1.2 Inductive Reasoning
... Technology Use a calculator to explore the pattern. Write a conjecture based on what you observe. 10. 101 25 __?__ ...
... Technology Use a calculator to explore the pattern. Write a conjecture based on what you observe. 10. 101 25 __?__ ...
1.2 Inductive Reasoning
... Technology Use a calculator to explore the pattern. Write a conjecture based on what you observe. 10. 101 3 25 5 __?__ ...
... Technology Use a calculator to explore the pattern. Write a conjecture based on what you observe. 10. 101 3 25 5 __?__ ...
Some remarks on Euler`s totient function - HAL
... Clearly α0 . . . αr is k written in binary form. Therefore there can be at most one odd number n such that φ(n) = 2k . If αi = 1 only when Fi is prime, then there exists n odd such that φ(n) = 2k and O(2k ) = 1. On the other hand, if there is an αi such that Fi is not prime, then there does not exis ...
... Clearly α0 . . . αr is k written in binary form. Therefore there can be at most one odd number n such that φ(n) = 2k . If αi = 1 only when Fi is prime, then there exists n odd such that φ(n) = 2k and O(2k ) = 1. On the other hand, if there is an αi such that Fi is not prime, then there does not exis ...
RECURSIVE REAL NUMBERS 784
... procedures formalized as recursive functions. We content ourselves with indicating what adjustments must be made where other than rational operations are to be performed. We suppose a polynomial P(z) given by its degree and by r.e., r.e. sequences of rational numbers converging to the real and imagi ...
... procedures formalized as recursive functions. We content ourselves with indicating what adjustments must be made where other than rational operations are to be performed. We suppose a polynomial P(z) given by its degree and by r.e., r.e. sequences of rational numbers converging to the real and imagi ...
Math 230 E Fall 2013 Homework 5 Drew Armstrong
... Since n is odd (why?), the prime 2 does not appear in this factorization, so each prime factor is either qi ≡ 1 mod 4 or qi ≡ 3 mod 4. We claim that at least one prime factor is ≡ 3 mod 4. Suppose not, i.e., suppose that every prime factor is ≡ 1 mod 4. Then n is a product of numbers ≡ 1 mod 4, henc ...
... Since n is odd (why?), the prime 2 does not appear in this factorization, so each prime factor is either qi ≡ 1 mod 4 or qi ≡ 3 mod 4. We claim that at least one prime factor is ≡ 3 mod 4. Suppose not, i.e., suppose that every prime factor is ≡ 1 mod 4. Then n is a product of numbers ≡ 1 mod 4, henc ...
Lecture Notes for MA 132 Foundations
... it complicates many statements. So we will not count 1 as a prime. I would say that Theorem 1.2 is less obvious than Theorem 1.1. But in fact 1.2 will be the first theorem we prove in the course, whereas you won’t meet a proof of 1.1 until a lot later. Theorem 1.1 becomes a little less obvious when ...
... it complicates many statements. So we will not count 1 as a prime. I would say that Theorem 1.2 is less obvious than Theorem 1.1. But in fact 1.2 will be the first theorem we prove in the course, whereas you won’t meet a proof of 1.1 until a lot later. Theorem 1.1 becomes a little less obvious when ...
primitive roots: a survey
... depends on α, the limiting value of the fractional part of the base-2 logarithm of n. That is, it looks like lim n→∞ Pn does not exist! And this is indeed the case, though the oscillation in P n is very gentle. We have lim sup Pn ≈ 0.72135465 which is achieved when α ≈ 0.139, and lim inf Pn ≈ 0.7213 ...
... depends on α, the limiting value of the fractional part of the base-2 logarithm of n. That is, it looks like lim n→∞ Pn does not exist! And this is indeed the case, though the oscillation in P n is very gentle. We have lim sup Pn ≈ 0.72135465 which is achieved when α ≈ 0.139, and lim inf Pn ≈ 0.7213 ...
The quadratic formula
... quartic a4 x 4 + a3 x 3 + a2 x 2 + a1 x + a0 (where the ai are all integers) there are similar (but more complicated) formulas. For degree 5, there are no such formulas. This is called the insolubility of the quintic and it is a famous result proved by Abel and Galois in the early 19th century. Howe ...
... quartic a4 x 4 + a3 x 3 + a2 x 2 + a1 x + a0 (where the ai are all integers) there are similar (but more complicated) formulas. For degree 5, there are no such formulas. This is called the insolubility of the quintic and it is a famous result proved by Abel and Galois in the early 19th century. Howe ...
Proof of a theorem of Fermat that every prime number of the form 4n
... not divisible by the prime number 4n + 1, from which it follows in turn that certainly not all first differences are divisible by 4n + 1. 6. By virtue of which, the power of this proof is better observed, it is to be noted that the difference of order 2n is produced from 2n + 1 terms of the given s ...
... not divisible by the prime number 4n + 1, from which it follows in turn that certainly not all first differences are divisible by 4n + 1. 6. By virtue of which, the power of this proof is better observed, it is to be noted that the difference of order 2n is produced from 2n + 1 terms of the given s ...
