1) Solve the following system of equations: −3y2 + 2xy + x + 9 = 0 (1
... −3y 2 + 2xy − 4y = 0. What can you do with this? Factor — every term has a y in it. Get y(−3y + 2x − 4) = 0. Use the fact that this implies either y = 0 or −3x + 2x − 4 = 0. • If y = 0, plug that in to either original equation to get x = −9. • If −3y + 2x − 4 = 0, make a system out of that and equat ...
... −3y 2 + 2xy − 4y = 0. What can you do with this? Factor — every term has a y in it. Get y(−3y + 2x − 4) = 0. Use the fact that this implies either y = 0 or −3x + 2x − 4 = 0. • If y = 0, plug that in to either original equation to get x = −9. • If −3y + 2x − 4 = 0, make a system out of that and equat ...
Core 1
... the point (3, 7) lie on the line y = 2x + 3, or above it, or below it? Justify the form by joining the points (0, c) and (x, y) by a line of gradient m. The one exception is a line parallel to the y-axis. Discuss aliases such as 3x + 4y = 24 and find the gradient and y-intercept. Now cover the formu ...
... the point (3, 7) lie on the line y = 2x + 3, or above it, or below it? Justify the form by joining the points (0, c) and (x, y) by a line of gradient m. The one exception is a line parallel to the y-axis. Discuss aliases such as 3x + 4y = 24 and find the gradient and y-intercept. Now cover the formu ...
Full text
... We shall show that the Lucas numbers may be defined by a particularly simple Diophantine equation and thus exhibit them as the positive numbers in the range of a very simple polynomial of the 9th degree. Our results are based upon the following identity (D ...
... We shall show that the Lucas numbers may be defined by a particularly simple Diophantine equation and thus exhibit them as the positive numbers in the range of a very simple polynomial of the 9th degree. Our results are based upon the following identity (D ...
An Upper Bound on the nth Prime - Mathematical Association of
... In 1845, J. Bertrand conjectured that for any integer n > 3, there exists at least one prime p between n and 2n − 2 [1]. In 1852, P. Tchebychev offered the first demonstration of this now-famous theorem. Today, Bertrand’s Postulate is often stated as, “for any positive integer n ≥ 1, there exists a ...
... In 1845, J. Bertrand conjectured that for any integer n > 3, there exists at least one prime p between n and 2n − 2 [1]. In 1852, P. Tchebychev offered the first demonstration of this now-famous theorem. Today, Bertrand’s Postulate is often stated as, “for any positive integer n ≥ 1, there exists a ...
s02.1
... throughout the life of a data structure or procedure. Each change to the data structure maintains the correctness of the invariant ...
... throughout the life of a data structure or procedure. Each change to the data structure maintains the correctness of the invariant ...
Lecture 2: Irrational numbers
... We want to appreciate one of the great moments of mathematics: the insight that there are numbers which are irrational. It was the Pythagoreans, who realized this first and - according to legend - tried even to ”cover the discovery up” and kill Hippasus, one of the earlier discoverers. We have seen ...
... We want to appreciate one of the great moments of mathematics: the insight that there are numbers which are irrational. It was the Pythagoreans, who realized this first and - according to legend - tried even to ”cover the discovery up” and kill Hippasus, one of the earlier discoverers. We have seen ...
A sample from this course
... of lengths of given segments. It follows that every length that can be constructed is an algebr aic number , a number that can be the solution to a polynomial equation with integer coefficients. In 1882, Ferdinand Lindemann proved that π is a tr anscendental number, not an algebr aic number. ...
... of lengths of given segments. It follows that every length that can be constructed is an algebr aic number , a number that can be the solution to a polynomial equation with integer coefficients. In 1882, Ferdinand Lindemann proved that π is a tr anscendental number, not an algebr aic number. ...
Document
... The index number tells us how many times the base number is multiplied by itself. e.g. 34 means 3 x 3 x 3 x 3 = 81 ...
... The index number tells us how many times the base number is multiplied by itself. e.g. 34 means 3 x 3 x 3 x 3 = 81 ...
Solution
... (8) If an−1 ≡ 1 (mod n), then we say n passes Fermat’s primality test for a. Which one of the following statements is TRUE (A) If a number n passes Fermat’s test for some a, then it must be a prime number. (B) If a number n passes Fermat’s test for every a coprime with n, then n must be a prime numb ...
... (8) If an−1 ≡ 1 (mod n), then we say n passes Fermat’s primality test for a. Which one of the following statements is TRUE (A) If a number n passes Fermat’s test for some a, then it must be a prime number. (B) If a number n passes Fermat’s test for every a coprime with n, then n must be a prime numb ...
Section 3 - Divisibility
... Unique Factorization Theorem • Theorem: Given any integer n > 1, there exist positive integer k; prime numbers p1,p2,...,pk; and positive integers e1,e2,...,ek, with n = (p1)e1 ⋅ (p2)e2 ⋅ (p3)e3...(pk)ek, and any other expression of n as a product of prime numbers is identical to this except, perhap ...
... Unique Factorization Theorem • Theorem: Given any integer n > 1, there exist positive integer k; prime numbers p1,p2,...,pk; and positive integers e1,e2,...,ek, with n = (p1)e1 ⋅ (p2)e2 ⋅ (p3)e3...(pk)ek, and any other expression of n as a product of prime numbers is identical to this except, perhap ...
Generating Anomalous Elliptic Curves
... of this form. Then, using Section 3, we compute a curve E over Fp with modular invariant jD . This curve is provided by the reduction modulo p of the equation (3), with j = jD . To decide which one between the two curves E or Ẽ is anomalous, one simply takes a point P ∈ E(Fp ) − {O} at random, and ...
... of this form. Then, using Section 3, we compute a curve E over Fp with modular invariant jD . This curve is provided by the reduction modulo p of the equation (3), with j = jD . To decide which one between the two curves E or Ẽ is anomalous, one simply takes a point P ∈ E(Fp ) − {O} at random, and ...
The Partition Function and Ramanujan`s 5k + 4 Congruence
... MacMahon, a mathematician known for his lists and tables of values, used this function to help construct his list of values of p(n). To see why the product given above generates values of the partition function, one first expresses each term of the product as a geometric series, then multiplies thes ...
... MacMahon, a mathematician known for his lists and tables of values, used this function to help construct his list of values of p(n). To see why the product given above generates values of the partition function, one first expresses each term of the product as a geometric series, then multiplies thes ...
Full-text PDF - American Mathematical Society
... The two outer groups are repetitions and have been omitted from the line above. Since [27P2; S3]=Z2 and [28P2; S3]=0, both homomorphisms /* are trivial, and #[S6aP2aP2; 53]=2. On the other hand, #[29P2; 53]=2 and #[512; S3]=4. Thus a=25. ...
... The two outer groups are repetitions and have been omitted from the line above. Since [27P2; S3]=Z2 and [28P2; S3]=0, both homomorphisms /* are trivial, and #[S6aP2aP2; 53]=2. On the other hand, #[29P2; 53]=2 and #[512; S3]=4. Thus a=25. ...
Introduction to Proofs
... Theorem: (For integers n) If n is the sum of two prime numbers, then either n is odd or n is even. Proof: Any integer n is either odd or even. So the conclusion of the implication is true regardless of the truth of the hypothesis. Thus the implication is true trivially. This kind will be discuss in ...
... Theorem: (For integers n) If n is the sum of two prime numbers, then either n is odd or n is even. Proof: Any integer n is either odd or even. So the conclusion of the implication is true regardless of the truth of the hypothesis. Thus the implication is true trivially. This kind will be discuss in ...