Math Club (September 29) Real Numbers Paul Cazeaux

... Or how I started to love Analysis. How well do you know your R?

1. Who is the ancient Greek mathematician that believed all numbers to be rational, and when one of his stu- dents discovered the existence of irrational numbers, that student was drowned at sea as a sacrifice to the gods? A bit of history: At the beginning, there were only the natural integers Ratios , used for counting stuff. then ap- 2. Pick the irrational and rational numbers out of the list: peared early in Antiquity, with the first incommensurable ( √ ) numbers being discovered in the VIth century BC. in An- 3 1 + 5 √ 0, √ , cos(π), eπ, √ , −3, 0.19191919 ..., cient Greece. Eudoxus of Cnidus formalized a theory of com- 2 1 − 5 mensurable and incommensurable ratios, which we would call nowadays the positive real numbers. Negative numbers had to wait many centuries, because they are so scary... 3. Saying a number is transcendental means... Although they were used for many centuries, a proper theory or construction of Real Numbers had to wait until the end of (a) a synonym for irrational, the 19th century, when Peano, Dedekind, Cantor, Weiestrass (b) it is the root of some nonzero polynomial with in- and other mathematicians finally formalized the classical re- teger coefficients, lation (c) it is not the root of any nonzero polynomial with integer coefficients, N ⊂ Z ⊂ Q ⊂ R. (d) it transcends dental functions.

4. Is R the only ordered and complete commutative ﬁeld? Rational Numbers: the ﬁeld Q

The rationals numbers p/q are built out of the ring Z of Some properties of Q : relative integers, which is itself built out of the associative totally ordered field monoid N of natural integers. • Q is a . In each case, we plug ’holes’ in our ability to solve equations: • Q is countable. • Equations like n + 1 = 0 have no solution in N. • Every totally ordered field contains a ring isomorphic → we introduce opposites −n of nonzero natural integers. to Z: Z = {n × 1, n ∈ Z} and a sub-field isomorphic to Q, • Equations like 2x − 1 = 0 have no solution in Z. Q = {pq−1, p ∈ Z, q ∈ Z \{0}}. → we introduce inverses 1/n of nonzero relative integers. • has the Archimedean property: for any x, y ∈ , Formal definition: Q Q ∗ x, y > 0 =⇒ ∃n ∈ , nx > y. • Equivalence relation on pairs of integers (a, b) ∈ Z×Z : N

(a, b) ∼ (c, d) iff ad = cb. • We have an absolute value: ∗ Denote equivalence classes a/b ∈ Q := (Z × Z )/∼. |x| = max(x, −x). • Compatible addition and multiplication operations: a c ad + cb a c ac + = , × = b d bd b d bd Activities • Operations are commutative, associative, distributive. 1. Show that any totally ordered field has the • Every element has an opposite: −a/b = (−a)/b. Archimidean property if, and only if, Q is dense in K: −1 • Every nonzero rational has an inverse: (a/b) = b/a. For any x < y in K, there is r ∈ Q such that x < r < y. • Total order on Q: a/b is positive iff ab > 0 and 2. Show that Q is countable by coming up with a way to x ≥ y iff x − y ≥ 0, x, y ∈ Q. enumerate elements of Q.

• Inclusion Z ⊂ Q: n 7→ n/1. 3. Check the classical properties of the absolute value. The holes in Q

Why are we not happy with Q? • Algebra! Some simple equations have no solution: x2 = 2. • Analysis! Some sequences that should morally converge do not: take the partial sums of the classical series ∞ X 1 u = . n k! k=1 • Topology! Some non-empty parts of Q are bounded from above, yet have no maximum: take 2 A = {x ∈ Q, x ≤ 3}.

