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Mathematics in Context Sample Review Questions
... From this triangle construct two squares with sides of length a + b, also shown above. The two squares have the same lengths for their sides, so their areas must be equal. a. Calculate the area of the left-hand square by adding up the areas of the triangles and squares that compose it. (4) ...
... From this triangle construct two squares with sides of length a + b, also shown above. The two squares have the same lengths for their sides, so their areas must be equal. a. Calculate the area of the left-hand square by adding up the areas of the triangles and squares that compose it. (4) ...
Note on a conjecture of PDTA Elliott
... stated some questions and conjectures related to shifted-prime factorizations. One of these questions is the following ([11, Question 2]): Rosen's question. Give an upper bound for the minimum number of primes required to represent all positive integers not exceeding n by shifted-prime factorization ...
... stated some questions and conjectures related to shifted-prime factorizations. One of these questions is the following ([11, Question 2]): Rosen's question. Give an upper bound for the minimum number of primes required to represent all positive integers not exceeding n by shifted-prime factorization ...
8Mathstandards unit 5
... 1. Know that numbers that are not rational are called irrational. Understand informally that every number has a decimal expansion; for rational numbers show that the decimal expansion repeats eventually, and convert a decimal expansion which repeats eventually into a rational number. 2. Use rational ...
... 1. Know that numbers that are not rational are called irrational. Understand informally that every number has a decimal expansion; for rational numbers show that the decimal expansion repeats eventually, and convert a decimal expansion which repeats eventually into a rational number. 2. Use rational ...
[Part 2]
... I mentioned this result to Dr. P.M. Lee of York University and he has pointed out to me that Lemma 3 can be derived from H. Siebeck's work on recurring series (L.E. Dickson, History of the Theory of Numbers, p. 394f). A colleague of his has also discovered a non-elementary proof of the above theorem ...
... I mentioned this result to Dr. P.M. Lee of York University and he has pointed out to me that Lemma 3 can be derived from H. Siebeck's work on recurring series (L.E. Dickson, History of the Theory of Numbers, p. 394f). A colleague of his has also discovered a non-elementary proof of the above theorem ...
Conditions Equivalent to the Existence of Odd Perfect
... In recent years, this M AGAZINE has published several interesting articles examining the abundancy index of a number [1, 3, 4]. In [1], R. Laatsch provided a comprehensive summary of what is known about the abundancy index, including a proof that the image of I (n) is dense in the interval (1, ∞). H ...
... In recent years, this M AGAZINE has published several interesting articles examining the abundancy index of a number [1, 3, 4]. In [1], R. Laatsch provided a comprehensive summary of what is known about the abundancy index, including a proof that the image of I (n) is dense in the interval (1, ∞). H ...
Lacunary recurrences for Eisenstein series
... The most striking difference between the recurrences (1.2) and (1.3) is that in (1.3), only about a third of the previous Eisenstein series are needed, while in (1.2), all Eisenstein series occur. In the end of [1], Romik asked for a direct proof of (1.3) using the theory of modular forms. Here, we ...
... The most striking difference between the recurrences (1.2) and (1.3) is that in (1.3), only about a third of the previous Eisenstein series are needed, while in (1.2), all Eisenstein series occur. In the end of [1], Romik asked for a direct proof of (1.3) using the theory of modular forms. Here, we ...
Name: Exam 2 Directions: You must show all of your work for full
... and 13x12 + 50x4 + 3 is always > 0, because all the terms in it are positive ( numbers raised to an even power will always be positive). Adding all positive terms can never give you zero. Since there’s no c with f 0 (c) = 0, the conditions for Rolle’s theorem aren’t met. It’s a continuous differenti ...
... and 13x12 + 50x4 + 3 is always > 0, because all the terms in it are positive ( numbers raised to an even power will always be positive). Adding all positive terms can never give you zero. Since there’s no c with f 0 (c) = 0, the conditions for Rolle’s theorem aren’t met. It’s a continuous differenti ...
Numbers: Fun and Challenge
... 4t2 = (u − s2 )(u + s2 ), there exist positive integers a, b such that u − s2 = 2b2 , u + s2 = 2a2 , t2 = ab, gcd(a, b) = 1. From t2 = ab we see that there exist integers x1 , y1 such that a = x21 , b = y12 and t = x1 y1 . It follows that u = x41 + y14 and s2 = x41 − y14 . Let z1 = s, so (x1 , y1 , ...
... 4t2 = (u − s2 )(u + s2 ), there exist positive integers a, b such that u − s2 = 2b2 , u + s2 = 2a2 , t2 = ab, gcd(a, b) = 1. From t2 = ab we see that there exist integers x1 , y1 such that a = x21 , b = y12 and t = x1 y1 . It follows that u = x41 + y14 and s2 = x41 − y14 . Let z1 = s, so (x1 , y1 , ...
The Impossibility of Trisecting an Angle with Straightedge and
... and compass. That is, given an angle θ , give a procedure using only straightedge and compass that will construct the angle θ 3 in a finite number of steps. This problem, which dates to around 400 B.C., fascinated mathematicians and amateurs over the centuries and many “solutions” have been proposed ...
... and compass. That is, given an angle θ , give a procedure using only straightedge and compass that will construct the angle θ 3 in a finite number of steps. This problem, which dates to around 400 B.C., fascinated mathematicians and amateurs over the centuries and many “solutions” have been proposed ...
