aa r X i v : . [ m a t h . G M ] J un The Product eπ Is Irrational
N. A. Carella
Abstract:
This note shows that the product eπ of the natural base e and the circle number π is an irrationalnumber. Contents
Irrational number; Natural base e ; Circle number π . AMS Mathematical Subjects Classification: he product eπ is irrational The natural base e and the circle number π are irrational numbers, but the arithmetic properties of the sum e + π and the product eπ seemto be unknown. The known information on the continued fractions and the convergents of the two irrationalnumbers e and π are used here toconstruct an infinite subsequence of rational approximations for the product eπ . Theorem 1.1.
The product eπ is an irrational number.Proof. Let e = [ a , a , a , . . . ] be the continued fraction of the irrational number e . By Lemma 2.3, thereexists a sequence ofconvergents { p n /q n : n ∈ N } such that (cid:12)(cid:12)(cid:12)(cid:12) e − p n q n (cid:12)(cid:12)(cid:12)(cid:12) < a n +1 q n . (1)Similarly, let π = [ b , b , b , . . . ] be the continued fraction of the irrational number π , and let { u m /v m : m ∈ N } be thesequence of convergents such that (cid:12)(cid:12)(cid:12)(cid:12) π − u m v m (cid:12)(cid:12)(cid:12)(cid:12) < b m +1 v m . (2)Now, suppose that the product eπ = r/s = 8 . . . . ∈ Q − Z is a rational number. Then1 s q n v m ≤ (cid:12)(cid:12)(cid:12)(cid:12) eπ − p n u m q n v m (cid:12)(cid:12)(cid:12)(cid:12) < πa n +1 q n + 2 eb m +1 v m + 1 a n +1 b m +1 q n v m . (3)The left side follows from Lemma 2.1, and the right side follows from Lemma 2.4.Multiplying through by the integer sq n v m leads to1 ≤ | eπsq n v m − sp n u m | < πsv m a n +1 q n + 2 esq n b m +1 v m + sa n +1 b m +1 q n v m . (4)By Lemma 3.1, there exists a pair of infinite subsequences of rational approximations p k − q k − and u m k v m k (5)that satisfies the inequality (2 k ) − ε q k − ≪ v m k ≪ k ) − ε q k − , (6)where ε > k, m k −→ ∞ . Replacing (6) into (4) yields1 ≤ | eπsq k − v m k − sp k − u m k | (7) < πsv m k a k − q k − + 2 esq k − b m k +1 v m k + sa k − b m k +1 q k − v m k ≪ − ε πsk ε + 2 es (2 k ) − ε b m k +1 + s kb m k +1 (2 k ) − ε q k − ≪ − ε πsk ε + 2 es (2 k ) − ε + s (2 k ) − ε q k − , where the last line uses b m ≥
1. By hypothesis, the quantities e · π · s · p k − · v m k ≥ and s · q k − · u m k ≥ k → ∞ . This data implies that the number e · π · s · q k − · v m k (9)can not be an integer. Ergo, the product eπ is not a rational number. (cid:4) he product eπ is irrational The structure of the proof, in equations (3), (4), and (7), is similar to some standard proofs of irrationalnumbers. Among these well known proofs are the Fourier proof of the irrationality of e , see [2, p. 35], [19,p. 13], the proofs for ζ (2), and ζ (3) in [4], et alii.A sample of numerical data is compiled in Section 4 to demonstrate the practicality of this technique. Except for Lemmas 2.4, and 3.1, all the materials covered in this section are standard results in the literature,see [9], [10], [12], [15], [17], [19], et alii.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 root of an irreduciblepolynomial f ( x ) ∈ Z [ x ] of degree deg( f ) >
1, otherwise it is transcendental . Lemma 2.1.
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) (10) 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 rational number α ∈ Q by other rationalnumbers. On the other hand, irrational numbers α ∈ R − Q have effective approximations by rationalnumbers. If thecomplementary inequality | α − p/q | < c/q holds for infinitely many rational approximations p/q , then italready shows that thereal number α ∈ R is irrational, so it is sufficient to prove the irrationality of real numbers. Lemma 2.2 (Dirichlet) . Suppose α ∈ R is an irrational number. Then there exists an infinite sequence ofrational 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 (11) for all integers n ∈ N . Lemma 2.3.
Let α = [ a , a , a , . . . ] be the continued fraction of a real number, and let { p n /q n : n ≥ } bethesequence 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 (12) for all integers n ∈ N . This is standard in the literature, the proof appears in [9, Theorem 171], [17, Corollary 3.7], and similarreferences.Lemma 2.2 and Lemma 2.3 suggest that a nonrational product αβ should have an inequality of the form0 < (cid:12)(cid:12)(cid:12)(cid:12) αβ − p n u m q n v m (cid:12)(cid:12)(cid:12)(cid:12) < ca n +1 b m +1 q n v m , (13)where c > he product eπ is irrational Lemma 2.4.
