Complex Numbers
... not hard to parametrize such a chain, so it determines is a curve.) As an example, let G be the open disk with center 0 and radius 2. Then any two points in G can be connected by a chain of at most 2 segments in G, so G is connected. Now let G0 = G \ {0}; this is the punctured disk obtained by remov ...
... not hard to parametrize such a chain, so it determines is a curve.) As an example, let G be the open disk with center 0 and radius 2. Then any two points in G can be connected by a chain of at most 2 segments in G, so G is connected. Now let G0 = G \ {0}; this is the punctured disk obtained by remov ...
Homework 3
... Proof of the division algorithm: Let S be the set of nonnegative integers of the form b − ka where k is an integer. That is S = {b − ka : k ∈ Z and b − ka ≥ 0}. Then S is non-empty. So see this we consider two cases. Case1: b ≥ 0. Let k = 0 to get that b = b − 0a ∈ S, which implies S 6= ∅. Case 2: b ...
... Proof of the division algorithm: Let S be the set of nonnegative integers of the form b − ka where k is an integer. That is S = {b − ka : k ∈ Z and b − ka ≥ 0}. Then S is non-empty. So see this we consider two cases. Case1: b ≥ 0. Let k = 0 to get that b = b − 0a ∈ S, which implies S 6= ∅. Case 2: b ...
MATH 115, SUMMER 2012 LECTURE 5 Last time:
... The following two theorems are very useful; the second follows from the first. To motivate these two results, recall an important caution: you cannot substitute congruent numbers as exponents! For example, even though 1 ≡ 5 mod 4, it is not true that 21 ≡ 25 mod 4. So when can we use congruence to s ...
... The following two theorems are very useful; the second follows from the first. To motivate these two results, recall an important caution: you cannot substitute congruent numbers as exponents! For example, even though 1 ≡ 5 mod 4, it is not true that 21 ≡ 25 mod 4. So when can we use congruence to s ...
Bertrand`s Conjecture: At least one Prime between n and 2n *
... We have found enough pieces and can put them together. The decisive is: Will °2nquestion ...
... We have found enough pieces and can put them together. The decisive is: Will °2nquestion ...
A Theory of Theory Formation
... Invents concept of pairs (a,b) for which there exists an element c such that: a*b=c & b*a=c ...
... Invents concept of pairs (a,b) for which there exists an element c such that: a*b=c & b*a=c ...
1.1 Introduction. Real numbers.
... an understanding of sequences and their limits. These appear in analysis whenever you get an answer not at once, but rather by making closer and closer approximations to it. Since they give a quick insight into some of the most important ideas in analysis, they will be our starting point, beginning ...
... an understanding of sequences and their limits. These appear in analysis whenever you get an answer not at once, but rather by making closer and closer approximations to it. Since they give a quick insight into some of the most important ideas in analysis, they will be our starting point, beginning ...
9.6 Mathematical Induction
... In 1852, Francis Guthrie conjectured that any map on a flat surface could be colored in at most four colors so that no two bordering regions would share the same color. Mathematicians tried unsuccessfully for almost 150 years to prove (or disprove) the conjecture, until Kenneth Appel and Wolfgang Ha ...
... In 1852, Francis Guthrie conjectured that any map on a flat surface could be colored in at most four colors so that no two bordering regions would share the same color. Mathematicians tried unsuccessfully for almost 150 years to prove (or disprove) the conjecture, until Kenneth Appel and Wolfgang Ha ...
Lagrange`s Four Square Theorem
... integer squares. We now have that 1, 2, and all odd primes can be written as a sum of four squares. By the Four Squares identity, every natural number can be written as a sum of four squares. ...
... integer squares. We now have that 1, 2, and all odd primes can be written as a sum of four squares. By the Four Squares identity, every natural number can be written as a sum of four squares. ...
Full text
... each entry of the powers of the matrices in the theorem is plus or minus a Fibonacci number. This completes the proof of Theorem 2. • In [1], the problem is posed to find all four-by-four Fibonacci matrices. This can be attacked by the above method. One difficulty is proving the analog of Lemma 4 fo ...
... each entry of the powers of the matrices in the theorem is plus or minus a Fibonacci number. This completes the proof of Theorem 2. • In [1], the problem is posed to find all four-by-four Fibonacci matrices. This can be attacked by the above method. One difficulty is proving the analog of Lemma 4 fo ...
