1}{10}$, the tenths position,

${10}^{-2}=\frac{1}{100}$, the hundredths position.

...

In 230 the 23 is in the tens position. In $\frac{23}{100}=0.23$ the 23 is in the hundredths position: "zero point two three" or "twenty-three one-hundredths". In $\frac{12345}{1000}=12.345$ the 12345 is in the thousandths position.

$$\begin{array}{l}23\u2064\frac{45}{100}=\frac{23\times 100}{100}+\frac{45}{100}\\ \phantom{23\u2064\frac{45}{100}}=\frac{2345}{100}\\ \phantom{23\u2064\frac{45}{100}}=23.45\end{array}$$

Between the ones position and the tenths position or, in other words, between the *integer part* and the *fractional part* of a number,
a *decimal mark* or *decimal separator* is inserted.
In English-speaking countries (and parts of Asia) a period is used as decimal separator: 2.063, pronounced as
"two point zero six three".
In most European countries a comma is used as decimal separator.
International standard (SI/ISO 31-0) is the use of a period as decimal separator.

Sometimes a zero integer part is omitted: 0.001 ("zero point zero zero one") is the same as .001 ("point zero zero one").

A *decimal* or *decimal numeral* or *decimal number* generally refers to a number expressed in decimal notation.
However, in some sources "decimal" refers to any digit after the decimal separator. In these cases, for example,
23.001 has 3 "decimals" and the third "decimal" is 1.

Some examples

$$\begin{array}{l}23.45=23\u2064\frac{45}{100}=23+\frac{9\times 5}{2\times 5\times 2\times 5}\\ \phantom{23.45}=23+\frac{9}{2\times 2\times 5}=23\u2064\frac{9}{20}\end{array}$$

$$\begin{array}{l}0.125=\frac{125}{1000}=\frac{5\times 5\times 5}{2\times 5\times 2\times 5\times 2\times 5}\\ \phantom{0.125}=\frac{1}{2\times 2\times 2}=\frac{1}{8}\end{array}$$

$$\frac{1}{2}=\frac{5}{10}=0.5$$

$$\frac{1}{5}=\frac{2}{10}=0.2$$

$$\begin{array}{l}\frac{3}{80}=\frac{3}{2\times 2\times 2\times 2\times 5}\\ \phantom{\frac{3}{80}}=\frac{3\times 5\times 5\times 5}{2\times 2\times 2\times 2\times 5\times 5\times 5\times 5}\\ \phantom{\frac{3}{80}}=\frac{375}{\mathrm{10\; 000}}=0.0375\end{array}$$

$$\begin{array}{l}\frac{7}{125}=\frac{7}{5\times 5\times 5}\\ \phantom{\frac{7}{125}}=\frac{7\times 2\times 2\times 2}{1000}=\frac{56}{1000}=0.056\end{array}$$

A fraction can also be converted to a decimal notation by applying long division:

0. | 0 | 5 | 6 | |||

125 | ) | 7. | 0 | 0 | ||

0 | ↓ | |||||

7 | 0 | (*) | ||||

0 | ↓ | |||||

7 | 0 | 0 | ||||

6 | 2 | 5 | ↓ | |||

7 | 5 | 0 | ||||

7 | 5 | 0 | ||||

0 |

(*), One digit (zero) *after* the decimal point is moved down. Now a dot must be placed in the result as well.

Learn the next decimal fractions by heart:

$$\frac{1}{2}=0.5,\frac{1}{4}=0.25,\frac{1}{5}=0.2,\frac{1}{8}=0.125,\frac{1}{25}=0.04,\frac{1}{125}=0.008$$

$$\frac{1}{20}=0.05,\frac{1}{40}=0.025,\frac{1}{50}=0.02,\frac{1}{80}=0.0125$$

Now write $\frac{3}{4}$ , $\frac{4}{5}$ , $\frac{3}{8}$ and $\frac{1}{16}$ in decimal notation (you can check your answers by long division or with a calculator). And how would you calculate $0.4+\frac{3}{5}$ if you had no calculator?

