solutions to problem set seven
... Roughly by looking at part (b) we estimate that it takes ln2 b steps. This is not quite correct, though. Perhaps this error was within their expectations of “estimate”, but to me it seems closer to the sum of the base two logs of the non-zero place values when the number is expressed as a binary num ...
... Roughly by looking at part (b) we estimate that it takes ln2 b steps. This is not quite correct, though. Perhaps this error was within their expectations of “estimate”, but to me it seems closer to the sum of the base two logs of the non-zero place values when the number is expressed as a binary num ...
1 - Columbia Math Department
... A few of these questions can be answered with completely elementary methods. For example, it is easy to see that sequences of the form (n! + 2, n! + 3, n! + 4, ..., n! + n) show that there are arbitrarily long gaps between prime numbers. In contrast is the following result due to Euclid: Theorem 1.1 ...
... A few of these questions can be answered with completely elementary methods. For example, it is easy to see that sequences of the form (n! + 2, n! + 3, n! + 4, ..., n! + n) show that there are arbitrarily long gaps between prime numbers. In contrast is the following result due to Euclid: Theorem 1.1 ...
irrationality and transcendence 4. continued fractions.
... Now as in the quadratic case, the first term here will vanish and the last two will be bounded. However the second term will be, roughly, a constant times qk−1 , which is unbounded, and this will wreck the entire argument. • In view of the theorem just proved and earlier results, we can sum up the c ...
... Now as in the quadratic case, the first term here will vanish and the last two will be bounded. However the second term will be, roughly, a constant times qk−1 , which is unbounded, and this will wreck the entire argument. • In view of the theorem just proved and earlier results, we can sum up the c ...
Extra Examples — Page references correspond to locations of Extra
... to show that neither root can be written as a quotient of integers. It might be easier to try a proof by contradiction because we can make the assumption that the roots are rational and hence can be written as fractions. That is, there is a very specific form in which the roots can be written. Suppos ...
... to show that neither root can be written as a quotient of integers. It might be easier to try a proof by contradiction because we can make the assumption that the roots are rational and hence can be written as fractions. That is, there is a very specific form in which the roots can be written. Suppos ...
Proofs
... be written elegantly in a formal mathematical style. Complete sentences of explanation are required. Do not simply write an equation; you must explain what the equation is giving and/or why it is being used. Moreover, all equations must be properly aligned with no scratch outs. Always give a conclus ...
... be written elegantly in a formal mathematical style. Complete sentences of explanation are required. Do not simply write an equation; you must explain what the equation is giving and/or why it is being used. Moreover, all equations must be properly aligned with no scratch outs. Always give a conclus ...
Доказательство великой теоремы Ферма - ferma-gold
... In contrast to existing variant of proof of the Ferma theorem presented by U. Waylls the present proof is a direct proof and doesn’t use special cases of earlier proofs of this theorem. In our opinion a successful approach was found which allows to prove the Ferma theorem sufficiently simply and thu ...
... In contrast to existing variant of proof of the Ferma theorem presented by U. Waylls the present proof is a direct proof and doesn’t use special cases of earlier proofs of this theorem. In our opinion a successful approach was found which allows to prove the Ferma theorem sufficiently simply and thu ...
CHAP11 Cryptography
... it from his celebrated “Last Theorem”. Fermat's Last Theorem states that for all integers n ≥ 3 there are no solutions to the equation xn + yn = zn for non-zero integers x, y and z. We all know that 32 + 42 = 52 and 52 + 122 = 132. There infinitely many such integer solutions to the equation x2 + y2 ...
... it from his celebrated “Last Theorem”. Fermat's Last Theorem states that for all integers n ≥ 3 there are no solutions to the equation xn + yn = zn for non-zero integers x, y and z. We all know that 32 + 42 = 52 and 52 + 122 = 132. There infinitely many such integer solutions to the equation x2 + y2 ...
A note on induced Ramsey numbers
... The Ramsey number rpF ; qq of a k-uniform hypergraph F is the smallest natural number n such that every q-coloring of the edges of Knpkq , the complete k-uniform hypergraph on n vertices, contains a monochromatic copy of F . In the particular case when q “ 2, we simply write rpF q. The existence of ...
... The Ramsey number rpF ; qq of a k-uniform hypergraph F is the smallest natural number n such that every q-coloring of the edges of Knpkq , the complete k-uniform hypergraph on n vertices, contains a monochromatic copy of F . In the particular case when q “ 2, we simply write rpF q. The existence of ...
Chapter 10 - The Exponential and Logarithm Functions
... The values of '2:" as x gets closer to ..J3 seem to be converging to some defmite number. By doing more and more calculations, we could approximate this number to as high a degree of accuracy as we wished. We thus have a method for genera~ the decimal expansion of a number which could be called 2if3 ...
... The values of '2:" as x gets closer to ..J3 seem to be converging to some defmite number. By doing more and more calculations, we could approximate this number to as high a degree of accuracy as we wished. We thus have a method for genera~ the decimal expansion of a number which could be called 2if3 ...
