is irrational

UNIVERSITY OF NATURAL SCIENCES HO CHI MINH City http://www.hcmuns.edu.vn/

Introduction to Diophantine methods:

irrationality and transcendence

Michel Waldschmidt,

Professeur, Universit´e P. et M. Curie (Paris VI) http://www.math.jussieu.fr/∼miw/coursHCMUNS2007.html

Examen – second session 3 hours

Exercise 1. Forn≥1 define a_n=

(1 ifnis a power of 2, 0 otherwise.

Hence

(a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, . . .) = (1, 1, 0, 1, 0, 0, 0, 1, 0, 0, . . .).

Prove that the number written in binary notation

0.a1a2a3a4a5a6a7a8a9a10· · ·= 0.1 1 0 1 0 0 0 1 0 0. . . is irrational.

Exercise 2. Which are the correct sentences? Explain your answer.

(i) The sum of two rational numbers is (A) always rational (B) always irrational

(C) sometimes rational, sometimes irrational.

(ii) The sum of two irrrational numbers is (A) always rational

(B) always irrational

(C) sometimes rational, sometimes irrational.

(iii) The sum of rational number and an irrational number is (A) always rational

(B) always irrational

(C) sometimes rational, sometimes irrational.

(iv), (v), (vi) Same questions with the product instead of the sum.

(vii) If (an)_n≥0 is an infinite sequence withan∈ {−1,1}for alln≥0, then the number

X

n≥0

an2⁻ⁿ is irrational.

Exercise 3. Defineu0 = 0, u1 = 1, and by induction un = 2u_n−1+u_n−2 for n≥2.

a) Check, for anyn≥1,

u²_n−2unun−1−u²_n−1= (−1)ⁿ⁻¹.

b) Show that the sequence (u_n/u_n−1)_n≥1 converges asn → ∞. What is the limit?

c) Prove that there exists a sequence (p_n/q_n)_n≥1 of rational numbers such that

n→∞lim qn

qn

√ 2−pn

= 1

2√ 2· d) Prove that for anyκ >2√

2, there are only finitely manyp/q∈Qsatisfying

√ 2−p

q

≤ 1 κq²·

Exercise 4. Letαbe a complex number. Show that the following properties are equivalent.

(i)αis root of a polynomial of degree≤3 with rational coefficients.

(ii) TheQ–vector space spanned by 1, α, α², α³. . . has dimension≤3.

(iii) For any integerm ≥1, the numberα^m is root of a polynomial of degree

≤3 with rational coefficients.

Exercise 5. Recall the next Theorem due to Hermite and Lindemann: for any non–zero complex number z, one at least of the two numbersz,e^z is transcen- dental. Deduce the following results.

a) For any non–zero algebraic numberα, the numberse^α, cos(α) and sin(α) are transcendental.

b) Letλ∈C,λ6= 0. Assume e^λis algebraic. Thenλis transcendental.

c) The numbers log 2 andπare transcendental.

Examen – Second session Solutions

Solution exercise 1.

Form≥1 betweena₂m anda₂m+1 there are 2^m−1 consecutive zeroes, a₂m+1=a₂m+2=· · ·=a₂m+1−1= 0.

Therefore the sequence (a_n)_n≥0 is not ultimately periodic, and it follows that the given number is irrational.

Solution exercise 2.

(i) The sum of two rational numbers is (A) always rational because

a b + c

d =ad+bc bd · The set of rational numbers is a field.

(ii) The sum of two irrrational numbers is

(C) sometimes rational, sometimes irrational.

Ifxis irrational andris rational theny=r−xis irrational, while the sum of xandy isr, hence is rational.

Ifxis irrational thenx+x= 2xis also irrational.

(iii) The sum of rational number and an irrational number is (B) always irrational.

Ifris rational andr+xis also rational thenx= (r+x)−ris rational. Hence ifris rational andxis irrational thenr+xis irrational.

(iv) The product of two rational numbers is (A) always rational

because

a b · c

d= ac bd· The set of rational numbers is a field.

(v) The product of two irrrational numbers is

(C) sometimes rational, sometimes irrational.

Ifxis irrational andris rational then y=r/x is irrational, while the product ofxandy isrhence is rational.

Ift is rational thent² is also rational. Therefore if x is irrational then√ xis also irrational. Now the product of√

xwith itself isx, hence irrational.