## Non-terminating decimal expansions

If a fully reduced fraction's denominator has any prime factors other than 2 or 5, it cannot be precisely expressed as a decimal number. However, such a fraction can be approximated to an arbitrary precision by a decimal number. As an example we use long division to determine the decimal expansion of 1/3:

0. | 3 | 3 | |||

3 | ) | 1. | 0 | 0 | |

0 | ↓ | ||||

1 | 0 | (*) | |||

9 | ↓ | ||||

1 | 0 | ||||

9 | |||||

etc. |

(*), One digit (zero) *after* the decimal point is moved down. Now a dot must be placed in the result as well.

The example above shows that 1/3 requires an endless sequence of threes after the decimal point in order to be precisely represented by a decimal notation. The long division never ends, and the same step repeats over and over again, adding more and more threes to the result. The longer the sequence of threes in the decimal expansion, the more accurately 1/3 is approximated.

0. | 0 | 4 | ||

25 | ) | 1 | ||

0 | ↓ | |||

1 | 0 | |||

0 | ↓ | |||

1 | 0 | 0 | ||

1 | 0 | 0 | ||

0 |

0. | 1 | 2 | 3 | ||||

333 | ) | 4 | 1 | ||||

0 | ↓ | ||||||

4 | 1 | 0 | |||||

3 | 3 | 3 | ↓ | ||||

7 | 7 | 0 | |||||

6 | 6 | 6 | ↓ | ||||

1 | 0 | 4 | 0 | ||||

9 | 9 | 9 | ↓ | ||||

4 | 1 | 0 | |||||

etc. |

$\frac{1}{25}=0.04$ and $\frac{41}{333}=0.123123123\mathrm{\dots}$

In the examples above, 1/25 can be converted to a decimal fraction
(4/100) and thus can be expressed precisely using a decimal representation (0.04).
The decimal expansion *terminates* eventually.
Of course the long division could be continued, resulting in endlessly repeating trailing zeros in the decimal expansion (0.04000...).
But trailing zeros do not affect the value of the number and are generally omitted,
except when they express *significant figures*, indicating "precision".

In the examples above, 41/333 is non-terminating. The long division does not end with remainder zero. The long division can be continued as long as you wish, resulting in an arbitrary long sequence of trailing digits. Because it is impossible to write infinitely many digits, the decimal representation never precisely represents the rational number. However, we see that a same sequence of digits eventually starts repeating endlessly: the decimal expansion eventually becomes periodic. With 1/3 this recurring sequence is 3, with 41/333 the recurring sequence is 123.

There are several notations to indicate an infinitely repeating or recurring part in a decimal expansion. Common use is a
*vinculum*, a horizontal line, over the first repeating digit(s).

$$\frac{1}{3}=0.\overline{)3},\frac{41}{333}=0.\overline{)123}$$

$$\frac{11}{60}=0.18\overline{)3},\frac{9001}{9000}=1.000\overline{)1}$$

1. | 0 | 0 | 0 | 1 | 1 | |||||

9000 | ) | 9 | 0 | 0 | 1 | |||||

9 | 0 | 0 | 0 | ↓ | ↓ | ↓ | ↓ | |||

1 | 0 | 0 | 0 | 0 | ||||||

9 | 0 | 0 | 0 | ↓ | ||||||

1 | 0 | 0 | 0 | 0 | ||||||

9 | 0 | 0 | 0 | |||||||

etc. |

In a long division, every subtraction step results in a "remainder" that must be smaller than the denominator. So, there is a finite number of those possible "remainders" and thus a same "remainder" must eventually appear for the second time. That is when the repeating must start because the digit brought down is always zero and a step identical to a step that appeared before, starts a new sequence identical to a sequence that appeared before. This proves that every rational number can be expressed as a decimal number with either a terminating decimal expansion (repeating sequence is zero), or with a non-terminating decimal expansion that eventually starts to endlessly repeat a finite sequence of digits.