Let α = [ a , a , a , . . . ] and β = [ b , b , b , . . . ] be distinct continued fractions for two distinctirrational numbers α, β ∈ R such that αβ = m ∈ Z , is not an integer, respectively. Then < (cid:12)(cid:12)(cid:12)(cid:12) αβ − p n u m q n v m (cid:12)(cid:12)(cid:12)(cid:12) < βa n +1 q n + 2 αb m +1 v m + 1 a n +1 b m +1 q n v m , (14) where { p n /q n : n ≥ } and { u m /v m : m ≥ } are the sequences of convergents respectively.Proof. By Lemma 2.3, there exists a sequence of convergents { p n /q n : n ∈ N } such that (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 , (15)and the corresponding long form is p n q n − a n +1 q n < α < p n q n + 1 a n +1 q n . (16)Similarly, there exists a sequence of convergents { u m /v m : m ∈ N } such that (cid:12)(cid:12)(cid:12)(cid:12) β − u m v m (cid:12)(cid:12)(cid:12)(cid:12) < b m +1 v m , (17)and the corresponding long form is u m v m − b m +1 v m < β < u m v m + 1 b m +1 v m . (18)The product of the last two long forms returns (cid:18) p n q n − a n +1 q n (cid:19) (cid:18) u m v m − b m +1 v m (cid:19) < αβ < (cid:18) p n q n + 1 a n +1 q n (cid:19) (cid:18) u m v m + 1 b m +1 v m (cid:19) . (19)Expanding these expressions produces p n u m q n v m − a n +1 q n u m v m − b m +1 v m p n q n + 1 a n +1 b m +1 q n v m < αβ (20) < p n u m q n v m + 1 a n +1 q n u m v m + 1 b m +1 v m p n q n + 1 a n +1 b m +1 q n v m . This can be rearranging as the equivalent inequality p n u m q n v m − a n +1 q n u m v m − b m +1 v m p n q n − a n +1 b m +1 q n v m < αβ (21) < p n u m q n v m + 1 a n +1 q n u m v m + 1 b m +1 v m p n q n + 1 a n +1 b m +1 q n v m . The previous inequality is rewritten as a standard absolute value inequality0 < (cid:12)(cid:12)(cid:12)(cid:12) αβ − p n u m q n v m (cid:12)(cid:12)(cid:12)(cid:12) < a n +1 q n u m v m + 1 b m +1 v m p n q n + 1 a n +1 b m +1 q n v m . (22)To complete the proof, use the trivial upper bound p n q n ≤ α and u m v m ≤ β, (23)for all large integers n, m ≥
1, confer (16) and (18). (cid:4) he product eπ is irrational For distinct irrationals α, β ∈ (0 , < (cid:12)(cid:12)(cid:12)(cid:12) αβ − p n u m q n v m (cid:12)(cid:12)(cid:12)(cid:12) < a n +1 q n + 1 b m +1 v m + 1 a n +1 b m +1 q n v m (24)can be used to stream line the proof of a result such as Theorem 1.1. Lemma 2.5 (Euler) . The continued fraction e = [2 , , , , , , , , , , , . . . ] of the natural base hasunbounded quotients and the subsequence of convergents p n /q n satisfies the inequality (cid:12)(cid:12)(cid:12)(cid:12) e − p n q n (cid:12)(cid:12)(cid:12)(cid:12) < a n +1 q n . (25)The quotients have the precise form a = 2 , a k = a k − = 1 , a k − = 2 k, (26)for k ≥
1. The derivation appears in [13], [10, Theorem 2], [17, Theorem 3.10], [6], and other.
The correlation of a pair of convergents { p n /q n : n ≥ } and { u m /v m : n ≥ } provides information on thedistribution of nearly equal values of the continuants q n and v m .The regular pattern and unbounded properties of the partial quotients a n = a k − = 2 k of the continued frac-tion of e , see Lemma 2.5, are used here to generate a pair of infinite subsequences of rational approximations { p k − /q k − : k ≥ } and { u m k /v m k : k ≥ } , for which the product p k − u m k q k − v m k −→ eπ as k, m k −→ ∞ . (27)Furthermore, the values q k − ≍ v m k are sufficiently correlated. The notation f ( x ) ≍ g ( x ) is defined by g ( x ) ≪ f ( x ) ≪ g ( x ).The recursive relations p − = 1 , p = a , p n = a n p n − + p n − ,q − = 0 , q = 1 , q n = a n q n − + q n − , (28)for all n ≥
1, see [9], [12], [15], are used to estimate the rate of growth of the subsequences of continuants { q n : n ≥ } and { v m : m ≥ } . Lemma 3.1.
