aa r X i v : . [ m a t h . G M ] J u l Linear Independence Of Some Irrational Numbers
N. A. Carella
Abstract : This note presents an analytic technique for proving the linear independence ofcertain small subsets of real numbers over the rational numbers. The applications of this testproduce simple linear independence proofs for the subsets of triples { , e, π } , { , e, π − } , and { , π r , π s } , where 1 ≤ r < s are fixed integers. The algebra and number theory literature has many elementary techniques used to verify thelinear independence of small finite subsets of algebraic numbers { α , α , . . . , α d } ⊂ Q over therational numbers Q . A few examples of these algebraic subsets are1. { , √ , √ , √ } ⊂ Q ,2. { , √ , √ , √ } ⊂ Q , 3. { , α, α , . . . , α d } ⊂ R , where f ( α ) = 0,4. { , ω, ω , . . . ω ϕ ( n ) } ⊂ R , where ω n = 1.But, these techniques are intrinsically algebraic, and do not seem to be applicable to smallsubsets of nonalgebraic numbers { α , α , . . . , α d } ⊂ R . This note presents an analytic tech-nique for proving the linear independence of certain small subsets of nonalgebraic numbers { α , α , α } ⊂ R over the rational numbers Q . The applications of this test produce simpleproofs for the followings subsets of triples. Theorem 1.1.
The real numbers , e and π are rationally independent. Theorem 1.2.
The real numbers , e and π − are rationally independent. Theorem 1.3.
For any pair of integers ≤ r < s , the real numbers , π r and π s are rationallyindependent. The proofs are presented in Section 6, Section 8, and Section 10 respectively.Since both e and π are irrational numbers, it is immediate that at least one, the trace T r ( α ) = e + π or the norm N ( α ) = eπ of the polynomial f ( x ) = ( x + e )( x + π ), is irrational. Simpleapplications of the above linear independence results demonstrate that both of these numbersare irrational, see Corollary 7.1, and Corollary 9.1 respectively. Similar application demonstratesthat the real number π + π (1)is irrational, see Corollary 11.1. July 31, 2020
AMS MSC :Primary 11J72; Secondary 11K06
Keywords : Irrational number; Irrationality criteria; Linear independence; Uniform distribution. The Irrational Limit Test
The irrational limit test converts some apparently intractable decision problems in the realdomain R to simpler decision problems in the finite field domain F = { , } . Definition 2.1.
Let α ∈ R be a real number. The irrational limit test is a map I : R −→ F = { , } defined by I ( α ) = lim x →∞ x X − x ≤ n ≤ x e iαn . (2)The normalization is intrinsic to the number π . But, it can be modified as needed. The irrationallimit test is a point map or equivalently a class map, and it is not invertible. But, inversion isnot required in the applications to decision problems. Lemma 2.1.
For any real number α ∈ R , the irrational limit test satisfies the followings. I (2 πmα ) = ( if and only if α ∈ Q , for some m ∈ Z × , if and only if α / ∈ Q , for any m ∈ Z × . (3) Proof.
Given any rational number α ∈ Q , there is an integer m ∈ Z such that αm ∈ Z , and thelimit is I (2 πmα ) = lim x →∞ x X − x ≤ n ≤ x e i παmn = lim x →∞ x X − x ≤ n ≤ x . (4)The above proves that for any rational number α ∈ Q , and any integer m , the sequence { αmn : n ∈ Z } (5)is not uniformly distributed. While for any irrational number α / ∈ Q , and any integer m = 0,the sine function sin( απm ) = 0. Hence, applying Lemma 5.1, the evaluation of the limit is I (2 παm ) = lim x →∞ x X − x ≤ n ≤ x e i παmn ≤ lim x →∞ x | sin ( απm ) | (6)= 0 . The above proves that for any irrational number α ∈ Q , and any integer m = 0, the sequence { αmn : n ∈ Z } , (7)is uniformly distributed. This proof is equivalent to the Weil criterion, see [9, Theorem 2.1]. (cid:4) As it is evident, the class function I maps the class of rational numbers Q to 1 and the class ofirrational numbers I = R − Q to 0. The irrational limit test induces an equivalence relation onthe set of real numbers R : • A pair of real numbers a and b are equivalent a ∼ b if and only if I (2 πa ) = I (2 πb ). Thisoccurs if either both a and b are rational numbers or both a and b are irrational numbers. • A pair of real numbers a and b are not equivalent a b if and only if I (2 πa ) = I (2 πb ).This occurs if either a or b is a rational numbers but not both.Some standard irrationality tests, criteria, and proofs are given in [1, Chapter 7], [3], [7], [11],[16], [17], et alii. 2 Approximation By Lattice Points
A handful of elementary integer relations and integers points approximations are considered inthis Section.
