Real and Complex Number Systems

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yedlu, Winter 2025

Introduction and Rationale

The introduction of real number set \(\BB{R}\) arises from the discovery that the rational number system (\(\BB{Q}\)) is inadequate in many scenarios. “Irrational numbers” is then introduced to fill the gap and is often approximated by rational numbers.

(Complex number set \(\BB{C}\) is introduced to fill the gap of algebraic calculations on real numbers.)

Here’s an example that shows the need of extension from the rational number system.

Example 1 - Algebraic Imcompleteness

We now show that

\[\begin{aligned} p^2 &= 2 \end{aligned}\]

is not satisfied by any \(p \in \BB{Q}\).

Example 1 Solution

Proof. Assume such \(p\) exists. Since \(p \in \BB{Q}\), we can denote this as \(p = m/n\) where \(m, n\) are integers that are not both even and \(n \neq 0\).

Rewrite the equation:

\[\begin{aligned} m^2 &= 2 n^2 \end{aligned}\]

This shows that \(m^2\) is even.

Then, our first conjecture would be that \(m\) is even:

If \(m\) is odd, then we can use

\[\begin{aligned} m &= 2k + 1, k \in \BB{N} \\ \implies m^2 &= 4k^2 + 4k + 1 \end{aligned}\]

to show the contrapositive statement (if \(m^2\) is even then \(m\) is even) is true.

It follows that both sides are divisible by \(4\) (if \(m = 2k\) is divisible by 2 then \(m^2 = 4k^2\) is divisible by 4), hence \(n^2\) (divisible by 2) and eventually \(n\) is even (using conjecture above).

It all leads to a contradiction! Hence the equation is not satisfied by any \(p \in \BB{Q}\). \(\blacksquare\)

Example 2 - Analytic Imcompleteness

The set

\[\begin{aligned} A = \left\{p \in \BB{Q} \mid p > 0, p^2 < 2\right\} \end{aligned}\]

contains no largest number.

Example 2 Solution

Proof. Given any \(p \in A\), let

\[\begin{aligned} q &= p - \frac{p^2 - 2}{p + 2} \end{aligned}\]

We first show that \(q > p\) holds when \(p \in A\). Observe that \(p^2 - 2 < 0\) and \(p + 2 > 0\). Therefore \(\frac{p^2 - 2}{p + 2}\) < 0 and \(q > p\) holds.

Now we show that \(q^2 < 2\):

\[\begin{aligned} q^2 - 2 &= (p^2 - \frac{2p(p^2 - 2)}{p+2} + \frac{(p^2 - 2)^2}{(p+2)^2}) - 2 \\ &= \frac{2(p^2 - 2)}{(p+2)^2} < 0 \end{aligned}\]

Thus \(q \in A\) and \(q > p\) always hold.

\(\blacksquare\)

Conceptual Background

We have shown that the real number set \(\BB{R}\) is needed to solve some problems in rational number “calculations”. Nevertheless, we would need some building blocks in order to rigorously define what real number set is.

Set Notations

Let \(A, B\) be sets.

Notation	Explanation
\(A = \emptyset\)	Empty Set: \(A\) has no elements.
\(x \in A\)	\(x\) is an element of \(A\).
\(x \notin A\)	\(x\) is not an element of \(A\).
\(A \subset B\)	\(\forall x \in A \implies x \in B\)
\(A = B\)	\(A \subset B \wedge B \subset A\)

Ordered Sets

This is a basic concept that can been seen anywhere (and is usually assumed to be a common fact). We will show how to define ordered sets rigorously.

Definition - Ordered Sets

Let \(S\) be a set and \(<\) be a binary relation. If

1. \(\forall x, y \in S\), one and only one of the following statements is true:
   • \(x < y\)
   • \(x = y\)
   • \(y < x\);
2. \(\forall x, y, z \in S\), \(x < z\) is true when:
   • \(x < y\) and
   • \(y < z\).

We call \(<\) an order of \(S\) and \((S, <)\) an ordered set.

Boundness

We here define several (upper) bound properties in a rigorous way. Same arguments for lower bound properties.

Definition - Bounds

Let \(E \subset S\). If there is \(\beta \in S\) such that

\[\begin{aligned} \beta \geq x, \quad \forall x \in E, \end{aligned}\]

we say that \(\beta\) is an upper bound of \(E\). Set \(E\) is bounded above by \(\beta\).

Least Upper Bound

Say \(\beta\) is an upper bound of \(E \subset S\). If for any upper bound of \(E\), \(\gamma\), we have

\[\begin{aligned} \beta \leq \gamma, \end{aligned}\]

then \(\beta\) is the least upper bound of set \(E\).