Let e = [ a , a , a , . . . ] and π = [ b , b , b , . . . ] be the continued fractions of this pair of irrationalnumbers. Let δ > and ε > be a pair of arbitrary small numbers. Then, the followings hold. (i) If b m = o ( m ) , then the exists a pair of subsequences of convergents p k − /q k − and u m k /v m k suchthat (2 k ) − ε q k − ≪ v m k ≪ k ) − ε q k − . (29)(ii) If b m = O ( m ) , then the exists a pair of subsequences of convergents p k − /q k − and u m k /v m k suchthat (2 k ) − ε q k − ≪ v m k ≪ k ) − ε q k − . (30)(iii) If b m = O ( m δ ) , then the exists a pair of subsequences of convergents p k − /q k − and u m k /v m k suchthat ( m k ) − ε v m k ≪ q n k ≪ m k ) − ε v m k . (31)5 he product eπ is irrational Proof. Case (i): The partial quotients b m = o ( m ) are bounded or unbounded. Make the change of index n ≡ −→ k = ( n + 2) / p k − /q k − of the number e as k → ∞ , see Lemma 2.5 for more details. Observe that... ... ... q k − = q k − + q k − = q k − q k − = 2 kq k − + q k − ≍ kq k − q k = q k − + q k − ≍ kq k − q k +1) − = q k +1) − + q k +1) − ≍ kq k − q k +1) − = 2( k + 1) q k +1) − + q k +1) − ≍ k ( k + 1) q k − q k +1) = q k +1) − + q k +1) ≍ k ( k + 1) q k − q k +2) − = q k +2) − + q k +2) − ≍ k ( k + 1) q k − q k +2) − = 2( k + 2) q k +1) − + q k +2) − ≍ k ( k + 1)( k + 2) q k − ... ... ... (32)This verifies that these numbers has exponential rate of growth in k of the form q k + t ) − ≍ (4 k ) t +1 q k − , (33)for some t ≥
0, as k → ∞ . Next, consider the sequence of convergents u m /v m of the number π . Byhypothesis, the partial quotients b m = o ( m ) are bounded or unbounded. Furthermore, to simplify thenotation, assume that b m k ≍ m − δk for infinitely many integers m k = m k , m k , m k , m k , . . . ≥
1. Then, thisimplies the existence of a subsequence of convergents u m k /v m k such that... ... ... v m k = b m k v m k − + v m k − = v m k v m k = b m k v m k − + v m k − ≍ (2 k − k m − δk ) v m k v m k = b m k v m k − + v m k − ≍ (2 k − k m − δk ) v m k v m k = b m k v m k − + v m k − ≍ (2 k − k m − δk ) v m k ... ... ... (34)Moreover, the existence of a single value v m k >
1, see Tables 1 and 2, such that(2 k ) − ε q k − ≪ v m k ≪ k ) − ε q k − , (35)implies the existence of an infinite subsequence of lower bounds v m ks = (2 k s − k m − δk ) s v m k (36) ≫ (2 k s − k m − δk ) s (2 k ) − ε q k − ≫ (2 k ) − ε q k + t ) − , where q k + t ) − = (2 k s − k m − δk ) s q k − , (37)and use (33) to identify the relation (2 k s − k m − δk ) s = o ((4 k ) t +1 ) (38)for some s ≥ t ≥
1. The corresponding subsequence of upper bounds satisfies v m ks = (2 k s − k m − δk ) s v m k (39) ≪ k s − k m − δk ) s (2 k ) − ε q k − ≪ k ) − ε q k + t ) − . he product eπ is irrational Combining the last two inequalities yields the required relation(2 k ) − ε q k + t ) − ≪ v m ks ≪ k ) − ε q k + t ) − (40)for some s, t ≥ k → ∞ . Case (ii): The the partial quotients b m are bounded or unbounded, and b m = O ( m ) . The proof for this caseis similar to Case (i).
Case (ii): The partial quotients b m are unbounded, and b m = O ( m δ ). In this case, (40) can fail, but sincethe inequality in Lemma 2.4 is symmetric in q n and v m , the proof is almost the same as Case (i), but thesubsequences of convergents are switched to obtain the required relation(2 k t − k m − δk ) t +1 v m kt ≪ q n k + s ≪ k t − k m − δk ) t +1 v m kt (41)for some s, t ≥ k → ∞ . (cid:4) The distribution of all the continuants { q n : n ≥ } associated with a subset of continued fractions of boundedpartial quotients is the subject of Zeremba conjecture, see [5] for advanced details. For any continued fraction,the numbers { q n : n ≥ } have exponential growth q n = a n q n − + q n − ≥ √ ! n , (42)which is very sparse subsequence of integers. The least asymptotic growth occurs for the (1 + √ / , , , . . . ]. But the combined subset of continuants for a subset of continued fractions of bounded partialfractions has positive density in the sunset of integers N = { , , , . . . } . For each fixed pair of index ( n, m ), the basic product inequality0 < (cid:12)(cid:12)(cid:12)(cid:12) eπ − p n q n u m v m (cid:12)(cid:12)(cid:12)(cid:12) < πa n q n + 2 eb m v m + 1 a n +1 b m +1 q n v m (43)is used to test each rational approximation. It is quite easy to find the pairs ( n, m ) to construct a subsequenceof rational approximations p n u m q n v m → eπ (44)as n, m → ∞ . The subsequence of rational approximations is generated by the following algorithm. Algorithm 1.
