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Chapter 5
Completeness I
Completeness is the key property of the real numbers that the rational numbers lack. Before examining this property we explore the rational and irrational numbers, discovering that both sets populate the real line more densely than you might imagine, and that they are inextricably entwined.
5.1 Rational Numbers
Definition A real number is rational if it can be written in the form p q , where p and q are integers with q 6 = 0. The set of rational numbers is denoted by Q. A real number that is not rational is termed irrational.
Example 12 , − 56 , 100 , (^567877) − 1239 , 82 are all rational numbers.
Exercise 1
- What do you think the letter Q stands for?
- Show that each of the following numbers is rational: 0, -10, 2.87, 0.0001, − 8 −^9 , 0.6666....
- Prove that between any two distinct rational numbers there is another rational number.
- Is there a smallest positive rational number?
- If a is rational and b is irrational, are a + b and ab rational or irrational? What if a and b are both rational? Or both irrational?
Historical Roots The proof that
√ 2 is irrational is attributed to Pythagoras ca. 580 − 500 BC who is well known to have had a triangle fetish.
What does
√ 2 have to do with triangles?
A sensible question to ask at this point is this: are all real numbers rational? In other words, can any number (even a difficult one like π or e) be expressed as a simple fraction if we just try hard enough? For good or ill this is not the case, because, as the Greeks discovered:
46 CHAPTER 5. COMPLETENESS I
Theorem√ 2 is irrational.
Euler’s constant
Euler’s γ constant is defined by
γ = lim n→∞
[ (^) ∑n
k=
1 k − log n
]
= 0. 5772 ...
It is not known whether γ is ra- tional or irrational. It is only known that, if γ = p q , q is
larger than 10^10
′ 000 .
This theorem assures us that at least one real number is not rational. You will meet the famous proof of this result in the Foundations course. Later in the course you will prove that e is irrational. The proof that π is irrational is also not hard but somewhat long and you will probably not meet it unless you hunt for it. We now discover that, despite the fact that some numbers are irrational, the rationals are spread so thickly over the real line that you will find one wherever you look. Exercise 2
- Illustrate on a number line those portions of the sets
{m| m ∈ Z}, {m/ 2 | m ∈ Z}, {m/ 4 | m ∈ Z}, {m/ 8 |m ∈ Z}
that lie between ±3. Is each set contained in the set which follows in the list? What would an illustration of the set {m/ 2 n| m ∈ Z} look like for some larger positive integer n?
- Find a rational number which lies between 57/65 and 64/73 and may be written in the form m/ 2 n, where m is an integer and n is a non-negative integer.
Integer Part
If x is a real number then [x], the integer part of x, is the unique integer such that
[x] ≤ x < [x] + 1.
For example
[3.14] = 3 and [− 3 .14] = − 4.
Theorem Between any two distinct real numbers there is a rational number.
I.e. if a < b, there is a rational p q with a < p q < b.
Open Interval
For a < b ∈ R, the open inter- val (a, b) is the set of all num- bers strictly between a and b: (a, b) = {x ∈ R : a < x < b}
Proof. Consider the set of numbers of the form p q with q fixed, and p any integer. Assume that there are no such numbers between a and b. Let p q be the number immediately before a. Then p+1 q is the number immediately after b. We necessarily have
p + 1 q
p q
≥ b − a ⇐⇒
q
≥ b − a.
If we choose q sufficiently large, then the above inequality is wrong. Then there is at least one rational number between a and b.
Chalk and Cheese
Though the rationals and ir- rationals share certain prop- erties, do not be fooled into thinking that they are two-of- a-kind. You will learn later that the rationals are “count- able”, you can pair them up with the natural numbers. The irrationals, however, are mani- festly “uncountable”
Corollary Let a < b. There is an infinite number of rational numbers in the open interval (a, b).
