From Rainbow to the Lonely Runner
... keeps a constant speed and all runners have different speeds. A runner is called “lonely” at some moment if he or she has (circular) distance at least r/k apart from all other runners. Conjecture: For each runner, there exists some time that he or she is lonely. ...
... keeps a constant speed and all runners have different speeds. A runner is called “lonely” at some moment if he or she has (circular) distance at least r/k apart from all other runners. Conjecture: For each runner, there exists some time that he or she is lonely. ...
From Rainbow to the Lonely Runner
... keeps a constant speed and all runners have different speeds. A runner is called “lonely” at some moment if he or she has (circular) distance at least r/k apart from all other runners. Conjecture: For each runner, there exists some time that he or she is lonely. ...
... keeps a constant speed and all runners have different speeds. A runner is called “lonely” at some moment if he or she has (circular) distance at least r/k apart from all other runners. Conjecture: For each runner, there exists some time that he or she is lonely. ...
Day 10: Precious Conjectures Grade 7
... disprove each of the remaining conjectures. Students complete each of the three questions for each conjecture, as modelled for Goldbach’s Conjecture. Whole Class Æ Debrief Students present their findings. Lead a discussion about what would be enough to prove a conjecture true or to disprove or refut ...
... disprove each of the remaining conjectures. Students complete each of the three questions for each conjecture, as modelled for Goldbach’s Conjecture. Whole Class Æ Debrief Students present their findings. Lead a discussion about what would be enough to prove a conjecture true or to disprove or refut ...
1.5 Methods of Proof
... Two important questions that arise in the study of mathematics are: (1) When is a mathematical argument correct? (2) What methods can be used to construct mathematical arguments? This section helps answer these questions by describing various forms of correct and incorrect mathematical arguments. A ...
... Two important questions that arise in the study of mathematics are: (1) When is a mathematical argument correct? (2) What methods can be used to construct mathematical arguments? This section helps answer these questions by describing various forms of correct and incorrect mathematical arguments. A ...
I1 Pythagoras` Theorem and Introduction Trigonometric Ratios
... Very little is known of the life of Pythagoras, but he was born on the island of Samos and is credited with the founding of a community at Crotona in Southern Italy by about 530 BC. The community had religious and political purposes, but also dealt with mathematics, especially the properties of whol ...
... Very little is known of the life of Pythagoras, but he was born on the island of Samos and is credited with the founding of a community at Crotona in Southern Italy by about 530 BC. The community had religious and political purposes, but also dealt with mathematics, especially the properties of whol ...
Engaging Students in Proof and Reasoning in High School Non
... them.##She#identified#student#work#that#had#the#potential#to#help#her#students#discover#and#make#sense# of#an#important#mathematics#concept.##Specifically,#when#Pretty#counted#on#from#the#larger#number,#the# teacher#understood#Pretty’s#strategy#was#based#on#the#commutative#property.##The#teacher#als ...
... them.##She#identified#student#work#that#had#the#potential#to#help#her#students#discover#and#make#sense# of#an#important#mathematics#concept.##Specifically,#when#Pretty#counted#on#from#the#larger#number,#the# teacher#understood#Pretty’s#strategy#was#based#on#the#commutative#property.##The#teacher#als ...
Full text
... The first assertion we shall disprove states that there are infinitely many pairs of positive coprime integers x, y such that 2\y, x2 + y2 E D, and ...
... The first assertion we shall disprove states that there are infinitely many pairs of positive coprime integers x, y such that 2\y, x2 + y2 E D, and ...
Solution 9
... Sol. Assume for a contradiction that 10 = a/b for some a, b ∈ Z with b 6= 0. We can assume that the fraction a/b is in lowest terms. Then we have 10b2 = a2 which implies that a2 , and therefore a, is even. Write a = 2a0 for some a0 ∈ Z. Then we have 10b2 = 4a20 i.e. 5b2 = 2a20 and so 5b2 is even. It ...
... Sol. Assume for a contradiction that 10 = a/b for some a, b ∈ Z with b 6= 0. We can assume that the fraction a/b is in lowest terms. Then we have 10b2 = a2 which implies that a2 , and therefore a, is even. Write a = 2a0 for some a0 ∈ Z. Then we have 10b2 = 4a20 i.e. 5b2 = 2a20 and so 5b2 is even. It ...
![[Part 1]](http://s1.studyres.com/store/data/008795785_1-ac97e8bcbccd58a3457a6ac11081a546-300x300.png)
[Part 1]
... Other possibilities suggest themselves. Letting n - 3k | (n - k + 1, 2k + 1) gives again n - 3 k | n - k + l whence also n - 3k j 3(n - k + 1) - (n - 3k) o r n - 3k | 2n + 3. If we then take 2n + 3 = prime we again obtain a useful theorem. It seems clear from just these samples that the theorems of ...
... Other possibilities suggest themselves. Letting n - 3k | (n - k + 1, 2k + 1) gives again n - 3 k | n - k + l whence also n - 3k j 3(n - k + 1) - (n - 3k) o r n - 3k | 2n + 3. If we then take 2n + 3 = prime we again obtain a useful theorem. It seems clear from just these samples that the theorems of ...
Cyclic Groups
... • Theorem 4.5 In a finite group, the number of elements of order d is a multiple of (d). • Proof: Let G be a finite group with n elements of order d. Let b be the number of cyclic subgroups G with order d. Each element of order d belongs to exactly one cyclic subgroup of order d. Thus n = b•(d). ...
