2  Real Numbers

2.1 Fields

Definition 1: Binary operation
A binary operation on a set \(K\) is a function \[ \circ \ \colon K \times K \to K \] which maps the ordered pair \((x,y)\) into \(x \circ y\).

Definition 2: Properties of binary operations

Let \(K\) be a set and \(\circ \ \colon K \times K \to K\) be a binary operation on \(K\). We say that:

  1. \(\circ\) is commutative if \[ x \circ y = y \circ x \,, \quad \, \forall \, x,y \in K \]
  2. \(\circ\) is associative if \[ (x \circ y) \circ z = x \circ (y \circ z) \,, \quad \, \forall \, x,y,z \in K \]
  3. An element \(e \in K\) is called neutral element of \(\circ\) if \[ x \circ e = e \circ x = x \,, \quad \, \forall \, x \in K \]
  4. Let \(e\) be a neutral element of \(\circ\) and let \(x \in K\). An element \(y \in K\) is called an inverse of \(x\) with respect to \(\circ\) if \[ x \circ y = y \circ x = e \,. \]
Example 3

Question. Let \(K=\{0,1\}\) be a set with binary operation \(\circ\) defined by the table \[ \begin{array}{c|cc} \circ & 0 & 1 \\ \hline 0 & 1 & 1 \\ 1 & 0 & 0 \\ \end{array} \]

  1. Is \(\circ\) commutative? Justify your answer.

  2. Is \(\circ\) associative? Justify your answer.

Solution.

  1. The operation \(\circ\) is not commutative, since \[ 0 \circ 1 = 1 \neq 0 = 1 \circ 0 \,. \]

  2. The operation \(\circ\) is not associative, since \[ (0 \circ 1) \circ 1 = 1 \circ 1 = 0 \,, \] while \[ 0 \circ (1 \circ 1) = 0 \circ 0 = 1 \,, \] so that \[ (0 \circ 1) \circ 1 \neq 0 \circ (1 \circ 1)\,. \]

2.2 Fields

Definition 4: Field

Let \(K\) be a set with binary operations of addition \[ +\ \colon K \times K \to K \,, \quad (x,y) \mapsto x + y \] and multiplication \[ \cdot\ \colon K \times K \to K \,, \quad (x,y) \mapsto x \cdot y = xy \,. \] We call the triple \((K, + , \cdot)\) a field if:

  1. The addition \(+\) satisfies: \(\,\forall \, x,y,z \in K\)
    • (A1) Commutativity and Associativity: \[ x+y = y+x \] \[ (x+y)+z = x+(y+z) \]
    • (A2) Additive Identity: There exists a neutral element in \(K\) for \(+\), which we call \(0\). It holds: \[ x + 0 = 0 + x = x \]
    • (A3) Additive Inverse: There exists an inverse of \(x\) with respect to \(+\). We call this element the additive inverse of \(x\) and denote it by \(-x\). It holds \[ x + (-x) = (-x) + x = 0 \]
  2. The multiplication \(\cdot\) satisifes: \(\,\forall \, x,y,z \in K\)
    • (M1) Commutativity and Associativity: \[ x \cdot y = y \cdot x \] \[ (x \cdot y) \cdot z = x \cdot (y \cdot z) \]
    • (M2) Multiplicative Identity: There exists a neutral element in \(K\) for \(\cdot\), which we call \(1\). It holds: \[ x \cdot 1 = 1 \cdot x = x \]
    • (M3) Multiplicative Inverse: If \(x \neq 0\) there exists an inverse of \(x\) with respect to \(\cdot\). We call this element the multiplicative inverse of \(x\) and denote it by \(x^{-1}\). It holds \[ x \cdot x^{-1} = x^{-1} \cdot x = 1 \]
  3. The operations \(+\) and \(\cdot\) are related by
    • (AM) Distributive Property: \(\,\forall \, x,y,z \in K\) \[ x \cdot (y + z) = (x \cdot y) + (y \cdot z) \,. \]

Theorem 5
Let \(K\) with \(+\) and \(\cdot\) defined by \[ \begin{array}{c|cc} + & 0 & 1 \\ \hline 0 & 0 & 1 \\ 1 & 1 & 0 \\ \end{array} \qquad \qquad \begin{array}{c|cc} \cdot & 0 & 1 \\ \hline 0 & 0 & 0 \\ 1 & 0 & 1 \\ \end{array} \] Then \((K,+,\cdot)\) is a field.

