Full text
... for a natural algebraic and geometric setting for their analysis. In this way many known results are unified and simplified and new results are obtained. Some of the results extend to Fibonacci representations of higher order, but we do not present these because we have been unable to extend the the ...
... for a natural algebraic and geometric setting for their analysis. In this way many known results are unified and simplified and new results are obtained. Some of the results extend to Fibonacci representations of higher order, but we do not present these because we have been unable to extend the the ...
Sieving and the Erdos-Kac theorem
... One way of using Proposition 3 is to take P to be the set of primes below z where z is suitably small so that the error term arising from the |rd |’s is negligible. If the numbers a in A are not too large, then there cannot be too many primes larger than z that divide a, and so Proposition 3 furnish ...
... One way of using Proposition 3 is to take P to be the set of primes below z where z is suitably small so that the error term arising from the |rd |’s is negligible. If the numbers a in A are not too large, then there cannot be too many primes larger than z that divide a, and so Proposition 3 furnish ...
16(4)
... Finally, the diagonals of Table 2.1 can be obtained as follows. The tth member of the main diagonal is found by evaluating the tth-order determinant with leading terms given by the tth "knight's move" from the apex of the triangle. The first super-diagonal is achieved using the same principle, but s ...
... Finally, the diagonals of Table 2.1 can be obtained as follows. The tth member of the main diagonal is found by evaluating the tth-order determinant with leading terms given by the tth "knight's move" from the apex of the triangle. The first super-diagonal is achieved using the same principle, but s ...
Public Key Cryptography and RSA Review: Number Theory Basics
... • Security relies on the difficulty of factoring large composite numbers • Essentially the same algorithm was discovered in 1973 by Clifford Cocks, who works for the British intelligence ...
... • Security relies on the difficulty of factoring large composite numbers • Essentially the same algorithm was discovered in 1973 by Clifford Cocks, who works for the British intelligence ...
(pdf)
... Proposition 2.1. The order, which is the least positive power of an element that equals 1, of any a ∈ F∗q divides q − 1. Proof. In F∗q , we list all q − 1 elements and multiply each of them by a. We get a permutation of the same elements, because any two distinct elements remain distinct after multi ...
... Proposition 2.1. The order, which is the least positive power of an element that equals 1, of any a ∈ F∗q divides q − 1. Proof. In F∗q , we list all q − 1 elements and multiply each of them by a. We get a permutation of the same elements, because any two distinct elements remain distinct after multi ...
AN ANALOGUE OF THE HARER-ZAGIER FORMULA FOR
... +(2n − 3) (10n2 − 9n) ηv (n − 2) + 8 ηv−1 (n − 2) − 8 ηv−2 (n − 2) +5(2n − 3)(2n − 4)(2n − 5) (ηv (n − 3) − 2 ηv−1 (n − 3)) −2(2n − 3)(2n − 4)(2n − 5)(2n − 6)(2n − 7) ηv (n − 4), ...
... +(2n − 3) (10n2 − 9n) ηv (n − 2) + 8 ηv−1 (n − 2) − 8 ηv−2 (n − 2) +5(2n − 3)(2n − 4)(2n − 5) (ηv (n − 3) − 2 ηv−1 (n − 3)) −2(2n − 3)(2n − 4)(2n − 5)(2n − 6)(2n − 7) ηv (n − 4), ...
Solutions 7
... (b) Show that f is its own inverse function, meaning that f (f (x)) = x for every x ∈ A. (c) Show that, if x ∈ A and x is not equal to 1 or p − 1, then f (x) is not 1 or −1. Therefore, when searching by trial and error for the inverse of, let’s say, 3 (mod 11), one need only check the numbers form 2 ...
... (b) Show that f is its own inverse function, meaning that f (f (x)) = x for every x ∈ A. (c) Show that, if x ∈ A and x is not equal to 1 or p − 1, then f (x) is not 1 or −1. Therefore, when searching by trial and error for the inverse of, let’s say, 3 (mod 11), one need only check the numbers form 2 ...
Full text
... The first-named author in his book [8] put forward the following problems: Does there exist a pseudoprime of the form 2 " - 2 ? (problem #22) and: Do there exist infinitely many even pseudoprime numbers which are the products of three primes? (problem #51). In 1989 McDaniel [4] gave an example of a ...
... The first-named author in his book [8] put forward the following problems: Does there exist a pseudoprime of the form 2 " - 2 ? (problem #22) and: Do there exist infinitely many even pseudoprime numbers which are the products of three primes? (problem #51). In 1989 McDaniel [4] gave an example of a ...
8(4)
... (p,q) the next stage will be of the correct form, Case 2. Suppose N + Z would give a carry; that i s , 10 - p + q> 10 or q ^ p. This carry is ignored, so the next number will be (10 - p) + (q) ...
... (p,q) the next stage will be of the correct form, Case 2. Suppose N + Z would give a carry; that i s , 10 - p + q> 10 or q ^ p. This carry is ignored, so the next number will be (10 - p) + (q) ...
Lecture 56 - TCD Maths
... and thus p = pi for some integer i between 1 and r . But then p = p1 , since p1 is the smallest of the prime numbers p1 , p2 , . . . , pr . Similarly p = q1 . Therefore p = p1 = q1 . Let m = n/p. Then m = p2 p3 · · · pr = q2 q3 · · · qs . But then r = s and pi = qi for all integers i between 2 and r ...
... and thus p = pi for some integer i between 1 and r . But then p = p1 , since p1 is the smallest of the prime numbers p1 , p2 , . . . , pr . Similarly p = q1 . Therefore p = p1 = q1 . Let m = n/p. Then m = p2 p3 · · · pr = q2 q3 · · · qs . But then r = s and pi = qi for all integers i between 2 and r ...
