Wikipedia
binary operations : In mathematics, associativity is a property that a binary operation can have. It means that, within an expression containing two or more of the same associative operators in a row, the order that the operations are performed does not matter as long as the sequence of the operands is not changed... More from Wikipedia
binary operations : In mathematics, Light's associativity test is a procedure invented by F W Light for testing whether a binary operation defined in a finite set by a Cayley multiplication table is associative. Direct verification of the associativity of a binary operation specified by a Cayley table is.. More from Wikipedia
Binary Operations
Binary Operations - Binary Operations are as given below, Commutative Law Associative Law Let S be any non-empty set. An operation * is called a binary operation on S if " a, b Î S a * b Î SLet S be any non-empty..
Binary Operations
binary operation: Let S be any non-empty set. An operation * is called a binary operation on S if " a, b S a * b S Commutative law: Let * be a binary operation on the set S. * is said to be associative in ..
demonstrations.wolfram.com The Wolfram Demonstrations Project contains thousands of free interactive visualizations, with new entries added daily. It is possible to use binary operators such as And, Or, or Xor as rules for cellular automata (CA). The binary operations are applied to the neighbor cells. Contributed by: Daniel de Souza Carvalho
GIMPLE memory representation into at form [4]. This will couple several datastructures into single structure. For usual binary operation this means replacing statement list (48 bytes), statement annotation (96 bytes), GIMPLE modify statement (48 bytes), binary expression (64 bytes) into attened form of roughly 90 bytes (for operation with no memory operands). Somewhat overestimating the savings by assuming that all 20565 statements of combine.c are such a binary operations, ...
Question : Number of binary operations on the set {a,b} are?
The answer is 16. How ? - If you know please explain how?
Answer : like the first guy said, its set up aa ab ba bb that is four sets. in binary, everything is base two, so it's 2^4 or 16.. More from Yahoo Answers
Answer : like the first guy said, its set up aa ab ba bb that is four sets. in binary, everything is base two, so it's 2^4 or 16.. More from Yahoo Answers
Question : According the following problem:
Let S be a set having exactly one element. How many different binary operations can be defined on S? Answer the question if S has exactly 2 elements exactly 3 elements exactly n elements.
there are 16 possible binary operations on a set of 2 elements. How many of these give a structure of a group? How many of the 19,683 possible binary operations on a set of 3 elements give a group structure.
Answer : Well, a binary operation is a function b:S S S. How do you figure out how many of these there are? The total number of functions F:A B is going to be counted quite simply. For each a in A, you need to pick a value b from B. So you have |B| choices for each |A|. So: Pick a1 in A, and choose b1 in B so that f(a1) = b1 (there are |B| choices). Pick a2 in A, and choose b1 in B so that f(a2) = b2 (there are |B| choices). etc. You will have |B| choices, and you will choose |A| times. So there are |B|^|A| choices. In this case, for A = S S and B=S, you get: The number of B:S S S is |S|^|S S| = S^(S^2) For S = 1, you get 1^(1^2) = 1. For S = 2, you get 2^(2^2) = 2^4 = 16. For S = 3, you get 3^(3^2) = 3^9 = 19683 Now, how many groups of order 2 and 3 are there? Only one of each type. So among the 16 choices, you have to have the same "pattern" of multiplication as only a single specific group. For example, if S={a,b} then the only possibilities are these two tables: .... More from Yahoo Answers
Answer : Well, a binary operation is a function b:S S S. How do you figure out how many of these there are? The total number of functions F:A B is going to be counted quite simply. For each a in A, you need to pick a value b from B. So you have |B| choices for each |A|. So: Pick a1 in A, and choose b1 in B so that f(a1) = b1 (there are |B| choices). Pick a2 in A, and choose b1 in B so that f(a2) = b2 (there are |B| choices). etc. You will have |B| choices, and you will choose |A| times. So there are |B|^|A| choices. In this case, for A = S S and B=S, you get: The number of B:S S S is |S|^|S S| = S^(S^2) For S = 1, you get 1^(1^2) = 1. For S = 2, you get 2^(2^2) = 2^4 = 16. For S = 3, you get 3^(3^2) = 3^9 = 19683 Now, how many groups of order 2 and 3 are there? Only one of each type. So among the 16 choices, you have to have the same "pattern" of multiplication as only a single specific group. For example, if S={a,b} then the only possibilities are these two tables: .... More from Yahoo Answers
Looking for More Help!