Definition 6: Subtraction and division

Let \((K,+,\cdot)\) be a field. We define:

  1. Subtraction as the operation \(-\) defined by \[ x - y := x + (-y) \,, \quad \forall \, x , y \in K \,, \] where \(-y\) is the additive inverse of \(y\).

  2. Division as the operation \(/\) defined by \[ x/y := x \cdot y^{-1}\,, \quad \forall \, x , y \in K \,, \,\, y \neq 0 \,, \] where \(y^{-1}\) is the multiplicative inverse of \(y\).

Proposition 7: Uniqueness of neutral elements and inverses

Let \((K,+,\cdot)\) be a field. Then

  1. There is a unique element in \(K\) with the property of \(0\).
  2. There is a unique element in \(K\) with the property of \(1\).
  3. For all \(x \in K\) there is a unique additive inverse \(-x\).
  4. For all \(x \in K\), \(x \neq 0\), there is a unique multiplicative inverse \(x^{-1}\).
Proof
  1. Suppose that \(0 \in K\) and \(\widetilde{0} \in K\) are both neutral element of \(+\), that is, they both satisfy (A2). Then \[ 0 + \widetilde{0} = 0 \] since \(\widetilde{0}\) is a neutral element for \(+\). Moreover \[ \widetilde{0} + 0 = \widetilde{0} \] since \(0\) is a neutral element for \(+\). By commutativity of \(+\), see property (A1), we have \[ 0 = 0 + \widetilde{0} = \widetilde{0} + 0 = \widetilde{0} \,, \] showing that \(0 = \widetilde{0}\). Hence the neutral element for \(+\) is unique.
  2. Exercise.
  3. Let \(x \in K\) and suppose that \(y, \widetilde{y} \in K\) are both additive inverses of \(x\), that is, they both satisfy (A3). Therefore \[ x + y = 0 \] since \(y\) is an additive inverse of \(x\) and \[ x + \widetilde{y} = 0 \] since \(\widetilde{y}\) is an additive inverse of \(x\). Therefore we can use commutativity and associativity and of \(+\), see property (A1), and the fact that \(0\) is the neutral element of \(+\), to infer \[\begin{align*} y & = y + 0 = y + (x + \widetilde{y}) \\ & = (y + x) + \widetilde{y} = (x + y) + \widetilde{y} \\ & = 0 + \widetilde{y} = \widetilde{y} \,, \end{align*}\] concluding that \(y = \widetilde{y}\). Thus there is a unique additive inverse of \(x\), and \[ y = \widetilde{y} = -x \,, \] with \(-x\) the element from property (A3).
  4. Exercise.
Theorem 8

Consider the sets \(\mathbb{N}\), \(\mathbb{Z}\), \(\mathbb{Q}\) with the usual operations \(+\) and \(\cdot\). We have:

  • \((\mathbb{N}, + , \cdot)\) is not a field.

  • \((\mathbb{Z}, + , \cdot)\) is not a field.

  • \((\mathbb{Q}, + , \cdot)\) is a field.