Fix an arbitrary small number ε > n ≡ k = ( n + 2) / a n +1 = a k − = 2 k , and fix the convergent p n /q n = p k − /q k − of the natural base e .4. Choose a convergent u m k /v m k of π such that(2 k ) − ε q k − ≤ v m k ≤ k ) − ε q k − . (45)5. If Step 4 fail, then increment n ≡ he product eπ is irrational ( n, m ) = (19,10)( a n , b m ) = (12 , p n = 13580623 q n = 4996032 u m = 5419351 v m = 1725033 (cid:12)(cid:12)(cid:12) eπ − p n q n u m v m (cid:12)(cid:12)(cid:12) = 0.00000000000012256862192 a n q n u m v m + b m v m p n q n + a n +1 b m +1 q n v m = 0.000000000000136378880Table 1: A Rational Approximation Of The Product eπ .( n, m ) = (31,21)( a n , b m ) = (20 , p n = 22526049624551 q n = 8286870547680 u m = 3587785776203 v m = 1142027682075 (cid:12)(cid:12)(cid:12) eπ − p n q n u m v m (cid:12)(cid:12)(cid:12) = 8 . × − a n q n u m v m + b m v m p n q n + a n +1 b m +1 q n v m = 1 . × − Table 2: A Rational Approximation Of The Product eπ .6. Output a k − = 2 k , p k − /q k − and u m k /v m k .The parameters for two small but very accurate approximations are listed in the Tables 1 and 2. Theseexamples demonstrate the practicality of the algorithm. Remark 4.1.
Step 5 makes the algorithm independent of the rate of growth of the partial quotients of thenumber π . Various versions of this algorithm are possible, for example, by modifying the interval in (45). The continued fraction α = [ a , a , a , . . . ] of an irrational number α ∈ R − Q and its inverse1 α = ( [0 , a , a , · · · ] if α > , [ a , a , a , · · · ] if < α < , (46)see [10, p. 14], are the same up to a partial quotient a . Thus, the same result as in Theorem 1.1 is valid forthe ratios πe and eπ . (47)Furthermore, since 1 ± πe − is irrational, the sum e + π = e (1 + πe − ), and difference e − π = e (1 − πe − ),are also irrationals.In general, the products of irrational numbers are not irrationals, for example, the products of the irrationalalgebraic integers α = a + b √ d and β = a − b √ d , with d = m , are rationals. However, the products ofalmost every pair of distinct, and nonconjugates irrationals seems to be irrationals. This can be proved usingKhintchin theorem. 8 he product eπ is irrational The rationality or irrationality of a product is heavily determined by the rate of growth of the partial quo-tients of the continued fractions of these numbers. The next result covers a large collection of such numbers.
Theorem 5.1.
Let α = [ a , a , a , . . . ] and β = [ b , b , b , . . . ] be distinct continued fractions for two distinctirrational numbers α, β ∈ R such that αβ = m ∈ Z , is not an integer, respectively. Suppose that at least onehas unbounded quotients a n → ∞ or b m → ∞ . Then, the product αβ is irrational.Proof. Same as Theorem 1.1 mutatis mutandis. (cid:4)
The simplest cubic irrationals such as 2 / = [1 , , , , , , , , , , , , . . . ] have unknown patterns of par-tial quotients, and there are no information on the magnitude a n = O ( n r ) with r ≥ R = { α , α , . . . , α d } ⊂ R be the set of real roots of the irreduciblepolynomial f ( x ) = a d x d + · · · + a x + a ∈ Z [ x ], and let A ⊂ R , and B = R − A . Then, the product ofirrational real numbers α = Y α i ∈A α i and β = Y β i ∈B β i (48)is an integer αβ = a ∈ Z . There lots of algebraic relationships. In particular, all the symmetric polynomials S ( x , x , · · · , x d ) ∈ Z [ x , x , · · · , x d ] are integer valued at the roots of any irreducible polynomial f ( x ) ∈ Z [ x ]. Theorem 6.1.
Let α = [ a , a , a , . . . ] be a root of an irreducible polynomial f ( x ) ∈ Z [ x ] of deg( f ) = d > .Then, the partial quotients a n = O (1) are bounded for all n ≥ .Proof. Let α = [ a , a , a , . . . ] and β = [ b , b , b , . . . ] be distinct continued fractions for two distinct irrationalnumbers α = α , and β = α α · · · α d − ∈ R , where f ( α i ) = 0 for i = 0 , , , ..., d −
1. Assume that thepartial quotients a n → ∞ as n → ∞ are unbounded. Then, a modified version of the proof of Theorem 1.1can be used to prove that the product αβ is irrational.But, αβ = f (0) ∈ Z is an integer, which contradicts the assumption a n → ∞ as n → ∞ . Therefore, thepartial quotients a n = O (1) are bounded. (cid:4) The converse of this theorem for the products of algebraic irrational numbers is not valid. For example, theproduct of the two algebraic irrational numbers √ · √ √ √ √ e m · e n = e m + n for n = ± m ,is irrational and has unbounbed partial quotients, The theory of uniform distribution is intrinsically linked to irrational numbers, confer [11]. Given this fact,it is not surprising that some of its technique can be used to study the properties of irrational numbers.
Definition 7.1.
A pair of real numbers x, y ∈ R are equivalent over the rational numbers if there exists r ∈ Q such that x = ry . Definition 7.2.