Lemma 3.1.
The numbers e and π are not integer multiple. Specifically, ek = mπ for anyintegers k, m ≥ .Proof. Numerically sin( e ) = 0 . . . . = 0. Computing it using the infinite product yields0 = sin ( e ) = 1 e Y n ≥ (cid:18) − e π n (cid:19) . (8)Ergo, for each integer n ≥
1, the local factor1 − e π n = 1 − (cid:16) eπn (cid:17) = 0 (9)cannot vanish. This proves that e/πn = 1. Equivalently, these numbers are not integer multiple: ek = mπ for any integers k, m ≥ (cid:4) The discrete lines L ( r ) = { (2 r + 1) π/ r ∈ Z × } and L ( r ) = { (2 r + 1) π : r ∈ Z × } (10)never intercept the discrete lattice L ( k, m ) = { ke + m : k, m ∈ Z × } , (11)but comes arbitrarily close. A proof, based on the simplest form of the Kronecker approximationtheorem, see Theorem 12.1, is given below. Lemma 3.2. If k and m are nonzero integers, and let r ∈ Z , then1. ke + m = (2 r + 1) π/ . ke + m = rπ. Proof. (i) It is sufficient to verify the inequality (3.2)-i on the first quadrant, which is specifiedby k ≥ m ≥
1. The verification in any quadrant is almost the same. Let { p n /q n : n ≥ } be the sequence of convergents of the irrational number e . The Diophantine approximationinequalities 12 q n +1 ≤ | q n e − p n | ≤ q n (12)and | q n e − p n | ≤ | ke − m | (13)for k ≤ q n , see Lemma 12.4, and Lemma 12.5, lead to the lattice points approximation | ke + m − (2 r + 1) π/ | ≥ || ke − m | − (2 r + 1) π/ | (14) ≥ || q n e − p n | − (2 r + 1) π/ | , where r ∈ Z , and | r + 1 | ≥
1. Rearranging it, and applying the reverse triangle inequality | X − Y | ≥ || X | − | Y || , yield (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) e − p n q n (cid:12)(cid:12)(cid:12)(cid:12) − (cid:12)(cid:12)(cid:12)(cid:12) (2 r + 1) π q n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q n +1 (cid:12)(cid:12)(cid:12)(cid:12) − (cid:12)(cid:12)(cid:12)(cid:12) (2 r + 1) π q n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (15) ≥ q n +1 > . Therefore, relation ke + m = (2 r + 1) π/ k, m ) = ( k =0 , m = 0) and r ∈ Z . (ii) The proof for this case is similar. (cid:4) e = [ a , a , a , . . . ] has arbitrary long arithmetic progres-sions, very visibly, see Theorem 12.1, the relation ke + m = (2 r + 1) π/ π = [ b , b , b , . . . ] has arbitrary long arithmetic progressions. But, this isunknown. These elementary results seem to be implied by a more advanced technique given in[14] about certain equivalence of irrational numbers. Lemma 3.3.
