Full text
... KIRCHHOFF8S MATRIX TREE THEOREM For any graph with two or more vertices, all the cofactors of M are equal, and the value of each cofactor equals t(G). Clearly, the matrix tree theorem solves the problem of finding the number of spanning trees of a graph. Furthermore, we note that this is an effectiv ...
... KIRCHHOFF8S MATRIX TREE THEOREM For any graph with two or more vertices, all the cofactors of M are equal, and the value of each cofactor equals t(G). Clearly, the matrix tree theorem solves the problem of finding the number of spanning trees of a graph. Furthermore, we note that this is an effectiv ...
Lucas` square pyramid problem revisited
... This assertion of Lucas, made first in 1875, amounts to the statement that the only solutions in positive integers (s, t) to the Diophantine equation ...
... This assertion of Lucas, made first in 1875, amounts to the statement that the only solutions in positive integers (s, t) to the Diophantine equation ...
CIS 260 Recitations 3 Feb 6 Problem 1 (Complete proof of example
... this leads to a contradiction. So our assumptions is that (p q ) is true, or equivalently that p q is true.] Proof of claim 1: (by contradiction) Assume a is odd (and a 2 is even). Then by definition a 2k 1 for some integer k . When we square both sides, we obtain that a 2 (2k 1) 2 ...
... this leads to a contradiction. So our assumptions is that (p q ) is true, or equivalently that p q is true.] Proof of claim 1: (by contradiction) Assume a is odd (and a 2 is even). Then by definition a 2k 1 for some integer k . When we square both sides, we obtain that a 2 (2k 1) 2 ...
FUNCTIONS WHICH REPRESENT PRIME NUMBERS
... real number A with property (1). Furthermore, given any real number A >1, there exists a value c with property (1). The proof of the first part of the theorem is a slight rearrangement of Mills' proof, and the proof of the second part employs the same basic idea in a different setting. ...
... real number A with property (1). Furthermore, given any real number A >1, there exists a value c with property (1). The proof of the first part of the theorem is a slight rearrangement of Mills' proof, and the proof of the second part employs the same basic idea in a different setting. ...
Acta Mathematica et Informatica Universitatis Ostraviensis - DML-CZ
... Let k be the minimal positive integer, if such exists, for which the statement of Theorem 1 is not valid. Then results of Brown and Kedlaya imply that k > 4. Since neither Ck nor 2ck is a square we see that Q(s/c~k) and Q(\/2ck) are real quadratic number fields. Moreover 2c* - 1 + 2sk\/c~k = (sk + y ...
... Let k be the minimal positive integer, if such exists, for which the statement of Theorem 1 is not valid. Then results of Brown and Kedlaya imply that k > 4. Since neither Ck nor 2ck is a square we see that Q(s/c~k) and Q(\/2ck) are real quadratic number fields. Moreover 2c* - 1 + 2sk\/c~k = (sk + y ...
Teacher Guide-Rational vs Irrational
... The instructor should meticulously state the Main question of the lesson and afterwards he/she is recommended to invite the students to think of a probable answer and write it on a piece of paper. Then the instructor must make sure that the student must understand the representation and notation of ...
... The instructor should meticulously state the Main question of the lesson and afterwards he/she is recommended to invite the students to think of a probable answer and write it on a piece of paper. Then the instructor must make sure that the student must understand the representation and notation of ...
1 Introduction 2 History 3 Irrationality
... look up the Lindemann- Weierstrass Theorem. There is a good proof of the theorem there. What is important to note is that this theorem gives us a method to prove certain numbers are transcendental. We will use this fact to prove a few corollaries about e and π. ...
... look up the Lindemann- Weierstrass Theorem. There is a good proof of the theorem there. What is important to note is that this theorem gives us a method to prove certain numbers are transcendental. We will use this fact to prove a few corollaries about e and π. ...
THE PRIME FACTORS OF CONSECUTIVE, INTEGERS II by P
... fI(pq - 1) > min (p, q), the primes p and q cannot both be used for f2 but can be used for fl . Assume now f2 (n) = k and assume that the set n + 1, . . ., n + k contains no power of a prime . Then f2 (n) > f3 W . Since f2 (n) _ k there must be r numbers n + 111 . . .1 n + it in the set which togeth ...
... fI(pq - 1) > min (p, q), the primes p and q cannot both be used for f2 but can be used for fl . Assume now f2 (n) = k and assume that the set n + 1, . . ., n + k contains no power of a prime . Then f2 (n) > f3 W . Since f2 (n) _ k there must be r numbers n + 111 . . .1 n + it in the set which togeth ...
