![How to Guess What to Prove Example](http://s1.studyres.com/store/data/019741163_1-43550415244519a54d7e071705908981-300x300.png)
How to Guess What to Prove Example
... Proof: By strong induction. Let P (n) be the statement that n has a unique factorization. We prove P (n) for n > 1. Basis: P (2) is clearly true. Induction step: Assume P (2), . . . , P (n). We prove P (n + 1). If n + 1 is prime, we are done. If not, it factors somehow. Suppose n + 1 = rs r, s > 1. ...
... Proof: By strong induction. Let P (n) be the statement that n has a unique factorization. We prove P (n) for n > 1. Basis: P (2) is clearly true. Induction step: Assume P (2), . . . , P (n). We prove P (n + 1). If n + 1 is prime, we are done. If not, it factors somehow. Suppose n + 1 = rs r, s > 1. ...
Irrational and Algebraic Numbers, IVT, Upper and Lower Bounds
... Now suppose f (c) > 0, then f (x) > 0 ∀x ∈ (c − ², c + ²) for some ² > 0, and so c − ²/2 is an upper bound for S. If x ∈ S and x > c − ²/2, then f (c − ²/2) < 0. So c − ²/2 is an upperbound for S. But c − ²/2 < c, contradicting c being the least upper bound of S. Thus we can’t have f (c) > 0. Theref ...
... Now suppose f (c) > 0, then f (x) > 0 ∀x ∈ (c − ², c + ²) for some ² > 0, and so c − ²/2 is an upper bound for S. If x ∈ S and x > c − ²/2, then f (c − ²/2) < 0. So c − ²/2 is an upperbound for S. But c − ²/2 < c, contradicting c being the least upper bound of S. Thus we can’t have f (c) > 0. Theref ...
SOLUTIONS TO QUIZ
... 2. No calculators, electronic watches, cellphones etc. will be allowed in this quiz. 3. This is a ”closed book” quiz, meaning that, looking at notes or computers or taking the help of cheat sheets is not permitted. 4. As per Penn’s Academic Integrity, any form of cheating is strongly discouraged and ...
... 2. No calculators, electronic watches, cellphones etc. will be allowed in this quiz. 3. This is a ”closed book” quiz, meaning that, looking at notes or computers or taking the help of cheat sheets is not permitted. 4. As per Penn’s Academic Integrity, any form of cheating is strongly discouraged and ...
arXiv:math/0008222v1 [math.CO] 30 Aug 2000
... f (n) if n ≡ 0, 3 (mod 4), and f (−1 − n) = −f (n) if n ≡ 1, 2 (mod 4). John and Sachs [JS] have independently investigated the 2-adic behavior of f , and explicitly determined it modulo 26 . Their methods, as well as ours, can be used to write formulas for f modulo any power of 2, but no closed for ...
... f (n) if n ≡ 0, 3 (mod 4), and f (−1 − n) = −f (n) if n ≡ 1, 2 (mod 4). John and Sachs [JS] have independently investigated the 2-adic behavior of f , and explicitly determined it modulo 26 . Their methods, as well as ours, can be used to write formulas for f modulo any power of 2, but no closed for ...
G:\stirling primes\slides stirl - College of Science and Mathematics
... result as we did with Corollary 1. Such theorems would allow us to throw out more known composite values of nk extend a search with larger values of k. ...
... result as we did with Corollary 1. Such theorems would allow us to throw out more known composite values of nk extend a search with larger values of k. ...
Mat 2345 Student Responsibilities — Week 5 Week 5 Overview 2.4
... (It doesn’t matter if there is more than one expansion for a number as long as our construction takes this into account.) ...
... (It doesn’t matter if there is more than one expansion for a number as long as our construction takes this into account.) ...
Test 2A (Sec. 2-1, 2-2, 2-5, 2-6) Multiple Choice Identify the choice
... 6. Laisha’s Internet Services designs web sites and recently began a weekly advertising campaign. Laisha noticed an increase in her customers over a period of five consecutive weeks. Based on the pattern shown in the graph, make a conjecture about the number of customers Laisha will have in the seve ...
... 6. Laisha’s Internet Services designs web sites and recently began a weekly advertising campaign. Laisha noticed an increase in her customers over a period of five consecutive weeks. Based on the pattern shown in the graph, make a conjecture about the number of customers Laisha will have in the seve ...
Irrationality of ratios of solutions to tanx = x and related matter
... Theorem 2. Suppose that h(x) is a non-constant rational function with algebraic coefficients such that h(x) is real valued (exculding poles) for real x, then equation (9) has countably infinitely many positive isolated solutions. Each non-zero solution is transcendental. If x and y are any two disti ...
... Theorem 2. Suppose that h(x) is a non-constant rational function with algebraic coefficients such that h(x) is real valued (exculding poles) for real x, then equation (9) has countably infinitely many positive isolated solutions. Each non-zero solution is transcendental. If x and y are any two disti ...
