Here
... and choose the most appropriate topics from the remaining four chapters. As summarized in the chart below, Chapters 3 – 6 are completely independent of each other. Flexibility is further enhanced by optional sections and appendices, by weaving some topics into the exercise sets of multiple sections, ...
... and choose the most appropriate topics from the remaining four chapters. As summarized in the chart below, Chapters 3 – 6 are completely independent of each other. Flexibility is further enhanced by optional sections and appendices, by weaving some topics into the exercise sets of multiple sections, ...
Introduction to Number Theory 2
... by powers of two. Clearly, a division (rule 6) reduces the “numerator” by a factor of two. A modular reduction (using rule 7 and then rule 1), reduces the number by at least two: as if a > b then a = qb + r ≥ b + r > r + r, thus r < a/2, i.e, a mod b < a/2. Therefore, at most O(log n) modular reduct ...
... by powers of two. Clearly, a division (rule 6) reduces the “numerator” by a factor of two. A modular reduction (using rule 7 and then rule 1), reduces the number by at least two: as if a > b then a = qb + r ≥ b + r > r + r, thus r < a/2, i.e, a mod b < a/2. Therefore, at most O(log n) modular reduct ...
On the independence numbers of the powers of graph
... C52n K 2 , which is an independent extension for some subgraph of C5. It’s sufficient to indicate corresponding two components of the subgraph in C52 n graph, since vertex set of C52n K 2 is a combination of vertices of two C52 n graphs. As such components consider subgraphs in C52 n induced by ...
... C52n K 2 , which is an independent extension for some subgraph of C5. It’s sufficient to indicate corresponding two components of the subgraph in C52 n graph, since vertex set of C52n K 2 is a combination of vertices of two C52 n graphs. As such components consider subgraphs in C52 n induced by ...
Not Always Buried Deep
... infinitely many integers n for which each fi (n) is prime, unless there is a “local obstruction,” i.e., a prime p dividing the product f1 (n) . . . fk (n) for every value of n. We adapt our heuristic to motivate a stronger, quantitative form of this conjecture put forward by Bateman & Horn [BH62], s ...
... infinitely many integers n for which each fi (n) is prime, unless there is a “local obstruction,” i.e., a prime p dividing the product f1 (n) . . . fk (n) for every value of n. We adapt our heuristic to motivate a stronger, quantitative form of this conjecture put forward by Bateman & Horn [BH62], s ...
Cryptography Lecture 1: Remainders and Modular Arithmetic Spring
... This procedure works when a is is not negative. (1.) Divide a by n. If the result is a whole number without a decimal then STOP. The remainder is 0! If the result has a decimal, go to step (2.) (2.) Remove the number that precedes the decimal (the “whole number part”). Do this by subtracting the pre ...
... This procedure works when a is is not negative. (1.) Divide a by n. If the result is a whole number without a decimal then STOP. The remainder is 0! If the result has a decimal, go to step (2.) (2.) Remove the number that precedes the decimal (the “whole number part”). Do this by subtracting the pre ...
Set Theory for Computer Science (pdf )
... properties such as being a natural number, or being irrational, but it was rare to think of say the collection of rational numbers as itself an object. (There were exceptions. From Euclid mathematicians were used to thinking of geometric objects such as lines and planes and spheres which we might to ...
... properties such as being a natural number, or being irrational, but it was rare to think of say the collection of rational numbers as itself an object. (There were exceptions. From Euclid mathematicians were used to thinking of geometric objects such as lines and planes and spheres which we might to ...
1 (mod n)
... ab=1 (mod (n)), we have ab = t(n)+1, for t>=1 Suppose that x in Zn*; then we have (xb)a = xt(n)+1 (mod n) = (x(n))tx = 1tx (mod n) = x (mod n) As desired. For x in Zn but not in Zn*, (do exercise) ...
... ab=1 (mod (n)), we have ab = t(n)+1, for t>=1 Suppose that x in Zn*; then we have (xb)a = xt(n)+1 (mod n) = (x(n))tx = 1tx (mod n) = x (mod n) As desired. For x in Zn but not in Zn*, (do exercise) ...