... • Theorem 4.5 In a finite group, the number of elements of order d is a multiple of (d). • Proof: Let G be a finite group with n elements of order d. Let b be the number of cyclic subgroups G with order d. Each element of order d belongs to exactly one cyclic subgroup of order d. Thus n = b•(d). ...
(A B) (A B) (A B) (A B)
... Proof: We must show that when AB AB is true then A=B is true. (Proof by contradiction) Assume that AB AB is true but AB. If AB then this means that either xA but xB, or xB but xA. If xA but xB, then x AB but x AB so AB is not a subset of AB and we have a contradiction t ...
... Proof: We must show that when AB AB is true then A=B is true. (Proof by contradiction) Assume that AB AB is true but AB. If AB then this means that either xA but xB, or xB but xA. If xA but xB, then x AB but x AB so AB is not a subset of AB and we have a contradiction t ...
(A B) (A B) (A B) (A B)
... Proof: We must show that when AB AB is true then A=B is true. (Proof by contradiction) Assume that AB AB is true but AB. If AB then this means that either xA but xB, or xB but xA. If xA but xB, then x AB but x AB so AB is not a subset of AB and we have a contradiction t ...
... Proof: We must show that when AB AB is true then A=B is true. (Proof by contradiction) Assume that AB AB is true but AB. If AB then this means that either xA but xB, or xB but xA. If xA but xB, then x AB but x AB so AB is not a subset of AB and we have a contradiction t ...
PPT - School of Computer Science
... standing. Domino k-1 ≥ 0 did fall, but k-1 will knock over domino k. Thus, domino k must fall and remain standing. Contradiction. ...
... standing. Domino k-1 ≥ 0 did fall, but k-1 will knock over domino k. Thus, domino k must fall and remain standing. Contradiction. ...
Notes for week 11.
... An informal proof of this result could be the observation that either a or (a+1) must be divisible by 2, and therefore the product a(a+1) must also be divisible by 2. However, in our studies we saw a very similar example that provides a “template” for proving the result. That is: Let a ∈ ℕ show that ...
... An informal proof of this result could be the observation that either a or (a+1) must be divisible by 2, and therefore the product a(a+1) must also be divisible by 2. However, in our studies we saw a very similar example that provides a “template” for proving the result. That is: Let a ∈ ℕ show that ...
ON THE FRACTIONAL PARTS OF LACUNARY SEQUENCES
... 5. Fractional parts of powers of 10 Proof of Theorem 5. Fix u in {0, 1, .√. . , 9} and set tn = 10n/2 , ν = −u/10 in Theorem 1. Note that 0.4625 > 1/( 10 − 1). It follows from Theorem 1 that there is a positive ξ such that {ξ 10n/2 −√u/10} < 0.4625 for each integer n ≥ 0. Hence {ξ 10k − u/10} < 1/2 ...
... 5. Fractional parts of powers of 10 Proof of Theorem 5. Fix u in {0, 1, .√. . , 9} and set tn = 10n/2 , ν = −u/10 in Theorem 1. Note that 0.4625 > 1/( 10 − 1). It follows from Theorem 1 that there is a positive ξ such that {ξ 10n/2 −√u/10} < 0.4625 for each integer n ≥ 0. Hence {ξ 10k − u/10} < 1/2 ...
A Basis Theorem for Perfect Sets
... integer; we carry the values assigned to Gx , Gy , and t during one step to the beginning of the next step, but these variables do not achieve a limit over the course of the recursion. When we speak of the least Gk , we are referring to the Gk with minimal index. Initially X and Y are not defined an ...
... integer; we carry the values assigned to Gx , Gy , and t during one step to the beginning of the next step, but these variables do not achieve a limit over the course of the recursion. When we speak of the least Gk , we are referring to the Gk with minimal index. Initially X and Y are not defined an ...
Methods of Proof
... Thus, any other numbers that differ by 1 cannot both be perfect squares Thus, a non-perfect square must exist in any set that contains two numbers that differ by 1 Note that we didn’t specify which one it was! ...
... Thus, any other numbers that differ by 1 cannot both be perfect squares Thus, a non-perfect square must exist in any set that contains two numbers that differ by 1 Note that we didn’t specify which one it was! ...
No. 10/2016: Prime Tuples in Function Fields
... the Twin Prime Conjecture, and a quantitative version of it known as the Hardy–Littlewood Conjecture. We will also see that these and other questions about prime numbers can be extended to questions about function fields, and discuss recent progress which has been made to answer them in this context ...
... the Twin Prime Conjecture, and a quantitative version of it known as the Hardy–Littlewood Conjecture. We will also see that these and other questions about prime numbers can be extended to questions about function fields, and discuss recent progress which has been made to answer them in this context ...
algebraic numbers and topologically equivalent measures in the
... Let r be an algebraic integer of degree 2 and s be binomially related to r. Then r, s satisfy (1) and (2). Therefore s is an algebraic integer of degree 2 and s = (pr) and r = (qs) for some integers p, q. Applying the preceding Lemma, we obtain s = r or s = Í - r. There is another proof for Theorem ...
... Let r be an algebraic integer of degree 2 and s be binomially related to r. Then r, s satisfy (1) and (2). Therefore s is an algebraic integer of degree 2 and s = (pr) and r = (qs) for some integers p, q. Applying the preceding Lemma, we obtain s = r or s = Í - r. There is another proof for Theorem ...