Size and degree anti-Ramsey numbers, Graphs and Combinatorics
... where c1 is an absolute constant to be chosen later. Define a (random) edge-coloring of G in two steps as follows. First, let C = [10d] = {1, 2, . . . , 10d} be a palette of 10d colors and let f : E → C be a random coloring of the edges of G obtained by picking, for each edge e of G, randomly and in ...
... where c1 is an absolute constant to be chosen later. Define a (random) edge-coloring of G in two steps as follows. First, let C = [10d] = {1, 2, . . . , 10d} be a palette of 10d colors and let f : E → C be a random coloring of the edges of G obtained by picking, for each edge e of G, randomly and in ...
Math 554 - Fall 08 Lecture Note Set # 1
... (−1) ∈ IP and therefore 0 < −1. But (−1) · (−1) = −(−1) = 1 by part (i) and the HW Problem that additive inverses are unique. This shows that 1 ∈ IP by property (b) of the positive cone and the assumption that (−1) ∈ IP . Contradiction, by the trichotomy property (c). For part (iii), observe that 0 ...
... (−1) ∈ IP and therefore 0 < −1. But (−1) · (−1) = −(−1) = 1 by part (i) and the HW Problem that additive inverses are unique. This shows that 1 ∈ IP by property (b) of the positive cone and the assumption that (−1) ∈ IP . Contradiction, by the trichotomy property (c). For part (iii), observe that 0 ...
PPT
... both change and hence the odd parity count changes by 2 – and remains even If 2 people of different parities shake, then they both swap parities and the odd parity count is unchanged ...
... both change and hence the odd parity count changes by 2 – and remains even If 2 people of different parities shake, then they both swap parities and the odd parity count is unchanged ...
Circular sets of prime numbers and p
... The pro-p-group GS (k)(p) = G(kS (p)/k), i.e. the Galois group of the maximal p-extension of k which is unramified outside S, contains valuable information on the arithmetic of the number field k. If all places dividing p are in S, then we have some structural knowledge on GS (k)(p), in particular, ...
... The pro-p-group GS (k)(p) = G(kS (p)/k), i.e. the Galois group of the maximal p-extension of k which is unramified outside S, contains valuable information on the arithmetic of the number field k. If all places dividing p are in S, then we have some structural knowledge on GS (k)(p), in particular, ...
Cyclic Compositions of a Positive Integer with Parts Avoiding an
... If in a cyclic 0-1 sequence [(δ1 , . . . , δn )]R we identify 1 with a black bead and 0 with a white bead, then we get a (fixed) necklace with n beads; e.g., see Graham et al. [9, Section 4.9]. In Knopfmacher and Robbins [14], a bijection is given between necklaces of n beads with k black and n − k ...
... If in a cyclic 0-1 sequence [(δ1 , . . . , δn )]R we identify 1 with a black bead and 0 with a white bead, then we get a (fixed) necklace with n beads; e.g., see Graham et al. [9, Section 4.9]. In Knopfmacher and Robbins [14], a bijection is given between necklaces of n beads with k black and n − k ...
P I ROOF BY
... A way to say that something is surprisingly different from usual is to exclaim “Now, that’s a horse of a different color!” The well-known mathematician George Pólya posed the following false “proof” showing through mathematical induction that actually, all horses are of the same color. Base case: If ...
... A way to say that something is surprisingly different from usual is to exclaim “Now, that’s a horse of a different color!” The well-known mathematician George Pólya posed the following false “proof” showing through mathematical induction that actually, all horses are of the same color. Base case: If ...
Notes - People @ EECS at UC Berkeley
... The latter expression is clearly divisible by 3, so (n + 1)3 − (n + 1) must also be divisible by 3. The theorem follows, by the principle of induction. 2 The next example we will look at is an inequality between two functions of n. Such inequalities are useful in computer science when showing that ...
... The latter expression is clearly divisible by 3, so (n + 1)3 − (n + 1) must also be divisible by 3. The theorem follows, by the principle of induction. 2 The next example we will look at is an inequality between two functions of n. Such inequalities are useful in computer science when showing that ...
Some Proofs of the Existence of Irrational Numbers
... invokes the Prime Factorization theorem in order to have the phrase “reducedform fraction” be well-defined; the two proofs presented above, however, do not require this secondary result, and so are somewhat stronger than the most common proof. ...
... invokes the Prime Factorization theorem in order to have the phrase “reducedform fraction” be well-defined; the two proofs presented above, however, do not require this secondary result, and so are somewhat stronger than the most common proof. ...
