Introduction#


Some sets, structures, and the well-ordering principle#

Some sets#

\( \begin{aligned} \mathbb{Z} = \{ \dots, -2, -1, 0, 1, 2, \dots \} && \text{the set of integers} \\ \mathbb{P} = \{ 1, 2, \dots \} && \text{the set of positive integers} \\ \mathbb{N} = \mathbb{P} \cup \{ 0 \} = \{ 0, 1, 2, \dots \} && \text{the set of natural numbers (nonnegative integers)} \\ \mathbb{Q} = \left\{ \frac{p}{q} \mid p, q \in \mathbb{Z}, q \ne 0 \right\} && \text{the set of rational numbers} \end{aligned} \)

\(\{ \mathbb{Z}, +, \cdot, \le \}\) The set of integers under the arithmetical operations of addition and multiplication, and the binary order relation less-than-or-equal-to \(\le\) on the set \(\mathbb{Z}\).

Well-Ordering Principle#

Well-Ordering Principle [Humphreys & Prest]

Any non-empty set \(X \subseteq \mathbb{P}\) of positive integers has a smallest element (i.e., an element that is less than or equal to every member of the set \(X\)).

\[ \color{red}\boxed{ \exists k \in X \,\, \forall x \in X \,\, (k \le x) } \]

In other words, we have a set \(X\) of positive integers which, for some reason, we know is non-empty (i.e., it contains at least one element). Then the well-ordering principle guarantees a least element \(k\) of \(X\).
Equivalently, there is no unending strictly-decreasing sequence of positive integers.


Extended Integers#

An extended integer has the form

\(m + n\sqrt{10}\)

where \(m\) and \(n\) are ordinary integers.

Extended Integer Addition

Extended Integer Subtraction

Extended Integer Multiplication

Extended Integer Division

\( \begin{aligned} \frac{a + b\sqrt{10}}{m + n\sqrt{10}} &= \frac{a + b\sqrt{10}}{m + n\sqrt{10}} \times \frac{m - n\sqrt{10}}{m - n\sqrt{10}} \\ &= \frac{a \times m - 10b \times n + (b \times m - a \times n) \sqrt{10}}{m^2 - 10n^2} \\ \end{aligned} \)

\(m + n\sqrt{10}\) is a divisor of \(a + b\sqrt{10}\) iff their ratio is another extended integer.

\( \begin{aligned} 2 + \sqrt{10} \mid 12 + 3\sqrt{10} \leftrightarrow & \frac{12 + 3\sqrt{10}}{2 + \sqrt{10}} = 1 + \sqrt{10} \end{aligned} \)

Indivisible Extended Numbers are not primes because they do not satisfy Theorem 1.1.

[EXAMPLE]

Show that the extended integer \(2\) is not prime according to Theorem 1.1.

Let \(4 + \sqrt{10}\) and \(4 - \sqrt{10}\) be extended integers.

\((4 + \sqrt{10}) \times (4 - \sqrt{10}) = 6\)

\(2 \mid 6 = 3\)

Theorem 1.1 says

\(p \mid ab \rightarrow (p \mid a \lor p \mid b)\)

However

\( (2 \nmid 4 + \sqrt{10}) \land (2 \nmid 4 - \sqrt{10}) \)

Therefore, the extended integer \(2\) is not prime.


More first-order statements#

Lagrange’s Theorem#

Every natural number is the sum of four squares.

\[ \color{red} \forall n \in \mathbb{N} \,\, \exists a, b, c, d \in \mathbb{N} \,\, (n = a^2 + b^2 + c^2 + d^2) \]