Axioms of Real Number System

In this article, we shall look at some very basic ideas about the Real Analysis, i.e. the study of the structure of Real Number System. We shall discuss the three axioms that are considered to be satisfied by the set of Real Numbers, R

The three axioms are : 

1. Field Axioms
2. Order Axioms
3. Completeness Axiom

Field Axioms : The set R    is represented as a field (R, +, .)     where +    and .   are the binary operations of addition and multiplication respectively. It consists of 4 axioms for addition and multiplication each and one distributive law.

(i) Axioms for addition : 

• a+b = b+a \ ∀ \ a,b \ ∈ R
• (a+b)+c = a+(b+c) \ ∀ \ a,b,c \ ∈ R
• R contains an element 0 such that a + 0 = a \ ∀ \ a ∈ R 
• For each a ∈ R    there corresponds an element -a ∈ R    such that a+(-a) = 0

(ii) Axioms for multiplication :

• ab = ba \ ∀\ a,b  ∈ R
• (ab)c = a(bc) \ ∀ \ a,b,c  ∈ R
• R    contains an element 1    such that  1.a = a \ ∀ \ a ∈ R    and  1 ≠ 0
• If a ∈ R \ and \ a≠0    then there exists an element \frac{1}{a} ∈ R     such that a. (\frac{1}{a} ) = 1

(iii) The distributive law :

• a(b+c) = ab+ac \ ∀ \ a,b,c ∈ R

Order Axioms : We define >    (Greater Than) as the order relation, and it satisfies the following axioms -

• Law of Trichotomy - For a,b∈ R    only one of the expressions can be true : a>b , a=b , b>a
• Transitivity - For a,b,c∈R \ a>b,\ b>c ⇒ a>c
• Monotone Property for addition - For a,b,c∈R, \ a>b ⇒ a+c > b+c 
• Monotone Property for multiplication - For a,b,c∈R,\ a>b, c>0 ⇒ ac > bc

We call >   linear order and R    is called a linearly ordered field.

Before defining the Completeness Axiom, we shall look at the concept of Boundedness. Here, we shall define a few terms before stating the Completeness Axiom.

Aggregate : Any non empty subset, say A   , of R   is known as an aggregate. For example, the set Z^+   is an aggregate. Similarly, the set B = {1,2,4,8} is also an aggregate since B ⊆ R   But, the set A = {x,y,z} and the empty set ∅   are not aggregates.

Upper bound : A subset S   of R   is said to be bounded above if ∃ \ k_1\ ∈ R   such that x ∈ S ⇒ x \leq k_1  . This number k_1   is called an upper bound of S  . For example, the set R^-    of negative real numbers is bounded above and 0   is an upper bound. Similarly, the set Z^-    of negative integers is bounded above and -1   is the upper bound. But, the set R^+   of positive real numbers is not bounded above.

Lower Bound : A subset S   of R   is said to be bounded below if ∃ \ k_2\ ∈ R   such that x ∈ S ⇒ x \geq k_2   This number k_2   is called a lower bound of S. For example, the set R^+   is bounded below and 0   is a lower bound. Similarly, the set Z^+   is bounded below and 1   is the upper bound. But, the set R^-   is not bounded below.

Least Upper Bound : Consider an upper bound u   of an aggregate S   and any real number less than u   is not an upper bound of S  , then we say u   is the least upper bound(lub) or supremum(sup) of S.

Greatest Lower bound : Consider a lower bound v   of an aggregate S   and any real number greater than v   is not a lower bound of S  , then we say v   is the greatest lower bound(glb) or infimum(inf) of S.

Example : Let S = [0,1]  . For S, we see that 1 is an upper bound and any number less than 1 is not an upper bound of S, hence, 1 is supremum of S. Also, 0 is a lower bound and any number greater than 0 is not a lower bound, so, 0 is infimum of S.

Boundedness : An aggregate S is bounded if it is both bounded above and bounded below. That is, it must have both an upper bound and a lower bound. For example, any finite set is bounded, the empty set ∅   is bounded. But, the sets Q   and R   are not bounded.

Note : An aggregate need not have a greatest and a least member to be bounded above or bounded below respectively.

Now being done with the required definition, we state the Completeness Axiom(also called the least upper bound axiom).

" Every non-empty set of Real numbers which is bounded above has a supremum."

The set R satisfies the Field Axioms, Order Axioms, and the Completeness Axiom. Hence the set of real numbers R   is called a complete ordered field.

