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 ...
CHAP02 Numbers
... But then came the computer age with its digital foundations. The need for security in data transmission gave rise to the need for secure codes and the techniques for this were discovered to be lying dormant in the theory of prime numbers. Throughout the history of mathematics this story has been rep ...
... But then came the computer age with its digital foundations. The need for security in data transmission gave rise to the need for secure codes and the techniques for this were discovered to be lying dormant in the theory of prime numbers. Throughout the history of mathematics this story has been rep ...
Difficulties in Factoring a Number: Prime Numbers
... ATM and ATM seem to be reducible to one another (a solution to either one could be used to solve the other by simply reversing the answer). However, ATM is not mapping reducible to ATM because it is not Turingrecognizable (find a solution to map each unacceptable
to an acceptable is ...
... ATM and ATM seem to be reducible to one another (a solution to either one could be used to solve the other by simply reversing the answer). However, ATM is not mapping reducible to ATM because it is not Turingrecognizable (find a solution to map each unacceptable
Generating Prime Numbers
... one composite image. In [1] they improve the result by proving the following theorem. Theorem 2. Given a positive integer n, f (x) takes an infinite number of values that are divisible by at least n distinct primes, and an infinite number of values that are divisible by pn for some prime p. In [4] t ...
... one composite image. In [1] they improve the result by proving the following theorem. Theorem 2. Given a positive integer n, f (x) takes an infinite number of values that are divisible by at least n distinct primes, and an infinite number of values that are divisible by pn for some prime p. In [4] t ...
Keys GEO SY14-15 Openers 2-5
... 12. Create a BIG coordinate plane. 13. Take out a protractor. 14. Put the straight edge of the inner semicircle along the positive xaxis. 15. Place the left ‘vertex’ of the semicircle on the origin. 16. Trace it and label it with 3 non-collinear points: 2 on the vertices and 1 at the top of the semi ...
... 12. Create a BIG coordinate plane. 13. Take out a protractor. 14. Put the straight edge of the inner semicircle along the positive xaxis. 15. Place the left ‘vertex’ of the semicircle on the origin. 16. Trace it and label it with 3 non-collinear points: 2 on the vertices and 1 at the top of the semi ...
Enumeration in Algebra and Geometry
... this question for a generic collection of hyperplanes. But for some special hyperplane arrangements the answer can be much more interesting than for the generic case, and yet not so easy to gain. The second task of this thesis belongs to the area of enumerative geometry and is similar to the (no les ...
... this question for a generic collection of hyperplanes. But for some special hyperplane arrangements the answer can be much more interesting than for the generic case, and yet not so easy to gain. The second task of this thesis belongs to the area of enumerative geometry and is similar to the (no les ...
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 ...
Of course
... whose removal disconnects G . Then C(G) = 2 C(G1) C(G2) . Prove or rather see this. Note: Let e be the distinguished edge between vertices vj and vj . Any composition of G can be obtained in exactly two ways.: either e is included to supply the component vi in G1 and the component vj in G2 or not. T ...
... whose removal disconnects G . Then C(G) = 2 C(G1) C(G2) . Prove or rather see this. Note: Let e be the distinguished edge between vertices vj and vj . Any composition of G can be obtained in exactly two ways.: either e is included to supply the component vi in G1 and the component vj in G2 or not. T ...
Full text
... surprising, it is actually very natural as Benford’s law is equivalent to the logarithms of the set being equidistributed modulo 1. For more on Benford’s law see [15, 16, 21, 24], as well as [20] for a compilation of articles on its theory and applications. Obviously, we would not be discussing Benf ...
... surprising, it is actually very natural as Benford’s law is equivalent to the logarithms of the set being equidistributed modulo 1. For more on Benford’s law see [15, 16, 21, 24], as well as [20] for a compilation of articles on its theory and applications. Obviously, we would not be discussing Benf ...
PRIME NUMBERS We denote by N the set of natural numbers: 1,2
... Indeed, one can use Euclid’s algorithm to give another proof of Theorem 6. We will now return back to factorisations of natural numbers into primes. We start with the following: Theorem 9. (Euclid’s first theorem) Let p be a prime number and let a1 , a2 ∈ N. If p | a1 a2 then p | a1 or p | a2 . More ...
... Indeed, one can use Euclid’s algorithm to give another proof of Theorem 6. We will now return back to factorisations of natural numbers into primes. We start with the following: Theorem 9. (Euclid’s first theorem) Let p be a prime number and let a1 , a2 ∈ N. If p | a1 a2 then p | a1 or p | a2 . More ...
