Description Details
...table by elementary methods; others are partial solutions of certain unsolved problems. An important method is based on the famous Euler-Bell-Kalm ar lemma, with many applications. Article 20 may be considered also as an introduction to Chapter 3 on Arithmetic functions. Here many papers study the famous Smarandache function, the source of inspiration of so many mathematic...

Full Text Search Details
...d problems. An important method is based on the famous Euler-Bell-Kalm ar lemma, with many applications. Article 20 may be considered also as an intr... ...a 2 1 4 X a 2 : (2) First we need the following auxiliary proposition. Lemma. Let p and p 0 denote the semi-perimeters of triangles ABC and A 0 B ... ...ven, see also [4], pp.47-48). First we extend (2) in the following manner: Lemma 1. In any triangle ABC holds the following inequality: b +c a sin A ... ...90 , then (3) implies (2). One has equality only when b A = 90 ; b =c. Lemma 2. In any triangle ABC one has r 8 > < > : b +c a 2 if b A 90 ... ...B @ 1 sin A 2 1 1 C A 1 1 sin A 2 1 : (6) Proof. By R = a 2 sinA and Lemma 2, one can write R r a 2 sinA 0 B @ a 2 sin A 2 a 2 1 C A which g... ...es of a right triangle, we shall prove this result separately, as follows: Lemma. Let XYZ be a right triangle of legs y;z and hypothenuse x. Suppose t... ...< p 2y. Now, y x = sinY > sin 45 = 1 p 2 , and we are done. Applying the Lemma to the particular triangle KBC, we obtain the inequality (5). (X K, ... ...otations, as the equation in the title. We shall use, among others the two lemmas of another article [4] (see also [3]). Lemma 1. If x> 0, y> 0, (x;y)... ...[4] (see also [3]). Lemma 1. If x> 0, y> 0, (x;y) = 1, then (x +y;xy) = 1. Lemma 2. If k> 0 is not a perfect square, then p k62Q. Another solution wil...

Full Text Search Details
....F.ofP.R.China(60472068) 10 RESEARCHONSMARANDACHEPROBLEMSINNUMBERTHEORYII Lemma Forn≥ 1,wehave ϕ(n) =n Y p|n 1− 1 p . Proof. SeeTheorem2.4inrefer... ... p|n 1− 1 p . Proof. SeeTheorem2.4inreference[2]. Now we use the above lemma to complete the proof of Theorem. We will discussitintwocases. I. (n,... ... . §2. Somelemmas Tocompletetheproofofthetheorem,weneedfollowingLemmas: Lemma1. Foranyrealnumberx> 1andpositiveintegerm,wehavethe asymptoticformula... ... (n) = x 2 2 Y p|m p p+1 +O(x 3 2 + ). (7) ThiscompletestheproofofLemma1. Lemma 2. For any real number x ≥ 3 and any nonnegative arithmetical functio... ...ion, wewillcompletetheproofofthetheorem. Fromthedefini tionofδ m (a k (n)),Lemma1andLemma2,wehave X n≤x δ m (a k (n)) = h x 1 k i −1 X t=1 X n≤g(t) δ ... ... +O(x 2− 2 k ). For the cases of k = 2,3, we can also prove the results by Lemma 2. For example, X n≤x δ m (a 2 (n)) Ontheadditivek thpowercomplement... ... Somesimplelemmas Beforetheproofofthetheorem,somesimplelemmaswillbeuseful. Lemma 1. If ord(2,(2n)!) < 2n− 2, then T n is not an integer, where ord(2,(... ...here ord(2,(2n)!)denotestheorderofprime 2in (2n)!. Proof. Seereference[2]. Lemma2. Foranypositiveintegern≥ 1,wehavetheidentity: α 2 (2n)≡ +∞ X i=1 2... ...+ 2 α 4 withα 4 > α 3 > α 2 > α 1 ≥ 0, then a(2n,2) ≥ 3. For these n, from Lemma 1 we know that T n is not an integer. Since there are infinite positiv...

Full Text Search Details
...ific spaces (using scientific: symbols, meta-language, matrices, theorems, lemmas etc.). E) Third Paradoxist Manifesto. Therefore, don’t enforce ... ...ific spaces (using scientific: symbols, meta-language, matrices, theorems, lemmas etc.). 4. New literary terms Trough the “paradoxism”, the followi...