Let ≤ u < v be a pair of integers, and let k , m , and r be any nonzero integers,then1. kπ u + mπ v = rπ. ke u + me v = rπ. Proof. (i) It is sufficient to verify the inequality (3.3)-i on the first quadrant, which is specifiedby k ≥ m ≥
1. The verification in any quadrant is almost the same. Let { p n /q n : n ≥ } be the sequence of convergents of the irrational number π v − u . The Diophantine approximationinequalities 12 q n +1 ≤ (cid:12)(cid:12) q n π v − u − p n (cid:12)(cid:12) ≤ q n (16)and (cid:12)(cid:12) q n π v − u − p n (cid:12)(cid:12) ≤ (cid:12)(cid:12) kπ v − u − m (cid:12)(cid:12) (17)for k ≤ q n , see Lemma 12.4, and Lemma 12.5, lead to the lattice points approximation | kπ u + mπ v − rπ | = (cid:12)(cid:12) π u (cid:0) kπ v − u + m (cid:1) − rπ (cid:12)(cid:12) (18) ≥ (cid:12)(cid:12) π u (cid:12)(cid:12) kπ v − u − m (cid:12)(cid:12) − rπ (cid:12)(cid:12) ≥ (cid:12)(cid:12) π u (cid:12)(cid:12) q n π v − u − p n (cid:12)(cid:12) − rπ (cid:12)(cid:12) , where r ∈ Z , and π u ≥
1. Rearranging it, and applying the reverse triangle inequality | X − Y | ≥|| X | − | Y || , yield (cid:12)(cid:12)(cid:12)(cid:12) π u (cid:12)(cid:12)(cid:12)(cid:12) π v − u − p n q n (cid:12)(cid:12)(cid:12)(cid:12) − (cid:12)(cid:12)(cid:12)(cid:12) rπq n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) π u q n +1 (cid:12)(cid:12)(cid:12)(cid:12) − (cid:12)(cid:12)(cid:12)(cid:12) rπq n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (19) ≥ q n +1 > . Therefore, relation kπ u + mπ v = rπ is false for any nontrivial integer point ( k, m ) = ( k = 0 , m =0) and r ∈ Z . (ii) The proof for this case is similar. (cid:4) The nonvanishing of the sine function at certain real numbers are required in the proofs ofcertain results. These are verified using either the irrationality of the real number π or viathe infinite product sin ( x ) = x − Q n ≥ (cid:0) − ( xπ − n − ) (cid:1) for any real number x ∈ R or thereflection formula Γ (1 − z ) Γ ( z ) = π/ sin πz of the gamma function Γ( z ) for z ∈ C . Lemma 4.1.
For any pair of integers r = 1 , and m = 0 , the sine function satisfies thefollowings.1. sin ( m ) = 0 . sin ( π r m ) = 0 . Proof. (i) The verification, using the reflection formula of the gamma function, yieldssin( m ) = sin ( π · m/π ) (20)= π Γ (1 − m/π ) Γ ( m/π ) = 0 4or any integer m ≥ z ) has its poles at the negative integers z = n ≤
0, and 1 − mπ and mπ (21)are irrational numbers, not negative integers. (ii) Similar to the previous case. (cid:4) Lemma 4.2. If k and m are nonzero integers, then sin( ke + m ) = 0 . (22) Proof.
The task to prove that the set of nontrivial integer solutions ( k, m ) = (0 ,
0) of theequation sin( ke + m ) = 0 (23)is empty splits into three different cases. Case 1. k = 0, and m = 0. The relationsin ( ke + m ) = sin ( m ) (24) = 0is true, see Lemma 4.1. Case 2. k = 0, and m = 0. By Lemma 3.1, e = aπ for any integer a ≥
1. Thus, the multiple ke = akπ = nπ . This implies that the equationsin( ke + m ) = sin( ke ) = sin( nπ ) = 0 (25)is impossible. Case 3. k = 0, and m = 0. By Lemma 3.2, ke + m = rπ for any integers k , m , and r ∈ Z .This implies that the equation sin( ke + m ) = sin ( rπ ) (26)is impossible. (cid:4) Lemma 4.3. If k and m are nonzero integers, then sin( keπ + m ) = 0 . (27) Proof.
The task to prove that the set of nontrivial integer solutions ( k, m ) = (0 ,
0) of theequation sin( keπ + m ) = 0 (28)is empty splits into three different cases. Case 1. k = 0, and m = 0. The relationsin ( keπ + m ) = sin ( m ) (29) = 0is true, see Lemma 4.1. Case 2. k = 0, and m = 0. Since e is irrational, the relation keπ = nπ , where n = 0, isimpossible. This implies that the equationsin( keπ + m ) = sin( keπ ) = sin( nπ ) = 0 (30)5s impossible. Case 3. k = 0, and m = 0. By Lemma 3.2, keπ + m = (2 r + 1) π for any integers k , m , and r ∈ Z . This implies that the equationsin( keπ + m ) = sin ((2 r + 1) π ) (31)is impossible. (cid:4) Lemma 4.4.
Let ≤ r < s be a pair of integers, and let k and m be nonzero integers, then sin (cid:0) kπ r +1 + mπ s +1 (cid:1) = 0 . (32) Proof.