A pair of real sequences { S n ∈ R : n ≥ } and { T n ∈ R : n ≥ } are equivalent over therational numbers if there exists r ∈ Q such that S n = rT n for almost all integers n ∈ N .9 he product eπ is irrational Lemma 7.1. (Equivalent Number Test)
If the real sequences { S n ∈ R : n ≥ } and { T n ∈ R : n ≥ } areequivalent over the rational numbers, then the limits relations lim x →∞ x X − x ≤ n ≤ x e iS n = lim x →∞ x X − x ≤ n ≤ x e iT n (49) holds. The
Equivalent Number Test in Lemma 7.1 effectively performs the followings tasks.1. Classifies the two rational numbers as a single class.2. Classifies a known irrational number α ∈ R − Q and a rational number r ∈ Q as two distinct equivalentclasses. That is, as not being equivalent. For example, set S n = 2 παn and R n = 2 πrn and applyLemma 7.2.3. Classifies a known irrational number α ∈ R − Q and π as two distinct equivalent classes. That is, asnot being equivalent. Thisfollows from Theorem 7.1.However, the converse of Lemma 7.1 is not valid: it cannot distinguish between two known irrational numbers,which do not involve the number π is some effective way, as being inequivalent. For example, it cannotestablish that √ √ Theorem 7.1.
The following numbers are linearly independent over the rational numbers Q . (i) The nonnegative numbers π and α ∈ R − π Q . (ii) The nonnegative numbers ln π and α ∈ R − ln π Q . Proof. (i) Suppose that the equation a · α + b · π = c has a nontrivial rational solution ( a, b, c ) = (0 , , bπn = 2( c − aα ) n for all n ∈ Z . Accordingly, Lemma 7.2 implies that the limitslim x →∞ x X − x ≤ n ≤ x e i bπn = lim x →∞ x X − x ≤ n ≤ x e − i aα − c ) n . (50)are equivalent. These limits are evaluated in two distinct ways.I. Based on the properties of the number π . Use the identity e i πn = 1 to evaluate of the left side limit aslim x →∞ x X − x ≤ n ≤ x e i bπn = lim x →∞ x X − x ≤ n ≤ x . (51)II. Based on the properties of the number aα − c . Since the number t = aα − c = kπ for any irrationalnumber α ∈ R − ln π Q , and any integer k ∈ Z , the value sin( t ) = sin( aα − c ) = 0. Applying Lemma 7.2, theright side has the limit lim x →∞ x X − x ≤ n ≤ x e − i aα − c ) n = lim x →∞ x sin((2 x + 1) t )sin( t ) ≤ lim x →∞ x aα − c ) (52)= 0 . Clearly, these distinct limits contradict equation (54). Hence, the Diophantine linear equation a · α + b · π = c ∈ Q [ α, π ] has no solutions in rational numbers. 10 he product eπ is irrational (ii) Assume a · α + b · ln π = c has a nontrivial rational solution ( a, b, c ) = (0 , , πn = 2 e ( c − aα ) /b n, (53)where n ∈ Z . Procceed as in the previous case, and evaluate the limitslim x →∞ x X − x ≤ n ≤ x e i πn = lim x →∞ x X − x ≤ n ≤ x e − i e ( c − aα ) /b n . (54) (cid:4) Corollary 7.1.
The followings statements are valid. (i)
The numbers e and π are linearly independent over the rational numbers. (ii) The trace π + e and norm π · e of the polynomial f ( x ) = ( x − π )( x − e ) are irrational numbers.Proof. (i) To prove the linear independence assume that a · e + b · π = c, where a, b, c ∈ Q , has a nontrivialsolution in rationalnumbers ( a, b, c ) = ((0 , , b · π · n = 2( c − a · e ) n for all n ∈ Z . Repeat the argument in theproof of Theorem 7.1. (ii) The irrationality of the trace e + π follows from linear independence, Theorem7.1. To prove the irrationality of the norm, assume that e · π = r ∈ Q is rational. Set 2 πn = 2 e − rn , andapply the Equivalent number test, Lemma 7.1. (cid:4) Corollary 7.2.
The followings statement are valid. (i)
The number π and ln 2 are linearly independent over the rational numbers. (ii) The trace π + ln 2 and norm π · ln 2 of the polynomial f ( x ) = ( x − π )( x − ln 2) are irrational numbers.Proof. The same as the previous Corollary 7.1. (cid:4)
Theorem 7.2.
The followings statement are valid. (i)
The number e π a is an irrational number for all rational a ∈ Q . (ii) The number e π a + π b is an irrational number for any rational pair a, b ∈ Q .Proof. (i) It is sufficient to consider a = 0 since the case a = 0 reduces to e π a = e . On the contrary, let e = e π a = r ∈ Q be a rational number. Rewrite it as π = π − a ln r . Now apply the equivalent number test,Lemma 7.1, to the sequences S n = 2 πn = T n = 2 nπ − a ln r , where n ≥ (cid:4) Corollary 7.3.
The followings statement are valid. (i)
The real number i iπ a is irrational number for any rational a ∈ Q . (ii) The real number i i ( π a + π b ) is irrational number for any rational pair a, b ∈ Q .Proof. (i) For a = 0, is the implest i i = e − π/ = r ∈ Q , and taking logarithm yield the equation − π = 2 ln r .Now apply theequivalent number test, Lemma 7.1, to the sequences S n = − πn = T n = 2 n ln r , where n ∈ Z . In fact, byTheorem 7.1, the twonumbers − π/ r are linearly independent over the rational numbers. (cid:4) Lemma 7.2.