Full Text Search Details
...ific spaces (using scientific: symbols, meta-language, matrices, theorems, lemmas etc.). E) Third Paradoxist Manifesto. Therefore, don’t enforce ... ...ific spaces (using scientific: symbols, meta-language, matrices, theorems, lemmas etc.). 4. New literary terms Trough the “paradoxism”, the followi...

Full Text Search Details
...f each permutation R a in π is the same for all a ∈ L n (m) ( a ≠ e). LEMMA [65]: Let L be a right alternative loop of even order say 2s (s ≥ 3) ... ... Thus no two different permutations have any transposition in common. LEMMA [65]: Let L = {1, 2, … , n + 1} n odd and S n +1 be the set of all ... ...EM [65]: Let π ∈ S n+1 , satisfy the conditions (i) and (ii) of the above lemma. If f(n) is the number of distinct possible choices of π, then any ri... ...ve loop is a S-loop. So all ARIF loops are S- loops. Further using the Lemma 2.4 of Kinyon [41] we have another interesting result. 50 THE... ...M 3.1.9: Every RIF-loop is a S-loop. Proof: According to Kinyon's [41] Lemma 2.4. every RIF loop is an ARIF loop. But by our Theorem 3.1.8 every ...

Full Text Search Details
...llary. For any real number , we have the asymptotic formula 2. One simple lemma To complete the proof of the theorem, we need the following simple le... ... To complete the proof of the theorem, we need the following simple lemma: Lemma. For any xed prime and real number , we have Proof. First we come ... ... power 3 Then we can immediately obtain This completes the proof of the Lemma. 3. Proof of the Theorem In this section, we shall use the above Lem... ...hall complete the proof of the theorem. First we need following one simple Lemma. That is, Lemma. Let be a prime and be any positive integer, then we... ... (1) and (2), we can easily get if otherwise This completes the proof of Lemma. Now we use above Lemma to complete the proof of Theorem. For any rea... ...e numbers of power and -power roots 7 then from the de nition of and the Lemma we have (3) (4) where . Note that if , then we have (see referenc... ...er , we have the asymptotic formula where is the Euler constant. 2. Some lemmas Before the proof of the theorems, some lemmas will be usefull. 14 R... ...mas will be usefull. 14 RESEARCH ON SMARANDACHE PROBLEMS IN NUMBER THEORY Lemma 1. For any real number , we have the asymptotic formula: where is th... ...symptotic formula: where is the Euler constant. Proof. See reference [2]. Lemma 2. For any real number and any nonnegative arithmetical function with...

Full Text Search Details
... in a given set of embeddings of a graph on surfaces by using the Burnside Lemma as the following. Theorem 1.3.4 For a given graph Γ, let E ⊂E L (Γ), ... ...E and P g =P}. Proof Define the groupH = Aut1 2 Γ×<α>. Then by the Burnside Lemma and the Theorem 1.3.3, we get that n(E,Γ)= 1 |H| g∈H |Φ 1 (g)|, whe... ... with g[Γ](x), r[Γ](x) with g[Γ](x) and R[Γ](x) with G[Γ](x) as follows. Lemma 1.4.1([25][45]) For a given map M, the number of non-isomorphic roote... ...ooted maps on an orientable surfaceS with genus γ(Γ)+k−1. According to the Lemma 1.4.1, we know that r k (Γ) = M∈M k (Γ,S) 4ε(M) |AutM| = 4ε(Γ) |Aut... ...by π −1 (x) ={ x| x∈( X α,β and π( x) =x}. We have the following result. Lemma 2.1.1 Let π : X α,β → X α,β be a covering mapping. Then for any two... ...e following numerical result is obvious by the definition of a lifting map. Lemma 2.1.2Thenumbersof verticesand edgesinthe lifting mapM ϑ arerespective... ...esinthe lifting mapM ϑ arerespective ν(M ϑ ) =ν(M)|G| and ε(M ϑ ) =ε(M)|G| Lemma 2.1.3 Let F = (C ∗ )(αC ∗ α −1 ) be a face in the map M. Then there a... ...tion on Ω is called fixed-free if G x =1 G for∀x∈ Ω. We have the following. Lemma 2.2.1 Any automorphism group G of a map M = (X α,β ,P) is fixed-free o... ...group of a lifting map has a obvious subgroup, determined by the following lemma. Lemma 2.2.2 Let M ϑ be a lifting map by the voltage assignment ϑ : X...