Discrete Mathematics
... “How many ways can you guys be matched with each other?” she wonders. “This is clearly the same problem as seating you on six chairs; it does not matter whether the chairs are around the dinner table of at the three boards. So the answer is 720 as before.” “I think you should not count it as a diff ...
... “How many ways can you guys be matched with each other?” she wonders. “This is clearly the same problem as seating you on six chairs; it does not matter whether the chairs are around the dinner table of at the three boards. So the answer is 720 as before.” “I think you should not count it as a diff ...
Version 1.0 of the Math 135 course notes - CEMC
... 39.3 Infinite Sets Are Even Weirder Than You Thought 39.4 Not All Infinite Sets Have the Same Cardinality . . ...
... 39.3 Infinite Sets Are Even Weirder Than You Thought 39.4 Not All Infinite Sets Have the Same Cardinality . . ...
Grade 7/8 Math Circles Modular Arithmetic The Modulus Operator
... Visualizing the Mod Operation To find A mod N , we can either do it algebraically/with a calculator, or we can do it visually, by using a “clock” of numbers 1. Construct a clock with N hours with 0 at the top, going clockwise until N − 1 2. Start at 0 and move around the clock A hours (a) If A is ne ...
... Visualizing the Mod Operation To find A mod N , we can either do it algebraically/with a calculator, or we can do it visually, by using a “clock” of numbers 1. Construct a clock with N hours with 0 at the top, going clockwise until N − 1 2. Start at 0 and move around the clock A hours (a) If A is ne ...
Products of random variables and the first digit phenomenon
... numbers whose mantissae are small are more numerous than those whose mantissae are large. This fact is called the First Digit Phenomenon. He also noticed that this phenomenon seems independent of the units. This led him to make a scale-invariance hypothesis (more or less satisfied in real life) from ...
... numbers whose mantissae are small are more numerous than those whose mantissae are large. This fact is called the First Digit Phenomenon. He also noticed that this phenomenon seems independent of the units. This led him to make a scale-invariance hypothesis (more or less satisfied in real life) from ...
Modular Arithmetic
... - A number and its negative are usually not congruent: 2 6≡ (−2) (mod 9), since 2 − (−2) = 4 is not a multiple of 9. This is the source of many mistakes. - Suppose a ≡ b and c ≡ d (mod n). Then a + c ≡ b + d (mod n) and a · c ≡ b · d (mod n) - Dividing is not so simple: 6 ≡ 36 (mod 10), but dividing ...
... - A number and its negative are usually not congruent: 2 6≡ (−2) (mod 9), since 2 − (−2) = 4 is not a multiple of 9. This is the source of many mistakes. - Suppose a ≡ b and c ≡ d (mod n). Then a + c ≡ b + d (mod n) and a · c ≡ b · d (mod n) - Dividing is not so simple: 6 ≡ 36 (mod 10), but dividing ...
MIDTERM 1 TUESDAY, FEB 23 SOLUTIONS 1.– (15 points
... Euler’s theorem by simply squaring our result, while there is no obvious way to deduce our result form Euler’s theorem. It is not possible to improve further, and get an even smaller exponent, by question c. RemarkL” Hence the smallest natural number u such that au ≡ 1 (mod 2n+2 ) for all odd a is 2 ...
... Euler’s theorem by simply squaring our result, while there is no obvious way to deduce our result form Euler’s theorem. It is not possible to improve further, and get an even smaller exponent, by question c. RemarkL” Hence the smallest natural number u such that au ≡ 1 (mod 2n+2 ) for all odd a is 2 ...
3. Mathematical Induction 3.1. First Principle of
... induction hypothesis. Here we assume not just P (n), but P (j) for all the integers j between k and n (inclusive). We use this assumption to show P (n + 1). This method of induction is also called strong mathematical induction. It is used in computer science in a variety of settings such as proving ...
... induction hypothesis. Here we assume not just P (n), but P (j) for all the integers j between k and n (inclusive). We use this assumption to show P (n + 1). This method of induction is also called strong mathematical induction. It is used in computer science in a variety of settings such as proving ...
31(1)
... be due to erosion. The Egyptian architect, two thousand years before Herodotus was born, might well have aimed at a perfect phi pyramid. Maybe the architect's plans will eventually be found entombed with his mummy. ...
... be due to erosion. The Egyptian architect, two thousand years before Herodotus was born, might well have aimed at a perfect phi pyramid. Maybe the architect's plans will eventually be found entombed with his mummy. ...
3 Congruence arithmetic
... However, by defining addition and multiplication on Z/nZ (equivalence classes), we can use the tools of modern algebra to study congruence arithmetic. In particular, we have made Z/nZ into a what is called a ring (formal definition later). In this way we would write our above equation as (7Z + 5) + ...
... However, by defining addition and multiplication on Z/nZ (equivalence classes), we can use the tools of modern algebra to study congruence arithmetic. In particular, we have made Z/nZ into a what is called a ring (formal definition later). In this way we would write our above equation as (7Z + 5) + ...
(0.4) K -f, - American Mathematical Society
... supply of known continued fractions with known tails increases, then the possibility of using the method also increases. A complete, or even a partial answer to the question raised in the present paper obviously makes it possible to construct continued fractions with known tails. (A different type o ...
... supply of known continued fractions with known tails increases, then the possibility of using the method also increases. A complete, or even a partial answer to the question raised in the present paper obviously makes it possible to construct continued fractions with known tails. (A different type o ...