We always have |a| ≥ 0. We can change the sign "within" the absolute value. Often we make use of the symmetry |a| = |-a| and therefore also of |a - b| = |b - a|. If a and b are both positive or both negative then their product a · b > 0 and we have |a · b| = |a| · |b|. If a and b have different signs, the a · b < 0 and again we have |a · b| = |a| · |b|, also if a or b or both are zero, we always have |a · b| = |a| · |b| and |a/b| = |a|/|b|, if b ≠ 0.

Proof. If a ≥ 0 then |a| = a and -a ≤ 0, and thus is -a ≤ a ≤ |a|. If a ≤ 0 then |a| = -a and -a ≥ 0, and thus is a ≤ -a ≤ |a|. ☐

Proof. (i) ⇒: From |x| ≤ ε and 1.) we have that x ≤ |x| ≤ ε and -x ≤ |x| ≤ ε. From the last inequality it follows that -ε ≤ x and thus -ε ≤ x ≤ ε.

(ii) ⇐: Supposing -ε ≤ x ≤ ε. Mutiplying by -1 we get also ε ≥ -x ≥ -ε. Is x ≥ 0 then is |x| = x und thus |x| ≤ ε. If x ≤ 0 then is |x| = -x and it follows likewise that |x| ≤ ε. ☐

**3**.) The *triangle inequality*: For all x, y ∈ **R**, |x + y| ≤ |x| + |y|.

Proof. Because of 1.) we have x ≤ |x| and y ≤ |y|. By Addition
we obtain x + y ≤ |x| + |y|. In the same manner we obtain -x - y ≤ |-x| + |-y|, i. e., -(x + y) ≤ |x| + |y|. Multiplying the last inequality
by -1, we get also x + y ≥ -(|x| + |y|). So we have -(|x| + |y|) ≤ x + y ≤ |x| + |y| and making use of 2.) it follows |x + y| ≤ |x| + |y|. ☐

**4**.) The *reverse triangle inequality*: |x - y| ≥ ||x| - |y||.

Proof. In the triangle inequality we plug in the number x - y intstead of x and obtain
|x - y + y| ≤ |x - y| + |y| or |x| - |y| ≤ |x - y|. By interchanging x and y in the last inequality, we obtain |y| - |x| ≤ |y - x|,
i. e., -(|x| - |y|) ≤ |x - y|, since |y - x| = |x - y|. This is equivalent to |x| - |y| ≥ -|x - y|. Thus we have because of -|x - y| ≤ |x| - |y| ≤ |x - y| and
2.), that |ΙxΙ - ΙyΙ| ≤ |x - y| or |x - y| ≥ |ΙxΙ - ΙyΙ|. ☐

Remark: Alternative method to show **3**.): From 1.) we conclude -|x| ≤ x ≤ |x| and likewise -|y| ≤ y ≤ |y|. Adding these two inequalities leads us to
-(|x| + |y|) ≤ x + y ≤ |x| + |y| and thus according to 2.), to |x + y| ≤ |x| + |y|. ☐

Only on account of having a negative sign, an algebraic expression need not be negative;
(-(-x) = x). Sometimes the signum function comes in useful: sign(x) = 1 if x > 0, sign(x) = -1 if x < 0 und sign(x) = 0 if x = 0.

For every number x we have: x = |x| · sign(x).

In chapter 7 we found the structure (**R**, +, ⋅) to be a field. Both operations (**R** x **R**) → **R** (addition and multiplication) are both commutative and associative, multiplication is
distributive with regard to addition, for each operation there exists an identity element, for every element in **R** there exists an additive inverse element and for each element except zero there exists a
multiplicative inverse element.
Reminder:

**The field axioms**

Let *a, b, c* be real numbers.

**A1**. *a + b = b + a* and *a⋅ b = b⋅ a*. - (Commutativity)

**A2**. *a + (b + c) = (a + b) + c* and *a⋅ (b⋅ c) = (a⋅ b)⋅ c*. - (Associativity)

**A3**. *a⋅ (b + c) = a⋅ b + a⋅ c*. - (Multiplication is distributive with regard to addition)

**A4**. *ꓱ 0, 1 ∈ R, 0 ≠ 1 | ∀ a ∈ R a + 0 = a* and

The field axioms A1 to A5 are also satisfied by the complex numbers.

With a ninth axiom, the field

Let a, b ∈

The set {x ∈

x

The intervals with center x

A sequence of closed intervals I

A set Ω of open intervals is called an

This characterisation A9 of the real numbers (it's actually the Borel-Lebesgue theorem about compactness) we were taught at the Universiteit van Suid-Afrika by Hanno Rund, who was also leading in writing [60], wherein completeness is introduced in the same way. This wiskunde (mathematics) book for undergraduates was published in 1970. It was the first book written in Afrikaans (in the past also called cape Dutch or colonial Dutch), in the new spirit of 20

Here are some more propositions which are equivalent to the completeness axiom A9:

⋅ A9a. The supremum principle: Every non-empty set of real numbers which is bounded from above has a least upper bound.

⋅ A9b. Every Dedekind cut has one and only one dividing number.

⋅ A9c. The principle of nested intervals.

⋅ A9d. Every Cauchy sequence converges.

Remark: Whereas the rational numbers are enumerable (countably infinite), the real numbers are non-denumerable (uncountably infinite). According to the

We want to demonstrate the equivalence of A9 and the supremum principle A9a. For doing so, we need again some new tools and definitions.

