Lecture 4: Combinations, Subsets and Multisets
... Partition X into subsets X1 , X2 , . . . , Xn where X j has the sets {i, j} with i < j as its elements. Then |X j | = j − 1 for each j, since i can be chosen only from the set {1, 2, . . . , j − 1}. Using the sum rule |X1 | + |X2 | + · · · + |Xn | = |X| we have a proof of 0 + 1 + 2 + · · · + (n − 1) ...
... Partition X into subsets X1 , X2 , . . . , Xn where X j has the sets {i, j} with i < j as its elements. Then |X j | = j − 1 for each j, since i can be chosen only from the set {1, 2, . . . , j − 1}. Using the sum rule |X1 | + |X2 | + · · · + |Xn | = |X| we have a proof of 0 + 1 + 2 + · · · + (n − 1) ...
Section 2.2
... In this section we consider the notions of finite and infinite sets, and the cardinality of a set. Reasonable goals for this section are to become familiar with these ideas and to practice interpreting descriptions of sets that are presented in terse mathematical notation (this means, amongst other th ...
... In this section we consider the notions of finite and infinite sets, and the cardinality of a set. Reasonable goals for this section are to become familiar with these ideas and to practice interpreting descriptions of sets that are presented in terse mathematical notation (this means, amongst other th ...
Math 3390 Introduction to topology, Assignment 2. Due October 26
... are all singletons, except for the equivalence class {0, 1}. (So 0 is identified with 1, and there are no other identifications). Show that X is homeomorphic to the subspace {(x, y) | x2 + y 2 = 1} of R2 . Question 6: Quotient maps need not be open maps or closed maps. Here is an example that shows ...
... are all singletons, except for the equivalence class {0, 1}. (So 0 is identified with 1, and there are no other identifications). Show that X is homeomorphic to the subspace {(x, y) | x2 + y 2 = 1} of R2 . Question 6: Quotient maps need not be open maps or closed maps. Here is an example that shows ...
DIOPHANTINE APPROXIMATION OF COMPLEX NUMBERS
... has infinitely many non-trivial solutions x, y ∈ Z[i D]. 2. Geometry of Numbers Hilde Gintner [5] mainly refers to two theorems of Minkowksi [15] coming from his famous theory of ’Geometry of Numbers’. Those theorems can also be found in [2], where J.W.S. Cassels so to say translated them to a more ...
... has infinitely many non-trivial solutions x, y ∈ Z[i D]. 2. Geometry of Numbers Hilde Gintner [5] mainly refers to two theorems of Minkowksi [15] coming from his famous theory of ’Geometry of Numbers’. Those theorems can also be found in [2], where J.W.S. Cassels so to say translated them to a more ...
4(x + 2)
... We can also use the Distributive Property in reverse. This is called Factoring. When we factor an expression, we find all numbers or variables that divide into all of the parts of an expression. Example: 7x + 35 Both the 7x and 35 are divisible by 7 7(x + 5) By removing the 7 we have factored the p ...
... We can also use the Distributive Property in reverse. This is called Factoring. When we factor an expression, we find all numbers or variables that divide into all of the parts of an expression. Example: 7x + 35 Both the 7x and 35 are divisible by 7 7(x + 5) By removing the 7 we have factored the p ...
Homework 2 January 19, 2006 Math 522 Direction: This homework
... in M2 (Z), then XY 6= Y X. I have selected the above matrices in such a way that no matter which way you multiply them you will get a zero matrix. 6. Let Z be the set of integers. Let P(Z) be the power set of Z, that is the collection of all subsets of Z. Let ⊕ and be two binary operations on P(Z) ...
... in M2 (Z), then XY 6= Y X. I have selected the above matrices in such a way that no matter which way you multiply them you will get a zero matrix. 6. Let Z be the set of integers. Let P(Z) be the power set of Z, that is the collection of all subsets of Z. Let ⊕ and be two binary operations on P(Z) ...
Coursework 6
... 5 Without using a calculator, write [280 ]19 as a class [a]19 where 0 ≤ a < 19. Try to be efficient in your calculations. 6 How many of the elements of Z12 have multiplicative inverses? What about Z30 ? Describe a general method that would allow you to calculate how many elements of Zm have multipl ...
... 5 Without using a calculator, write [280 ]19 as a class [a]19 where 0 ≤ a < 19. Try to be efficient in your calculations. 6 How many of the elements of Z12 have multiplicative inverses? What about Z30 ? Describe a general method that would allow you to calculate how many elements of Zm have multipl ...
Math 322, Fall Term 2011 Final Exam
... (c) If a group of order 25 acts on a set X, then there is an element x ∈ X such that its orbit has 8 elements. (d) The number of conjugacy classes in Sn is equal to the number of nonisomorphic abelian groups of order pn , for any prime p. (e) The group S6 has a subgroup H of order 360 whose normalis ...