Table of Contents Details
... hybrid DSm rule . . . 134 6.5 Generalization of DSm rules for sets . . . . 135 6.5.1 General DSm rules for imprecise beliefs . . . 136 6.5.2 Some lemmas and a theorem . . . . 136 6.5.3 An example with multiple-interval masses . . . 138 6.6 Conclusion . . . . . 140 6.7 References . . . . . 141 7 A Generalized Pignistic Transformation 143 7.1 A short introduction t...

Full Text Search Details
...in C. First, we determine the minimal nonnegative s for which the inequality s 2 +5s+4¸k holds. We denote it by s(k). Directly it is seen the followin... ...)´1, as a corollary from (12) we obtain Proposition 7. For n¸1theidentity n Y k=1 P d (k)= n Y k=1 k [ n k ] (13) holds. 53 Now, we need the followin... ...he above se- quence, using the third de¯nition of it. Initially, we shall prove the 8 The results in this section are taken from [37] 65 following tw... ...ndhavinginmindthatv = C n +4 and C 1 =7, the smallest of them satis¯es the inequality v¸11. Corollary. For every natural number n¸1: C n+1 ·C n +6: (1... ...s the set of all<t;w> ; 10 The results in this section are taken from [36] 73 K o (t;w) denotes the set of all<t;w>2N 2 ; K 0 (t;w)=K o (t;w)¡f<2;1>g... ...;w)=K o (t;w)¡f<2;1>g. Lemma 1. If<t;w>2N 2 and<x;y> 6 =< 2;1 >,thenthere exists<l;m>,suchthat<l;m>2N 2 and the equalities x=l+3m and y =l+2m (4) hold... ... 2 and the equalities x=l+3m and y =l+2m (4) hold. Lemma 2. Let<l;m>2N 2 .Ifx and y are given by (1), then x and y satisfy (4) and<x;y>2N 2 . Note tha... ...of (5) and (8). ProofofLemma2. Let<l;m>2N 2 .T h e nw ec h e c kt h e equality 2(l+3m) 2 ¡3(l+2m) 2 =5; under the assumption of validity of (2) and th...

Full Text Search Details
...olution iff all its equations of form (4) have a common solution. The following lemma can be established. Let a o t r = b a FRE of form (4). We hav... ...ation equations (1). This value changing is called an equivalence operation. Lemma 1.6.1: For j 1 , j 2 ∈ {1, 2,…,n}, if 2 1 2 1 , j ij j j b a ... ...alled an equivalence operation. The reader is expected to prove the following lemma: Lemma 1.8.1: For j 1 , j 2 ∈ J, if 1 j b > 2 j b , and ... ...n the corresponding a ij of matrix A is called a critical element for x ˆ . Lemma 1.8.2: If a ij is a critical element, then p ij ≥ 0. It ... ...hen p ij ≥ 0. It is left for the reader to supply proofs for the following lemmas: Lemmas 1.8.3: For column, j if there is only one i ∈ I su... ...ith equation in A ⊗ x = b. Of course, it does not mean that x is a solution. Lemma 1.13.1: Let i ∈ M, j ∈ N. If x ∈ S (A, b)-F ij , then a ij ⊗ x... ...: Let i ∈ M, j ∈ N. If x ∈ S (A, b)-F ij , then a ij ⊗ x j < b i . 83 Lemma 1.13.2: Let x ∈ S (A, b). Then the following statements are eq... ...s are equivalent: (i) x ∈ S (A, b) (ii) (∃ϕ: M →N) (∀i∈ M) x ∈ F iϕ(i) . Lemma 1.13.3: Let i ∈ M, j ∈ N. Then (i) F ij = {x ∈ S (A, b); x ... ... b) ∈ B(n). If this is the case, then x = x (A, b) is the greatest solution. Lemma 1.13.4: Let x ∈ S(A, b), then x x ≤ . THEOREM 1.13.7: Let A ...

