course Mth 151 «m{X|rassignment #022
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19:17:20 `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 @?
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RESPONSE --> 1 2 1 2 1 2 1 2 This system is closed. The identity is 2, because any number @ 2 remains the same. Inverse: 1@2=2@1=1 This system is also associative, so it is a group. confidence assessment: 3
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19:17:24 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.
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RESPONSE --> OK self critique assessment: 3
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19:19:30 `q002. Which of the properties closure, identity, commutativity, inverse, does the standard addition operation + have on the set {-1, 0, 1}?
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RESPONSE --> -1 0 1 -1 -2 -1 0 0 -1 0 1 1 0 1 2 This system is not closed. 0 is the identity, because any number +0 stays the same. This system does have the property of inverse: -1 + 1 =0 so -1 and 1 are inverse 0 + 0 =0 so 0 is its own inverse confidence assessment: 3
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19:19:42 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.
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RESPONSE --> OK self critique assessment: 3
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19:20:58 `q003. Does the operation * of standard multiplication on the set {-1, 0, 1} have the properties of closure, identity and inverse?
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RESPONSE --> -1 0 1 -1 1 0 -1 0 0 0 0 1 -1 0 1 This system is closed. The identity is 1, because any number *1 stays the same. The number 0 has no inverse, so this is not a group. confidence assessment: 3
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19:21:04 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.
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RESPONSE --> OK self critique assessment: 3
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19:22:17 `q004. Does the operation * of standard multiplication on the set {-1, 1} have the properties of closure, identity and inverse?
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RESPONSE --> -1 1 -1 1 -1 1 -1 1 This system is closed. The identity is 1, because any number *1 stays the same. -1 is its own inverse becuase -1*-1=1 1 is its own inverse because 1*1=1 confidence assessment: 3
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19:22:27 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.
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RESPONSE --> OK self critique assessment: 3
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19:23:07 `q005. Is the operation * of standard multiplication on the set {-1, 1} a group. Note that the operation does have the property of associativity.
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RESPONSE --> Yes, it is a group, because it has the properties of closure, inverse, and identity. confidence assessment: 3
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19:23:15 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.
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RESPONSE --> OK self critique assessment: 3
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19:23:59 `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).
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RESPONSE --> (3+4) + 5 = 3 + (4+5) 7 + 5 = 3 + 9 12 = 12 confidence assessment: 3
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19:24:03 (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.
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RESPONSE --> OK self critique assessment: 3
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19:26:23 `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).
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RESPONSE --> 2 @ (0@1) = (2@0) @1 0@1= 0 2@0 = 0 2@0 = 0 1 @0=0 Both sides equal 0. (2@1) @ 1 = 2 @ (1@1) 2@1 = 1 1 @1 = 2 1@1 = 2 2 @ 2 = 2 Both sides equal 2. confidence assessment: 3
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19:26:33 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).
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RESPONSE --> OK self critique assessment: 3
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19:28:06 `q008. Does the result of the preceding exercise prove that the @ operation is associative on the set {0, 1, 2}?
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RESPONSE --> The two operations that we worked were associative, but in order to prove that the operation is associative on the entire set, we would have to perform the operation on every possible combination of the elements of the set. confidence assessment: 3
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19:28:15 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.
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RESPONSE --> OK self critique assessment: 3
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19:32:30 `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.
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RESPONSE --> Combinations of a,b,c from the set {-1,1): (1,1,1), (-1, -1, -1) , (1,1,-1), (1,-1,1), (1,-1,-1), (-1,-1,1), (-1,1,1) Verify 3: (1*1) *1 = 1 * (1*1) 1 *1 = 1 * 1 1 = 1 (-1*-1) * -1 = -1 * (-1*-1) 1 * -1 = -1 * 1 -1 = -1 (-1 *-1) * 1 = -1 (-1*1) 1 * 1 = -1 * -1 1 = 1 confidence assessment: 3
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19:32:37 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.
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RESPONSE --> OK self critique assessment: 3
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