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Solutions to polynomials in two variables
... For degree 3 polynomials in two-variables, things are still more complicated. Without going into great detail, professor Tucker explained that given 2 rational solutions to such a polynomial, one can construction a third by “adding” the first 2 together. Because of this, it is possible in many cases ...
... For degree 3 polynomials in two-variables, things are still more complicated. Without going into great detail, professor Tucker explained that given 2 rational solutions to such a polynomial, one can construction a third by “adding” the first 2 together. Because of this, it is possible in many cases ...
Modular Arithmetic
... EXAMPLE 1 The final digit of a 12-digit Universal Product Code (UPC) is a check digit. Suppose the digits of a UPS are represented by X1X2X3X4X5X6-X7X8X9X10X11Xc To calculate the check digit Xc, first compute 3(X1+X3+X5+X7+X9+X11) + (X2+X4+X6+X8+X10) (mod 10). If this number is zero, then Xc=0, If ...
... EXAMPLE 1 The final digit of a 12-digit Universal Product Code (UPC) is a check digit. Suppose the digits of a UPS are represented by X1X2X3X4X5X6-X7X8X9X10X11Xc To calculate the check digit Xc, first compute 3(X1+X3+X5+X7+X9+X11) + (X2+X4+X6+X8+X10) (mod 10). If this number is zero, then Xc=0, If ...
Why Do All Composite Fermat Numbers Become
... It has been proved that any prime number p satisfies Fermat’s little theorem, which includes Fermat primes. But there are some composite numbers also satisfy Fermat’s little theorem, in which the smallest such composite number is 341=11×31, so that such composite numbers are called pseudoprimes to b ...
... It has been proved that any prime number p satisfies Fermat’s little theorem, which includes Fermat primes. But there are some composite numbers also satisfy Fermat’s little theorem, in which the smallest such composite number is 341=11×31, so that such composite numbers are called pseudoprimes to b ...
Normal numbers without measure theory - Research Online
... expansion. A similar definition to the above may be made for a number to be simply normal to other bases. The following result was proved by Émile Borel in 1904. Borel’s Theorem. There is a subset Z of [0, 1) which has measure zero and is such that every number in [0, 1) which is not in Z is simply ...
... expansion. A similar definition to the above may be made for a number to be simply normal to other bases. The following result was proved by Émile Borel in 1904. Borel’s Theorem. There is a subset Z of [0, 1) which has measure zero and is such that every number in [0, 1) which is not in Z is simply ...
ON A PROBLEM OF SIDON IN ADDITIVE NUMBER THEORY, AND
... less than n would be greater than cn2, and so there would be a number m < 2n for which f(sn) > cn2/2n, which is contrary to hypothesis. Therefore, by Fabry’s gap theorem, the power series circle as its natural ...
... less than n would be greater than cn2, and so there would be a number m < 2n for which f(sn) > cn2/2n, which is contrary to hypothesis. Therefore, by Fabry’s gap theorem, the power series circle as its natural ...
A Nonlinear Expression for Fibonacci Numbers and Its Consequences
... Denote by N and Z the natural number set (including O) and the set of integers, respectively. It is known that Fibonacci numbers could be defined on N as well as on Z (cf. Graham-KnuthPatashnik [1]). More percisely, the two sets of numbers {Fn } (n ∈ N) and {fn } (n ∈ Z) are generated by the followi ...
... Denote by N and Z the natural number set (including O) and the set of integers, respectively. It is known that Fibonacci numbers could be defined on N as well as on Z (cf. Graham-KnuthPatashnik [1]). More percisely, the two sets of numbers {Fn } (n ∈ N) and {fn } (n ∈ Z) are generated by the followi ...
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... By "doubling diagram mod m" we mean the directed graph whose vertices are 0 and the natural numbers less than m, with directed arcs (arrows) from each vertex x to 2x reduced modulo m. ...
... By "doubling diagram mod m" we mean the directed graph whose vertices are 0 and the natural numbers less than m, with directed arcs (arrows) from each vertex x to 2x reduced modulo m. ...
Three Connections to Continued Fractions
... More generally, we can look at so-called generalized continued fractions, i.e. expressions of the form x = b0 + ...
... More generally, we can look at so-called generalized continued fractions, i.e. expressions of the form x = b0 + ...
1 The Principle of Mathematical Induction
... using the PMI. The work involved in prove that (1) holds for Z(n) is called the base case. The work is to prove that Z(1) is true. Exercise 6. For each of the following statements, suppose you want to prove them true for all natural numbers using the PMI. Write the proof from the beginning until you ...
... using the PMI. The work involved in prove that (1) holds for Z(n) is called the base case. The work is to prove that Z(1) is true. Exercise 6. For each of the following statements, suppose you want to prove them true for all natural numbers using the PMI. Write the proof from the beginning until you ...
Mathematical Diversions
... The statement that every even integer greater than 4 can be expressed as the sum of two odd prime numbers (for example, 6 = 3+3, 8 =3+5, 10 = 5+5 or 3+7, 12 = 5+7, etc.) Was proved by Christian Goldbach in 1742. Was proved by Leonhard Euler in 1737. Has never been proved. Is false for n = 2,467,892, ...