... (c) If a group of order 25 acts on a set X, then there is an element x ∈ X such that its orbit has 8 elements. (d) The number of conjugacy classes in Sn is equal to the number of nonisomorphic abelian groups of order pn , for any prime p. (e) The group S6 has a subgroup H of order 360 whose normalis ...
The same paper as word document
... where the X is an arbitrary set ,are called the transfinite, real ,complex, quaternion Fuzzy subsets of X. Remark 12. The norm of the transfinite real ,complex ,quaternion numbers of some Archimedean base λ , is defined as in the minimal case of the real, complex ,and quaternion numbers. Only that t ...
... where the X is an arbitrary set ,are called the transfinite, real ,complex, quaternion Fuzzy subsets of X. Remark 12. The norm of the transfinite real ,complex ,quaternion numbers of some Archimedean base λ , is defined as in the minimal case of the real, complex ,and quaternion numbers. Only that t ...
Section 5.1: Basic Definitions of Set Theory
... No, the set {1} is not in this set (though the number 1 is). Remark 2.4. Note that there is a big difference between the symbols ∈ and ⊆ as illustrated in the last example. Example 2.5. Determine which of the following sets are equal, subsets of another or not subsets of each other: (i ) S = {x ∈ R| ...
... No, the set {1} is not in this set (though the number 1 is). Remark 2.4. Note that there is a big difference between the symbols ∈ and ⊆ as illustrated in the last example. Example 2.5. Determine which of the following sets are equal, subsets of another or not subsets of each other: (i ) S = {x ∈ R| ...
Math 153: Course Summary
... • G = Z/p, the group of hours on a clock with p hours. Here ∗ is addition. For instance, 5 + 9 = 3 in Z/11, because 5 hours after 9oclock is 3-oclock on a clock with 11 hours. • When p is prime, the nonzero hours of Z/p form a group, with ∗ being multiplication. This group is called (Z/p)∗ . For ins ...
... • G = Z/p, the group of hours on a clock with p hours. Here ∗ is addition. For instance, 5 + 9 = 3 in Z/11, because 5 hours after 9oclock is 3-oclock on a clock with 11 hours. • When p is prime, the nonzero hours of Z/p form a group, with ∗ being multiplication. This group is called (Z/p)∗ . For ins ...
1 Cardinality and the Pigeonhole Principle
... there is no injection A → B. In particular f is not injective. But not being injective means precisely that two different elements of the domain have the same image, as desired. This is a simple statement, but often gets used in surprising ways. In any problem involving sets you should be open to th ...
... there is no injection A → B. In particular f is not injective. But not being injective means precisely that two different elements of the domain have the same image, as desired. This is a simple statement, but often gets used in surprising ways. In any problem involving sets you should be open to th ...
Solutions.
... we take the quotient by x2 + x + 1, we are essentially setting x2 + x + 1 = 0, or equivalently x2 = −(x + 1), but −(x+1) = x+1 in F2 [x], so x2 = x + 1. Imposing this condition, we see that every polynomial an xn + · · · + a1 x + a0 can be reduced to a linear polynomial after repeated applications o ...
... we take the quotient by x2 + x + 1, we are essentially setting x2 + x + 1 = 0, or equivalently x2 = −(x + 1), but −(x+1) = x+1 in F2 [x], so x2 = x + 1. Imposing this condition, we see that every polynomial an xn + · · · + a1 x + a0 can be reduced to a linear polynomial after repeated applications o ...
MATH 161, SHEET 1: SETS, FUNCTIONS and CARDINALITY Sets
... of A is n, and we write |A| = n. If A is not finite, we say that A is infinite. In either case, if there exists a bijection between two sets A and B, then we say that A and B have the same cardinality and write |A| = |B|. ...
... of A is n, and we write |A| = n. If A is not finite, we say that A is infinite. In either case, if there exists a bijection between two sets A and B, then we say that A and B have the same cardinality and write |A| = |B|. ...
Birkhoff's representation theorem
This is about lattice theory. For other similarly named results, see Birkhoff's theorem (disambiguation).In mathematics, Birkhoff's representation theorem for distributive lattices states that the elements of any finite distributive lattice can be represented as finite sets, in such a way that the lattice operations correspond to unions and intersections of sets. The theorem can be interpreted as providing a one-to-one correspondence between distributive lattices and partial orders, between quasi-ordinal knowledge spaces and preorders, or between finite topological spaces and preorders. It is named after Garrett Birkhoff, who published a proof of it in 1937.The name “Birkhoff's representation theorem” has also been applied to two other results of Birkhoff, one from 1935 on the representation of Boolean algebras as families of sets closed under union, intersection, and complement (so-called fields of sets, closely related to the rings of sets used by Birkhoff to represent distributive lattices), and Birkhoff's HSP theorem representing algebras as products of irreducible algebras. Birkhoff's representation theorem has also been called the fundamental theorem for finite distributive lattices.