Full Text Search Details
...olution iff all its equations of form (4) have a common solution. The following lemma can be established. Let a o t r = b a FRE of form (4). We hav... ...ation equations (1). This value changing is called an equivalence operation. Lemma 1.6.1: For j 1 , j 2 ∈ {1, 2,…,n}, if 2 1 2 1 , j ij j j b a ... ...alled an equivalence operation. The reader is expected to prove the following lemma: Lemma 1.8.1: For j 1 , j 2 ∈ J, if 1 j b > 2 j b , and ... ...n the corresponding a ij of matrix A is called a critical element for x ˆ . Lemma 1.8.2: If a ij is a critical element, then p ij ≥ 0. It ... ...hen p ij ≥ 0. It is left for the reader to supply proofs for the following lemmas: Lemmas 1.8.3: For column, j if there is only one i ∈ I su... ...ith equation in A ⊗ x = b. Of course, it does not mean that x is a solution. Lemma 1.13.1: Let i ∈ M, j ∈ N. If x ∈ S (A, b)-F ij , then a ij ⊗ x... ...: Let i ∈ M, j ∈ N. If x ∈ S (A, b)-F ij , then a ij ⊗ x j < b i . 83 Lemma 1.13.2: Let x ∈ S (A, b). Then the following statements are eq... ...s are equivalent: (i) x ∈ S (A, b) (ii) (∃ϕ: M →N) (∀i∈ M) x ∈ F iϕ(i) . Lemma 1.13.3: Let i ∈ M, j ∈ N. Then (i) F ij = {x ∈ S (A, b); x ... ... b) ∈ B(n). If this is the case, then x = x (A, b) is the greatest solution. Lemma 1.13.4: Let x ∈ S(A, b), then x x ≤ . THEOREM 1.13.7: Let A ...

Full Text Search Details
...f the form ) 1 2 ( 2 1 − ⋅ = − k k n , 1 2 − = k p being a prime number. Lemma 1: Let p n i ⋅ = 2 when p is an odd prime number and i an integer ... ..., p k = −1 2 prime, one has that p n S = ) ( . Demonstration: For the Lemma it is sufficient to prove that ) ! ( 1 2 p e k ≤ − . If we can pro... ...n takes for n p p m = with n integer, 2 ≥ n and p prime number. To do so a Lemma required. Lemma 1 2 , , ≥ Ν ∈ ∀ n m n m         + − ... ... p E p p k E p k E p k E p k E k e log 3 2 ) ( L 10 And using the lemma we have () [] n p n p n p p E n n n n n n n n p p p p p p E p ...

Full Text Search Details
... its relation to the subspace generated by the characteristic bivectors. LEMMA 2.6.1: Let T be a linear bioperator on the finite dimensional bive... ...ar relation which may exist between bivectors in the various bispaces. This lemma or result however states that the characteristic bispaces associat... ...∪ W 2 i.e., if T(W) is contained in W. Now we just give an interesting lemma about invariant bivector subspaces of V relative to a linear biope... ...out invariant bivector subspaces of V relative to a linear bioperator T. LEMMA 2.6.2: Let V = V 1 ∪ V 2 be a finite dimensional bivector space ... ...' rrrr 1122 1k1k ' xC xC xC xC −−∪ −− by repeated applications of the above lemma 2.6.2 we shall arrive at an ordered basis B = ( ) ( ) { } 1122 1k... ...αα . To find ( ) ( ) 1122 1n1m , ..., , ,... , α ααα we start by applying lemma to the subspace W = W 1 ∪ W 2 to obtain the vector 12 11 α∪ α, t... ...he subspace W = W 1 ∪ W 2 to obtain the vector 12 11 α∪ α, then apply the lemma to 12 111 WWW =∪, the space spanned by α 1 = 12 11 α ∪α and obt... ...e spanned by α 1 = 12 11 α ∪α and obtain α 2 = 12 11 α ∪α . Next apply lemma to W 2 = 12 22 WW ∪ the space spanned by α 1 and α 2 . Contin...