For any real number t = 0 and a large integer x ≥ , the finite sum X − x ≤ n ≤ x e i tn = sin((2 x + 1) t )sin( t ) . (55)11 he product eπ is irrational Proof.
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 . (56)Lastly, use the geometric series to determine the closed form. (cid:4) Lemma 7.3.
The number ln r ∈ Q is an irrational for all rational r ∈ Q Very few arithmetic properties of the zeta constants are known. Recently, it was proved that1 , ζ (5) , ζ (5) , . . . , ζ (25) , (57)are linear independent over the rational numbers, confer [20]. An improment of earlier works appears in [8],it proves that many consecutive odd zeta constants are irrationals. In section, it is demonstrated that thezeta constants are pairwise linear independent. Theorem 8.1.
The following numbers are linearly independent over the rational numbers Q . (i) The nonnegative numbers , π r and π s , where r, s ∈ N are distinct nonzero integers. (ii) The nonnegative numbers , ζ (2 r + 1) and ζ (2 s + 1) , where r, s ∈ N are distinct nonzero integers.Proof. (i) Suppose that the equation 1 · a + π r · b + π s · c = 0 has a nontrivial rational solution ( a, b, c ) = (0 , , a, b, c ∈ Z are integers. Multiplying by 2 πn across the board, and rewriteit in the equivalent form 2 aπn = − (cid:0) π r +1 b + π s +1 c (cid:1) πn (58)for all n ∈ Z . Accordingly, Lemma 7.2 implies that the limitslim x →∞ x X − x ≤ n ≤ x e i aπn = lim x →∞ x X − x ≤ n ≤ x e − i ( π r +1 b + π s +1 c ) πn . (59)are equivalent. These limits are evaluated in two distinct ways.I. Based on the properties of the number π . Use the identity e i πn = 1 to evaluate of the left side limit aslim x →∞ x X − x ≤ n ≤ x e i aπn = lim x →∞ x X − x ≤ n ≤ x . (60)II. Based on the properties of the number π r +1 b + π s +1 c . Since r, s ≥
1, and r = s , the number t = π r +1 b + π s +1 c = πk for any nontrivial pair ( b, c ) = (0 , k ∈ Z , the value sin( t ) =sin (cid:0) π r +1 b + π s +1 c (cid:1) = 0. Applying Lemma 7.2, the right side has the limitlim x →∞ x X − x ≤ n ≤ x e − i ( π r +1 b + π s +1 c ) n = lim x →∞ x sin((2 x + 1) t )sin( t ) ≤ lim x →∞ x π r +1 b + π s +1 c ) (61)= 0 . Clearly, these distinct limits contradict equation (54). Hence, the Diophantine linear equation 1 · a + π r · b + π s · c = 0 has no solutions in rational numbers ( a, b, c ) = (0 , , he product eπ is irrational (ii) Suppose that the equation 1 · a + ζ (3) · b + ζ (5) · c = 0 has a nontrivial rational solution ( a, b, c ) = (0 , , a, b, c ∈ Z are integers. Multiplying by 2 πn across the board, and rewriteit in the equivalent form 2 aπn = − ζ (3) b + ζ (5) c ) πn (62)for all n ∈ Z . Accordingly, Lemma 7.2 implies that the limitslim x →∞ x X − x ≤ n ≤ x e i aπn = lim x →∞ x X − x ≤ n ≤ x e − i ζ (3) b + ζ (5) c ) πn . (63)are equivalent. These limits are evaluated in two distinct ways.I. Based on the properties of the number π . Use the identity e i πn = 1 to evaluate of the left side limit aslim x →∞ x X − x ≤ n ≤ x e i aπn = lim x →∞ x X − x ≤ n ≤ x . (64)II. Based on the properties of the number ( ζ (3) b + ζ (5) c ) π . Since the number t = ( ζ (3) b + ζ (5) c ) π = kπ forany nontrivial pair ( b, c ) = (0 , k ∈ Z , the value sin( t ) = sin ( ζ (3) πb + ζ (5) πc ) = 0. Notethat since ζ (3) is irrational, the expressionsin ( ζ (3) πb + ζ (5) πc ) = cos ( ζ (3) πb ) sin ( ζ (5) πc ) + cos ( ζ (5) πc ) sin ( ζ (3) πb ) = 0 (65)independently of the value of ζ (5) πc = 0. Applying Lemma 7.2, the right side has the limitlim x →∞ x X − x ≤ n ≤ x e − i ζ (3) b + ζ (5) c ) πn = lim x →∞ x sin((2 x + 1) t )sin( t ) ≤ lim x →∞ x ζ (3) πb + ζ (5) πc ) (66)= 0 . Clearly, these distinct limits contradict equation (54). Hence, the Diophantine linear equation 1 · a + ζ (3) · b + ζ (5) · c = 0 has no solutions in rational numbers ( a, b, c ) = (0 , , (cid:4) Another way to verify that sin( t ) = sin ( ζ (3) πb + ζ (5) πc ) = 0, is to note that ζ (3) is irrational, so theexpression sin ( ζ (3) πb + ζ (5) πc ) = cos ( ζ (3) πb ) sin ( ζ (5) πc ) + cos ( ζ (5) πc ) sin ( ζ (3) πb ) = 0 (67)independently of the value of ζ (5) πc . Corollary 8.1.