Proof by Induction
... from the contradiction that our assumption that the statement we want to prove is false is incorrect, so the statement we want to prove must be true. But that’s just silly. Why do we need the first and third steps? After all, the second step is a proof all by itself! Unfortunately, this redundant st ...
... from the contradiction that our assumption that the statement we want to prove is false is incorrect, so the statement we want to prove must be true. But that’s just silly. Why do we need the first and third steps? After all, the second step is a proof all by itself! Unfortunately, this redundant st ...
BSc-Mathematics-Syll..
... UNIT – I (12 Hours), Differential Equations of first order and first degree : Linear Differential Equations; Differential Equations Reducible to Linear Form; Exact Differential Equations; Integrating Factors; Change of Variables. UNIT – II (12 Hours), Orthogonal Trajectories. Differential Equations ...
... UNIT – I (12 Hours), Differential Equations of first order and first degree : Linear Differential Equations; Differential Equations Reducible to Linear Form; Exact Differential Equations; Integrating Factors; Change of Variables. UNIT – II (12 Hours), Orthogonal Trajectories. Differential Equations ...
RISES, LEVELS, DROPS AND - California State University, Los
... A composition of n consists of an ordered sequence of positive integers whose sum is n. A palindromic composition (or palindrome) is one for which the sequence reads the same forwards and backwards. We derive results for the number of “+” signs, summands, levels (a summand followed by itself), rises ...
... A composition of n consists of an ordered sequence of positive integers whose sum is n. A palindromic composition (or palindrome) is one for which the sequence reads the same forwards and backwards. We derive results for the number of “+” signs, summands, levels (a summand followed by itself), rises ...
Document
... Prove the base case: state what P(0) is, then prove it using any technique you'd like. Prove the inductive step: ...
... Prove the base case: state what P(0) is, then prove it using any technique you'd like. Prove the inductive step: ...
PROBLEM SET 7
... The pigeonhole principle is the following observation: Theorem 1. Suppose that > kn marbles are distributed over n jars, then one jar will contain at least ≥ k + 1 marbles. (It can also be formulated in terms of pigeons and pigeonholes, hence the name.) The proof of this pigeonhole principle is easy ...
... The pigeonhole principle is the following observation: Theorem 1. Suppose that > kn marbles are distributed over n jars, then one jar will contain at least ≥ k + 1 marbles. (It can also be formulated in terms of pigeons and pigeonholes, hence the name.) The proof of this pigeonhole principle is easy ...
Full text
... "1967 as the Sum of Three Squares," by Brother Alfred Brousseau (April, 1967 Fibonacci Quarterly), the general result is as follows: ...
... "1967 as the Sum of Three Squares," by Brother Alfred Brousseau (April, 1967 Fibonacci Quarterly), the general result is as follows: ...
Two statements that are equivalent to a
... one may take θ = 32999/33000. This result has been subsequently improved, and in 2001 Baker, Harman, and Pintz [2] proved that in (1) the constant θ may be taken to be 0.525. In other words, these authors showed that the interval [x, x + x0.525 ] contains at least one prime number for sufficiently l ...
... one may take θ = 32999/33000. This result has been subsequently improved, and in 2001 Baker, Harman, and Pintz [2] proved that in (1) the constant θ may be taken to be 0.525. In other words, these authors showed that the interval [x, x + x0.525 ] contains at least one prime number for sufficiently l ...
Solutions of the Pell Equations x2 − (a2b2 + 2b)y2 = N when N ∈ {±1,±4}
... by Theorem 3.4 and Corollary 2.8, it follows that the fundamental solution to the equation x2 − dy 2 = −4 is (2(a2 b + 1), 2a). Then the proof follows. Theorem 3.6 Let d = a2 b2 + 2b. Then the equation x2 − dy 2 = −4 has no positive integer solutions. Proof Assume that b is odd. If a is odd, then d ...
... by Theorem 3.4 and Corollary 2.8, it follows that the fundamental solution to the equation x2 − dy 2 = −4 is (2(a2 b + 1), 2a). Then the proof follows. Theorem 3.6 Let d = a2 b2 + 2b. Then the equation x2 − dy 2 = −4 has no positive integer solutions. Proof Assume that b is odd. If a is odd, then d ...
Notes on Calculus by
... Lemma 6.5.4 Let f be a differentiable function on an open interval I such that f (x) is > 0 (resp. < 0) for all x in I. Then f is a strictly increasing function, resp. strictly decreasing function, on I. Proof of Lemma. Assume f (x) > 0 for all x ∈ I. Pick any two points a, b ∈ I with a < b. We h ...
... Lemma 6.5.4 Let f be a differentiable function on an open interval I such that f (x) is > 0 (resp. < 0) for all x in I. Then f is a strictly increasing function, resp. strictly decreasing function, on I. Proof of Lemma. Assume f (x) > 0 for all x ∈ I. Pick any two points a, b ∈ I with a < b. We h ...