The task to prove that the set of nontrivial integer solutions ( k, m ) = (0 ,
0) of theequation sin (cid:0) kπ r +1 + mπ s +1 (cid:1) = 0 (33)is empty splits into three different cases. Case 1. k = 0, and m = 0. Since s ≥ (cid:0) kπ r +1 + mπ s +1 (cid:1) = sin (cid:0) mπ s +1 (cid:1) (34)= 0 , where π s +1 ≥ π , is false, see Lemma 4.1. Case 2. k = 0, and m = 0. Since r ≥ (cid:0) kπ r +1 + mπ s +1 (cid:1) = sin( kπ r +1 ) = 0 (35)where π r ≥ π , is false, see Lemma 4.1. Case 3. k = 0, and m = 0. By Lemma 3.2, kπ r +1 + mπ s +1 = aπ for any integers k = 0, m = 0,and a ∈ Z . This implies that the equationsin (cid:0) kπ r +1 + mπ s +1 (cid:1) = sin ( aπ ) (36)is impossible. (cid:4) Lemma 5.1.
For any real number t = kπ , k ∈ Z , and a large integer x ≥ , the finite sum1. X − x ≤ n ≤ x e i tn = sin((2 x + 1) t )sin( t ) . (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X − x ≤ n ≤ x e i tn (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ | sin( t ) | . Proof. (i) Expand the complex exponential sum into two subsums: X − x ≤ n ≤ x e i tn = e − i t X ≤ n ≤ x − e − i tn + X ≤ n ≤ x e i tn . (37)Lastly, use the geometric series to determine the closed form. (cid:4) Linear Independence Of , e , and π Proof. (Theorem 1.1:) On the contrary, the numbers 1, e and π are linearly dependent over therational numbers, and the equation1 · A + e · B + π · C = 0 , (38)where A, B, C ∈ Z × are integers, has a nontrivial rational solution ( A, B, C ) = (0 , , − πC = 2 ( eB + A ) . (39)Take the irrational limit test , see Lemma 2.1, in both sides to obtain I ( − πC ) = I (2( eB + A )) . (40)The left side and the right side are evaluated separately. Left Side.
The verification is based on the identity e − i πC = 1, where C is an integer. Theevaluation of the limit is I ( − πC ) = lim x →∞ x X − x ≤ n ≤ x e − i πCn = lim x →∞ x X − x ≤ n ≤ x . (41) Right Side.
The verification is based on the nonvanishing of the sine function sin ( eB + A ) = 0,see Lemma 4.2. An application of Lemma 5.1 yields I (2( eB + A )) = lim x →∞ x X − x ≤ n ≤ x e i (2( eB + A )) n (42) ≤ lim x →∞ x | sin ( eB + A ) | = 0 . Clearly, these distinct evaluations1 = I ( − πC ) = I (2( eB + A )) = 0 (43)contradict equation (40). This implies that equation (38) does not have a nontrivial rationalsolution ( A, B, C ) ∈ Z × × Z × × Z × . Hence, the numbers 1, e and π are linearly independentover the rational numbers Q × . (cid:4) e + π The continued fraction of the number understudy is e + π = [5; 1 , , , , , , , , , , , , , , , , , , , , , , , . . . ] . (44)The previous result immediately implies that this continued fraction is infinite. Corollary 7.1.
The real number e + π = 5 . . . . is irrational number.Proof. By Theorem 1.1, the equation1 · A + e · B + π · C = 0 , (45)has no nontrivial integer solutions ( A, B, C ) = (0 , , (cid:4) Conjecture 7.1.
The real number e + π is transcendental. Conjecture 7.2.
The irrationality measure of the real number e + π is µ ( e + π ) = 2 . A few values were computed to illustrate the prediction in this conjecture, see Table 1. Thefourth column displays the numerical approximation µ ( e + π ) of the actual value µ ( e + π ).7able 1: Numerical Data For Irrationality Measure | p n /q n − e − π | ≥ q µ ( e + π ) n . n p n q n µ ( e + π )1 5 12 6 13 41 7 3 . . . . . . . . , e , And π − Proof. (Theorem 1.2:) On the contrary, the numbers 1, e and π − are linearly dependent overthe rational numbers, and the equation1 · A + e · B + π − · C = 0 , (46)where A, B, C ∈ Z × are integers, has a nontrivial rational solution ( A, B, C ) = (0 , , − πA = 2 ( eπB + C ) . (47)Take the irrational limit test , see Lemma 2.1, in both sides to obtain I ( − πA ) = I (2( eπB + C )) . (48)The left side and the right side are evaluated separately. Left Side.
The verification is based on the identity e − i πA = 1, where A is an integer. Theevaluation of the limit is I ( − πA ) = lim x →∞ x X − x ≤ n ≤ x e − i πAn = lim x →∞ x X − x ≤ n ≤ x . (49) Right Side.
