Real Numbers

We know that a field is a commutative division ring.

The Order axiom states that there exists a subset of ${\mathbb{R}}^{+}$ of $\mathbb{R}$ such that for all $a,b\in {\mathbb{R}}^{+},a+b\in {\mathbb{R}}^{+},a\cdot b\in {\mathbb{R}}^{+}.$ and for all $a\in \mathbb{R}$ exactly one of the following is true: (i) $a\in {\mathbb{R}}^{+}$, (ii) $-a\in {\mathbb{R}}^{+}$, or (iii) $a=0$.

A more familiar and equivalent way of stating this can be achieved by defining to mean $b+(-a)\in {\mathbb{R}}^{+}$, and to mean $a+(-b)\in {\mathbb{R}}^{+}$, for $a,b,c\in \mathbb{R}$,

• Exactly one of the following is true: or $a=b$.

• If and , we have

• If and then (transitivity).

Note that rational numbers also satisfy the field axioms and the Order axiom. Complex numbers satisfy the field axioms but not the Order axiom because there is no total ordering that can be defined on complex numbers.

To give the Completeness axiom, we must first define what a least upper bound is.

The Completeness axiom states that every nonempty subset $A$ of $\mathbb{R}$ that's bounded above has a least upper bound.

One consequence of the Completeness axiom is that the real numbers contain numbers that are not rational. For example, the subset of $\mathbb{R}$ that is the finite decimal approximations of $\sqrt{5}$ is,

$$\{2,2.2,2.23,2.236,\dots \},$$

must have a supremum due to the Completeness axiom. That supremum is $\sqrt{5}$, but $\sqrt{5}$ is not rational, therefore, the reals contain non-rational numbers.

The absolute value of a real number is defined as follows:

One fact that follow from this definition is that for $a,b\in \mathbb{R}$, if , then . Proof: We have two cases to consider:

• If , then $\Vert a\Vert =a$, so . Since $a$ is positive, $b$ must also be positive, and therefore $-b$ must be negative, and so .

• If , then $\Vert a\Vert =-a$, so . Since $a$ is negative, $-a$ must be positive, so $b$ must also be positive, and therefore . Now, , and therefore $◻$

Another fact is that the absolute value function is a norm, and thus satisfies the triangle inequality. For real $a,b$:

$$|a+b|\le |a|+|b|$$