Proof. One can think of many proofs. One could proceed as above, but proving that there are more than N numbers between a and b, for arbitrarily large N. But we can also use the theorem directly. We know that there must
48 CHAPTER 5. COMPLETENESS I
Exercise 4 For each of the following sets of real numbers decide whether the set is bounded above, bounded below, bounded or none of these:
- {x : x^2 < 10 } 2. {x : x^2 > 10 } 3. {x : x^3 > 10 } 4. {x : x^3 < 10 }
Definition A number u is a least upper bound of A if
- u is an upper bound of A and
- if U is any upper bound of A then u ≤ U. A number l is a greatest lower bound of A if
- l is a lower bound of A and
- if L is any lower bound of A then l ≥ L.
The least upper bound of a set A is also called the supremum of A and is denoted by sup A, pronounced “soup A”. The greatest lower bound of a set A is also called the infimum of A and is denoted by inf A. Example Let A = { (^1) n : n = 2, 3 , 4 ,... }. Then sup A = 1/2 and inf A = 0.
Exercise 5 Check that 0 is a lower bound and 2 is an upper bound of each of these sets
- {x| 0 ≤ x ≤ 1 } 2. {x| 0 < x < 1 } 3. {1 + 1/n| n ∈ N}
- { 2 − 1 /n| n ∈ N} 5. {1 + (−1)n/n| n ∈ N} 6. {q| q^2 < 2 , q ∈ Q}.
For which of these sets can you find a lower bound greater than 0 and/or an upper bound less that 2? Identify the greatest lower bound and the least upper bound for each set. Can a least upper bound or a greatest lower bound for a set A belong to the set? Must a least upper bound or a greatest lower bound for a set A belong to the set?
We have been writing the least upper bound so there had better be only one. Exercise 6 Prove that a set A can have at most one least upper bound.
5.3 Axioms of the Real Numbers
Despite their exotic names, the following fundamental properties of the real numbers will no doubt be familiar to you. They are listed below. Just glimpse through them to check they are well known to you.
- For x, y ∈ R, x + y is a real number closure under addition
- For x, y, z ∈ R, (x + y) + z = x + (y + z) associativity of addition
5.3. AXIOMS OF THE REAL NUMBERS 49
- For x, y ∈ R, x + y = y + x commutativity of addition
- There exists a number 0 such that for x ∈ R, x + 0 = x = 0 + x existence of an additive identity
- For x ∈ R there exists a number −x such that x + (−x) = 0 = (−x) + x existence of additive inverses
- For x, y ∈ R, xy is a real number closure under multiplication
- For x, y, z ∈ R, (xy)z = x(yz) associativity of multiplication
- For x, y ∈ R, xy = yx commutativity of multiplication
- There exists a number 1 such that x · 1 = x = 1 · x for all x ∈ R. existence of multiplicative identity
- For x ∈ R, x 6 = 0 there exists a number x−^1 such that x · x−^1 = 1 = x−^1 · x existence of multiplicative inverses
- For x, y, z ∈ R, x(y + z) = xy + xz distributive law
- For x, y ∈ R, exactly one of the following statements is true: x < y, x = y or x > y trichotomy
- For x, y, z ∈ R, if x < y and y < z then x < z transitivity
- For x, y, z ∈ R, if x < y then x + z < y + z adding to an inequality
- For x, y, z ∈ R, if x < y and z > 0 then zx < zy multiplying an inequality
There is one last axiom, without which the reals would not behave as ex- pected:
Completeness Axiom Every non-empty subset of the reals that is bounded above has a least upper bound.
If you lived on a planet where they only used the rational numbers then all the axioms would hold except the completeness axiom. The set {x ∈ Q : x^2 ≤ 2 } has rational upper bounds 1. 5 , 1. 42 , 1. 415 ,... but no rational least upper bound.
5.4. BOUNDED MONOTONIC SEQUENCES 51
5.4 Consequences of Completeness - Bounded
Monotonic Sequences
The mathematician Weierstrass was the first to pin down the ideas of com- pleteness in the 1860’s and to point out that all the deeper results of analysis are based upon completeness. The most immediately useful consequence is the following theorem:
Theorem Increasing sequence version Every bounded increasing sequence is convergent.