The verification is based on the nonvanishing sin ( eπB + C ) = 0 of the sinefunction, see Lemma 4.3. An application of Lemma 5.1 yields I (2( eπB + C )) = lim x →∞ x X − x ≤ n ≤ x e i (2( eπB + C )) n (50) ≤ lim x →∞ x | sin ( eπB + C ) | = 0 . Clearly, these distinct evaluations1 = I ( − πA ) = I (2( eπB + C )) = 0 (51)contradict equation (48). This implies that equation (46) does not have a nontrivial rationalsolution ( A, B, C ) ∈ Z × × Z × × Z × . Hence, the numbers 1, e and π − are linearly independentover the rational numbers Q × . (cid:4) The Real Number eπ The continued fraction of the number understudy is eπ = [8; 1 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , . . . ] . (52)The previous result immediately implies that this continued fraction is infinite. Corollary 9.1.
The real number eπ = 8 . . . . is irrational.Proof. By Theorem 1.2, the equation1 · A + e · B + π − · C = 0 , (53)has no nontrivial integer solutions ( A, B, C ) = (0 , , (cid:4) Conjecture 9.1.
The real number eπ is transcendental. Conjecture 9.2.
The irrationality measure of the real number eπ is µ ( eπ ) = 2 . A few values were computed to illustrate the prediction in this conjecture, see Table 2. Thefourth column displays the numerical approximation µ ( eπ ) of the actual value µ ( eπ ).Table 2: Numerical Data For Irrationality Measure | p n /q n − eπ | ≥ q µ ( eπ ) n . n p n q n µ ( eπ )1 8 12 9 13 17 2 4 . . . . . . . .
10 Linear Independence Of , π r , And π s Proof. (Theorem 1.3:) On the contrary, the numbers 1, π r and π s are linearly dependent overthe rational numbers, and the equation1 · A + π r · B + π s · C = 0 , (54)where A, B, C ∈ Z × are integers, has a nontrivial rational solution ( A, B, C ) = (0 , , − πA = 2 (cid:0) π r +1 B + π s +1 C (cid:1) . (55)Take the irrational limit test , see Lemma 2.1, in both sides to obtain I ( − πA ) = I (cid:0) π r +1 B + π s +1 C ) (cid:1) . (56)The left side and the right side are evaluated separately.9 eft Side. The verification is based on the identity e − i πA = 1, where A is an integer. Theevaluation of the limit is I ( − πA ) = lim x →∞ x X − x ≤ n ≤ x e − i πAn = lim x →∞ x X − x ≤ n ≤ x . (57) Right Side.
The verification is based on the nonvanishing sin (cid:0) π r +1 B + π s +1 C (cid:1) = 0 of thesine function, see Lemma 4.4. An application of Lemma 5.1 yields I (cid:0) π r +1 πB + π s +1 C ) (cid:1) = lim x →∞ x X − x ≤ n ≤ x e i ( π r +1 B + π s +1 C ) ) n (58) ≤ lim x →∞ x | sin ( π r +1 B + π s +1 C ) | = 0 . Clearly, these distinct evaluations1 = I ( − πA ) = I (cid:0) π r +1 B + π s +1 C ) (cid:1) = 0 (59)contradict equation (56). This implies that equation (54) does not have a nontrivial rationalsolution ( A, B, C ) ∈ Z × × Z × × Z × . Hence, the numbers 1, π r and π s are linearly independentover the rational numbers Q × . (cid:4)
11 The Real Number π + π The continued fraction of the number understudy is π + π = [13 , , , , , , , , , , , , , , , , , , , , , , , , , , , . . . ] . (60)The previous result immediately implies that this continued fraction is infinite. Corollary 11.1.
The real number π + π = 13 . . . . is irrational.Proof. Set r = 1 and s = 2. By Theorem 1.3, the equation1 · A + π · B + π · C = 0 , (61)has no nontrivial integer solutions ( A, B, C ) = (0 , , (cid:4) Conjecture 11.1.
The real number π + π is transcendental. Conjecture 11.2.
The irrationality measure of the real number π + π is µ ( π + π ) = 2 . A few values were computed to illustrate the prediction in this conjecture, see Table 3. Thefourth column displays the numerical approximation µ ( π + π ) of the actual value µ ( π + π ).
12 Basic Diophantine Approximations Results
All the materials covered in this section are standard results in the literature, see [7], [10], [12],[15], [16], [17], et alii. 10able 3: Numerical Data For Irrationality Measure | p n /q n − eπ | ≥ q µ ( π + π ) n . n p n q n µ ( π + π )1 13 12 1158 89 2 . . . . . . . . . A real number α ∈ R is called rational if α = a/b , where a, b ∈ Z are integers. Otherwise, thenumber is irrational . The irrational numbers are further classified as algebraic if α is the rootof an irreducible polynomial f ( x ) ∈ Z [ x ] of degree deg( f ) >
1, otherwise it is transcendental . Lemma 12.1.