The number ζ (5) is irrational.Proof. By Theorem 8.1 the Diophantine equation 1 · a + ζ (3) · b + ζ (5) · c = 0 has a nontrivial rational solution( a, b, c ) = (0 , , ζ (5) = a/c ∈ Q . (cid:4) A zeta constant at the even integer argument has an exact Euler formula ζ (2 n ) = ( − n +1 (2 π ) n B n n )! (68)13 he product eπ is irrational in terms of the Bernoulli numbers B n , for n ≥
1. This formula expresses each zeta constant ζ (2 n ) as arational multiple of π n . The formula for the evaluation of the first even zeta constant ζ (2), known as theBasel problem, was proved by Euler, later it was generalized to all the even integer arguments. Today, thereare dozens of proofs, see [16] for an elementary introduction. In contrast, the evaluation of a zeta constantat an odd integer argument has one or two complicated transcendental power series. The general forms ofthese formulas are ζ ( s ) = a n π n − − b n X n ≥ n n − ( e πn −
1) if s = 4 n − ,a n π n − − b n X n ≥ n n − ( e πn − − c n X n ≥ n n − ( e πn + 1) if s = 4 n − , (69)where a n , b n , c n ∈ Q are rational numbers. These formulas express each zeta constat ζ (2 n + 1) as a nearlyrational multiple of π n +1 . The derivations involve the Ramanujan series for the zeta function, and appearin [18], [7], et alii. The first few are1. ζ (3) = 7 π − X n ≥ n ( e πn −
1) ,2. ζ (5) = π − X n ≥ n ( e πn − − X n ≥ n ( e πn + 1) , ζ (7) = 19 π − X n ≥ n ( e πn − , et cetera. These analysis are summarized in a compact formula. Definition 9.1.
Let s ≥ π -representation of the zeta constant ζ ( s ) is defined by theformula ζ ( s ) = ( rπ s if s = 4 n, n + 2 ,rπ s − u if s = 4 n − , n − , (70)where r ∈ Q is a rational number and u ∈ R is a real number.The irrationality of the first even zeta constant ζ (2) was proved by the technique of Lambert, see [3, p. 129].The irrationality of the first odd zeta constant ζ (3) was proved by Apery in 1978, [1]. The irrationality of theother odd zeta constants ζ ( s ) remain unknown for s ≥
5, see [8] and [20]. The π -representation in Definition9.1 offers a recursive method for proving the irrationality of ζ (2 n + 1) from the known irrationality of π n +2 for n ≥
1. For example, the irrationality of ζ (3) , ζ (5) , ζ (7) , ζ (9) , . . . , (71)can be derived from the known irrationality of the numbers π , π , π , π , . . . . More generally, this idea canbe used to recursively prove the irrationality of ζ ( s ) from the known irrationality of π s +1 for s ≥
2. Theinner working of this technique is demonstraded here for s = 5. Theorem 9.1.
The number ζ (5) is irrational.Proof. Use the π -representation of this number to rewrite this equation as ζ (5) = π
294 + U, (72)where U = 0 is real number, refer to Definition 9.1 for more information. The proof is split into two cases. Case 1. U = r π + r , with r , r ∈ Q × rational numbers. This immediately implies that ζ (5) = π
294 + U = π
294 + r π + r = r π + r , (73)14 he product eπ is irrational where r ∈ Q × a rational number. Ergo, ζ (5) is irrational. Case 2. U = r π + r , with r , r ∈ Q × rational numbers. Suppose that the equation ζ (5) = a/b ∈ Q is arational number. Proceed to rewrite the pi - representation as ab = π
294 + U, (74)where U = 0 is real number, refer to Definition 9.1 for more information. Multiply by 2 πn and the lowestcommon multiple to obtain the equation2 Aπn = 2 (cid:0) Bπ + CU π (cid:1) n, (75)where A = 294 a, B = b, C = 294 b ∈ Z and U ∈ R .Next, it will be demonstrated that left side sequence { Aπn : n ∈ Z } is not uniformily distriduted, but theright side sequence { (cid:0) Bπ + CU π (cid:1) n : n ∈ Z } is uniformly distributed. Toward this end, consider the limitslim x →∞ x X − x ≤ n ≤ x e i Aπn = lim x →∞ x X − x ≤ n ≤ x e i ( Bπ + CUπ ) n . (76)These limits are evaluated in two distinct ways.I. Based on the properties of the number 2 Aπ . Use the identity e i Aπ = 1, where A is a fixed integer, toevaluate the limit of the left side of equation (76) aslim x →∞ x X − x ≤ n ≤ x e i Aπn = lim x →∞ x X − x ≤ n ≤ x . (77)II. Based on the properties of the number Bπ + CU π , where U = r π + r . By Lemma 9.1 sin( t ) =sin (cid:0) Bπ + CU π (cid:1) = 0. Thus, Lemma 7.2 is applicable, and the limit of the left side of equation (76) islim x →∞ x X − x ≤ n ≤ x e i ( Bπ + CUπ ) n = lim x →∞ x sin((2 x + 1) t )sin( t ) ≤ lim x →∞ x Bπ + CU π ) (78)= 0 . Clearly, these distinct limits contradict equation (76). Hence, the number ζ (5) ∈ R is not a rational number. (cid:4) Lemma 9.1.
