Lectures 1-31 - School of Mathematical Sciences
... induction. Exercise: write down this proof ! Remark 4. Note that the number of sequences of length k which we can make using elements of X without repetition is just the same as the number of injections from {1, 2, . . . , k} to X. This should be ‘obvious’ by now. But it is worth bearing in mind tha ...
... induction. Exercise: write down this proof ! Remark 4. Note that the number of sequences of length k which we can make using elements of X without repetition is just the same as the number of injections from {1, 2, . . . , k} to X. This should be ‘obvious’ by now. But it is worth bearing in mind tha ...
On Cantor`s First Uncountability Proof, Pick`s Theorem
... In 1874, two years before the publication of his famous diagonalization argument, Georg Cantor’s first proof of the uncountability of the real numbers appeared in print [1]. Surprisingly, a small twist on Cantor’s line of reasoning shows that the Golden Ratio is irrational, as we shall demonstrate h ...
... In 1874, two years before the publication of his famous diagonalization argument, Georg Cantor’s first proof of the uncountability of the real numbers appeared in print [1]. Surprisingly, a small twist on Cantor’s line of reasoning shows that the Golden Ratio is irrational, as we shall demonstrate h ...
Machine Models - Columbia University
... Cantor’s Diabolical Diagonal If R were countable could list all numbers in (0,1) by a sequence: r1 , r2 , r3 , r4 , r5 , r6 , r7, … Cantor’s diabolical diagonalization creates a number “revil” between 0 and 1 which is not on the list, contradicting countability assumption. Let ri,j be the j ’th deci ...
... Cantor’s Diabolical Diagonal If R were countable could list all numbers in (0,1) by a sequence: r1 , r2 , r3 , r4 , r5 , r6 , r7, … Cantor’s diabolical diagonalization creates a number “revil” between 0 and 1 which is not on the list, contradicting countability assumption. Let ri,j be the j ’th deci ...
handout - inst.eecs.berkeley.edu
... Or (a − b)x = km for some integer k . gcd(x, m) = 1 =⇒ Prime factorization of m and x do not contain common primes. =⇒ (a − b) factorization contains all primes in m’s factorization. So (a − b) has to be multiple of m. =⇒ (a − b) ≥ m. But a, b ∈ {0, ...m − 1}. Contradiction. ...
... Or (a − b)x = km for some integer k . gcd(x, m) = 1 =⇒ Prime factorization of m and x do not contain common primes. =⇒ (a − b) factorization contains all primes in m’s factorization. So (a − b) has to be multiple of m. =⇒ (a − b) ≥ m. But a, b ∈ {0, ...m − 1}. Contradiction. ...
A Course on Number Theory - School of Mathematical Sciences
... Note that this is a slightly different use of the word “mod” from the one we used earlier to denote the remainder. But it is closely connected; two numbers are congruent modulo n if and only if they leave the same remainder when they are divided by n. Congruence modulo n is an equivalence relation; ...
... Note that this is a slightly different use of the word “mod” from the one we used earlier to denote the remainder. But it is closely connected; two numbers are congruent modulo n if and only if they leave the same remainder when they are divided by n. Congruence modulo n is an equivalence relation; ...
Examples of Conics
... For example, the even integers 2Z form a subgroup in the group Z of integers. A homomorphism between groups (G, ◦) and (H, ∗) is a map f : G −→ H that respects the group laws in the sense that we have f (g ◦g 0 ) = f (g)∗f (g 0 ). An isomorphism is a bijective homomorphism. Here are some examples: 1 ...
... For example, the even integers 2Z form a subgroup in the group Z of integers. A homomorphism between groups (G, ◦) and (H, ∗) is a map f : G −→ H that respects the group laws in the sense that we have f (g ◦g 0 ) = f (g)∗f (g 0 ). An isomorphism is a bijective homomorphism. Here are some examples: 1 ...
The Abundancy Index of Divisors of Odd Perfect Numbers
... is an integer (because gcd(qi αi , σ(qi αi )) = 1). Suppose ρi = 1. Then σ(N/qi αi ) = qi αi and σ(qi αi ) = 2N/qi αi . Since N is an odd perfect number, qi is odd, whereupon we have an odd αi by considering parity conditions from the last equation. But this means that qi is the Euler prime q, and w ...
... is an integer (because gcd(qi αi , σ(qi αi )) = 1). Suppose ρi = 1. Then σ(N/qi αi ) = qi αi and σ(qi αi ) = 2N/qi αi . Since N is an odd perfect number, qi is odd, whereupon we have an odd αi by considering parity conditions from the last equation. But this means that qi is the Euler prime q, and w ...
CS1231 - Lecture 09
... We have proven that f is a bijection. Let’s pause for a moment to consider this propositoin that we have just proven. Z is infinite. Z+ is infinite. Which is more infinite? Most would answer intuitively that Z contains more elements than Z+. But your intuition must be in subjection to logic. And log ...
... We have proven that f is a bijection. Let’s pause for a moment to consider this propositoin that we have just proven. Z is infinite. Z+ is infinite. Which is more infinite? Most would answer intuitively that Z contains more elements than Z+. But your intuition must be in subjection to logic. And log ...
101 Illustrated Real Analysis Bedtime Stories
... numbers in the zig-zag order indicated by the arrows. A small simplification made in this illustration is that it omits the nonpositive rational numbers. ...
... numbers in the zig-zag order indicated by the arrows. A small simplification made in this illustration is that it omits the nonpositive rational numbers. ...
