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A story of numbers

Map of the journey ahead

  • Set theoretic construction of numbers
  • How numbers relate to the physical world
  • Sets of numbers
  • Comparing the "size" of these sets of numbers

Notations

As we go along, we will "invent" some objects, so we shall "create" words to talk about those objects. The following is a list of all the words we shall see. You should definitely skip this list in your first reading.

Note

  • \( ℕ = \{0, 1, 2, … \} \)
  • \( ℤ = \{…, -2, -1, 0, 1, 2, … \} \)

What is a number?

prime

A prime number is an natural number which has only \( 1 \) and itself as divisors.

Theorem

A number is even iff its square is even.

even

A number \( n \) is called even if it can be written as \( n = 2k \) for some integer \( k \).

Euclid's theorem

\( \sqrt2 \) is not rational.

Proof

We prove this by contradiction.

Suppose \( \sqrt2 \) is rational. Then it can be written in the form \( \frac{p}{q} \), where \( p, q \) are integers with \( q ≠ 0 \). Assume that \( p \) and \( q \) have no common factors, for if they do, we can reduce the fraction to its lowest terms and then call the numerator \( p \) and the denominator \( q \).

Squaring and simplifying, we get \begin{equation} \label{eq:sqrt2-notin-Q} p^2 = 2 q^2 . \end{equation} This means \( p^2 \) is even. By the previous lemma, \( p \) is also even. Therefore, there exists an integer \( r \) such that \( p = 2 r \), and so \( p^2 = 4 r^2 \).

Putting this in equation \eqref{eq:sqrt2-notin-Q}, we get \( 4 r^2 = 2 q^2 \), which is the same as \( 2 r^2 = q^2 \). This means that \( q^2 \), and thus \( q \), is even.

But we had assumed that \( p \) and \( q \) have no common factors. Thus we have a contradiction. Therefore, our supposition must be wrong, and it must be that \( \sqrt2 \) is not rational.

\( \sqrt2 \) is not rational.
We prove this by contradiction. Suppose \( \sqrt2 \) is rational. Then it can be written in the form \( \frac{p}{q} \), where \( p, q \) are integers with \( q ≠ 0 \). Assume that \( p \) and \( q \) have no common factors, for if they do, we can reduce the fraction to its lowest terms and then call the numerator \( p \) and the denominator \( q \). Squaring and simplifying, we get \begin{equation} \label{eq:sqrt2-notin-Q-v1} p^2 = 2 q^2 . \end{equation} This means \( p^2 \) is even. By the previous lemma, \( p \) is also even. Therefore, there exists an integer \( r \) such that \( p = 2 r \), and so \( p^2 = 4 r^2 \). Putting this in equation \eqref{eq:sqrt2-notin-Q-v1}, we get \( 4 r^2 = 2 q^2 \), which is the same as \( 2 r^2 = q^2 \). This means that \( q^2 \), and thus \( q \), is even. But we had assumed that \( p \) and \( q \) have no common factors. Thus we have a contradiction. Therefore, our supposition must be wrong, and it must be that \( \sqrt2 \) is not rational.
Last update: 2020-07-30 15:59:07

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