Advanced Internet Technologies
... A3: There is an element 0 in R such that a + 0 = 0 + a = a for all a in S A4: For each a in S there is an element –a in S such that a + (-a) = (-a) + a = 0 A5: a + b = b + a for all a,b in A M1: If a and b belong to S, then ab is also in S M2: a (bc) = (ab) c for all a, b, c in S ...
... A3: There is an element 0 in R such that a + 0 = 0 + a = a for all a in S A4: For each a in S there is an element –a in S such that a + (-a) = (-a) + a = 0 A5: a + b = b + a for all a,b in A M1: If a and b belong to S, then ab is also in S M2: a (bc) = (ab) c for all a, b, c in S ...
[Chap. 2] Pythagorean Triples (b) The table suggests that in every
... This gives a primitive Pythagorean triple with the right value of b provided that B > 2r−1 . On the other hand, if B < 2r−1 , then we can just take a = 22r−2 − B 2 instead. (c) This part is quite difficult to prove, and it’s not even that easy to make the correct conjecture. It turns out that an odd ...
... This gives a primitive Pythagorean triple with the right value of b provided that B > 2r−1 . On the other hand, if B < 2r−1 , then we can just take a = 22r−2 − B 2 instead. (c) This part is quite difficult to prove, and it’s not even that easy to make the correct conjecture. It turns out that an odd ...
What every computer scientist should know about floating
... The IEEE standard goes further than just requiring the use of a guard digit. It gives an algorithm for addition, subtraction, multiplication, division, and square root and requires that implementations produce the same result as that algorithm. Thus, when a program is moved from one machine to anoth ...
... The IEEE standard goes further than just requiring the use of a guard digit. It gives an algorithm for addition, subtraction, multiplication, division, and square root and requires that implementations produce the same result as that algorithm. Thus, when a program is moved from one machine to anoth ...
Floating-Point Arithmetic Goldberg CS1991
... The IEEE standard goes further than just requiring the use of a guard digit. It gives an algorithm for addition, subtraction, multiplication, division, and square root and requires that implementations produce the same result as that algorithm. Thus, when a program is moved from one machine to anoth ...
... The IEEE standard goes further than just requiring the use of a guard digit. It gives an algorithm for addition, subtraction, multiplication, division, and square root and requires that implementations produce the same result as that algorithm. Thus, when a program is moved from one machine to anoth ...
Chapter 6 Sequences and Series of Real Numbers
... Sequences and Series of Real Numbers We often use sequences and series of numbers without thinking about it. A decimal representation of a number is an example of a series, the bracketing of a real number by closer and closer rational numbers gives us an example of a sequence. We want to study these ...
... Sequences and Series of Real Numbers We often use sequences and series of numbers without thinking about it. A decimal representation of a number is an example of a series, the bracketing of a real number by closer and closer rational numbers gives us an example of a sequence. We want to study these ...
Text (PDF format)
... Solving linear congruences, which have the form ax ≡ b (mod m), is an essential task in the study of number theory and its applications, just as solving linear equations plays an important role in calculus and linear algebra. To solve linear congruences, we employ inverses modulo m. We explain how t ...
... Solving linear congruences, which have the form ax ≡ b (mod m), is an essential task in the study of number theory and its applications, just as solving linear equations plays an important role in calculus and linear algebra. To solve linear congruences, we employ inverses modulo m. We explain how t ...
HARMONIOUS PAIRS Let σ(n) denote the sum of the divisors of the
... We are not aware of any previous work on harmonious pairs, as such. However, the following result can K be read out of a paper of Borho [2]: If a, b form a harmonious pair and Ω(ab) = K, then ab ≤ K 2 . Borho states this for amicable pairs, but only the harmonious property of the pair is used in the ...