FP3: Complex Numbers - Schoolworkout.co.uk
... This suggests that z n De Moivre’s theorem states that if ...
... This suggests that z n De Moivre’s theorem states that if ...
On the digits of prime numbers
... These concepts are strongly related with Sarnak’s conjecture if f is produced by a zero topological entropy dynamical system. For f = 1 these properties are equivalent: ...
... These concepts are strongly related with Sarnak’s conjecture if f is produced by a zero topological entropy dynamical system. For f = 1 these properties are equivalent: ...
title goes here - Stetson University
... 1. Preliminaries ----------------------------------------------------------------------------------------1.1. Introduction -----------------------------------------------------------------------------------1.1.1. A Brief History of Calculus -------------------------------------------------------1.1. ...
... 1. Preliminaries ----------------------------------------------------------------------------------------1.1. Introduction -----------------------------------------------------------------------------------1.1.1. A Brief History of Calculus -------------------------------------------------------1.1. ...
RESEARCH PROJECTS 1. Irrationality questions
... 1.2. Irrationality of π 2 and the infinitude of primes. Let π(x) count the number of primes at most x. The celebrated Prime Number Theorem R x states that π(x) ∼ x/ log x for x large (even better, π(x) ∼ Li(x), where Li(x) = 2 dt/ log t, which to first order is x/ log x). As primes are the building ...
... 1.2. Irrationality of π 2 and the infinitude of primes. Let π(x) count the number of primes at most x. The celebrated Prime Number Theorem R x states that π(x) ∼ x/ log x for x large (even better, π(x) ∼ Li(x), where Li(x) = 2 dt/ log t, which to first order is x/ log x). As primes are the building ...
Integer Divisibility
... Prime numbers satisfy many strange and wonderful properties. Observation: ...
... Prime numbers satisfy many strange and wonderful properties. Observation: ...
n - Stanford University
... include all three of these steps Youmeans must include all pthree these in yourdivisor. This this that both and qof have 2 assteps a common proofs! contradicts proofs! our earlier assertion that their only common divisors are 1 and -1. We have reached a contradiction, so our assumption was incorrect ...
... include all three of these steps Youmeans must include all pthree these in yourdivisor. This this that both and qof have 2 assteps a common proofs! contradicts proofs! our earlier assertion that their only common divisors are 1 and -1. We have reached a contradiction, so our assumption was incorrect ...
Fundamental Theorem of Arithmetic
... We will show that there are infinitely many primes, in fact, the proof we give is Euclid’s original proof. Lemma. Every positive integer n > 1 has a prime divisor. Proof. Let S = {n ∈ Z | n > 1 and n has no prime divisors}. If S 6= ∅, since S is bounded below, by the well ordering property S has a ...
... We will show that there are infinitely many primes, in fact, the proof we give is Euclid’s original proof. Lemma. Every positive integer n > 1 has a prime divisor. Proof. Let S = {n ∈ Z | n > 1 and n has no prime divisors}. If S 6= ∅, since S is bounded below, by the well ordering property S has a ...
Lecture 1
... 3. theorems that have been previously established as true, and 4. statements that are logically implied by the earlier statements in the proof. When actually building the bridge, it may not be at all obvious which blocks to use or in what order to use them. This is where experience is helpful, toget ...
... 3. theorems that have been previously established as true, and 4. statements that are logically implied by the earlier statements in the proof. When actually building the bridge, it may not be at all obvious which blocks to use or in what order to use them. This is where experience is helpful, toget ...
Some remarks on iterated maps of natural numbers,
... It is unlikely that the problem originates from Steinhaus. It is equally unlikely that it comes from Reg Allenby’s daughter as described on page 234 of [2] who describes numbers that terminate at 1 as “happy numbers”. It is conjectured that about 1/7 of the set of natural numbers are ‘happy’ though ...
... It is unlikely that the problem originates from Steinhaus. It is equally unlikely that it comes from Reg Allenby’s daughter as described on page 234 of [2] who describes numbers that terminate at 1 as “happy numbers”. It is conjectured that about 1/7 of the set of natural numbers are ‘happy’ though ...
2E Numbers and Sets What is an equivalence relation on a set X? If
... (a) What is the highest common factor of two positive integers a and b? Show that the highest common factor may always be expressed in the form λa + µb, where λ and µ are integers. (a) The hcf of a and b is a positive integer c such that c | a and c | b and such that if d | a and d | b then d | c (c ...
... (a) What is the highest common factor of two positive integers a and b? Show that the highest common factor may always be expressed in the form λa + µb, where λ and µ are integers. (a) The hcf of a and b is a positive integer c such that c | a and c | b and such that if d | a and d | b then d | c (c ...