course math 151 ëß󦀀üjÄ¡±°¨_üÿÄûŽšÜÒ¹assignment #022 022. Groups Liberal Arts Mathematics I 04-26-2009
......!!!!!!!!...................................
19:57:44 `q001. There are nine questions in this assignment. A group is a set and an operation on that set which has the properties of closure, associativity, identity and inverse. The set {1, 2} on the operation @ of Assignment 21, defined by x @ y = remainder when the product x * y is divided by 3, does have the associative property. Is the set {1, 2} a group on @?
......!!!!!!!!...................................
RESPONSE --> yes it has the associative property confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:58:25 The table for @ on {1, 2} is @ 1 2 1 2 1 2 1 2 The table shows us that all the possible results of x @ y on the set {1, 2} are in the set, so the operations is closed. We see that 2 when combined with anything does not change that number number, so 2 is the identity for this operation on this set. We see also that 1 @ 2 = 2 @ 1 = 1 so 1 and 2 are inverses. Therefore both of the elements in the set {1, 2} have inverses in the set. Since it has already been stated that the set has the associative property, we conclude that the set is a group.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
19:59:52 `q002. Which of the properties closure, identity, commutativity, inverse, does the standard addition operation + have on the set {-1, 0, 1}?
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
20:15:07 The table for this operation would be + -1 0 2 -1 -2 -1 0 0 -1 0 1 1 0 1 2 .The table shows that the operation is not closed, since the necessary results 2 and -2 are not in the set {-1, 0, 1}. The number 0 is the identity, as we can see by looking at the row across from 0 and the column below 0. We see that -1 + 1 = 0 so that -1 and 1 are inverses, and that 0 + 0 = 0 so 0 is its own inverse. The operation therefore has the inverse property. The order of addition doesn't affect the result, which as we see makes the table symmetric about a diagonal line from upper left to lower right (copy the table, sketch the line and see how everything above and to the right of the line is 'reflected' in the corresponding below and to the left of the line. We therefore conclude that the operation is commutative. This operation has some important properties, but since it is not closed on this set it is not an interesting operation on this set.
......!!!!!!!!...................................
RESPONSE --> - the table shows that is not closed because we have elements that are no in the set{-1,0,1} -zero is the identity because the 0 row and column are the same as in the set{-1,0,1} - we have the inverse property, -1 + 1 = 0 and 0+0=0 the operation is commutative because when I copied the table and draw a diagonal line the posterior half was identical to the inferior part self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:45:15 `q003. Does the operation * of standard multiplication on the set {-1, 0, 1} have the properties of closure, identity and inverse?
......!!!!!!!!...................................
RESPONSE --> the table show * -1 0 1 -1 1 0 -1 0 0 0 0 1 -1 0 1 that it is commutative, draw a diagonal line and the two halves are reflecting each other. it is closed because all the results in the table are the same elements on the set{-1,0,1} 1 is the identity because the 1 column and row have the same elements then the set {-1,0,1} confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:50:58 The table for this operation is * -1 0 1 -1 1 0 -1 0 0 0 0 1 -1 0 1 We see that every possible result of the operation is in the set {-1, 0, 1}, and the row across from 1 and the column beneath 1 show how us that 1 is the identity. -1 * -1 = 1 so -1 is its own inverse, and 1 * 1 = 1 so 1 is its own inverse. However, anything 0 is combine with gives us 0, so 0 cannot by combined with anything to get 1 and 0 therefore has no inverse.
......!!!!!!!!...................................
RESPONSE --> becuase 0*0=0 can not be combined its not inverse self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:56:27 `q004. Does the operation * of standard multiplication on the set {-1, 1} have the properties of closure, identity and inverse?
......!!!!!!!!...................................
RESPONSE --> the table shows * -1 1 -1 1 -1 1 -1 1 it has the property of closure because all the result elements on the table are the same than in the set{-1,1} the identity is 1, because under the 1 column and row the elements are the same and in the same order as the set{-1,1} it is inverse becuase -1 * -1 = 1 confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:59:25 The table for this operation is easily written: * -1 1 -1 1 -1 1 -1 1 All the results come from the set {-1,1} so the operation is closed. The row across from and column beneath 1 show us that 1 is the identity. Since -1 * -1 = 1 and 1 * 1 = 1, both -1 and 1 are their own inverses. Thus both of the elements in the set {-1,1} have inverses and the operation has the inverse property.
......!!!!!!!!...................................
RESPONSE --> so both of the elements in the set are inverse, I wasn't very clear about the inverse property, but I understand now that both elements in the set have to be able to multiply by itself and if the product is the same the operation has the inverse property
.................................................
......!!!!!!!!...................................
21:00:09 `q005. Is the operation * of standard multiplication on the set {-1, 1} a group. Note that the operation does have the property of associativity.
......!!!!!!!!...................................
RESPONSE --> if it doesn't have the property of associativity it is not a group confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:04:56 Has seen in the preceding example the operation is closed and has the identity and inverse properties. Given that it is sensitive, it is therefore a group.
