Proposition Let $a \in \mathbb{R}$. Then $-(-a) = a$.
Proof \begin{align}
(-a) + (-(-a)) = 0 & \text{ by A4}
\\
(-a) + a = 0 & \text{ by A4 and A1}
\\
a = (-(-a)) & \text{ by uniqueness of additive inverses}
\end{align}
Proposition For any $a \in \mathbb{R}$: $a \cdot 0 = 0$.
Proof \begin{align}
a \cdot (0 + 0) = a \cdot 0 & \text{ by A3}
\\
a \cdot 0 + a \cdot 0 = (a \cdot 0) + 0 & \text{ by D and A3}
\\
a \cdot 0 = 0 & \text{ subtracting } a \cdot 0 \text{ (by A1 - A4)}
\end{align}
2 Order axioms
We will assume the existence of a subset $\mathbb{P}$ of $\mathbb{R}$ called the positive numbers which satisfies the following three axioms:
given $a, b \in \mathbb{P}$, then also $a + b \in \mathbb{P}$
given $a, b \in \mathbb{P}$, then also $ab \in \mathbb{P}$
(trichotomy axiom) for any $a \in \mathbb{R}$, precisely one of $a \in \mathbb{P}$, $a = 0$, $-a \in \mathbb{P}$ holds
Proposition $1 \in \mathbb{P}$
Proof $1 = 0$ is impossible by $Z$.
Also if $-1 \in \mathbb{P}$, then $(-1)(-1) = -(-1) = 1 \in \mathbb{P}$ by $P2$, but that contradicts the trichotomy axiom.
The only remaining possibility is $1 \in \mathbb{P}$.
3 The Completness Axiom
Proposition (Approximation Property) Let $B$ be a non-empty, bounded above subset of $\mathbb{R}$. Let $b = \sup{B}$ and $\varepsilon > 0$. Then there exists $x \in B$ with $\sup{B} - \varepsilon < x \leq \sup{B}$.
Theorem (Archimedean Property) $\mathbb{N}$ is not bounded above.
4 Complex Numbers
This chapter was made into the separate course Introduction to Complex Numbers.
Slovak
English
Matej Balog