Figure 5.1: Bounded increasing sequences must converge.
Figure 5.1 should make this reasonable. Plotting the sequence on the real line as the set A = {a 1 , a 2 , a 3 ,... } we can guess that the limit should be sup A. Proof. Let (an) be a bounded increasing sequence. We show that an → sup A. Let ε be any positive number. By the above lemma, there exists aN ∈ A such that sup A − ε < aN ≤ sup A. Since (an) is increasing, we have
sup A − ε < an ≤ sup A
for all n > N. Then |an − sup A| < ε. This holds for every ε > 0, so that an → sup A.
Check that your proof has used the completeness axiom, the fact that the sequence is increasing, and the fact that the sequence is bounded above. If you have not used each of these then your proof must be wrong!
Corollary Decreasing sequence version Every bounded decreasing sequence is convergent.
Proof. The sequence (−an) is bounded and increasing, then it converges to a number −a. Then an → a by the theorem of Section 2.6.
Example In Chapter 3, we considered a recursively defined sequence (an) where a 1 = 1 and an+1 =
an + 2. We showed by induction that an ≥ 1 for all n (because a 1 = 1 and ak ≥ 1 =⇒ ak+1 =
ak + 2 ≥
3 ≥ 1) and that an ≤ 2 for all n (because a 1 ≤ 2 and ak ≤ 2 =⇒ ak+1 =
ak + 2 ≤
4 = 2). So (an) is bounded.
52 CHAPTER 5. COMPLETENESS I
We now show that the sequence is increasing.
a^2 n − an − 2 = (an − 2)(an + 1) ≤ 0 since 1 ≤ an ≤ 2 ∴ a^2 n ≤ an + 2 ∴ an ≤
an + 2 = an+1.
Decreasing?
To see whether a sequence (an) is decreasing, try testing
an+1 − an ≤ 0
or, when terms are positive,
an+ an ≤ 1.
Hence (an) is increasing and bounded. It follows from Theorem 5.4 that (an) is convergent. Call the limit a. Then a^2 = limn→∞ a^2 n+1 = limn→∞ an+2 = a+ so that a^2 − a − 2 = 0 =⇒ a = 2 or a = −1. Since (an) ∈ [1, 2] for all n we know from results in Chapter 3 that a ∈ [1, 2], so the limit must be 2.
Exercise 9 Consider the sequence (an) defined by
a 1 =
and an+1 =
a^2 n + 6
Show by induction that 2 < ak < 3. Show that (an) is decreasing. Finally, show that (an) is convergent and find its limit.
Exercise 10 Explain why every monotonic sequence is either bounded above or bounded below. Deduce that an increasing sequence which is bounded above is bounded, and that a decreasing sequence which is bounded below is bounded.
Exercise 11 If (an) is an increasing sequence that is not bounded above, show that (an) → ∞. Make a rough sketch of the situation.
The two theorems on convergence of bounded increasing or decreasing se- quences give us a method for showing that monotonic sequences converge even though we may not know what the limit is.
5.5 * Application - k
th
Roots *
So far, we have taken it for granted that every positive number a has a unique positive kth^ root, that is there exists b > 0 such that bk^ = a, and we have been writing b = a^1 /k. But how do we know such a root exists? We now give a careful proof. Note that even square roots do not exist if we live just with the rationals - so any proof must use the Axiom of Completeness. Stop Press √ 2 exists!!! Mathematicians have at last confirmed that
√ 2 is really there.
Phew! What a relief.
Theorem Every positive real number has a unique positive kth^ root.
Suppose a is a positive real number and k is a natural number. We wish to show that there exists a unique positive number b such that bk^ = a. The idea of the proof is to define the set A = {x > 0 : xk^ > a} of numbers that are too big to be the kth^ root. The infimum of this set, which we will show to exist by the
54 CHAPTER 5. COMPLETENESS I