......!!!!!!!!...................................
RESPONSE --> on the textbook on page 210(green box) it says that for an operation to be a group needs to satisfy the closure, associative, identity, and inverse properties. Since this one doesn't satisfy the associative property i understand it is not a group
.................................................
......!!!!!!!!...................................
21:25:53 `q006. We've referred to the property of associativity, but we haven't yet defined it. Associativity essentially means that when an operation (technically a binary operation, but don't worry about that a terminology at this point) is performed on three elements of a set, for example a + b + c, it doesn't matter whether we first perform a + b then add c, calculating (a + b) + c, or group the b and c so we calculate a + (b + c). If + means addition on real numbers, show that (3 + 4) + 5 = 3 + ( 4 + 5).
......!!!!!!!!...................................
RESPONSE --> [3 + 4] + 5 = 3 + [4 + 5] 7 + 5 = 3 + 9 12 = 12 confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:26:50 (3 + 4) + 5 = 7 + 5 = 12. 3 + ( 4 + 5) = 3 + 9 = 12. Either way we do the calculation we get the same thing. This is a familiar property of addition, and everyone in this course has used it for years.
......!!!!!!!!...................................
RESPONSE --> because the order of the elements doesn't matter the result is the same self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:36:10 `q007. Verify that for the operation @ defined on {0, 1, 2} by x @ y = remainder when x * y is double then divided by 3, we have 2 @ (0 @ 1) = ( 2 @ 0 ) @ 1. Verify also that (2 @ 1) @ 1 = 2 @ ( 1 @ 1).
......!!!!!!!!...................................
RESPONSE --> 2*[0*1] = [2*0] * 1 2*0 = 0*1 0 = 0 [2*1]*1 = 2*[1*1] 2*1 = 2*1 2 = 2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:44:27 0 @ 1 = 0, so 2 @ ( 0 @ 1) = 2 @ 0 = 0. 2 @ 0 = 0 so (2 @ 0 ) @ 1 = 0 @ 1 = 0. Thus 2 @ (0 @ 1) = ( 2 @ 0 ) @ 1. (2 @ 1) = 1 so (2 @ 1) @ 1 = 1 @ 1 = 1. 1 @ 1 = 2 so 2 @ (1 @ 1) = 2 @ 2 = 2. Thus (2 @ 1) @ 1 = 2 @ ( 1 @ 1).
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:44:44 `q008. Does the result of the preceding exercise prove that the @ operation is associative on the set {0, 1, 2}?
......!!!!!!!!...................................
RESPONSE --> yes confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:45:24 Not quite. In order to prove that the operation is associative on the set we would need to prove that ( a @ b) @ c = a @ ( b @ c ) for all possible combinations all of a, b, c. We verified the property for (a, b, c) = (2, 0, 1) and (2, 1, 1). We would also have to verify it for every possible combination of (a, b, c). That would take a lot of work, but it could be done. The possible combinations are (0,0,0), (0,0,1), (0,0,2), (0,1,0), (0,1,1), (0,1,2), (0,2,0), (0,2,1), (0,2,2), (1,0,0), etc., etc.. There are 27 possible combinations.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:47:00 `q009. Earlier we verified the properties of closure, identity and inverse for the multiplication operation * on the set {-1, 1}. We asserted that this operation was associative, so that this set with this operation forms a group. It would still be too time-consuming to prove that * is associative on {-1, 1}, but list the possible combinations of a, b, c from the set and verify associativity for any three of them.
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 0
.................................................
......!!!!!!!!...................................
21:51:38 The possible combinations are (a, b, c) = (-1, -1, -1) or (-1, -1, 1) or (-1, 1, -1) or (-1, 1, 1) or (1, -1, -1) or (1, -1, 1) or (1, 1, -1) or (1, 1, 1). Verifying these combinations in order: (-1 * -1) * -1 = -1 * ( -1 * -1) because 1 * -1 = -1 * 1 and both sides give -1. (-1 * -1) * 1 = -1 * ( -1 * 1) because 1 * 1 = -1 * -1 and both sides give 1. (-1 * 1) * -1 = -1 * ( 1 * -1) because -1 * -1 = -1 * -1 and both sides give 1. (-1 * 1) * 1 = -1 * ( 1 * 1) because -1 * 1 = -1 * 1 and both sides give -1. (1 * -1) * -1 = 1 * ( -1 * -1) because -1 * -1 = 1 * 1 and both sides give 1. (1 * -1) * 1 = 1 * ( -1 * 1) because -1 * 1 = 1 * -1 and both sides give -1. (1 * 1) * -1 = 1 * ( 1 * -1) because 1 * -1 = 1 * -1 and both sides give -1. (1 * 1) * 1 = 1 * ( 1 * 1) because 1 * 1 = 1 * 1 and both sides give 1. You should have verified three of these in the manner shown.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3