0. | 8 | 5 | 7 | 1 | 4 | 2 | ||

7 | ) | 6 | ||||||

0 | ↓ | |||||||

6 | 0 | |||||||

5 | 6 | ↓ | ||||||

4 | 0 | |||||||

3 | 5 | ↓ | ||||||

5 | 0 | |||||||

4 | 9 | ↓ | ||||||

1 | 0 | |||||||

7 | ↓ | |||||||

3 | 0 | |||||||

2 | 8 | ↓ | ||||||

2 | 0 | |||||||

1 | 4 | |||||||

6 |

0. | 1 | 1 | 2 | 2 | |||

303 | ) | 3 | 4 | ||||

0 | ↓ | ||||||

3 | 4 | 0 | |||||

3 | 0 | 3 | ↓ | ||||

3 | 7 | 0 | |||||

3 | 0 | 3 | ↓ | ||||

6 | 7 | 0 | |||||

6 | 0 | 6 | ↓ | ||||

6 | 4 | 0 | |||||

6 | 0 | 6 | |||||

3 | 4 |

$\frac{6}{7}=0.\overline{)857142}$ and $\frac{34}{303}=0.\overline{)1122}$

The *period* of a repeating sequence, that is, the number of digits in a repeating sequence,
can never be more than the denominator minus one. In the first of two examples above the possible "remainders"
are 1,2,3,4,5,6 and they all appeared in the long division.

$$\frac{1}{29}=0.\overline{)0344827586206896551724137931}$$ The repeating sequence of $\frac{1}{97}$ contains 96 digits!

Next some examples. Can you check them using long division?

$$\frac{2}{3}=0.\overline{)6}$$ $$\frac{1}{6}=0.1\overline{)6}$$ $$\frac{1}{9}=0.\overline{)1}$$ $$\frac{252455}{33300}=7.58\overline{)123}$$

### Conversion

We can use long division to convert a fraction to a (non-terminating) decimal notation. But how to convert a non-terminating decimal notation to a fraction? We can try to reason to a conversion, for instance with 0.333...:

$$\begin{array}{l}0.3=\frac{3}{10}=\frac{9}{30}=\frac{10-1}{30}=\frac{1}{3}-\frac{1}{30}\\ 0.33=\frac{33}{100}=\frac{99}{300}=\frac{100-1}{300}=\frac{1}{3}-\frac{1}{300}\\ 0.333=\frac{333}{1000}=\frac{999}{3000}=\frac{1000-1}{3000}=\frac{1}{3}-\frac{1}{3000}\end{array}$$

The more trailing threes you add to the decimal expansion, the closer you get to 1/3. The fraction that is subtracted from 1/3 becomes smaller and smaller, approaching "infinitely small" as the number of trailing threes approaches "infinitely many".

Let's try a different approach and see if we can find a more general method:

$$\begin{array}{rl}0.333\mathrm{\dots}\times 9=& 0.333\mathrm{\dots}\times \left(10-1\right)\\ \phantom{0.333\mathrm{\dots}\times 9}=& 3.333\mathrm{\dots}-0.333\mathrm{\dots}\\ \phantom{0.333\mathrm{\dots}\times 9}=& 3\\ \frac{0.333\mathrm{\dots}\times 9}{9}=& \frac{3}{9}\\ 0.333\mathrm{\dots}=& \frac{3}{9}=\frac{1}{3}\end{array}$$

$$\begin{array}{rl}0.123123\mathrm{\dots}\times 999=& 0.123123\mathrm{\dots}\times \left(1000-1\right)\\ \phantom{0.123123\mathrm{\dots}\times 999}=& 123.123123\mathrm{\dots}-0.123123\mathrm{\dots}\\ \phantom{0.123123\mathrm{\dots}\times 999}=& 123\\ 0.123123\mathrm{\dots}=& \frac{123}{999}=\frac{41}{333}\end{array}$$