If a real number α ∈ R is a rational number, then there exists a constant c = c ( α ) such that cq ≤ (cid:12)(cid:12)(cid:12)(cid:12) α − pq (cid:12)(cid:12)(cid:12)(cid:12) (62) holds for any rational fraction p/q = α . Specifically, c ≥ /b if α = a/b . This is a statement about the lack of effective or good approximations for any arbitrary rationalnumber α ∈ Q by other rational numbers. On the other hand, irrational numbers α ∈ R − Q haveeffective approximations by rational numbers. If the complementary inequality | α − p/q | < c/q holds for infinitely many rational approximations p/q , then it already shows that the real number α ∈ R is irrational, so it is sufficient to prove the irrationality of real numbers. Lemma 12.2 (Dirichlet) . Suppose α ∈ R is an irrational number. Then there exists an infinitesequence of rational numbers p n /q n satisfying < (cid:12)(cid:12)(cid:12)(cid:12) α − p n q n (cid:12)(cid:12)(cid:12)(cid:12) < q n (63) for all integers n ∈ N . Lemma 12.3.
Let α = [ a , a , a , . . . ] be the continued fraction of a real number, and let { p n /q n : n ≥ } be the sequence of convergents. Then < (cid:12)(cid:12)(cid:12)(cid:12) α − p n q n (cid:12)(cid:12)(cid:12)(cid:12) < a n +1 q n (64) for all integers n ∈ N . This is standard in the literature, the proof appears in [7, Theorem 171], [16, Corollary 3.7], [8,Theorem 9], and similar references.
Lemma 12.4.
Let α = [ a , a , a , . . . ] be the continued fraction of a real number, and let { p n /q n : n ≥ } be the sequence of convergents. Then . q n +1 q n ≤ (cid:12)(cid:12)(cid:12)(cid:12) α − p n q n (cid:12)(cid:12)(cid:12)(cid:12) ≤ q n , 2. a n +1 q n ≤ (cid:12)(cid:12)(cid:12)(cid:12) α − p n q n (cid:12)(cid:12)(cid:12)(cid:12) ≤ q n ,for all integers n ∈ N . The recursive relation q n +1 = a n +1 q n + q n − links the two inequalities. Confer [13, Theorem3.8], [8, Theorems 9 and 13], et alia. The proof of the best rational approximation stated below,appears in [15, Theorem 2.1], and [16, Theorem 3.8]. Lemma 12.5.
Let α ∈ R be an irrational real number, and let { p n /q n : n ≥ } be the sequenceof convergents. Then, for any rational number p/q ∈ Q × ,1. | αq n − p n | ≤ | αq − p | , 2. (cid:12)(cid:12)(cid:12)(cid:12) α − p n q n (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) α − pq (cid:12)(cid:12)(cid:12)(cid:12) ,for all sufficiently large n ∈ N such that q ≤ q n . Theorem 12.1. (Kronecker approximation theorem)
Let α, β ∈ R × be real numbers, and α irrational. Given a small number ε > , there exists infinitely many pairs of integers p, q ∈ N such that | αq − p − β | < ε. (65)The n th dimensional version and related problems are studied in [2], [6], and similar references. The concept of measures of irrationality of real numbers is discussed in [17, p. 556], [5, Chapter11], et alii. This concept can be approached from several points of views.
Definition 12.1.
The irrationality measure µ ( α ) of a real number α ∈ R is the infimum of thesubset of real numbers µ ( α ) ≥ (cid:12)(cid:12)(cid:12)(cid:12) α − pq (cid:12)(cid:12)(cid:12)(cid:12) ≪ q µ ( α ) (66)has finitely many rational solutions p and q . Equivalently, for any arbitrary small number ε > (cid:12)(cid:12)(cid:12)(cid:12) α − pq (cid:12)(cid:12)(cid:12)(cid:12) ≫ q µ ( α )+ ε (67)for all large q ≥ References [1] Aigner, Martin; Ziegler, Gunter M.
Proofs from The Book . Fifth edition. Springer-Verlag,Berlin, 2014.[2] Apostol, P.
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An Introduction to Irrationality and Transcendence Methods, Lecture1. ∼∼