Odd prime values of the Ramanujan tau function
... Remark 4 Considering the list p = 11, 17, 29, 41, 47, 59, 79, 89, 97, . . . for which we know LR (probable) primes, it is remarkable that the six first values correspond exactly to the odd values in the sequence of the Ramanujan primes: 2, 11, 17, 29, 41, 47, 59, 67, 71, 97, . . . . This sequence wa ...
... Remark 4 Considering the list p = 11, 17, 29, 41, 47, 59, 79, 89, 97, . . . for which we know LR (probable) primes, it is remarkable that the six first values correspond exactly to the odd values in the sequence of the Ramanujan primes: 2, 11, 17, 29, 41, 47, 59, 67, 71, 97, . . . . This sequence wa ...
IDEAL CLASSES AND SL 1. Introduction (C) on the Riemann
... needed for fractional ideals in a number field appears as an exercise in several introductory algebraic number theory books, but it may seem like an isolated fact in such books (I thought so when I first saw it!). Its use in the proof of Theorem 1.1 shows it is not. 2. Transitivity and Class Number ...
... needed for fractional ideals in a number field appears as an exercise in several introductory algebraic number theory books, but it may seem like an isolated fact in such books (I thought so when I first saw it!). Its use in the proof of Theorem 1.1 shows it is not. 2. Transitivity and Class Number ...
F. Roberts: Applied Combinatorics, L. Lovász
... 18. In how many ways can we partition n into exactly k parts if the order counts? 19. A code is being written using the four letters a, b, c and d. How many 12 digit codewords are there which use exactly 3 of each letter? 20. Of 15 computer programs to be run in a day, 5 of them are short, 4 are lon ...
... 18. In how many ways can we partition n into exactly k parts if the order counts? 19. A code is being written using the four letters a, b, c and d. How many 12 digit codewords are there which use exactly 3 of each letter? 20. Of 15 computer programs to be run in a day, 5 of them are short, 4 are lon ...
Indecomposable permutations with a given number of cycles
... Abstract. A permutation a1 a2 . . . an is indecomposable if there does not exist p < n such that a1 a2 . . . ap is a permutation of {1, 2, . . . , p}. We compute the asymptotic probability that a permutation of Sn with m cycles is indecomposable as n goes to infinity with m/n fixed. The error term i ...
... Abstract. A permutation a1 a2 . . . an is indecomposable if there does not exist p < n such that a1 a2 . . . ap is a permutation of {1, 2, . . . , p}. We compute the asymptotic probability that a permutation of Sn with m cycles is indecomposable as n goes to infinity with m/n fixed. The error term i ...
Computing Fibonacci Numbers Fast using the Chinese Remainder
... The inspiration for trying to use the Chinese Remainder Theorem (CRT) to compute Fibonacci numbers was not an original idea I had. I was given a paper that claimed to be able to use the CRT to compute Fibonacci numbers. I say claimed due to the fact that it did not contain enough explaination and da ...
... The inspiration for trying to use the Chinese Remainder Theorem (CRT) to compute Fibonacci numbers was not an original idea I had. I was given a paper that claimed to be able to use the CRT to compute Fibonacci numbers. I say claimed due to the fact that it did not contain enough explaination and da ...
Full text
... The polynomials Gn are a special case of Brahmagupta polynomials (take t = 1) introduced and studied in [10], [11]. These are linked with Morgan-Voyce polynomials, see [8] and are the Fibonacci polynomials when we further specialize y to be equal to 1. At this level, our proof was a mystery to us, w ...
... The polynomials Gn are a special case of Brahmagupta polynomials (take t = 1) introduced and studied in [10], [11]. These are linked with Morgan-Voyce polynomials, see [8] and are the Fibonacci polynomials when we further specialize y to be equal to 1. At this level, our proof was a mystery to us, w ...
Odd Crossing Number and Crossing Number Are
... This argument proves Equation (7) for maps with unit weights. The next step is to extend this lemma to maps with arbitrary weights. Consider two curves γ1 , γ2 whose endpoints are adjacent and in the same order. In a drawing minimizing one of the crossing numbers we can always assume that the two cu ...
... This argument proves Equation (7) for maps with unit weights. The next step is to extend this lemma to maps with arbitrary weights. Consider two curves γ1 , γ2 whose endpoints are adjacent and in the same order. In a drawing minimizing one of the crossing numbers we can always assume that the two cu ...
THE SOLOVAY–STRASSEN TEST 1. Introduction
... can be made into a deterministic primality test if we assume the truth of one of the most difficult unsolved problems in mathematics, called the Generalized Riemann Hypothesis (for Dirichlet L-functions). We will not explain here the Generalized Riemann Hypothesis, often abbreviated to GRH, but here ...
... can be made into a deterministic primality test if we assume the truth of one of the most difficult unsolved problems in mathematics, called the Generalized Riemann Hypothesis (for Dirichlet L-functions). We will not explain here the Generalized Riemann Hypothesis, often abbreviated to GRH, but here ...
The Farey Sequence and Its Niche(s)
... numbers. For example, there are many ways to represent 2. In its decimal notation we get 1.41421356237..., now we can turn that into a fraction, however the denominator get quite large rather fast. The goal of rational approximation of irrational numbers is to represent an irrational number with a f ...
... numbers. For example, there are many ways to represent 2. In its decimal notation we get 1.41421356237..., now we can turn that into a fraction, however the denominator get quite large rather fast. The goal of rational approximation of irrational numbers is to represent an irrational number with a f ...