Also, the set of rational numbers, Q   does not satisfy the completeness axiom. Hence, Q   is not a complete field.

The completeness axiom is a really fundamental and important property of real number systems, as proofs various theorems of calculus, the concepts of maxima and minima, mean-value theorems etc. rely on the completeness property of real numbers.

Solved Examples :

1).Prove that for any real numbers a and b, (a + b)³ = a³ + 3a²b + 3ab² + b³.

(a + b)³ = a³ + 3a²b + 3ab² + b³

Solution: Use the distributive property repeatedly:

(a + b)³ = (a + b)(a + b)(a + b)

= (a + b)(a² + 2ab + b²)

= a³ + 2a²b + ab² + a²b + 2ab² + b³

= a³ + 3a²b + 3ab² + b³

2).Show that if a, b, and c are real numbers and a ≠ 0, then the equation ax² + bx + c = 0 has at most two distinct real solutions.

2).Show that if a, b, and c are real numbers and a ≠ 0, then the equation ax² + bx + c = 0 has at most two distinct real solutions.

ax² + bx + c = 0 has at most two distinct real solutions (a ≠ 0)

Solution outline: Use the quadratic formula. The discriminant (b² - 4ac) determines the nature of solutions:

If b² - 4ac < 0, no real solutions

If b² - 4ac = 0, one real solution

If b² - 4ac > 0, two distinct real solutions

3).Prove that for any real numbers a and b, |a + b| ≤ |a| + |b|. (This is known as the triangle inequality.)

Triangle inequality: |a + b| ≤ |a| + |b|

We'll consider all possible cases:

Case 1: a ≥ 0 and b ≥ 0

|a + b| = a + b = |a| + |b|

Case 2: a < 0 and b < 0

|a + b| = -(a + b) = (-a) + (-b) = |a| + |b|

Case 3: a ≥ 0 and b < 0 (or vice versa)

|a + b| ≤ |a| + |-b| = |a| + |b|

(This is because a + b could be positive or negative, but its absolute value will always be less than or equal to |a| + |b|)

In all cases, |a + b| ≤ |a| + |b|, proving the inequality.

4).Demonstrate that √2 is irrational using a proof by contradiction.

Assume √2 is rational. Then it can be expressed as a fraction in lowest terms:

√2 = a/b, where a and b are integers with no common factors.

Square both sides: 2 = a²/b²

Multiply both sides by b²: 2b² = a²

This means a² is even, so a must be even. Let a = 2k for some integer k.

Substitute: 2b² = (2k)² = 4k²

Divide by 2: b² = 2k²

This means b² is even, so b must be even.

But if both a and b are even, they have 2 as a common factor, contradicting our assumption that a/b was in lowest terms.

Therefore, our initial assumption that √2 is rational must be false. √2 is irrational.

5).Prove that between any two distinct real numbers, there is always a rational number.

Let a and b be distinct real numbers with a < b.

Consider the number c = (a + b)/2, the midpoint between a and b.

If c is rational, we're done.

If c is irrational, we can find a rational number between a and c as follows:

Let n be the smallest positive integer such that n(b - a) > 1.

Consider the numbers a, a + 1/n, a + 2/n, ..., b.

These form an arithmetic sequence with a step size of 1/n.

Since n(b - a) > 1, there must be at least one term of this sequence between a and b.

All terms in this sequence are rational (since a + k/n is rational for integer k).

Therefore, there is always a rational number between any two distinct real numbers.

6).Show that for any real numbers a, b, and c, max(a,b,c) = (a + b + c + |a - b| + |b - c| + |c - a|)

Detailed proof:

We'll consider all possible orderings of a, b, and c:

Case 1: a ≥ b ≥ c

max(a,b,c) = a

|a - b| = a - b, |b - c| = b - c, |c - a| = a - c

(a + b + c + |a - b| + |b - c| + |c - a|) / 3

= (a + b + c + (a - b) + (b - c) + (a - c)) / 3

= (3a) / 3 = a

Case 2: a ≥ c ≥ b

max(a,b,c) = a

|a - b| = a - b, |b - c| = c - b, |c - a| = a - c

(a + b + c + |a - b| + |b - c| + |c - a|) / 3

= (a + b + c + (a - b) + (c - b) + (a - c)) / 3

= (3a) / 3 = a

The other four cases (b ≥ a ≥ c, b ≥ c ≥ a, c ≥ a ≥ b, c ≥ b ≥ a) can be proven similarly.

In all cases, the formula gives the correct maximum.