The dawn of Analysis. • With order comes idea of upper bound: Activities M = sup(A) iﬀ (x ≥ a, ∀a ∈ A) =⇒ x ≥ M.

convergence 2 • With absolute value comes notion of : 1. Show that there is no x ∈ Q such that x = 2. l = lim un iﬀ ∀ε > 0, ∃N ∈ , ∀n ≥ N, |un − l| < ε. n→∞ N 2. Show that the sequence un = 1/n converges to 0 in a ﬁeld K which has the Archimidean property. • If we don’t know l, we check the Cauchy property: un is a Cauchy sequence if P∞ 1 3. Show that un = k=1 k! forms a Cauchy sequence. ∀ε > 0, ∃N ∈ N, ∀n ≥ N, |un − um| < ε Completeness

Let K be an ordered ﬁeld. Here are some desired properties:

• Property of the upper bound: Any non-empty part of K, bounded from above, admits an upper bound. • Cauchy property: Any Cauchy sequence converges. • Property of nested intervals: If In = [an, bn] with In+1 ⊂ In and bn − an → 0, then T n≥0 In contains exactly one element of K.

Theorem. The following properties are equivalent: • K has the property of the upper bound; • Any increasing sequence bounded from above converges in K; • K is Archimedean and has the Cauchy property; K • is Archimedean and has the property of nested in- Activities tervals. Under these conditions, we say K is complete. 1. Let K be a complete ordered ﬁeld. Show that for any y ∈ K, y ≥ 0 if and only if there exists x ∈ K such 2 Narrator: sadly, Q is not complete. that y = x . Building R, take one: Dedekind Cuts

We’ve seen that some parts of Q do not have upper bounds. Theorem (Dedekind) The set D of Dedekind cuts can be Let’s fix that by making these ’holes’ into new objects! equipped with a structure of totally ordered, complete com- Definition. A Dedekind cut is a partition of Q into two mutative field. nonempty parts (α, β) such that 1. α ∪ β = Q, 2. For all a ∈ α, b ∈ β, we have a < b: 3. The right part β has no minimum (i.e. a lower bound that also belongs to β). We call α = βc the left part and β the right part of the cut. Note how only the knowledge of β is needed to form the cut. We call D ⊂ P ( ) the set of Dedekind cuts. Q Proof: tedious. For the completeness property, let us take A ⊂ D non-empty, bounded from above. Then Example. Any rational r ∈ Q defines a unique Dedekind  c cut α = {x ∈ Q, x ≤ r}, β = {x ∈ Q, x > r}. [ µ = βc This allows us to identify Q ⊂ D.   β∈A

can be shown to be a Dedekind cut: Structure as an ordered ring µ µc µ β + β0 = {x + x0, x ∈ β, x0 ∈ β0}, • and are nonempty, and has no minimum; c c − β = {x ∈ Q, s.t. ∀b ∈ β, x > −b} \ {− max(β )}, • if x ∈ µ, y ∈ µ , then x > y;  0 0 0 0 + {xx , x ∈ β, x ∈ β } if β, β ⊂ Q , • µ is the upper bound of A.  −(β × (−β0)) if β ⊂ +, β0 6⊂ +, β × β0 = Q Q 0 + 0 + −((−β) × β ) if β 6⊂ Q , β ⊂ Q , Activities  ((−β) × (−β0)) β, β0 6⊂ +. if Q 1. Show that the same construction, repeated with D as a β ≤ β0 iﬀ β ⊂ β0. starting point, yields a ﬁeld which is isomorphic to D. Building R, take two: Cauchy sequences

A sequence un of rationals is a Cauchy sequence if: Theorem. Ce may be equipped with a structure of totally ordered, complete commutative field. ∀ε ∈ Q, ε > 0, ∃N ∈ N, ∀n ≥ N, |un − um| < ε. Proof: tedious. For the completeness property, one starts by showing that any element of Ce is the limit of a sequence of "rationals" π({r, r, . . . }). Given a Cauchy sequence un in Ce, C 1. First, we define the set of Cauchy sequences . define "rationals" rn such that |rn − un| < 1/n. Then for This is a sub-ring of the set of rational sequences: m, n big enough,