The above examples show that a decimal expansion with integer part zero and the repeating sequence first occurring right after the decimal point, can be expressed as a fraction with the repeating sequence as numerator and a denominator that consists only of the digits 9, as many as there are digits in the repeating sequence:

$$0.\overline{)6}=\frac{6}{9}=\frac{2}{3}$$ $$0.\overline{)18}=\frac{18}{99}=\frac{2}{11}$$ $$0.\overline{)012}=\frac{012}{999}=\frac{12}{999}=\frac{4}{333}$$

We can use this method to convert any decimal expansion to a fraction:

$$\begin{array}{l}0.00\overline{)4}=\frac{1}{100}\times 0.\overline{)4}\\ \phantom{0.00\overline{)4}}=\frac{1}{100}\times \frac{4}{9}=\frac{4}{900}=\frac{1}{225}\end{array}$$ $$\begin{array}{l}1.23\overline{)4}=\frac{1}{100}\times 123.\overline{)4}=\frac{1}{100}\times \left(123+0.\overline{)4}\right)\\ \phantom{1.23\overline{)4}}=\frac{1}{100}\times \left(123+\frac{4}{9}\right)=\frac{123}{100}+\frac{4}{900}\\ \phantom{1.23\overline{)4}}=\frac{1107}{900}+\frac{4}{900}=\frac{1111}{900}=1\u2064\frac{211}{900}\end{array}$$

So, any decimal expansion can be expressed as a fraction with a numerator and denominator, and thus represents a rational number. Any rational number can be expressed as a decimal expansion. Decimal fractions can be expressed with a terminating decimal expansion (or a non-terminating expansion) and all other rational numbers can only be expressed with a non-terminating repetitive decimal expansion.

### Rounding

Fractions are often represented in decimal notation, rather than as a fraction expressed with a numerator and denominator. It is often easier to compare fractions in decimal notation than with a numerator and denominator. And computers generally store and work with numbers in decimal (or binary) representation.

It is impossible to involve an infinite number of digits, so non-decimal fractions (with non-terminating decimal expansions) are generally approximated by decimal fractions (with terminating decimal expansions). This is usually not a problem, because in practice no one ever needs infinite precision. Nevertheless, it does introduce errors, and errors may add up or may be multiplied in further calculations.

Approximation by decimal fractions is done by *rounding*.
Non-terminating decimal expansions are rounded depending on the desired precision.

$\frac{1}{3}\approx 0.3$ or $\frac{1}{3}\approx 0.33$ or $\frac{1}{3}\approx 0.333$ etc.

"≈" means "approximately equal to".

It is up to the user to decide how many digits after the decimal point should be involved. The more digits, the more accurate the approximation is.

Earlier we rounded quantities to their significant figures.
The same procedures can be used to round a decimal fraction.
In the example above, the first 1/3 is *rounded to (whole) tenths* or
*rounded to the nearest tenth*:
$\frac{1}{3}\approx \frac{3}{10}$
.
The second is rounded to hundredths, the third is rounded to thousandths.

Rounding 2/3 to thousandths: $$\frac{2}{3}=0.66666\mathrm{\dots},\frac{2}{3}\approx 0.667$$

We have seen that 2.9349999... = 2.935. Hence:

$$2.934\overline{)9}\approx 2.94$$

## Arithmetic with decimal fractions

### Adding and subtracting

Adding and subtracting decimal fractions in decimal notation is pretty straightforward. The same algorithms can be used as before:

1 | ||||

2 | 0 | . | 4 | 3 |

3 | 3 | . | 6 | |

5 | 4 | . | 0 | 3 |

1 | 1 | 1 | 1 | ||

3 | 1 | . | 0 | 3 | |

2 | . | 7 | 0 | ||

3 | 0 | 7 | . | 0 | 0 |

4 | 3 | . | 2 | 5 | |

6 | 1 | . | 4 | 9 | |

4 | 4 | 5 | . | 4 | 7 |

9 | ||||

4 | 10 | 10 | ||

4 | 5 | . | 0 | 0 |

3 | . | 7 | 7 | |

4 | 1 | . | 2 | 3 |

3 | 5 | 6 | ||||

2 | 6 | 7 | . | 8 | 3 | 9 |

8 | 8 | . | 1 | 6 | 1 |

### Multiplication and division by powers of 10

Multiplying a number by a power of 10 moves the decimal point to the right an equal number of digits as zeros the power of 10 has.