2.3 Ordered fields

Definition 9

Let \(K\) be a set with binary operations \(+\) and \(\cdot\), and with an order relation \(\leq\). We call \((K,+,\cdot,\leq)\) an ordered field if:

  1. \((K,+,\cdot)\) is a field

  2. There \(\leq\) is of total order on \(K\): \(\, \forall \, x, y, z \in K\)

    • (O1) Reflexivity: \[ x \leq x \]
    • (O2) Antisymmetry: \[ x \leq y \, \mbox{ and } \, y \leq x \,\, \implies \,\, x = y \]
    • (O3) Transitivity: \[ x \leq y \,\, \mbox{ and } \,\, y \leq z \,\, \implies \,\, x = z \]
    • (O4) Total order:
      \[ x \leq y \,\, \mbox{ or } \,\, y \leq x \]

  3. The operations \(+\) and \(\cdot\), and the total order \(\leq\), are related by the following properties: \(\, \forall x, y, z \in K\)

    • (AM) Distributive: Relates addition and multiplication via \[ x \cdot (y + z) = x \cdot y + x \cdot z \]
    • (AO) Relates addition and order with the requirement: \[ x \leq y \,\, \implies \,\, x + z \leq y + z \]
    • (MO) Relates multiplication and order with the requirement: \[ x \geq 0, \, y \geq 0 \,\, \implies \,\, x \cdot y \geq 0 \]

Theorem 10
\((\mathbb{Q},+,\cdot,\leq)\) is an ordered field.

2.4 Supremum and infimum

In the following we assume that \((K,+,\cdot,\leq)\) is an ordered field.

Definition 11: Upper bound and bounded above

Let \(A \subseteq K\):

  1. We say that \(b \in K\) is an upper bound for \(A\) if \[ a \leq b \,, \quad \forall \, a \in A \,. \]
  2. We say that \(A\) is bounded above if there exists and upper bound \(b \in K\) for \(A\).

Definition 12: Supremum
Let \(A \subseteq K\). A number \(s \in K\) is called least upper bound or supremum of \(A\) if:

  1. \(s\) is an upper bound for \(A\),
  2. \(s\) is the smallest upper bound of \(A\), that is, \[ \mbox{If } \, b \in K \, \mbox{ is upper bound for } \, A \, \mbox{ then } \, s \leq b \,. \]

If it exists, the supremum is denoted by \[ s := \sup \ A \,. \]

Remark 13
Note that if a set \(A \subseteq K\) in NOT bounded above, then the supremum does not exist, as there are no upper bounds of \(A\).

Proposition 14: Uniqueness of the supremum
Let \(A \subseteq K\). If \(\sup A\) exists, then it is unique.

Definition 15: Maximum
Let \(A \subseteq K\). A number \(M \in K\) is called the maximum of \(A\) if: \[ M \in A \,\, \mbox{ and } \,\, a \leq M \,, \, \forall a \in A \,. \] If it exists, we denote the maximum by \[ M = \max A \,. \]

Proposition 16: Relationship between Max and Sup
Let \(A \subseteq K\). If the maximum of \(A\) exists, then also the supremum exists, and \[ \sup A = \max A \,. \]

Definition 17: Lower bound, bounded below, infimum, minimum

Let \(A \subseteq K\):

  1. We say that \(l \in K\) is a lower bound for \(A\) if \[ l \leq a \,, \quad \forall \, a \in A \,. \]

  2. We say that \(A\) is bounded below if there exists a lower bound \(l \in K\) for \(A\).

  3. We say that \(i \in K\) is the greatest lower bound or infimum of \(A\) if:

    • \(i\) is a lower bound for \(A\),
    • \(i\) is the largest lower bound of \(A\), that is, \[ \mbox{If } \, l \in K \, \mbox{ is a lower bound for } \, A \, \mbox{ then } \, l \leq i \,. \] If it exists, the infimum is denoted by \[ i = \inf A \,. \]

  4. We say that \(m \in K\) is the minimum of \(A\) if: \[ m \in A \,\, \mbox{ and } \,\, m \leq a \,, \, \forall a \in A \,. \] If it exists, we denote the minimum by \[ m = \min A \,. \]

Proposition 18

Let \(A \subseteq K\):