If the number U = r π + r , then sin (cid:0) Bπ + CU π (cid:1) = 0 , where r , r ∈ Q × are rationalnumbers, and B, C ∈ Z × are integers.Proof. The contrary statement satisfies the relation0 = sin (cid:0) Bπ + CU π (cid:1) = cos (cid:0) Bπ (cid:1) sin ( CU π ) + cos (
CU π ) sin (cid:0) Bπ (cid:1) . (79)Hence, tan( Bπ ) = − tan( CU π ). Since the tangent function is periodic and one-to-one on the interval( − π/ , π/ Bπ = − CU π + mπ , for some m ∈ Z . Equivalently U = r π + r with r , r ∈ Q × . But this contradicts the assumption. (cid:4) he product eπ is irrational
10 Problems
Exercise 10.1.
Let f ( x ) = c d x d + · · · + c x + c ∈ Z [ x ] be an irreducible polynomial of deg( f ) = d > a n = B ( c , c , . . . c d ), where B ( x , x , . . . x d ) ∈ R [ x , x , . . . x d ] is a bounded function. Exercise 10.2.
Given two continued fractions α = α , and β = [ b , b , b , . . . ] for two distinct irrational num-bers α = α , find a formula for computing the product αβ = [ c , c , c , . . . ] where c i = c i ( a , a , . . . ; b , b , . . . ) . Exercise 10.3.
Prove that γ = X n ≥ (cid:18) n − log (cid:18) n (cid:19)(cid:19) . Exercise 10.4.
Prove that π, π , π , π , π , . . . , π n is irrational for any rational number r = 0. Exercise 10.5.
Prove that π, π , π , π , π , . . . , π n are pairwise, and triplewise linearly independentover the ration numbers. Exercise 10.6.
Prove or disprove that π = rζ (3), where 0 = r ∈ Q . Exercise 10.7.
Show that π ∈ R is not algebraic. Exercise 10.8.
Show that i i ∈ R is not algebraic. Exercise 10.9.
Show that ln π ∈ R is irrational. Exercise 10.10.
Prove or disprove that a pair of algebraic numbers α, β ∈ R are linearly independent overthe rational numbers, that is, aα + bβ = c has no nontrivial rational solutions ( a, b, c ) = (0 , , Exercise 10.11.
Prove or disprove that power series S (3) = X n ≥ n ( e πn −
1) = r π , where r ∈ Q is rational number. Exercise 10.12.
Prove or disprove that power series S (5) = X n ≥ n ( e πn −
1) = r π , where r ∈ Q is rational number. Exercise 10.13.
Prove or disprove that power series S (5) = X n ≥ n ( e πn + 1) = r π , where r ∈ Q is rational number. Exercise 10.14.
Prove or disprove that power series S (7) = X n ≥ n ( e πn −
1) = r π , where r ∈ Q is rational number. 16 he product eπ is irrational References [1] Apery, Roger. Irrationalite de ζ (2) et ζ (3). Luminy Conference on Arithmetic. Asterisque No. 61 (1979),1-13.[2] Aigner, Martin; Ziegler, Gunter M. Proofs from The Book. Fifth edition. Including illustrations by KarlH. Hofmann. Springer-Verlag, Berlin, 2014.[3] Berggren, Lennart; Borwein, Jonathan; Borwein, Peter Pi: a source book . Third edition. Springer-Verlag,New York, 2004.[4] Beukers, F. A note on the irrationality of ζ (2) and ζ (3). Bull. London Math. Soc. 11 (1979), no. 3,268-272.[5] Jean Bourgain, Alex Kontorovich, On Zaremba’s Conjecture, arXiv:1107.3776.[6] Cohn, Henry. A short proof of the simple continued fraction expansion of e . Amer. Math. Monthly 113(2006), no. 1, 57-62.[7] Bruce C. Berndt, Armin Straub, Ramanujan’s Formula for zeta (2 n + 1), arXiv:1701.02964.[8] Stephane Fischler, Johannes Sprang, Wadim Zudilin, Many odd zeta values are irrationals,arXiv:1803.08905.[9] Hardy, G. H.; Wright, E. M. An introduction to the theory of numbers. Sixth edition. Oxford UniversityPress, Oxford, 2008.[10] Lang, Serge. Introduction to Diophantine approximations. Second edition. Springer-Verlag, New York,1995.[11] Kuipers, L.; Niederreiter, H. Uniform distribution of sequences ∼ miw/articles/pdf/AWSLecture1.pdf.[20] Wadim Zudilin, One of the Odd Zeta Values from ζ (5) to ζζ