Dividing a number by a power of 10 moves the decimal point to the left an equal number of digits as zeros the power of 10 has.

13 × 10 = 130, 13 in the tens position.

$\frac{13}{100}=0.13$, 13 in the hundredths position.

Thirteen times ten. Ten has one zero. The decimal point moves one position to the right, turning 13 into 130. Thirteen divided by a hundred. A hundred has two zeros. The decimal point moves two position to the left, turning 13 into 0.13.

Moving the decimal point works because we are simply adding or removing zeros to the denominator:

$$\begin{array}{l}0.13\times 10=\frac{13}{100}\times 10\\ \phantom{0.13\times 10}=\frac{13}{10}=1.3\end{array}$$

$$\begin{array}{l}0.13\xf710=\frac{13}{100}\times \frac{1}{10}\\ \phantom{0.13\xf710}=\frac{13}{1000}=0.013\end{array}$$

### Long multiplication

When we use the multiplication algorithm (long multiplication) we "remove" the decimal points first. In the result we finally insert a decimal point again. For instance 34.14 × 0.5: We use the algorithm to calculate 3414 × 5. This is equal to calculating 34.14 × 100 × 0.5 × 10 = 34.14 × 0.5 × 1000. So the result must be divided by 1000 to get the answer to 34.14 × 0.5. Or in other words:

$$\begin{array}{l}34.14\times 0.5=\frac{3414}{100}\times \frac{5}{10}\\ \phantom{34.14\times 0.5}=\frac{3414\times 5}{1000}\end{array}$$

Before using the algorithm, how much do you think a half of 34.14 is?

2 | 2 | |||

3 | 4 | 1 | 4 | |

5 | ||||

1 | 7 | 0 | 7 | 0 |

Then: 17070/1000 = 17.07 and therefore 34.14 × 0.5 = 17.07.

So what we can do is simply count the total number of positions after the decimal points in the original factors: 3 in the example above (the position of 1,4 and 5). Then, in the result, move the decimal point from the end this much places to the left: from 17070 to 17.070 in the example above.

An other example: 1.37 × 0.052 = ...

1 | 3 | 7 | |

5 | 2 | ||

2 | 7 | 4 | |

6 | 8 | 5 | |

7 | 1 | 2 | 4 |

Then: 7124/100 000 = 0.07124 or moving the decimal point 5 places to the left (a total of 5 positions after the decimal points in the original factors). Therefore: 1.37 × 0.052 = 0.07124.

An other example: write ${2.4}^{2}$ as a decimal fraction:

2 | 4 | |

2 | 4 | |

9 | 6 | |

4 | 8 | |

5 | 7 | 6 |

Thus ${2.4}^{2}=2.4\times 2.4=5.76$.

### Long division

When we use the division algorithm (long division) we "remove" the decimal point from the divisor i.e. the denominator first.

$$\frac{1.00275}{0.25}=\frac{1.00275\times 100}{0.25\times 100}=\frac{100.275}{25}$$

Before using long division, how much do you think 100.275/25 approximately is?

4 | . | 0 | 1 | 1 | ||||

25 | ) | 1 | 0 | 0 | . | 2 | 7 | 5 |

1 | 0 | 0 | ↓ | ↓ | ||||

0 | 2 | 7 | ||||||

2 | 5 | ↓ | ||||||

2 | 5 | |||||||

2 | 5 | |||||||

0 |

Thus: $$\begin{array}{l}\frac{1.00275}{0.25}=4.011\\ \phantom{\frac{1.00275}{0.25}}=4\u2064\frac{11}{1000}\end{array}$$