Lecture 11: the Euler φ-function In the light of the previous lecture
... In the light of the previous lecture, we shall now look in more detail at the function defined there. Let n ≥ 1 be a natural number. Recall that we defined φ(n) to be the number of natural numbers 1 ≤ m ≤ n and coprime to n. This function is called Euler’s φ-function. For example, let’s calculate φ( ...
... In the light of the previous lecture, we shall now look in more detail at the function defined there. Let n ≥ 1 be a natural number. Recall that we defined φ(n) to be the number of natural numbers 1 ≤ m ≤ n and coprime to n. This function is called Euler’s φ-function. For example, let’s calculate φ( ...
CS 173: Discrete Structures, Fall 2011 Homework 3
... 1. Proof by cases [8 points] Problem 5 from Homework 2 defined the “extended real numbers.” An extended real number has the form a + bǫ, where ǫ is a special new positive number whose square is zero. To compare the size of two extended real numbers, we use the definition: a + bǫ < c + dǫ whenever ei ...
... 1. Proof by cases [8 points] Problem 5 from Homework 2 defined the “extended real numbers.” An extended real number has the form a + bǫ, where ǫ is a special new positive number whose square is zero. To compare the size of two extended real numbers, we use the definition: a + bǫ < c + dǫ whenever ei ...
Full text
... so that checking somewhat fewer than p/2 Fibonacci numbers is guaranteed to find the first Fibonacci number divisible by p. Theorem 6. If P(p 2 ) f P(p) then P(p^) = p : i " 1 P(p). Thus, subject to a rather odd condition, if we know P(p) we know P(p J ). ...
... so that checking somewhat fewer than p/2 Fibonacci numbers is guaranteed to find the first Fibonacci number divisible by p. Theorem 6. If P(p 2 ) f P(p) then P(p^) = p : i " 1 P(p). Thus, subject to a rather odd condition, if we know P(p) we know P(p J ). ...
A NOTE ON THE SMARANDACHE PRIME PRODUCT
... the properties of the sequence (Pn ) have been studied by Prakash [3]. Theorem 2.1 improves one of the results of Prakash [3], while our proof of Theorem 2.2 is much simpler than that followed by Prakash [3]. The expressions for Ps , P r , Ps, Pg and P lO show that Theorem 2.1 is satisfied with tigh ...
... the properties of the sequence (Pn ) have been studied by Prakash [3]. Theorem 2.1 improves one of the results of Prakash [3], while our proof of Theorem 2.2 is much simpler than that followed by Prakash [3]. The expressions for Ps , P r , Ps, Pg and P lO show that Theorem 2.1 is satisfied with tigh ...
More about Permutations and Symmetry Groups
... disjoint as sets. Since cycles are permutations, we are allowed to multiply them. Theorem 10.1. Any permutation can be expressed as a product of disjoint cycles. We will omit the proof, but describe the conversion procedure in an informal way. Given a permutation p, start with 1, then compute p(1), ...
... disjoint as sets. Since cycles are permutations, we are allowed to multiply them. Theorem 10.1. Any permutation can be expressed as a product of disjoint cycles. We will omit the proof, but describe the conversion procedure in an informal way. Given a permutation p, start with 1, then compute p(1), ...
Triangular numbers and elliptic curves
... have only finitely many solutions in natural numbers (r, s, t). In this paper we change perspective and seek rational solutions of (1). In other words, we define ∆r = r(r + 1)/2 for any rational number r, and then consider (1) for fixed n, m. There exist 8 obvious solutions to these equations, namel ...
... have only finitely many solutions in natural numbers (r, s, t). In this paper we change perspective and seek rational solutions of (1). In other words, we define ∆r = r(r + 1)/2 for any rational number r, and then consider (1) for fixed n, m. There exist 8 obvious solutions to these equations, namel ...
Answers - Grade 6/ Middle School Math
... 4) A brand of perfume costs $98.60 for 20fl.oz. Find the unit rate. (3pts) 4) $4.93 per fl.oz 98.60/20 = $4.93 per fl.oz ...
... 4) A brand of perfume costs $98.60 for 20fl.oz. Find the unit rate. (3pts) 4) $4.93 per fl.oz 98.60/20 = $4.93 per fl.oz ...