7).Prove that for any real numbers a and b, (a + b)(a - b) = a² - b².

Solution:

We'll use the distributive property:

(a + b)(a - b) = a(a - b) + b(a - b)

= a² - ab + ba - b²

= a² - ab + ab - b² (commutative property of multiplication)

= a² - b²

8).Demonstrate that the set of rational numbers is dense in the real numbers.

We need to show that between any two distinct real numbers, there is a rational number.

Let x and y be two distinct real numbers with x < y.

Consider the number z = (x + y) / 2, the midpoint between x and y.

If z is rational, we're done.

If z is irrational, we can construct a rational number between x and z as follows:

Let n be a positive integer such that 1/n < y - x.

Consider the numbers of the form x + m/n where m is a non-negative integer.

There will be some smallest m such that x + m/n > z.

For this m, x + (m-1)/n ≤ z < x + m/n.

The number x + (m-1)/n is rational and lies between x and y.

This proves that between any two real numbers, there is a rational number, demonstrating the density of rational numbers in the real numbers.

9).Prove the AM-GM inequality: For positive real numbers a and b, (a + b)/2 ≥ √(ab), with equality if and only if a = b.

Start with (√a - √b)² ≥ 0 (this is true for all real numbers)

Expand: a - 2√(ab) + b ≥ 0

Add 2√(ab) to both sides: a + b ≥ 2√(ab)

Divide both sides by 2: (a + b)/2 ≥ √(ab)

This proves the inequality. For the equality condition:

If a = b, then (a + b)/2 = (a + a)/2 = a = √(a²) = √(ab)

If (a + b)/2 = √(ab), then going backwards through our proof, we must have (√a - √b)² = 0, which is only true when √a = √b, or a = b.

10).Show that for any real numbers a, b, and c, if a < b, then ac < bc if c > 0, and ac > bc if c < 0.

Solution :

Given: a < b

Case 1: c > 0

Multiply both sides of a < b by c: ac < bc (since multiplying an inequality by a positive number preserves the inequality direction)

Case 2: c < 0

Multiply both sides of a < b by c: ac > bc (since multiplying an inequality by a negative number reverses the inequality direction)

Case 3: c = 0

If c = 0, then ac = bc = 0, so the original statement doesn't apply.

This proves the given statement for all non-zero real numbers c.

Practice Questions

1).1. Find which of the variables x, y, z and u represent rational numbers and which irrational numbers:

(i) x2 = 5 (ii) y2 = 9 (iii) z2 = .04 (iv) u2 = 17/4

2).Find the value of a, if 6/(3√2 – 2√3) = 3√2 – a√3.

3).Prove that one of any three consecutive positive integers must be divisible by 3.

4).Calculate: (15 - 6) - 4 and 15 - (6 - 4). Are the results the same?

5).Prove that if a + x = b + x, then a = b, using the properties of real numbers.

6).How does the proof of (a + b)(a - b) = a² - b² demonstrate the use of multiple axioms of real numbers?

7).Explain why the formula max(a,b,c) = (a + b + c + |a - b| + |b - c| + |c - a|) / 3 works. What role do the absolute value terms play?

8).Explain intuitively why the triangle inequality (|a + b| ≤ |a| + |b|) makes sense geometrically.

9).Show that 5 + √3 is irrational.

10). If a = 5 + 2√6 and b = 1/a, what will be the value of a2 + b2?

FAQS

1).What are the axioms for the real number system?

The axioms of real numbers can be divided into three groups: field axioms, order axioms, and completeness axioms. and g : F ≠ F â F, g(x, y) = xy, called addition and multiplication respectively, satisfy the following axiom F1.

2).What is the completeness axiom of the real number system?

Completeness Axiom: a least upper bound of a set A is a number x such that x ≥ y for all y ∈ A, and such that if z is also an upper bound for A, then necessarily z ≥ x.

3).What is the completeness theorem of reals

The Completeness Axiom A fundamental property of the set R of real numbers : Completeness Axiom : R has “no gaps”. ∀S ⊆ R and S 6= ∅, If S is bounded above, then supS exists and supS ∈ R. (that is, the set S has a least upper bound which is a real number).

4).Why is the axiom of completeness important?

The completeness axiom of R, or equivalently the least upper bound property, is introduced at the start of any course in practical analysis. It is then shown that it can be used to prove Archimedean properties, that it is related to the notion of Cauchy sequences, etc.

Reference Links :

https://www.geeksforgeeks.org/engineering-mathematics/peano-axioms-number-system-discrete-mathematics/