Student Name: assignment #022
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22:51:44 `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 doubled and divided by 3, does have the associative property. Is the set {1, 2} a group on @?
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RESPONSE --> yes it is, because 2*1 is 4\3 is 1 with remainder of 1 which is in your number set
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22:54:10 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 --> gotcha
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22:59:06 `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 --> inverse cause it just flips the numbers
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23:00:39 The table for this operation would be + -1 0 1 -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 --> k
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23:02:35 `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 --> as a whole i would say inverse. zero has it's own identity, of course with multiplication you get zero automatically.
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23:03:13 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 --> right, gotcha
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23:08:13 `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 --> identity cause when you multiply you get one of the numbers you are multiplying with
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23:09:32 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 --> i didn't take -1 and -1 into consideration
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23:11:04 `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 --> no cause it's a closed set
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23:11:59 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 --> i got mixed up somewhere
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23:18:13 `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=7, then +5 which is 12. (4+5)=9, then +3 is 12.show is does equal each other
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23:18:43 (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 --> k
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23:24:08 `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 --> yes the left side is 2 and the right side is 2. cause 2@1=1 and that 1@1 is 2(left side). on the right side (1@1)=2 and 2@2 is 4 then 8 then 6 with 2 left over which is and equality
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23:24:17 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 --> right
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23:24:47 `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 --> yes it does
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23:25:20 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. In answer to a question by a student the instructor wrote a function to check associativity on this set and it did verify associativity, using a computer algebra system (DERIVE). You don't have to understand this, or even read it, but you might find it interesting: The function, in case you're interested: f(a, b) := 2·x·y/3 - INT(2·x·y/3) g(a, b, c) := f(f(a, b), c) - f(a, f(b, c)) VECTOR(VECTOR(VECTOR(g(a, b, c), a, 0, 2), b, 0, 2), c, 0, 2) should evaluate to all 0's. The result is as follows: [[0, 0, 0; 0, 0, 0; 0, 0, 0], [0, 0, 0; 0, 0, 0; 0, 0, 0], [0, 0, 0; 0, 0, 0; 0, 0, 0]] So the operation is associative.
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RESPONSE --> that was a guess, but gotcha
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23:29:32 `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 --> i'm not sure how to get three numbers for a,b,c with two numbers
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23:30:27 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 --> i wasn't sure where to get the numbers
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