Number sets

Each number has an exact position on the number line (a straight line). Conversely, each position on the number line corresponds to a number.

We can divide the numbers into different groups:

The natural numbers

These are the numbers you use to count things.

N={1,2,3,...}\mathbb{N}=\{1,2,3,...\}

The natural numbers can be further divided, e.g.:

Note: 00 is not counted as a natural number. However, we can use the notation

N0={0,1,2,3,...}\mathbb{N}_0=\{0, 1,2,3,...\}

to include the 00.

The integers

If 00 and the "negative" natural numbers are added to the natural numbers, we get the integers:

Z={...,3,2,1,0,1,2,3,...}\mathbb{Z}=\{...,-3,-2,-1,0,1,2,3,...\}

Note that we say "negative 2" and not "minus 2".

The rational numbers

If we add to the integers all the fractions 12,13,23,...\frac{1}{2}, -\frac{1}{3}, \frac{2}{3}, ..., we get the rational numbers:

Q={all fractions pq with pZ and qN}\mathbb{Q} = \text{\{all fractions $\frac{p}{q}$ with $p\in \mathbb{Z}$ and $q\in \mathbb{N}$\}}

Any integer can of course also be represented as a fraction: 1=11,2=21,2=21,...1=\frac{1}{1}, 2=\frac{2}{1}, -2=\frac{-2}{1}, .... The integers are therefore automatically included in the above definition.

Note that pp is called the numerator and qq the denominator, and the small horizontal line is called the fraction line. We say "pp over qq". But for small denominators we also say, for example,

Note that because of 1=11,2=21,2=21,...1=\frac{1}{1}, 2=\frac{2}{1}, -2=\frac{-2}{1}, ... it follows that each integer is also a rational number.

The "Q" stands for "quotient".

The irrational numbers

All numbers which are not rational numbers, that is, all numbers which we cannot write as a fraction. We will not introduce a symbol for this set. It can be shown that π\pi (say "pai") and 2\sqrt{2} are irrational numbers.

The real numbers

All numbers on the number line, that is, the rational and irrational numbers.

Exercise 1

Draw a Venn-diagram showing the relationship between the different number sets N,Z,Q\mathbb{N},\mathbb{Z},\mathbb{Q} and R\mathbb{R}.

Solution

We have NZQR\mathbb{N}\subset \mathbb{Z}\subset \mathbb{Q}\subset \mathbb{R}:

The decimal representation of numbers

We can also write the numbers with the help of a decimal point. For example:

Observe that the rational numbers have a terminating or repeating decimal representation, while this does not seem to be the case for the irrational numbers like π\pi or 2\sqrt{2}. Indeed, this is a general rule, which we will not prove:

Theorem 1

A number is rational exactly when its decimal representation is finite or repeating.

So, finding the decimal representation of an irrational number is impossible. Read also the following article about this topic.

In contrast, finding the decimal representation of rational numbers is straight forward using a division. Here is an example:

Example 1

Find the decimal representation of 23\frac{2}{3}:

It is also possible to find the fraction for a terminating or repeating decimal number.

Example 2

Express the following numbers as a fraction pq\frac{p}{q}:

  1. 1.321.32

  2. 0.30.\overline{3}

  3. 1.231.\overline{23}

Solution
  1. 1.32=13.210=1321001.32=\frac{13.2}{10}=\frac{132}{100}
  2. Multiply the number 0.30.\overline{3} by 1010, and subtract from it the number itself: 100.30.3=3.30.3=310\cdot 0.\overline{3} - 0.\overline{3} = 3.\overline{3}-0.\overline{3}=3 But we also have 100.30.3=90.310\cdot 0.\overline{3}- 0.\overline{3}=9\cdot 0.\overline{3} Thus, we have 90.3=39\cdot 0.\overline{3} = 3 and it follows that 0.3=39=130.\overline{3} = \frac{3}{9} = \frac{1}{3}
  3. Same trick, but this time we multiply by 100100: 1001.231.23=123.231.23=122100\cdot 1.\overline{23} - 1.\overline{23} = 123.\overline{23}-1.\overline{23}=122 But we also have that 1001.231.23=991.23100\cdot 1.\overline{23}- 1.\overline{23}=99\cdot 1.\overline{23} Thus, we have 991.23=12299\cdot 1.\overline{23} = 122 and it follows that 1.23=122991.\overline{23} = \frac{122}{99}
Exercise 2
  1. To which ("smallest") number set N,Z,Q,R\mathbb{N}, \mathbb{Z}, \mathbb{Q}, \mathbb{R} does the following number belong?

    1. 23-23
    2. 42\frac{4}{2}
    3. 0.30.3
    4. 0.30.\overline{3}
    5. 1.141.\overline{14}
    6. 0.101001000100001...0.101001000100001...
  2. Determine the decimal representation of the following numbers:

    1. 14\frac{1}{4}
    2. 108\frac{10}{8}
    3. 17\frac{1}{7}
  3. Is there a number that contains every natural number? If so, this would mean that this number contains every single book that was and will be written ... .

  4. Write 2.5132.\overline{513} as a fraction pq\frac{p}{q}

  5. Show that 0.9999....0.9999.... equals 11.

Solution
  1. It is
    1. 23Z-23 \in \mathbb{Z}
    2. 42=2N\frac{4}{2}=2 \in \mathbb{N}
    3. 0.3Q0.3 \in \mathbb{Q} (because it is a terminating decimal number)
    4. 0.3Q0.\overline{3} \in \mathbb{Q} (because it is a repeating decimal number)
    5. 1.14Q1.\overline{14} \in \mathbb{Q} (because it is a repeating decimal number)
    6. 0.101001000100001...R0.101001000100001...\in \mathbb{R}. This is an irrational number, because it is a non-terminating, not-repeating decimal number.
  2. See the picture below
  3. Yes, e.g. 0.123456789101112131415....0.1\,2\,3\,4\,5\,6\,7\,8\,9\,10\,11\,12\,13\,14\,15\,....
  4. It is 10002.5132.5139992.513=2513.5132.513=2511\underbrace{1000\cdot 2.\overline{513}-2.\overline{513}}_{999\cdot 2.\overline{513}}=2513.\overline{513}-2.\overline{513} = 2511 Thus 9992.513=2511999\cdot 2.\overline{513}=2511 and therefore 2.513=25119992.\overline{513}=\frac{2511}{999}
  5. Let's write 0.999...=0.90.999...=0.\overline{9} as a fraction: 100.90.990.9=9.90.9=9\underbrace{10\cdot 0.\overline{9}-0.\overline{9}}_{9\cdot 0.\overline{9}}=9.\overline{9}-0.\overline{9}=9 Thus we have 90.9=99\cdot 0.\overline{9} = 9 and therefore 0.9=99=10.\overline{9} = \frac{9}{9}=1