2  Real Numbers

2.1 Fields

Definition 1: Binary operation

A binary operation on a set $K$ is a function $$\circ :K\times K\to K$$ which maps the ordered pair $(x,y)$ into $x\circ y$.

Definition 2: Properties of binary operations

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

1. $\circ$ is commutative if $$x\circ y=y\circ x,\mathrm{\forall }x,y\in K$$
2. $\circ$ is associative if $$(x\circ y)\circ z=x\circ (y\circ z),\mathrm{\forall }x,y,z\in K$$
3. An element $e\in K$ is called neutral element of $\circ$ if $$x\circ e=e\circ x=x,\mathrm{\forall }x\in K$$
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 $$\begin{array}{ccc}\circ & 0& 1\\ 0& 1& 1\\ 1& 0& 0\end{array}$$

1. Is $\circ$ commutative? Justify your answer.

2. Is $\circ$ associative? Justify your answer.

Solution.

1. The operation $\circ$ is not commutative, since $$0\circ 1=1\ne 0=1\circ 0.$$

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 $$(0\circ 1)\circ 1\ne 0\circ (1\circ 1).$$

2.2 Fields

Definition 4: Field

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

1. The addition $+$ satisfies: $\mathrm{\forall }x,y,z\in K$
   • (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: $\mathrm{\forall }x,y,z\in K$
   • (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 $x\ne 0$ 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: $\mathrm{\forall }x,y,z\in K$ $$x\cdot (y+z)=(x\cdot y)+(y\cdot z).$$

Theorem 5

Let $K$ with $+$ and $\cdot$ defined by $$\begin{array}{ccc}+& 0& 1\\ 0& 0& 1\\ 1& 1& 0\end{array}\begin{array}{ccc}\cdot & 0& 1\\ 0& 0& 0\\ 1& 0& 1\end{array}$$ 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 $$x-y:=x+(-y),\mathrm{\forall }x,y\in K,$$ where $-y$ is the additive inverse of $y$.

2. Division as the operation $/$ defined by $$x/y:=x\cdot y^{-1},\mathrm{\forall }x,y\in K,y\ne 0,$$ 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 $x\in K$, $x\ne 0$, there is a unique multiplicative inverse $x^{-1}$.

Proof

1. Suppose that $0\in K$ and $\tilde{0}\in K$ are both neutral element of $+$, that is, they both satisfy (A2). Then $$0+\tilde{0}=0$$ since $\tilde{0}$ is a neutral element for $+$. Moreover $$\tilde{0}+0=\tilde{0}$$ since $0$ is a neutral element for $+$. By commutativity of $+$, see property (A1), we have $$0=0+\tilde{0}=\tilde{0}+0=\tilde{0},$$ showing that $0=\tilde{0}$. Hence the neutral element for $+$ is unique.
2. Exercise.
3. Let $x\in K$ and suppose that $y,\tilde{y}\in K$ 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 $$x+\tilde{y}=0$$ since $\tilde{y}$ 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 $$\begin{array}{rl}y& =y+0=y+(x+\tilde{y})\\ & =(y+x)+\tilde{y}=(x+y)+\tilde{y}\\ & =0+\tilde{y}=\tilde{y},\end{array}$$ concluding that $y=\tilde{y}$. Thus there is a unique additive inverse of $x$, and $$y=\tilde{y}=-x,$$ 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 $\le$. We call $(K,+,\cdot ,\le )$ an ordered field if:

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

2. There $\le$ is of total order on $K$: $\mathrm{\forall }x,y,z\in K$

   • (O1) Reflexivity: $$x\le x$$
   • (O2) Antisymmetry: $$x\le y\text{ and }y\le x⟹x=y$$
   • (O3) Transitivity: $$x\le y\text{ and }y\le z⟹x=z$$
   • (O4) Total order:
     $$x\le y\text{ or }y\le x$$

3. The operations $+$ and $\cdot$, and the total order $\le$, are related by the following properties: $\mathrm{\forall }x,y,z\in K$

   • (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: $$x\le y⟹x+z\le y+z$$
   • (MO) Relates multiplication and order with the requirement: $$x\ge 0,y\ge 0⟹x\cdot y\ge 0$$

Theorem 10

$(\mathbb{Q},+,\cdot ,\le )$ is an ordered field.

2.4 Supremum and infimum

In the following we assume that $(K,+,\cdot ,\le )$ 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 $$a\le b,\mathrm{\forall }a\in A.$$
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, $$\text{If }b\in K\text{ is upper bound for }A\text{ then }s\le b.$$

If it exists, the supremum is denoted by $$s:=supA.$$

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 $supA$ exists, then it is unique.

Definition 15: Maximum

Let $A\subseteq K$. A number $M\in K$ is called the maximum of $A$ if: $$M\in A\text{ and }a\le M,\mathrm{\forall }a\in A.$$ If it exists, we denote the maximum by $$M=maxA.$$

Proposition 16: Relationship between Max and Sup

Let $A\subseteq K$. If the maximum of $A$ exists, then also the supremum exists, and $$supA=maxA.$$

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 $$l\le a,\mathrm{\forall }a\in A.$$

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, $$\text{If }l\in K\text{ is a lower bound for }A\text{ then }l\le i.$$ If it exists, the infimum is denoted by $$i=infA.$$

4. We say that $m\in K$ is the minimum of $A$ if: $$m\in A\text{ and }m\le a,\mathrm{\forall }a\in A.$$ If it exists, we denote the minimum by $$m=minA.$$

Proposition 18

Let $A\subseteq K$:

1. If $infA$ exists, then it is unique.
2. If the minimum of $A$ exists, then also the infimum exists, and $$infA=minA.$$

Proposition 19

Let $A\subseteq K$. If $infA$ and $supA$ exist, then $$infA\le a\le supA,\mathrm{\forall }a\in A.$$

Proposition 20: Relationship between sup and inf

Let $A\subseteq K$. Define $$-A:=\{-a:a\in A\}.$$ They hold

1. If $supA$ exists, then $infA$ exists and $$inf(-A)=-supA.$$
2. If $infA$ exists, then $supA$ exists and $$sup(-A)=-infA.$$

2.5 Axioms of Real Numbers

Definition 21: Completeness

Let $(K,+,\cdot ,\le )$ 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 $$supA\in K.$$

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,
• $supA$ does not exist in $\mathbb{Q}$.

One of such sets is, for example, $$A=\{q\in \mathbb{Q}:q\ge 0,q^2<2\}.$$

Proposition 23

Let $(K,+,\cdot ,\le )$ be a complete ordered field. Suppose that $A\subseteq K$ is non-empty and bounded below. Then $$infA\in K.$$

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 $\le$, such that

• $(\mathbb{R},+,\cdot ,\le )$ 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 $1+1\ne 0$.

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 $$x\ge 1.$$