(vi) The product of rational number and an irrational number is (C) sometimes rational, sometimes irrational.

The product of 0 and any irrational number is 0, hence rational.

Ifr6= 0 is rational andrxis also rational thenx=rx/ris rational, hence ifr is rational6= 0 and xis irrational thenrxis irrational.

(vii) If the sequence (an)_n≥0 is ultimately periodic, then the number X

n≥0

a_n2⁻ⁿ

is rational. For instance fora_n= 1 for alln≥0 the sum is 2.

Solution exercise 3. Forn≥1 set

vn=un/un−1 and wn=u²_n−2unun−1−u²_n−1 so that

wn=u²_n−1(v²_n−2vn−1).

a) From the recurrence formulau_n+1= 2u_n+u_n−1one deduces

w_n+1=u²_n+1−2u_n+1u_n−u²_n=u_n+1(u_n+1−2u_n)−u²_n =u_n−1(2u_n+u_n−1)−u²_n=−w_n. Thereforew_n= (−1)ⁿw₀, and sincew₀= 1 we concludew_n= (−1)ⁿ⁻¹.

b) The roots of the polynomialX²−2X−1 areα= 1 +√

2 andα⁰ = 1−√ 2.

Notice thatα⁰<0< α and

wn = (un−αu_n−1)(un−α⁰u_n−1).

From the recurrence formula we deduceun >2u_n−1 forn≥2, henceun ≥2ⁿ forn≥0 and

un−α⁰u_n−1≥un≥2ⁿ. Using|wn|= 1 we obtain

|α−vn| ≤ 1 2²ⁿ⁻¹· Therefore the sequence (vn)_n≥1converges to α= 1 +√

2 asn→ ∞.

c) We havewn=u²_n−1(vn−α)(vn−α⁰) and the limit of the sequence (vn−α⁰)_n≥1 isα−α⁰ = 2√

2. Hence

n→∞lim u_n−1|u_n−1α−un|= 1 2√

2· Forn≥1 definepn=u_n−1−un and qn =u_n−1. Then

n→∞lim qn

qn

√ 2−pn

= 1

2√ 2· d) Forκ >2√

2, letp/q∈Qsatisfy

√ 2−p

q

≤ 1 κq²· We have

≤ |(p+ ²−2q(p+ −q²| |(p − − ⁰)| ≤ 1

− ⁰ 1

·

Hence q²κ(κ−2√

2) <1. Therefore q is bounded, and since p is the nearest integer toq√

2 there are only finitely many solutionsp/q.

Solution exercise 4.

The implication (iii)⇒(i) is trivial: take m= 1.

Proof of (i)⇒(ii). Sinceαis root of a polynomial of degree≤3 with rational coefficients, then it is also root of a monic polynomialX³−aX²−bX−c of degree 3 with rational coefficients (multiply byX orX²if necessary and divide by the leading coefficient). By induction, for each integerm≥1 we can write α^m=a_mα²+b_mα+c_m with a_m,b_mand c_m in Q. Hence the Q–vector space spanned by 1, α, α², α³. . . is also spanned by 1, α, α², hence has dimension

≤3.

Proof of (ii)⇒(iii). If the Q–vector space spanned by 1, α, α², α³. . . has dimension≤3, then the four numbers 1,α^m,α^2m,α^3mare linearly dependent overQ.

Solution exercise 5.

a) Let α be a non–zero algebraic number. Taking z = α in the Hermite–

Lindemann Theorem shows that e^α is transcendental. Also iα is a non–zero algebraic number, hencee^iα is transcendental. This means that it is not root of a polynomial with rational coefficients, and this implies that it is not root of a polynomial with algebraic coefficients. Sincee^iα is root of the polynomials

X²−2Xcos(α) + 1 and X²−2iXsin(α)−1, it follows that cos(α) and sin(α) are transcendental.

b) Letλ∈C,λ6= 0. Ife^λ is algebraic, then the Hermite–Lindemann Theorem withz=λshows thatλis transcendental.

c) Forλ= log 2 the numbere^λ = 2 is algebraic, hence log 2 is transcendental.

Forλ=iπ the numbere^λ =−1 is algebraic, hence iπ is transcendental. The product of two algebraic numbers is algebraic, and i is algebraic, hence π is transcendental.