PART II. SEQUENCES OF REAL NUMBERS
... Therefore, sk+1 ∈ S and {sn } is increasing. THEOREM 8. A monotone sequence is convergent if and only if it is bounded. Proof: Let {sn } be a monotone sequence. If {sn} is convergent, then it is bounded (Theorem 2). Now suppose that {sn } is a bounded, monotone sequence. In particular, suppose {sn} ...
... Therefore, sk+1 ∈ S and {sn } is increasing. THEOREM 8. A monotone sequence is convergent if and only if it is bounded. Proof: Let {sn } be a monotone sequence. If {sn} is convergent, then it is bounded (Theorem 2). Now suppose that {sn } is a bounded, monotone sequence. In particular, suppose {sn} ...
Number Theory - Abstractmath.org
... “ 3 / 6 “ is the name of a number (the number 1/ 2 ), and does not form a complete sentence in itself. An additional source of confusion is the fact that the numbers are reversed between the two notations: 3 | 6 because 6 / 3 is an integer. In the notation “ k | m ” and in the phrase “ m = q k ” i ...
... “ 3 / 6 “ is the name of a number (the number 1/ 2 ), and does not form a complete sentence in itself. An additional source of confusion is the fact that the numbers are reversed between the two notations: 3 | 6 because 6 / 3 is an integer. In the notation “ k | m ” and in the phrase “ m = q k ” i ...
双曲線暗号について
... The ring is a pair of a set G and operators defined on G, * for multiplication and + for addition, that is (G,*,+). And it also satisfy an associative law and commutative law for each operators. >At least the ring must have two operators for Add. and Multi. The residual ring Z/pZ has ‘the operation ...
... The ring is a pair of a set G and operators defined on G, * for multiplication and + for addition, that is (G,*,+). And it also satisfy an associative law and commutative law for each operators. >At least the ring must have two operators for Add. and Multi. The residual ring Z/pZ has ‘the operation ...
40(3)
... 3. THE BURGSTAHLER PROBLEM AS AN INVERSION PROBLEM The real problem considered in this paper is the following: Find the sequence 62n+1 given that u ...
... 3. THE BURGSTAHLER PROBLEM AS AN INVERSION PROBLEM The real problem considered in this paper is the following: Find the sequence 62n+1 given that u ...
2×2 handouts
... Problem and basic idea The problem of assigning messages to points on elliptic curves is difficult because there are no polynomial-time algorithms to write down points of an arbitrary elliptic curve. Fortunately, there is a fast randomized algorithm, to assign points of any elliptic curve to message ...
... Problem and basic idea The problem of assigning messages to points on elliptic curves is difficult because there are no polynomial-time algorithms to write down points of an arbitrary elliptic curve. Fortunately, there is a fast randomized algorithm, to assign points of any elliptic curve to message ...
the origins of the genus concept in quadratic forms
... product of two numbers represented by F will be a quadratic residue mod 8 or a quadratic residue mod 4, respectively. Hence if 8 | D, then exactly one of the following is true: all numbers represented by F are 1 (mod 8) , or all are 3 (mod 8) , or all are 5 (mod 8) , or all are 7 (mod 8) . L ...
... product of two numbers represented by F will be a quadratic residue mod 8 or a quadratic residue mod 4, respectively. Hence if 8 | D, then exactly one of the following is true: all numbers represented by F are 1 (mod 8) , or all are 3 (mod 8) , or all are 5 (mod 8) , or all are 7 (mod 8) . L ...
Slides - FI MUNI
... Elliptic curves version of ElGamal digital signatures has the following form for signing (a message) m, an integer, by Alice and to have the signature verified by Bob: Alice chooses a prime p, an elliptic curve Ep (a, b), a point P on Ep and calculates the number of points n on Ep – what can be done ...
... Elliptic curves version of ElGamal digital signatures has the following form for signing (a message) m, an integer, by Alice and to have the signature verified by Bob: Alice chooses a prime p, an elliptic curve Ep (a, b), a point P on Ep and calculates the number of points n on Ep – what can be done ...
Bases for Sets of Integers
... Finally we wish to point out that the size of m,, depends rather delicately on the arithmetical structure of the sequence A and not just on the coarse aspects of its "rate of growth ." The fact is that to every set, A, there is a fairly nearby set, A', which has a relatively small basis . This pertu ...
... Finally we wish to point out that the size of m,, depends rather delicately on the arithmetical structure of the sequence A and not just on the coarse aspects of its "rate of growth ." The fact is that to every set, A, there is a fairly nearby set, A', which has a relatively small basis . This pertu ...
Pythagorean Triples. - Doug Jones`s Mathematics Homepage
... We now know that one side of a Pythagorean triple right triangle must be odd. So give me any odd number, and I can fairly quickly give you back a Pythagorean triple with that odd number as a side.11 Here’s how it works. Let n be any odd number (odd positive integer). Then square it, subtract one, an ...
... We now know that one side of a Pythagorean triple right triangle must be odd. So give me any odd number, and I can fairly quickly give you back a Pythagorean triple with that odd number as a side.11 Here’s how it works. Let n be any odd number (odd positive integer). Then square it, subtract one, an ...