(Another way to define the least upper bound is to say any \(b < \beta\) is not an upper bound of \(E\).)

Supremum/Infimum

The least upper bound of set \(E\) can be also called as the supremum of \(E\):

\[\begin{aligned} \sup E \end{aligned}\]

The greatest lower bound of set \(E\) can be also called as the infimum of \(E\):

\[\begin{aligned} \inf E \end{aligned}\]

Least-Upper-Bound (LUB) Property

This property turns out to be useful in defining and understanding \(\BB{R}\).

Definition - LUB Property

Let \(S\) be an ordered set.

We say \(S\) has LUB property if for any non-empty \(E \subset S\) that has an upper bound, there is:

\[\begin{aligned} \sup E \in S \end{aligned}\]

Using our findings above, we can show that \(\BB{Q}\) does not have the LUB property:

Example: LUB Property and \(\BB{Q}\)

Let

\[\begin{aligned} A = \left\{p \in \BB{Q} \mid p > 0, p^2 < 2 \right\} \end{aligned}\]

The non-empty set \(A \subset \BB{Q}\) does not have a supremum in \(\BB{Q}\).

Symmetry in Least-Upper-Boundness

Theorem

\(S\) has the least-upper-bound property implies that \(S\) also has the greatest-lower-bound property.

This is equivalent to:

Suppose \(S\) is an ordered set with the LUB property. \(B\) is an non-empty subset of \(S\) which is bounded below (in \(S\)).

Let \(L\) be the set of all lower bounds of \(B\).

Then,

\[\begin{aligned} \alpha &= \sup L \end{aligned}\]

exists in \(S\), and \(\alpha = \inf B\). In particular, \(\inf B \in S\).

Proof of Theorem

Proof.

Since \(B\) is bounded below in \(S\), \(L \neq \emptyset\).

Every \(x \in B\) is an upper bound of \(L\), hence \(L\) is bounded above. By the LUB property, \(\exists \alpha = \sup L \in L \subset S\).

Suppose, for the sake of contradiction, that \(\alpha\) is not a lower bound of \(B\). Then \(\exists x \in B\) with \(x < \alpha\).

However, \(x \in B\) is an upper bound of \(L\) (from 1), so \(\alpha \leq x\). This contradicts \(x < \alpha\).

Hence, \(\alpha\) must be a lower bound of \(B\) (\(\alpha \in L\)).

Let \(\gamma\) be any lower bound of \(B\) (\(\gamma \in L\)). Since \(\alpha = \sup L\), we have \(\gamma \leq \alpha\) always true.

Therefore, \(\alpha\) is the greatest element in \(L\). By the definition, \(\alpha = \inf B\).

Since \(\alpha = \sup L \in S\) and \(\alpha = \inf B\), \(S\) also has the greatest-lower-bound property. \(\blacksquare\)

Fields

We’ve learnt rules on additions and multiplications since ages. But how to properly define operations and other related topics? The concept of fields comes in.

Definition - Fields

A field is a set \(F\) with two operations:

• addition (+),
• multiplication (\(\cdot\)),

which satisfy “field axioms” (A)(M)(D).

Field Axioms

(A) Addition

#	Name	Description
A1	Closure	\(x, y \in F \implies x+y \in F\)
A2	Commutativity	\((\forall x,y \in F) \quad x+y = y+x\)
A3	Associativity	\((\forall x,y,z \in F) \quad (x+y)+z = x+(y+z)\)
A4	Identity	\((\exists 0 \in F)(\forall x \in F) \quad 0+x = x\)
A5	Inverse	\((\forall x \in F)(\exists -x \in F)x+(-x) = 0\)

(M) Multiplication

#	Name	Description
M1	Closure	\(x, y \in F \implies xy \in F\)
M2	Commutativity	\((\forall x,y \in F) \quad xy=yx\)
M3	Associativity	\((\forall x,y,z \in F) \quad (xy)z = x(yz)\)
M4	Identity	\((\exists 1 \in F)(\forall x \in F) \quad 1x = x\)
M5	Inverse	\((\forall x \in F)(\exists \, 1/x \in F) \quad x \cdot 1/x = 1\)

(D) The Distributive Law

For all \(x,y,z \in F\):

\[\begin{aligned} x(y+z) &= xy + xz \end{aligned}\]

Consequences of the Field Axioms

• \(x + y = x + z \implies y = z\)
• \(x + y = x \implies y = 0\)
• \(x + y = 0 \implies y = -x\)
• \(-(-x) = x\)