For a pdf file
... The above equation implies that b2 is even and hence b is even. Since we know a is even this means that a and b have 2 as a common factor which contradicts the assumption that a and b have no common factors. ...
... The above equation implies that b2 is even and hence b is even. Since we know a is even this means that a and b have 2 as a common factor which contradicts the assumption that a and b have no common factors. ...
Algebraic Proof (H)
... Please also note that the layout in terms of fonts, answer lines and space given to each question does not reflect the actual papers to save space. These questions have been collated by me as the basis for a GCSE working party set up by the GLOW maths hub - if you want to get involved please get in ...
... Please also note that the layout in terms of fonts, answer lines and space given to each question does not reflect the actual papers to save space. These questions have been collated by me as the basis for a GCSE working party set up by the GLOW maths hub - if you want to get involved please get in ...
Algebraic Proof (H)
... Please also note that the layout in terms of fonts, answer lines and space given to each question does not reflect the actual papers to save space. These questions have been collated by me as the basis for a GCSE working party set up by the GLOW maths hub - if you want to get involved please get in ...
... Please also note that the layout in terms of fonts, answer lines and space given to each question does not reflect the actual papers to save space. These questions have been collated by me as the basis for a GCSE working party set up by the GLOW maths hub - if you want to get involved please get in ...
Look at notes for first lectures in other courses
... are both rational functions and satisfy G(x) = – F(1/x). Example 1: f(n) = c^n for fixed non-zero c. F(x) = 1 + cx + c^2 x^2 + ... = 1/(1-cx) G(x) = c^{-1} x + c^{-2} x^2 + ... = (x/c)/(1-x/c)) = 1/(c/x-1) = -1/(1-c/x) = -F(1/x). Example 2: f(n) = (n choose k) for fixed k. (n choose k) - (n-1 choose ...
... are both rational functions and satisfy G(x) = – F(1/x). Example 1: f(n) = c^n for fixed non-zero c. F(x) = 1 + cx + c^2 x^2 + ... = 1/(1-cx) G(x) = c^{-1} x + c^{-2} x^2 + ... = (x/c)/(1-x/c)) = 1/(c/x-1) = -1/(1-c/x) = -F(1/x). Example 2: f(n) = (n choose k) for fixed k. (n choose k) - (n-1 choose ...
Section.1.1
... A false statement is called a contradiction. For example, “S and not S” is a contradiction for any statement S. A truth table will show us that “if A then B,” is equivalent to “A and not B implies false.” So to prove “if A then B,” it suffices to assume A and also to assume not B, and then argue tow ...
... A false statement is called a contradiction. For example, “S and not S” is a contradiction for any statement S. A truth table will show us that “if A then B,” is equivalent to “A and not B implies false.” So to prove “if A then B,” it suffices to assume A and also to assume not B, and then argue tow ...
ON Prk SEQUENCES + k = b\ a2a3 + k = y2 axa3 + fe ,2 36 [Feb.
... When b2 - k is positive and square free, (5) has an infinite number of solutions. Henceforth, we concentrate on the solution X = a2(a1 + b±) , Y = a 2 + bx of (5). This gives a2a^ ...
... When b2 - k is positive and square free, (5) has an infinite number of solutions. Henceforth, we concentrate on the solution X = a2(a1 + b±) , Y = a 2 + bx of (5). This gives a2a^ ...
22 - AbstractAlgebra.net: The home of introductory abstract algebra
... The mapping φ(x) = x3 from (R, +) to itself is not an isomorphism because the homomorphism property is not satisfied. (Non)Example 5: U (10) is not isomorphic to U (12). Although both groups have order four, U (10) is cyclic and therefore has an element of order four. On the other hand, all non-iden ...
... The mapping φ(x) = x3 from (R, +) to itself is not an isomorphism because the homomorphism property is not satisfied. (Non)Example 5: U (10) is not isomorphic to U (12). Although both groups have order four, U (10) is cyclic and therefore has an element of order four. On the other hand, all non-iden ...
Geometry Name_____________________________________ 2.1
... 11. Carrie collected canned food for a homeless shelter in her area each day for one week. On day one, she collected 7 cans of food. On day two, she collected 8 cans. On day three she collected 10 cans. On day four, she collected 13 cans. If Carrie wanted to give at least 100 cans of food to the she ...
