Alg II 5-7 The Binomial Theorem
... The "coefficients only" column matches the numbers in Pascal's Triangle. Pascal's Triangle is a triangular array of numbers in which the first and last number of each row is 1. Each of the other numbers in the row is the sum of the two numbers above it. ...
... The "coefficients only" column matches the numbers in Pascal's Triangle. Pascal's Triangle is a triangular array of numbers in which the first and last number of each row is 1. Each of the other numbers in the row is the sum of the two numbers above it. ...
Chapter 8 Number Theory
... 8-3 The Pigeonhole Principle (鴿籠原理) The pigeonhole principle: If m pigeons occupy n pigeonholes and m>n, then at least one pigeonhole has two or more pigeons roosting in it. Eg. Let S ⊂ Z, and S has 37 elements. Then S contains two elements that have the same remainder upon division by 36. (Proof) ...
... 8-3 The Pigeonhole Principle (鴿籠原理) The pigeonhole principle: If m pigeons occupy n pigeonholes and m>n, then at least one pigeonhole has two or more pigeons roosting in it. Eg. Let S ⊂ Z, and S has 37 elements. Then S contains two elements that have the same remainder upon division by 36. (Proof) ...
Chapter 8 Number Theory 8-1 Prime Numbers and Composite N
... 8-3 The Pigeonhole Principle (鴿籠原理) The pigeonhole principle: If m pigeons occupy n pigeonholes and m>n, then at least one pigeonhole has two or more pigeons roosting in it. Eg. Let S Z, and S has 37 elements. Then S contains two elements that have the same remainder upon division by 36. (Proof) ...
... 8-3 The Pigeonhole Principle (鴿籠原理) The pigeonhole principle: If m pigeons occupy n pigeonholes and m>n, then at least one pigeonhole has two or more pigeons roosting in it. Eg. Let S Z, and S has 37 elements. Then S contains two elements that have the same remainder upon division by 36. (Proof) ...
Approximation of irrational numbers. Let α be an irrational number
... value κ and whose y-component is the denominator q. Now draw a rectangle with vertices at (− N1 , 0), ( N1 , 0), (− N1 , N + 1), ( N1 , N + 1). The theorem says that for any choice of N , the rectangle must contain a lattice point in its interior. Even if N is very very large, i.e. the rectangle is ...
... value κ and whose y-component is the denominator q. Now draw a rectangle with vertices at (− N1 , 0), ( N1 , 0), (− N1 , N + 1), ( N1 , N + 1). The theorem says that for any choice of N , the rectangle must contain a lattice point in its interior. Even if N is very very large, i.e. the rectangle is ...
For a pdf file
... Clearly, n > p. Since p is the largest prime number, n cannot be a prime number. In other words, n is composite. Let q be any prime number. Because of the way n is constructed, when n is divided by q the remainder is 1. That is, n is not a multiple of q. This contradicts the Fundamental Theorem of A ...
... Clearly, n > p. Since p is the largest prime number, n cannot be a prime number. In other words, n is composite. Let q be any prime number. Because of the way n is constructed, when n is divided by q the remainder is 1. That is, n is not a multiple of q. This contradicts the Fundamental Theorem of A ...
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... is demanded in each of m partols, with a total quantity demanded of T units, (T ;> m)„ The number of different ways of partitioning T into m demands is 0T ...
... is demanded in each of m partols, with a total quantity demanded of T units, (T ;> m)„ The number of different ways of partitioning T into m demands is 0T ...
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... We shall represent points in the plane by column vectors with two components. The set C = {(V) | both a and j3 are non-negative and at least one of a and /3 is positive} will be called the positive cone. Our preseat objective is to associate with each vector in the positive cone an /.^-sequence. Def ...
... We shall represent points in the plane by column vectors with two components. The set C = {(V) | both a and j3 are non-negative and at least one of a and /3 is positive} will be called the positive cone. Our preseat objective is to associate with each vector in the positive cone an /.^-sequence. Def ...
On mathematical induction
... Example 4.2. Let A be a set with n elements. Then the set P(A) of all subsets of A (the set P(A) is called the power set of A) contains 2n elements. Proof. We prove the theorem by induction over n. First we check the case in which A contains 0 elements. In this case A is the empty set ∅ and his only ...
... Example 4.2. Let A be a set with n elements. Then the set P(A) of all subsets of A (the set P(A) is called the power set of A) contains 2n elements. Proof. We prove the theorem by induction over n. First we check the case in which A contains 0 elements. In this case A is the empty set ∅ and his only ...
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... Forty-two abstracts on all branches of mathematics and science related to the Fibonacci numbers and their generalizations have been received. All contributed papers will appear subject to approval by a referee in the Conference Proceedings, which are expected to be published in 1989. The program for ...
... Forty-two abstracts on all branches of mathematics and science related to the Fibonacci numbers and their generalizations have been received. All contributed papers will appear subject to approval by a referee in the Conference Proceedings, which are expected to be published in 1989. The program for ...
Math 261 Spring 2014 Final Exam May 5, 2014 1. Give a statement
... (B) Find the number of zeroes at the end of the normal (base ten) representation of 24!. By the FTOA, every factor in 24! can be written as a product of primes. In particular, we are interested in the number of times the primes 2 and 5 appear since a zero at the end of a number corresponds to a fact ...
... (B) Find the number of zeroes at the end of the normal (base ten) representation of 24!. By the FTOA, every factor in 24! can be written as a product of primes. In particular, we are interested in the number of times the primes 2 and 5 appear since a zero at the end of a number corresponds to a fact ...
On the Reciprocal of the Binary Generating Function for the Sum of
... If A is a set of natural numbers containing 0, then there is a unique nonempty “reciprocal” set A−1 of natural numbers such that every positive integer can be written in the form a + a0 , where a ∈ A and a0 ∈ A−1 , in an even number of ways. It is straightforward to see that the generating functions ...
... If A is a set of natural numbers containing 0, then there is a unique nonempty “reciprocal” set A−1 of natural numbers such that every positive integer can be written in the form a + a0 , where a ∈ A and a0 ∈ A−1 , in an even number of ways. It is straightforward to see that the generating functions ...
6.042J Lecture 02: Solutions
... Now we justify the claim that if an is even, so is a. There is a simple proof by contradiction: suppose to the contrary that a is odd. It’s a familiar (and easily verified1 ) fact that the product of two odd numbers is odd, from which it follows that the product of any finite number of odd numbers is ...
... Now we justify the claim that if an is even, so is a. There is a simple proof by contradiction: suppose to the contrary that a is odd. It’s a familiar (and easily verified1 ) fact that the product of two odd numbers is odd, from which it follows that the product of any finite number of odd numbers is ...
DEPARTMENT OF MATHEMATICS
... b. : Z R under addition given by = the greatest integer x. c. : Z6 Z2 given (x) = the remainder of x when divided by 2. ...
... b. : Z R under addition given by = the greatest integer x. c. : Z6 Z2 given (x) = the remainder of x when divided by 2. ...
Issues in Nonlinear Hyperperf ect Numbers
... perfect numbers; either such sets are empty, or their euclidean distance from zero is greater than some very large number, or no particularly unique prime decomposition ...
... perfect numbers; either such sets are empty, or their euclidean distance from zero is greater than some very large number, or no particularly unique prime decomposition ...