Prop. If n is an integer, then 3 | (n 3 − n). Proof. By the Division
... for all integers n. This would be correct, and a uniform proof (for all primes p) follows from Fermat’s Little Theorem. Remark. You may be tempted to conjecture the following. If a is not a prime number, then a - (na − n) for some integers n. You could check this conjecture for all composite numbers ...
... for all integers n. This would be correct, and a uniform proof (for all primes p) follows from Fermat’s Little Theorem. Remark. You may be tempted to conjecture the following. If a is not a prime number, then a - (na − n) for some integers n. You could check this conjecture for all composite numbers ...
Full text
... The possibility that 32 = ''' ~ $t ~ 1 (with 3i > 1) has also been considered. In this case, it is known (see [1]) that $i ^ 2 and (see [7]) that $i ^ 3; by a previously mentioned result [8], we also have that 3i t 1 (mod 3). We shall prove Theorem 2. If N as above is an odd perfect number and 3 2 = ...
... The possibility that 32 = ''' ~ $t ~ 1 (with 3i > 1) has also been considered. In this case, it is known (see [1]) that $i ^ 2 and (see [7]) that $i ^ 3; by a previously mentioned result [8], we also have that 3i t 1 (mod 3). We shall prove Theorem 2. If N as above is an odd perfect number and 3 2 = ...
On the b-ary Expansion of an Algebraic Number.
... where k0 0, a k0 6 0 if k0 > 0, the ak 's are integers from f0; 1; . . . ; b 1g and ak is non-zero for infinitely many indices k. The sequence (ak )k k0 is uniquely determined by u: it is its b-ary expansion. We then define the function nbdc, `number of digit changes', by nbdc(n; u; b) Cardf1 ...
... where k0 0, a k0 6 0 if k0 > 0, the ak 's are integers from f0; 1; . . . ; b 1g and ak is non-zero for infinitely many indices k. The sequence (ak )k k0 is uniquely determined by u: it is its b-ary expansion. We then define the function nbdc, `number of digit changes', by nbdc(n; u; b) Cardf1 ...
Graduate Texts in Mathematics 232
... finite group theory at several points to give alternative proofs. Often the resulting approaches simplify and promote generalization, as well as providing elegance. We also use this approach because we want to try to explain how different approaches to elementary results are worked out later in differe ...
... finite group theory at several points to give alternative proofs. Often the resulting approaches simplify and promote generalization, as well as providing elegance. We also use this approach because we want to try to explain how different approaches to elementary results are worked out later in differe ...
The Rational Numbers
... corresponding to a hypothetical solution to Ma,b ), analogously to what we did when we constructed the integers. And, just as then, we have to use an equivalence relation to relate two pairs that should correspond to the same hypothetical solution (which, in this case, means that they satisfy equati ...
... corresponding to a hypothetical solution to Ma,b ), analogously to what we did when we constructed the integers. And, just as then, we have to use an equivalence relation to relate two pairs that should correspond to the same hypothetical solution (which, in this case, means that they satisfy equati ...
Patterns and Inductive Reasoning
... 34. All prime numbers are odd. 35. The sum of two numbers is always greater than the larger number. 36. If the product of two numbers is even, then the two numbers must be even. 37. If the product of two numbers is positive, then the two numbers must both ...
... 34. All prime numbers are odd. 35. The sum of two numbers is always greater than the larger number. 36. If the product of two numbers is even, then the two numbers must be even. 37. If the product of two numbers is positive, then the two numbers must both ...
Chapter 2.1, 2.4
... states, the 2000 population of each state, and the number of people per 1000 residents who are licensed drivers in each state. Based on the table, which two states could be used as a counterexample for the following statement? The greater the population of a state, the lower the number of drivers pe ...
... states, the 2000 population of each state, and the number of people per 1000 residents who are licensed drivers in each state. Based on the table, which two states could be used as a counterexample for the following statement? The greater the population of a state, the lower the number of drivers pe ...
literature review
... which are essential in creating cryptosystems, are found using primality tests. Occasionally, large composite numbers are inadvertently selected in the place of prime numbers. Such mistakes occur when primality testing yields false positives. One large class of false positives to many primality test ...
... which are essential in creating cryptosystems, are found using primality tests. Occasionally, large composite numbers are inadvertently selected in the place of prime numbers. Such mistakes occur when primality testing yields false positives. One large class of false positives to many primality test ...