... 11. Carrie collected canned food for a homeless shelter in her area each day for one week. On day one, she collected 7 cans of food. On day two, she collected 8 cans. On day three she collected 10 cans. On day four, she collected 13 cans. If Carrie wanted to give at least 100 cans of food to the she ...
8.1
... The slope of a line is the ratio of the rise between any two points on the line to the run between to two points. ...
... The slope of a line is the ratio of the rise between any two points on the line to the run between to two points. ...
Surprising Connections between Partitions and Divisors
... “partitions” of a given number as sums of positive integers. For example, the seven partitions of 5 are 5, 4 + 1, 3 + 2, 3 + 1 + 1, 2 + 2 + 1, 2 + 1 + 1 + 1, and 1 + 1 + 1 + 1 + 1. The “partition function” p(n) is defined as the number of partitions of n. Thus, p(5) = 7. Prime numbers and divisors ...
... “partitions” of a given number as sums of positive integers. For example, the seven partitions of 5 are 5, 4 + 1, 3 + 2, 3 + 1 + 1, 2 + 2 + 1, 2 + 1 + 1 + 1, and 1 + 1 + 1 + 1 + 1. The “partition function” p(n) is defined as the number of partitions of n. Thus, p(5) = 7. Prime numbers and divisors ...
Full text
... conditions demand that the modulus of the leading weight coefficient a0 should be sufficiently large (see also §5). There are many sequences that satisfy these conditions. For example, take an arbitrary holomorphic function hx(z) defined in a neighborhood of zero. Then the sequence appearing as the ...
... conditions demand that the modulus of the leading weight coefficient a0 should be sufficiently large (see also §5). There are many sequences that satisfy these conditions. For example, take an arbitrary holomorphic function hx(z) defined in a neighborhood of zero. Then the sequence appearing as the ...
September 21: Math 432 Class Lecture Notes
... there is an integer a such that a2 ≡ −1 mod p. Thus p divides a2 + 1. One now checks that a + i is a nontrivial gcd with p, but that it is not divisible by p, and that P := gcd(p, a + i) is an element of norm p, so that p is the sum of two squares. As an exercise, the reader should work out that in ...
... there is an integer a such that a2 ≡ −1 mod p. Thus p divides a2 + 1. One now checks that a + i is a nontrivial gcd with p, but that it is not divisible by p, and that P := gcd(p, a + i) is an element of norm p, so that p is the sum of two squares. As an exercise, the reader should work out that in ...
Counting Your Way to the Sum of Squares Formula
... the elements of a suitably chosen set in two different ways and then equate the two resulting expressions. In the two-part Resonance article [1], many such examples were studied. In this article, which may be regarded as a continuation of that one, we do the same for the formulas for the sum of the ...
... the elements of a suitably chosen set in two different ways and then equate the two resulting expressions. In the two-part Resonance article [1], many such examples were studied. In this article, which may be regarded as a continuation of that one, we do the same for the formulas for the sum of the ...
section 2
... of algebra • 2 parts: 1) find all roots, how you do this depends on if you are given a real root or a complex root a) given a real root: do synthetic division, rewrite your answer, find the remaining zeros by doing quadratic formula (these will be complex numbers),write your answer in set notation.) ...
... of algebra • 2 parts: 1) find all roots, how you do this depends on if you are given a real root or a complex root a) given a real root: do synthetic division, rewrite your answer, find the remaining zeros by doing quadratic formula (these will be complex numbers),write your answer in set notation.) ...
Prove
... To prove a theorem (proposition, lemma) of the form p ⇒ q, we often attempt to build a bridge of implications p ⇒ p1 ⇒ p2 ⇒ · · · ⇒ q2 ⇒ q1 ⇒ q. To obtain such a bridge, we might try to use deductive reasoning (”what can we conclude from p?”) to get p1 from p, then again to get p2 from p1 , and so o ...
... To prove a theorem (proposition, lemma) of the form p ⇒ q, we often attempt to build a bridge of implications p ⇒ p1 ⇒ p2 ⇒ · · · ⇒ q2 ⇒ q1 ⇒ q. To obtain such a bridge, we might try to use deductive reasoning (”what can we conclude from p?”) to get p1 from p, then again to get p2 from p1 , and so o ...