A Brief Note on Proofs in Pure Mathematics
... You should note that induction can only be applied when the statements are indexed by the natural numbers. There has to be a first case (the base case), and you have to have be able to move to the ‘next’ statement (the induction step). If the statements are indexed by real numbers, then there is no ...
... You should note that induction can only be applied when the statements are indexed by the natural numbers. There has to be a first case (the base case), and you have to have be able to move to the ‘next’ statement (the induction step). If the statements are indexed by real numbers, then there is no ...
Class notes, rings and modules : some of 23/03/2017 and 04/04/2017
... to be “aligned” with a basis of M in order to compute the quotient. Thus, our goal now is to find such aligned bases for the submodule N from the first picture. Here is an algorithm for doing it. We write the coordinates of the generators of N as rows of a matrix (called relations matrix). In our ex ...
... to be “aligned” with a basis of M in order to compute the quotient. Thus, our goal now is to find such aligned bases for the submodule N from the first picture. Here is an algorithm for doing it. We write the coordinates of the generators of N as rows of a matrix (called relations matrix). In our ex ...
What is Euler`s Prime Generating Polynomial? Main Theorem:
... Lehmer in (1936), and will be the main item of my talk. ...
... Lehmer in (1936), and will be the main item of my talk. ...
A product of Gamma function values at fractions with the
... A BSTRACT. We give an exact formula for the product of the values of Euler’s Gamma function evaluated at all rational numbers between 0 and 1 with the same denominator in lowest terms; the answer depends on whether or not that denominator is a prime power. A consequence is a surprisingly nice formul ...
... A BSTRACT. We give an exact formula for the product of the values of Euler’s Gamma function evaluated at all rational numbers between 0 and 1 with the same denominator in lowest terms; the answer depends on whether or not that denominator is a prime power. A consequence is a surprisingly nice formul ...
Fibonacci numbers at most one away from a perfect power
... change signs so that x ≡ −1 (mod 4) and 2 | y. 2 The Modularity Theorem states that all elliptic curves are modular. Wiles proved this for semi-stable elliptic curves, which was enough for the proof of Fermat’s Last Theorem. Since then the proof of the Modularity Theorem has been completed in a seri ...
... change signs so that x ≡ −1 (mod 4) and 2 | y. 2 The Modularity Theorem states that all elliptic curves are modular. Wiles proved this for semi-stable elliptic curves, which was enough for the proof of Fermat’s Last Theorem. Since then the proof of the Modularity Theorem has been completed in a seri ...
Congruent Number Problem 1 Congruent number problem
... 10-th century in Arab manuscripts (Al-Kazin) but it is possibly much older. It turns out to be a beautiful example of the modern theory of the arithmetic of elliptic curves, but it is more accurate to say that this theory grew out of the study of this problem. In the 17-th century, Fermat gave a won ...
... 10-th century in Arab manuscripts (Al-Kazin) but it is possibly much older. It turns out to be a beautiful example of the modern theory of the arithmetic of elliptic curves, but it is more accurate to say that this theory grew out of the study of this problem. In the 17-th century, Fermat gave a won ...
CHAP07 Mersenne and Fermat Primes
... Then since xn + 1 = (x + 1)(xn−1 − xn−2 + ... − x + 1) for odd n, 2N + 1 = 2pQ + 1 is divisible by 2Q + 1, which is a contradiction. Hence the only prime divisors of N is 2 and so N is a power of 2. Example 3: The first five Fermat numbers are: n ...
... Then since xn + 1 = (x + 1)(xn−1 − xn−2 + ... − x + 1) for odd n, 2N + 1 = 2pQ + 1 is divisible by 2Q + 1, which is a contradiction. Hence the only prime divisors of N is 2 and so N is a power of 2. Example 3: The first five Fermat numbers are: n ...
G30 MATH SEMINAR 1 - PROOFS BY CONTRADICTION 1
... admits no (strictly) positive integer solution. (this theorem, known as Fermat’s Last Theorem has been proved by Andrew Wiles in 1994). 2. Proofs by contradiction - the method The example we saw in previous section is called a direct proof. We will see a useful tool when direct proofs do not work. T ...
... admits no (strictly) positive integer solution. (this theorem, known as Fermat’s Last Theorem has been proved by Andrew Wiles in 1994). 2. Proofs by contradiction - the method The example we saw in previous section is called a direct proof. We will see a useful tool when direct proofs do not work. T ...
Squares in arithmetic progressions and infinitely many primes
... 3. MORE HEAVY MACHINERY One day over lunch, in late 1989, Bombieri showed me a completely different proof of Theorem 2, this time relying on one of the most influential results in algebraic and arithmetic geometry, Faltings’ theorem [3]. Faltings’ theorem is not easy to state, requiring a general un ...
... 3. MORE HEAVY MACHINERY One day over lunch, in late 1989, Bombieri showed me a completely different proof of Theorem 2, this time relying on one of the most influential results in algebraic and arithmetic geometry, Faltings’ theorem [3]. Faltings’ theorem is not easy to state, requiring a general un ...
Irrationality of Square Roots - Mathematical Association of America
... Indeed, if α were a fraction with denominator q, then m + nα would also be fraction with denominator q. Such a fraction is either zero or at least 1/q in magnitude. Let us first note some previous proofs of irrationality based on this criterion. Arbitrarily small numbers m + nα have been constructed ...
... Indeed, if α were a fraction with denominator q, then m + nα would also be fraction with denominator q. Such a fraction is either zero or at least 1/q in magnitude. Let us first note some previous proofs of irrationality based on this criterion. Arbitrarily small numbers m + nα have been constructed ...