• For u, v ∈ C, 1 1 u + v = {u + v } ∈ C u × v = {u v } ∈ C |rn − rm| ≤ |rn − un| + |un − um| + |um − rn| ≤ + + ε, n n , n n , e e e e n m • We have neutral elements {0, 0,... } for the addition and {1, 1,... } for the multiplication. so {rn} is itself a Cauchy sequence, which corresponds to an element r of Ce. Finally one ﬁnds limn→∞ un − r = 0. 2. However, observe that C is too big:

• many sequences converge to the same number, Activities • there is no inverse for some nonzero sequences, such as {1, 0, 1,... }. 1. Show that a Cauchy sequence is bounded, then that the product uv of two Cauchy sequences is a Cauchy Define the equivalence relation sequence. u ∼ v iff lim u − v = 0. n→∞ 2. Show that if u ∼ u0 and v ∼ v0 then u + v ∼ u0 + v0 and uv ∼ u0v0. Definition. The ring Ce is defined as the quotient C/∼, iden- tifying sequences that have the same limit. 3. Show that if u 6∼ 0, then there exists v such that uv ∼ 1.

• This set can be ordered: we say that π(u) > 0 if for any 4. Show that the same construction, repeated with Ce as a representative sequence u, un > 0 for n big enough. starting point, yields a ﬁeld which is isomorphic to Ce. Counting the real numbers

• N, Z, Q are all countable. What about R?! Second proof. Any real number in (0, 1] admits a unique proper decimal representation

a = 0.d1d2 ...

Theorem. (Cantor) that does not end in 9999 ... . As before, assume numbers in The set R of real numbers is uncountable. (0, 1] can be ordered in a sequence a0, a1,... :

a0 = 0.d01d02d03 . . . d0p ...

a1 = 0.d11d12d13 . . . d1p ...

a2 = 0.d21d22d23 . . . d2p ... First proof. I start, you ﬁnish! Let us show that [0, 1] . is uncountable. Assume by contradiction that we can order . all its elements in a sequence {an}≥0. an = 0.dn1dn2dn3 . . . dnp ... . .

1. Divide I0 = [0, 1] in three pieces [0, 1/3], [1/3, 2/3], Using a diagonal procedure, create a number x ∈ (0, 1] with [2/3, 1]. Deﬁne I1 as the left-most piece that does not proper decimal representation 0.x1x2 . . . xp ... which is dif- contain a0. ferent from all the an. 2. Write an algorithm continuing this process, forming a sequence of nested intervals In such that an ∈/ In. Activities

3. Compute the length of the intervals In. 1. Show there is no surjection from N onto the set of func- 4. Conclude! tions from N into N. Rich numbers

• Real numbers are weird. Most of them are rich! A real number is rich if its decimal expansion contains every possible sequence of numbers. They are all irrational. We want to prove that almost all numbers in (0, 1] are rich.

Let’s start with: 1. Let A1 be the set of all real numbers in (0, 1] which Theorem. Let b ≥ 2 be an integer. A number x ∈ R is contains 0 as the ﬁrst digit of their decimal expansion. rational if and only if its developement in base b is periodic Show that A1 is an interval with length 1/10. after a long enough rank. 2. Let Ai be the set of all real numbers in (0, 1] which contains 0 as the ith digit of their decimal expansion. Prove this! Let 0 ≤ p/q < 1 a rational number, we get the Show that it is a collection of intervals with cumulative ﬁrst digit a1 by euclidean division: length 1/10i.

3. Compute the ’length’ of the set B0 of all real numbers bp = a q + r , 0 ≤ r < q. 1 1 1 that contains 0 anywhere in their decimal expansion.

• Why do we have 0 ≤ a1 < b? 4. Explain why the same should be true of any set Bn of all the real numbers that contain n’s decimal expansion as a sequence of digits anywhere in their decimal • How to get the next digits? expansion.

5. Assuming a countable union of sets of measure (length) • Why must this process repeat? (Pigeonhole...) zero has measure zero, you can conclude!

Activities 1. Now that you know this, write a few irrational numbers!