... The statement that every even integer greater than 4 can be expressed as the sum of two odd prime numbers (for example, 6 = 3+3, 8 =3+5, 10 = 5+5 or 3+7, 12 = 5+7, etc.) Was proved by Christian Goldbach in 1742. Was proved by Leonhard Euler in 1737. Has never been proved. Is false for n = 2,467,892, ...
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... and (6) a r e logically equivalent. F o r the proof of (4) it suffices to verify (6). Actually (6) can be verified by means of the principle of inclusion and exclusion in combinatorial analysis. ...
... and (6) a r e logically equivalent. F o r the proof of (4) it suffices to verify (6). Actually (6) can be verified by means of the principle of inclusion and exclusion in combinatorial analysis. ...
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... context, the Tchebycheff polynomials are distinguished among the family of Fibonacci-like polynomials defined by (2) and (6), as only for that case (i.e., for a = 1 and b = -1) the Fibonacci-like polynomials associate with standard organizations [3]. This can be seen easily after consulting Theorem ...
... context, the Tchebycheff polynomials are distinguished among the family of Fibonacci-like polynomials defined by (2) and (6), as only for that case (i.e., for a = 1 and b = -1) the Fibonacci-like polynomials associate with standard organizations [3]. This can be seen easily after consulting Theorem ...
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... Theorem 1, if n > / ( 7 ) = 4, then v2(S(2n, 7)) = v2{S{c-2n, 7)) = 2 for arbitrary positive integer c. Notice, however, that v2(S(%, 7)) = 2 also. We make the following Conjecture: For all k and 1 < k < 2n, we have v2 ($(2n ,k)) = d(k) -1. By Theorem 1, the Conjecture is true for all k~2m with m
... Theorem 1, if n > / ( 7 ) = 4, then v2(S(2n, 7)) = v2{S{c-2n, 7)) = 2 for arbitrary positive integer c. Notice, however, that v2(S(%, 7)) = 2 also. We make the following Conjecture: For all k and 1 < k < 2n, we have v2 ($(2n ,k)) = d(k) -1. By Theorem 1, the Conjecture is true for all k~2m with m
Math 416 – Introduction to Abstract Algebra
... Example: For any group, the identity map, id(x) = x, is always an automorphism. Example 9: : C C, with + as its operation, that maps a + bi to its complement, a – bi, is an automorphism. The same function is an automorphism on C*, with ...
... Example: For any group, the identity map, id(x) = x, is always an automorphism. Example 9: : C C, with + as its operation, that maps a + bi to its complement, a – bi, is an automorphism. The same function is an automorphism on C*, with ...
SOME HOMEWORK PROBLEMS Andrew Granville 1. Suppose that
... Use the abc-conjecture to prove that b) There are only finitely many triples of consecutive powerful numbers m − 1, m, m + 1. c) Show that if p1 = 1, p2 = 4, p3 = 8, p4 = 9, . . . is the sequence of powerful numbers then pn+2 − pn → ∞ as n → ∞. d) There are only finitely many powerful Fibonacci numb ...
... Use the abc-conjecture to prove that b) There are only finitely many triples of consecutive powerful numbers m − 1, m, m + 1. c) Show that if p1 = 1, p2 = 4, p3 = 8, p4 = 9, . . . is the sequence of powerful numbers then pn+2 − pn → ∞ as n → ∞. d) There are only finitely many powerful Fibonacci numb ...
[Part 1]
... the integers x and y. Lemma 3. Every positive integer m divides some Fibonacci number whose index does not exceed m 2 . Lemma 4. Let p be an odd prime and f- 5. Then p does not divide F . Proof of Lemma 4. According to [ 1 ] , p. 394, we have that either F divisible by p. F r o m the well known iden ...
... the integers x and y. Lemma 3. Every positive integer m divides some Fibonacci number whose index does not exceed m 2 . Lemma 4. Let p be an odd prime and f- 5. Then p does not divide F . Proof of Lemma 4. According to [ 1 ] , p. 394, we have that either F divisible by p. F r o m the well known iden ...
Transcendental values of the digamma function
... Lemma 12. (A. Baker [4]) If α1 , . . . , αn ∈ Q \ {0} and β1 , . . . , βn ∈ Q, then β1 log α1 + · · · + βn log αn is either zero or transcendental. Let us note that here and later, we interpret log as the principal value of the logarithm with argument in (−π, π]. A creative application of this theor ...
... Lemma 12. (A. Baker [4]) If α1 , . . . , αn ∈ Q \ {0} and β1 , . . . , βn ∈ Q, then β1 log α1 + · · · + βn log αn is either zero or transcendental. Let us note that here and later, we interpret log as the principal value of the logarithm with argument in (−π, π]. A creative application of this theor ...
PPT - School of Computer Science
... Four guys want to cross a bridge that can only hold two people at one time. It is pitch dark and they only have one flashlight, so people must cross either alone or in pairs (bringing the flashlight). Their walking speeds allow them to cross in 1, 2, 5, and 10 minutes, respectively. Is it possible ...
... Four guys want to cross a bridge that can only hold two people at one time. It is pitch dark and they only have one flashlight, so people must cross either alone or in pairs (bringing the flashlight). Their walking speeds allow them to cross in 1, 2, 5, and 10 minutes, respectively. Is it possible ...