Full text
... the equilateral hyperbola y = xa onto the hyperbola y = 4 xa , for each integer n > 0, that preserves isodecimal points. Finally, in section 7, we investigate those isodecimal points on parabola y = ax2 . Moreover, using the same technique, we construct a one-to-one correspondence, σ, between the pa ...
... the equilateral hyperbola y = xa onto the hyperbola y = 4 xa , for each integer n > 0, that preserves isodecimal points. Finally, in section 7, we investigate those isodecimal points on parabola y = ax2 . Moreover, using the same technique, we construct a one-to-one correspondence, σ, between the pa ...
Problem List 3
... Next are three problems involving congruences and Fermat’s little theorem. Let n be a positive integer. The Euler function φ(n) is equal to the number of positive integers k = 1, 2, . . . , n−1 relatively prime to n. The Euler function can be easily calculated using that φ(pk ) = pk −pk−1 for p prim ...
... Next are three problems involving congruences and Fermat’s little theorem. Let n be a positive integer. The Euler function φ(n) is equal to the number of positive integers k = 1, 2, . . . , n−1 relatively prime to n. The Euler function can be easily calculated using that φ(pk ) = pk −pk−1 for p prim ...
3-6 Fundamental Theorem of Algebra Day 1
... as the number of zeros. This is true for all polynomial functions. However, all of the zeros are not necessarily real zeros. Polynomials functions, like quadratic functions, may have complex zeros that are not real numbers. ...
... as the number of zeros. This is true for all polynomial functions. However, all of the zeros are not necessarily real zeros. Polynomials functions, like quadratic functions, may have complex zeros that are not real numbers. ...
CONGRUENCES Modular arithmetic. Two whole numbers a and b
... Modular arithmetic. Two whole numbers a and b are said to be congruent modulo n if they give the same remainders when divided by n. In other words, the difference a − b is divisible by n. Examples: 5 ≡ 1 (mod 2), 6 ≡ 2 (mod 4), etc. Congruences are very important because many of their properties are ...
... Modular arithmetic. Two whole numbers a and b are said to be congruent modulo n if they give the same remainders when divided by n. In other words, the difference a − b is divisible by n. Examples: 5 ≡ 1 (mod 2), 6 ≡ 2 (mod 4), etc. Congruences are very important because many of their properties are ...
CCGPS Advanced Algebra
... MCC9‐12.N.CN.9 (+) Know the Fundamental Theorem of Algebra; show that it is true for quadratic polynomials. Understand the relationship between zeros and factors of polynomials. MCC9‐12.A.APR.2 Know and apply the Remainder Theorem: For a polynomial p(x) and a number a, the remainder on division ...
... MCC9‐12.N.CN.9 (+) Know the Fundamental Theorem of Algebra; show that it is true for quadratic polynomials. Understand the relationship between zeros and factors of polynomials. MCC9‐12.A.APR.2 Know and apply the Remainder Theorem: For a polynomial p(x) and a number a, the remainder on division ...
Mathematical Statements and Their Proofs
... Let a and b be two even integers. Since they are even, they can be written in the forms a = 2x and b = 2y for some integers x and y, respectively. Then, a + b can be written in the form 2x + 2y, giving the following equation: a + b = 2x + 2y = 2(x + y) From this, we see that a + b is divisible by 2. ...
... Let a and b be two even integers. Since they are even, they can be written in the forms a = 2x and b = 2y for some integers x and y, respectively. Then, a + b can be written in the form 2x + 2y, giving the following equation: a + b = 2x + 2y = 2(x + y) From this, we see that a + b is divisible by 2. ...
Introduction to Discrete Mathematics
... A contradiction is a statement that is always false. (negation of a tautology) ...
... A contradiction is a statement that is always false. (negation of a tautology) ...
Lecture 4 - CSE@IIT Delhi
... We know that √2 is irrational, what about √2√2 ? Case 1: √2√2 is rational Then we are done, a=√2, b=√2. Case 2: √2√2 is irrational Then (√2√2)√2 = √22 = 2, a rational number So a=√2√2, b= √2 will do. So in either case there are a,b irrational and ab be rational. We don’t (need to) know which case is ...
... We know that √2 is irrational, what about √2√2 ? Case 1: √2√2 is rational Then we are done, a=√2, b=√2. Case 2: √2√2 is irrational Then (√2√2)√2 = √22 = 2, a rational number So a=√2√2, b= √2 will do. So in either case there are a,b irrational and ab be rational. We don’t (need to) know which case is ...