Methods of Proof
... If a direct proof of an assertion appears problematic, the next most natural strategy to try is a proof of the contrapositive. In Example 2.4.1 we use this method to prove that if the product of two integers, m and n, is even, then m or n is even. This statement has the form p → (r ∨ s). If you take ...
... If a direct proof of an assertion appears problematic, the next most natural strategy to try is a proof of the contrapositive. In Example 2.4.1 we use this method to prove that if the product of two integers, m and n, is even, then m or n is even. This statement has the form p → (r ∨ s). If you take ...
Exam II Review Sheet Solutions
... The second exam will be on Thursday, March 29. The syllabus will consist of Chapter NT from the text, together with the two number theory supplements passed out in class (Divisibility and Congruences). For reference, I will refer to these as Supplement D and Supplement C. You should be able to do al ...
... The second exam will be on Thursday, March 29. The syllabus will consist of Chapter NT from the text, together with the two number theory supplements passed out in class (Divisibility and Congruences). For reference, I will refer to these as Supplement D and Supplement C. You should be able to do al ...
ICS 251 – Foundation of Computer Science – Fall 2002
... The statement is true. We prove it using an indirect proof. Assume that x*y= z, where x is rational and y is irrational. Now if z is rational (negation of conclusion), we can write y= z/x = (a/b)/(c/d), where a,b,c,d are integers. Thus y=(ad)/(bc) = rational, which contradicts the fact that y is irr ...
... The statement is true. We prove it using an indirect proof. Assume that x*y= z, where x is rational and y is irrational. Now if z is rational (negation of conclusion), we can write y= z/x = (a/b)/(c/d), where a,b,c,d are integers. Thus y=(ad)/(bc) = rational, which contradicts the fact that y is irr ...
Full text
... Now, for there to exist a representation of n as a difference of two squares, one must be able to find a factorization ab = n for which (2) will yield a solution (x, y) in integers. Remark 2.1: We note that it is sufficient to consider only (2) since, if for a chosen factorization ab = n an integer ...
... Now, for there to exist a representation of n as a difference of two squares, one must be able to find a factorization ab = n for which (2) will yield a solution (x, y) in integers. Remark 2.1: We note that it is sufficient to consider only (2) since, if for a chosen factorization ab = n an integer ...
euler and the partial sums of the prime
... others), then he was indeed correct. In 1874, Mertens [5] showed that the di↵erence between the left and right-hand sides of (1) tends, as x ! 1, to the finite limit XX 1 ...
... others), then he was indeed correct. In 1874, Mertens [5] showed that the di↵erence between the left and right-hand sides of (1) tends, as x ! 1, to the finite limit XX 1 ...
CHECKING THE ODD GOLDBACH CONJECTURE UP TO 10 1
... investigate this conjecture numerically and prove it to be true for all integers less than 1020 . 2. Principle of the algorithm Because of the huge size of the set of odd integers considered, systematic verification for all integers is impossible. But it is in fact possible to use partial results of ...
... investigate this conjecture numerically and prove it to be true for all integers less than 1020 . 2. Principle of the algorithm Because of the huge size of the set of odd integers considered, systematic verification for all integers is impossible. But it is in fact possible to use partial results of ...
2 y x = − 6 y x = y x = − 2 y x = 5 y = − y x = 2 3 y x = 9 y x
... Decide what the left side would look like – think: Will there be a left side? Will it be odd symmetry? Will it be even symmetry? Adjust the graph if there is a reflection Do a little sketch beside each of the following, add in any notes for yourself within part a – if you find the above steps are no ...
... Decide what the left side would look like – think: Will there be a left side? Will it be odd symmetry? Will it be even symmetry? Adjust the graph if there is a reflection Do a little sketch beside each of the following, add in any notes for yourself within part a – if you find the above steps are no ...
Assignment # 3 : Solutions
... 33. Suppose n is an integer such that 2*3*4*5*n = 29*28*27*26*25. Does 29 | n? Why? Yes. Since 29 is one of the prime factors of the right hand side of the equation, it is also a prime factor of the left-hand side (by the unique factorization theorem). But 29 does not equal a prime factor of 2, 3, 4 ...
... 33. Suppose n is an integer such that 2*3*4*5*n = 29*28*27*26*25. Does 29 | n? Why? Yes. Since 29 is one of the prime factors of the right hand side of the equation, it is also a prime factor of the left-hand side (by the unique factorization theorem). But 29 does not equal a prime factor of 2, 3, 4 ...