... We are not aware of any previous work on harmonious pairs, as such. However, the following result can K be read out of a paper of Borho [2]: If a, b form a harmonious pair and Ω(ab) = K, then ab ≤ K 2 . Borho states this for amicable pairs, but only the harmonious property of the pair is used in the ...
Problems before the Semifinal 1 Solving equations of degree 3 and 4
... For example, if a collection contains x2 + 2y and x − y 3 , then one may apply the first operation in order to append the polynomial −5(x2 + 2y)2 + 3(x2 + 2y)(x − y 3 )6 ; moreover, if a collection already contains x2 − 2xy + y 2 , then applying the second operation one may append x − y and y − x. ...
... For example, if a collection contains x2 + 2y and x − y 3 , then one may apply the first operation in order to append the polynomial −5(x2 + 2y)2 + 3(x2 + 2y)(x − y 3 )6 ; moreover, if a collection already contains x2 − 2xy + y 2 , then applying the second operation one may append x − y and y − x. ...
Math 445 Homework 5 Solutions Due Wednesday, October 6 21. If
... 25. Find a primitive root modulo 23. Following our argument from class, since 23 − 1 = 22 = 2 · 11, we will find an a with a22/11 = a2 ≡ 1 (mod 23) and b with b22/2 = b11 ≡ 1 (mod 23) . Then c = ab will be a primitive root. But a = 2 works; 22 = 4 ≡ 1 . And since 11 is odd, b = 22 ≡ −1 works; 2211 ...
... 25. Find a primitive root modulo 23. Following our argument from class, since 23 − 1 = 22 = 2 · 11, we will find an a with a22/11 = a2 ≡ 1 (mod 23) and b with b22/2 = b11 ≡ 1 (mod 23) . Then c = ab will be a primitive root. But a = 2 works; 22 = 4 ≡ 1 . And since 11 is odd, b = 22 ≡ −1 works; 2211 ...
Lecture notes #5 - EECS: www
... (possibly zero or negative). However, x and m are relatively prime, so x cannot share any factors with m. This implies that a − b must be an integer multiple of m. This is not possible, since a − b ranges between 1 and m − 1. 2 Actually it turns out that gcd(m, x) = 1 is also a necessary condition f ...
... (possibly zero or negative). However, x and m are relatively prime, so x cannot share any factors with m. This implies that a − b must be an integer multiple of m. This is not possible, since a − b ranges between 1 and m − 1. 2 Actually it turns out that gcd(m, x) = 1 is also a necessary condition f ...
Chapter 4.6 - CS Course Webpages
... inverse of a modulo m and 0 is its own additive inverse. a +m (m− a ) = 0 and 0 +m 0 = 0 Distributivity: If a, b, and c belong to Zm , then a ∙m (b +m c) = (a ∙m b) +m (a ∙m c) and (a +m b) ∙m c = (a ∙m c) +m (b ∙m c). Exercises 42-44 ask for proofs of these properties. Multiplicatative in ...
... inverse of a modulo m and 0 is its own additive inverse. a +m (m− a ) = 0 and 0 +m 0 = 0 Distributivity: If a, b, and c belong to Zm , then a ∙m (b +m c) = (a ∙m b) +m (a ∙m c) and (a +m b) ∙m c = (a ∙m c) +m (b ∙m c). Exercises 42-44 ask for proofs of these properties. Multiplicatative in ...
x,y
... Lemma If aN-1 1 mod N for some a relatively prime to N, then it must hold for at least half the choices of a < N Proof Fix some value of a such that aN-1 1 mod N. Suppose b < N Satisfies the test, i.e., bN-1 1 mod N. Then, (a·b)N-1 aN-1·bN-1 aN-1 1 mod N Let S be the set of all b < N tha ...
... Lemma If aN-1 1 mod N for some a relatively prime to N, then it must hold for at least half the choices of a < N Proof Fix some value of a such that aN-1 1 mod N. Suppose b < N Satisfies the test, i.e., bN-1 1 mod N. Then, (a·b)N-1 aN-1·bN-1 aN-1 1 mod N Let S be the set of all b < N tha ...