9709/01 - TheAllPapers
... You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of ...
... You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of ...
Trees
... Proof by Construction Prove that a particular object exists Demonstrate how to construct it Alternatively, find a counterexample All shapes that have four sides of equal length are squares ...
... Proof by Construction Prove that a particular object exists Demonstrate how to construct it Alternatively, find a counterexample All shapes that have four sides of equal length are squares ...
Primes, Composites and Integer Division
... We will construct a number m > pn that is prime, thus contradicting the fact that pn is the largest prime number, and, consequently, the assumption that there are finitely many prime numbers (with pn being the largest). We may thus infer that the assumption that there are finitely many prime numbers ...
... We will construct a number m > pn that is prime, thus contradicting the fact that pn is the largest prime number, and, consequently, the assumption that there are finitely many prime numbers (with pn being the largest). We may thus infer that the assumption that there are finitely many prime numbers ...
Reciprocity Laws and Density Theorems
... SL2 (Z) and their action on the hyperbolic plane. SL2 (Z) denotes the group of 2 × 2 integer matrices with determinant 1. These groups act as isometries on the hyperbolic plane, a non-standard geometry in which there are many lines through a given point parallel to a given line. This geometry provid ...
... SL2 (Z) and their action on the hyperbolic plane. SL2 (Z) denotes the group of 2 × 2 integer matrices with determinant 1. These groups act as isometries on the hyperbolic plane, a non-standard geometry in which there are many lines through a given point parallel to a given line. This geometry provid ...
Chapter Two: Describing Distributions with Numbers Besides the
... The interquartile range IQR is the distance between the first and third quartiles: IQR = Q3 − Q1. The interquartile range is a measure of spread which is mainly used as the basis for identifying suspected outliers. The 1.5IQR Rule for outliers An observation, x, is called a suspected outlier if x < ...
... The interquartile range IQR is the distance between the first and third quartiles: IQR = Q3 − Q1. The interquartile range is a measure of spread which is mainly used as the basis for identifying suspected outliers. The 1.5IQR Rule for outliers An observation, x, is called a suspected outlier if x < ...
Full-Text PDF - EMS Publishing House
... to verify it empirically, but did not attain a demonstration. Fermat then continues 2. The eightfold multiple of an arbitrary number, diminished by 1, is composed of four squares – not only in integers – which perhaps others might have already seen – but also in fractions, as I promise to prove. The ...
... to verify it empirically, but did not attain a demonstration. Fermat then continues 2. The eightfold multiple of an arbitrary number, diminished by 1, is composed of four squares – not only in integers – which perhaps others might have already seen – but also in fractions, as I promise to prove. The ...
Discrete Math, Spring 2013 - Some Sample Problems
... 15. Let p be a prime number. a. Show that if x2 ≡ x (mod p) then x ≡ 0 (mod p) or x ≡ 1 (mod p). b. Suppose an integer a is not divisible by p. Show that ax ≡ 1 (mod p) has a solution x in Z. 16. Let a, b, c be positive integers such that a | c and b | c. a. Show that if (a, b) = 1 then ab | c. b. G ...
... 15. Let p be a prime number. a. Show that if x2 ≡ x (mod p) then x ≡ 0 (mod p) or x ≡ 1 (mod p). b. Suppose an integer a is not divisible by p. Show that ax ≡ 1 (mod p) has a solution x in Z. 16. Let a, b, c be positive integers such that a | c and b | c. a. Show that if (a, b) = 1 then ab | c. b. G ...
Fermat Numbers - William Stein
... Prime numbers are widely studied in the field of number theory. One approach to investigate prime numbers is to study numbers of a certain form. For example, it has been proven that there are infinitely many primes in the form a + nd, where d ≥ 2 and gcd(d, a) = 1 (Dirichlet’s theorem). On the other ...
... Prime numbers are widely studied in the field of number theory. One approach to investigate prime numbers is to study numbers of a certain form. For example, it has been proven that there are infinitely many primes in the form a + nd, where d ≥ 2 and gcd(d, a) = 1 (Dirichlet’s theorem). On the other ...
Number theory and proof techniques
... n and asks to find factors of it Integers r and s such that rs=n How would you do it? Can try all integers from 1 to n Can try all integers from 1 to sqrt(n) Can try all primes from 1 to sqrt(n) ...
... n and asks to find factors of it Integers r and s such that rs=n How would you do it? Can try all integers from 1 to n Can try all integers from 1 to sqrt(n) Can try all primes from 1 to sqrt(n) ...