Full Text Search Details
...m, ut ~~~g~f?k cuaci fecundun~ deterlninationem vis a malleo in ipfum LEMMA I. $&.dant~tutes, sit & gtiantitattim ratioties, pud ad a?palicat~m ... ...tu- do ejus A B in infinitum mitluitur, fit minus quovis dato. Ergo (per lemma I) figura inkripta Sr; circumfcripta & multo magis fi- gura curvilin... ...omnium, 8~ ita figura ad figa- ram i exiitente nimirum figura priore (per lemma\?~r) ad fummam priorem, & figura poileriore ad fummam. pofieriorem i... ... ad caxeras : erunt tota ad invicem in eadem illa data ratione. Nam ii in lemmatis hujus figuris fuman. tur parallelogramma inter fe ut partes, fumm... ...llelogrammum, idelt (per hypothefin) in ultima ratione partis ad partem. LEMMA V. EIBE& PRIMUS. Nam ii angulus ilIe non evanefcit, continebit ar... ...ident areze curvilinez Ab d, AC e cum re&ilineis Afdj Age ; ideoque (per lemma v,) erunt in duplicata ratione laterum Ad, Ae : Sed his areis propor... ...E funt ultimo in duplicata ratione laterum A,D, A&J. ,$&E. 2). ‘” * h LEMMA X, Exponantur tempera per lineas AD, AE, & velocitates genite per o... ...ne 23 2) ” J ad b J quam pro differentia quavis affignata. Eff ergo, per lemma rS ratio ultima AB qund. ad A6 q~d. eadem cum ratione ultima .,BD a... ... TvG ad $j$Tpad. ut T5: pad. a v f-j? aa SLr 2trad-w TV fplv, 81 (per lemma XIX.) BCx CApro Cz>x PF, net non (pun&is LJBER T & g coeuntibus) 2 ...

Full Text Search Details
...er will be 5, it too will be divisible by 5. b) We start this proof with a lemma, Lemma: For the Concatenated Odd Sequence, if the index is congruent ... ...o 3 and multiplication by a power of ten does not change that and from the lemma, K + 1 is congruent to 0 modulo 3. Therefore, the sum is congruent t... ... to 1 modulo 3, the product will also be congruent to 1 modulo 3. From the lemma, K + 2 is congruent to 2 modulo 3, so the sum is congruent to 0 modul...

Full Text Search Details
...mbined into one, as the results are similar. We start the proof with a lemma. Lemma: If A and B are elements of NL1, then: i) If A \/ B = ( ... ...n A = ( 0, 0, 1). vi) If ¬A = ( 0, 0, 1) then A = ( 1, 0, 0). Proof of lemma: Results i) through iv) are consequences of the definitions of \/ and... ...: Rule v) of the definition of a wff is applied to form A /\ B. From the lemma, we have that if this expression is an absolute tautology, then eithe... ...: Rule v) of the definition of a wff is applied to form A \/ B. From the lemma, we have that if this expression is an absolute tautology, then eithe... ...e terms of the triplets in A are zero. Again, we start the proof with a lemma. 95 Lemma: If A and B are elements of NL2, then: i) If A \/ ... ...n A = ( 0, 0, 1). vi) If ¬A = ( 0, 0, 1) then A = ( 1, 0, 0). Proof of lemma: (i) If either A or B had a middle term that was not zero, then by ...

Full Text Search Details
...fic spaces (using scientific: symbols, meta-language, matrices, theorems, lemmas, etc.). 2.6. Third Paradoxist Manifesto: Therefore, don't enforce... ... 2 ), ..., (a n ). Developing [P], we find many propositions (theorems, lemmas, etc.) (p 1 ), (p 2 ), ..., (p m ), by logical combinations of it... ...paces (using scientific: symbols, meta-language, matrices, theorems, lemmas, etc.). 83 b) New Types of 'Mathematical' Poetry with Fixed Form ...

Full Text Search Details
...r which reason, in geometry and mechanics, it is the equivalent of the fundamental lemmas which the Greeks have left us without proof. Note that ... ...required. He simple observes that it . . . is the equivalent of the fundamental lemmas (axioms) handed down from the ancient Greeks, and accepte... ...ed his view without proof, stating it . . is the equivalent of the fundamental lemmas which the Greeks have left us without proof , and therefore...

Full Text Search Details
...when the terminal numbers are known to be fixed. From this, the following lemma is obvious. Lemma 1. The only candidates for a solution to the equat...

Full Text Search Details
...h is Hamiltonian. The theorem of [26] can be proved using the following lemma. LEMMA [26]: Let S be a set of vertices of a nontrivial tree T a...