  1. If \(\inf A\) exists, then it is unique.
  2. If the minimum of \(A\) exists, then also the infimum exists, and \[ \inf A = \min A \,. \]

Proposition 19
Let \(A \subseteq K\). If \(\inf A\) and \(\sup A\) exist, then \[ \inf A \leq a \leq \sup A \,, \quad \forall \, a \in A \,. \]

Proposition 20: Relationship between sup and inf

Let \(A \subseteq K\). Define \[ - A := \{ - a \, \colon \,a \in A \} \,. \] They hold

  1. If \(\sup A\) exists, then \(\inf A\) exists and \[ \inf(-A) = - \sup A \,. \]
  2. If \(\inf A\) exists, then \(\sup A\) exists and \[ \sup(-A) = - \inf A \,. \]

2.5 Axioms of Real Numbers

Definition 21: Completeness

Let \((K,+,\cdot,\leq)\) be an ordered field. We say that \(K\) is complete if the following property holds:

  • (AC) For every \(A \subseteq K\) non-empty and bounded above \[ \sup A \in K \,. \]

Theorem 22
\(\mathbb{Q}\) is not complete. In particular, there exists a set \(A \subseteq \mathbb{Q}\) such that

  • \(A\) is non-empty,
  • \(A\) is bounded above,
  • \(\sup A\) does not exist in \(\mathbb{Q}\).

One of such sets is, for example, \[ A = \{ q \in \mathbb{Q}\, \colon \,q \geq 0 \,, \,\, q^2 < 2 \} \,. \]

Proposition 23
Let \((K,+,\cdot,\leq)\) be a complete ordered field. Suppose that \(A \subseteq K\) is non-empty and bounded below. Then \[ \inf A \in K\,. \]

Definition 24: System of Real Numbers \(\mathbb{R}\)

A system of Real Numbers is a set \(\mathbb{R}\) with two operations \(+\) and \(\cdot\), and a total order relation \(\leq\), such that

  • \((\mathbb{R},+,\cdot, \leq)\) is an ordered field

  • \(\mathbb{R}\) sastisfies the Axiom of Completeness

2.6 Inductive sets

Definition 25: Inductive set

Let \(S \subseteq \mathbb{R}\). We say that \(S\) is an inductive set if they are satisfied:

  • \(1 \in S\),
  • If \(x \in S\), then \((x + 1) \in S\).
Example 26

Question. Prove the following:

  1. \(\mathbb{R}\) is an inductive set.

  2. The set \(A=\{0,1\}\) is not an inductive set.

Solution.

  1. We have that \(1 \in \mathbb{R}\) by axiom (M2). Moreover \((x + 1) \in \mathbb{R}\) for every \(x \in \mathbb{R}\), by definition of sum \(+\).

  2. We have \(1 \in A\), but \((1 + 1) \notin A\), since \(1 + 1 \neq 0\).

Proposition 27
Let \(\mathcal{M}\) be a collection of inductive subsets of \(\mathbb{R}\). Then \[ S := \bigcap_{M \in \mathcal{M}} \, M \] is an inductive subset of \(\mathbb{R}\).

Definition 28: Set of Natural Numbers
Let \(\mathcal{M}\) be the collection of all inductive subsets of \(\mathbb{R}\). We define the set of natural numbers in \(\mathbb{R}\) as \[ \mathbb{N}:= \bigcap_{M \in \mathcal{M}} \, M \,. \]

Proposition 29: \({\mathbb{N}}_{\mathbb{R}}\) is the smallest inductive subset of \(\mathbb{R}\)
Let \(C \subseteq \mathbb{R}\) be an inductive subset. Then \[ \mathbb{N}\subseteq C \,. \] In other words, \(\mathbb{N}\) is the smallest inductive set in \(\mathbb{R}\).

Theorem 30
Let \(x \in \mathbb{N}\). Then \[ x \geq 1 \,. \]