Question

Q. Prove that set of reals ℝ is an ordered field.

Answer 1

Answer on Question #63839 – Math - Real Analysis

Question

Prove that set of reals $\mathbb{R}$ is an ordered field.

Solution

**Axioms of addition.** For any two real numbers $a, b$, there is an operation of addition which associates their sum denoted by $a + b$. The operation of addition satisfies the following axioms:

A1. (Associativity) $(x + y) + z = x + (y + z)$.

A2. (Existence of zero) There is a real number, called zero and denoted by 0, such that

$$x + 0 = 0 + x = x \text{ for all real numbers } x.$$

A3. (Existence of negative) For every $x \in \mathbb{R}$ there is $y \in \mathbb{R}$ such that

$$x + y = y + x = 0. \text{ The number } y \text{ is called the negative of } x \text{ and denoted by } -x.$$

A4. (Commutativity) $x + y = y + x$.

Hence $\mathbb{R}$ with the operation of addition is a commutative group.

**Axioms of multiplication.** There is an operation of multiplication which associates with any two real numbers $x, y$, the number $x \cdot y$. It satisfies the following axioms:

M1. (Associativity) $(x \cdot y) \cdot = x \cdot (y \cdot z)$.

M2. (Existence of identity) There is a real number, called identity and denoted by 1, such that $1 \neq 0$ and $1 \cdot x = x \cdot 1 = x$ for all real numbers $x$.

M3. (Existence of reciprocal) For every $x \in \mathbb{R} \setminus \{0\}$ there is $y \in \mathbb{R}$ such that $x \cdot y = y \cdot x = 1$. (the number $y$ is called the reciprocal of $x$ and denoted by $x - 1$ or $1x$).

M4. (Commutativity) $x \cdot y = y \cdot x$.

Hence $\mathbb{R} \setminus \{0\}$ together with the multiplication is a commutative group.

**The distributive law.** The next axiom defines the relation between the two operations.

$$D. x \cdot (y + z) = x \cdot y + x \cdot z$$

All computational rules for real numbers follow from the axioms A1–A4, M1–M4, D.

**Order** There is a subset $P$ of $\mathbb{R}$, called the set of positive numbers, satisfying the following two axioms:

O1. If $x, y \in P$, then $x + y$ and $x \cdot y \in P$.

O2. For every $x \in \mathbb{R}$, either $x \in P$ or $x = 0$ or $-x \in P$.

The axioms O1 and O2 imply that 1 is a positive number. Indeed, since $1 \neq 0$, the axiom O1 implies that either 1 or -1 is positive. Since $1 = 1 \cdot 1 = (-1) \cdot (-1)$, the axiom O1 implies that 1 is positive.

A set X with two operations, addition and multiplication, which satisfy axioms A1–A4, M1-M4, D, and O1-O2 is called an ordered field. The set of real numbers $\mathbb{R}$ is an ordered field.

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