Let

If there is a real number

If there is a real number

If

If

If

If

We must resort to infimum and/or supremum if a set has no minimum and/or no maximum. (Open or half-open intervals.)

A mathematician needs imagination, pencil and paper as well as ideas and methods to prove or disprove other ideas which jumped off his imagination or which are asserted by others.

Starting from the true statement A we show by algebraic rearrangement and inference that statement B follows from statement A.

In order to prove the assertion A, we assume A to be false and try to reach a contradiction. This method of proof is also called

Example: in A4 we have ∃ 1 ∈

A special case of an indirect proof is by

Instead of A ⇒ B we show ¬B ⇒ ¬A. (B is necessary for A).

(i) The assertion A(k) is true;

(ii) if A(n) is true, then A(n+1) is true.

We conclude that A(n) is true for all integers n ≥ k.

Give an example that contradicts the assumption.

Before we prove the equivalence of A9 and A9a, we need a formal characterisation of the notion supremum.

(i) x ≤ γ for all x ϵ

(ii) for every real number ε > 0 there exists an x ϵ

A ⇒ B(i): If γ = sup

A ⇒ B(ii): Suppose ¬B(ii) holds, i. e. there exists ε' > 0 such that for every x ϵ

B(i) ⇒ A: ∀ x ∈ X, x ≤ γ ⇒ γ ist eine obere Schranke von X gemäß Definition 8.2.

B(ii) ⇒ A: wir gehen von ¬A aus und führen einen Widerspruch herbei, indem wir B(ii) als wahr erachten: Angenommen γ', (γ' < γ), ist die
kleinste obere Schranke. Wir wählen ε = γ – γ' > 0. Dann, gemäß B(ii), ∃ x ∈ X | γ – ε = γ - (γ – γ') = γ' < x.
Das würde bedeuten, dass γ', im Gegensatz zu unserer Annahme, keine obere Schranke für X darstellt. ❑

B(ii) ⇒ A: we start out from ¬A and cause a contradiction by deeming (ii) to be true: Suppose γ', (γ' < γ), is the least upper bound. We choose ε = γ – γ' > 0. Thus, by B(ii), ∃ x ∈

Reminder:

This is the crucial point where analysis begins, and it is perhaps not immediately visible, what it's all about. So let's drop some more words on the situation. A non-empty set X contains at least one element x, that is a single number. If this is the case, x is both the maximim and supremum (also the minimum and the infimum) of X. In the case of a discrete set there exists also a maximum, and the same is true for an interval which is closed above. In any of these instances there is a least upper bound. So what we have to prove is actually, what happens when I demarcate a set X from the set of real numbers, without specifying a greatest element for X, like an interval which is open above. What happens at the boundary between the two ares? Is there a least upper bound for X? How can the seemingly dubious axiom A9 be applied to prove that X has a supremum?

Let's begin by assuming ¬A9a, that there is no least upper bound and try to derive a contradiction. At a distance > 0 on the upper side of X we choose an upper bound b of X. Then we choose any a ∈

P = {y | y ∈ [a, b] and y is no upper bound of

Q = {y | y ∈ [a, b] and y is an upper bound of

We have P ∪ Q = [a, b] and P ∩ Q = ∅. We construct an open cover of [a, b] by building an open interval about each y of [a, b]. For the extreme boundary poinst we take any a

(i) For y ∈ P we form the interval I(y) = (a

(ii) For y ∈ Q we form the interval J(y) = (q(y), b

The set of the open intervals I(y) and J(y) forms a cover of [a, b]. According to A9 there exists a finite subcover, say

{I(y

In view of J(y

Remark: The supremum of X, call this number c ∈ Q, is the minimum of Q. For c you cannot find a p(c) < c ∈ Q, so it is absurd to choose an interval (q(c), b

Suppose we have an infinite set Ω of open intervals that covers [a, b]. For the purpose of this proof we shall call a subset Ω' ⊂ Ω "admissible", if it meets the following two requirements:

(a) Ω' consists of only finitely many intervals I

(b) ⋃

There exist admissible subsets of Ω; For example, every interval I ∈ Ω for which a ∈ I, is admissible. If now Ω' = {I

We assume now, that c(Ω') ≤ b for every admissible Ω' and prove thereupon, that this assumption leads to a contradiction.

If c(Ω') ≤ b for every admissible Ω', then also r := sup{c(Ω')} ≤ b. According to our assumption A9a this supremum exists. Since r ∈ [a, b], there exists an interval ω ∈ Ω such that r ∈ ω. We choose now an admissible Ω'' such that c(Ω'') ∈ ω. This is always possible because of theorm 8.4 (ii). But the set Ω''' which consists of the intervals of Ω'' together with the interval ω, is also admissible and therefore holds that c(Ω''') > r, because r ∈ ω. This is in conflict with the assumption that c(Ω') ≤ r for every admissible Ω'. This completes the prove for A9 ⇔ A9a. ♫

We are now going to construct an infinite cover of the closed interval [1, 2] of

For every natural number n we form the open interval I

For every natural number n we form the open interval J

Now we prove that the two sets of intervals

{I

{J

We observe that a

Suppose now, that α ∈ [1, 2] is not covered, i. e. a

(a

and

α

Since we also have 2 - (a

α

Now is (b

This is only possible if α

α

From (1) follows furthermore that 2 - α

Since (b

2 - α

We have thus that 2 - α

α

From (4) and (6) it follows that

α

This is in conflict with the fact, that there is no rational number α which satisfies this equation. Thus, the sets of intervals {I

{I

r