The Binomial Theorem
... In this lesson, we study two ways to expand (a + b)n, where n is a positive integer. The first, which uses Pascal’s Triangle, is applicable if n is not too big, and if we want to determine all the terms in the expansion. The second method gives a general formula for the expansion of (a + b)n for any ...
... In this lesson, we study two ways to expand (a + b)n, where n is a positive integer. The first, which uses Pascal’s Triangle, is applicable if n is not too big, and if we want to determine all the terms in the expansion. The second method gives a general formula for the expansion of (a + b)n for any ...
Supplementary Notes
... there is a randomized polynomial-time algorithm A(x, r) such that for every input x, A(x, r) outputs the correct answer with probability at least 1 − 2−1000 . So when we discover an efficient randomized algorithm for a problem, it is reasonable to consider that problem to be solved for all practical ...
... there is a randomized polynomial-time algorithm A(x, r) such that for every input x, A(x, r) outputs the correct answer with probability at least 1 − 2−1000 . So when we discover an efficient randomized algorithm for a problem, it is reasonable to consider that problem to be solved for all practical ...
Reverse Mathematics and the Coloring Number of Graphs
... I would like to thank my advisors Reed Solomon and Joe Miller. I most certainly would not have made it this far without all of their guidance and support. Despite having three other Ph.D. students, two of which are also graduating this semester, for the past year Reed has always taken the time out o ...
... I would like to thank my advisors Reed Solomon and Joe Miller. I most certainly would not have made it this far without all of their guidance and support. Despite having three other Ph.D. students, two of which are also graduating this semester, for the past year Reed has always taken the time out o ...
Propositional Statements Direct Proof
... Given p → q, suppose that q is not true and p is true to deduce that this is impossible. In other words, we want to show that it is impossible for our hypothesis to occur but the result to not occur. We always begin a proof by contradiction by supposing that q is not true (¬q) and p is true. Example ...
... Given p → q, suppose that q is not true and p is true to deduce that this is impossible. In other words, we want to show that it is impossible for our hypothesis to occur but the result to not occur. We always begin a proof by contradiction by supposing that q is not true (¬q) and p is true. Example ...
sergey-ccc08
... • Are Mersenne primes essential to the method? • Has the method been pushed to its limit? ...
... • Are Mersenne primes essential to the method? • Has the method been pushed to its limit? ...
Notes for 11th Jan (Wednesday)
... Definition : Let (S, ≤) be a totally ordered set (where a ≥ b means that b ≤ a). S is said to satisfy where the “least upper bound” (sup) property if every non-empty subset A that is bounded above has a least upper bound, i.e., if ∃m ∈ S such that x ≤ m for every x ∈ A, then ∃n ≤ m ∈ S such that 1. ...
... Definition : Let (S, ≤) be a totally ordered set (where a ≥ b means that b ≤ a). S is said to satisfy where the “least upper bound” (sup) property if every non-empty subset A that is bounded above has a least upper bound, i.e., if ∃m ∈ S such that x ≤ m for every x ∈ A, then ∃n ≤ m ∈ S such that 1. ...
Note A Note on the Binomial Drop Polynomial of a Poset
... This is a consequence of very general results on "Tutte-like" polynomials on digraphs which will appear in [2]. 4. Our original motivation which led to (1) stemmed from encountering certain new classes of juggling pattern [1] where ( P , ~ ) was {1, 2..... n} (representing time) with the usual size ...
... This is a consequence of very general results on "Tutte-like" polynomials on digraphs which will appear in [2]. 4. Our original motivation which led to (1) stemmed from encountering certain new classes of juggling pattern [1] where ( P , ~ ) was {1, 2..... n} (representing time) with the usual size ...
FIELDS ON THE BOTTOM
... The following result is due to Landau [Lnd19]. Proposition 1.5: Every totally positive algebraic number a is a sum of finitely many squares of elements of Q(a). We mention that two years after Landau published his result, Carl Ludwig Siegel improved it by proving that every totally positive algebra ...
... The following result is due to Landau [Lnd19]. Proposition 1.5: Every totally positive algebraic number a is a sum of finitely many squares of elements of Q(a). We mention that two years after Landau published his result, Carl Ludwig Siegel improved it by proving that every totally positive algebra ...
