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course Mth 151 11/20/2013, 1:40 p.m. If your solution to stated problem does not match the given solution, you should self-critique per instructions at.............................................
Given Solution: In order to be a group the operation has to be closed, associative, the identity must be present, and each element in the set must have an inverse in the set. So the four questions that need to be answered are: Is the operation closed? Is the operation associative? Is the identity present? Is the inverse property satisfied? 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):3 ------------------------------------------------ Self-critique Rating:ok
********************************************* Question: `q002. Which of the properties closure, identity, commutativity, inverse, does the standard addition operation + have on the set {-1, 0, 1}? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: identity, 0, addition identity is 0 inverse, 0 is it's own inverse, and 1 and-1 are inverses. associative and commutative, the order in which the addition has does not matter. it is not closed, 2,-2 are not include in the set It is not a group. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: The table for this operation would be + -1 0 1 -1 -2 -1 0 0 -1 0 1 1 0 1 2 .The table shows that the operation is not closed, since the results 2 and -2 are not in the set {-1, 0, 1}. The number 0 is the identity, as we can see by looking at the row across from 0 and the column below 0. In other words, any number combined with 0 gives us that number (said another way, 0 doesn't change the number it's combined with). We see that -1 + 1 = 0 so that -1 and 1 are inverses, and that 0 + 0 = 0 so 0 is its own inverse. The operation therefore has the inverse property. The order of addition doesn't affect the result, which as we see makes the table symmetric about a diagonal line from upper left to lower right (copy the table, sketch the line and see how everything above and to the right of the line is 'reflected' in the corresponding below and to the left of the line. We therefore conclude that the operation is commutative. This operation has some important properties, but since it is not closed on this set it is not an interesting operation on this set. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):3 ------------------------------------------------ Self-critique Rating:ok
********************************************* Question: `q003. Does the operation * of standard multiplication on the set {-1, 0, 1} have the properties of closure, identity and inverse? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: *-1 0 1 -1 1 0-1 0 0 0 0 1-1 0 1 it has all but the inverse property. It does have an identity of 1, but 1 does not appear in the third row, which means nothing multiplied by 0 that equals 1. The far left row contains all the members, so it has closure. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: The table for this operation is * -1 0 1 -1 1 0 -1 0 0 0 0 1 -1 0 1 The numbers 0, -1 and 1 all occur as results of the operation, so every possible result of the operation is a member of the original set {-1, 0, 1}. The row across from 1 and the column beneath 1 show how us that 1 is the identity. We now check to see if the inverse property holds: -1 * -1 = 1 so -1 is its own inverse, and 1 * 1 = 1 so 1 is its own inverse. However, anything 0 is combined with gives us 0, so there is nothing that can be combined with 0 to get our identity 1. 0 therefore has no inverse, so we cannot say that every element of the set has an inverse in the set. The operation therefore does not have the inverse property. STUDENT COMMENT I see now how the -1*-1 = 1 is an inverse. I was only thinking about the 0 and how it never gives an inverse. INSTRUCTOR RESPONSE -1 does have an inverse. However since 0 doesn't have an inverse, the operation doesn't have the inverse property, which would imply that all elements of the set have inverses. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):3 ------------------------------------------------ Self-critique Rating:ok
********************************************* Question: `q004. Does the operation * of standard multiplication on the set {-1, 1} have the properties of closure, identity and inverse? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: * -1 1 -1 1-1 1-1 1 Yes, the set {-1,1} has closure, identity , and inverse properties. When removing 0 from the set it made the inverse property possible. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: The table for this operation is easily written: * -1 1 -1 1 -1 1 -1 1 All the results come from the set {-1,1} so the operation is closed. The row across from and column beneath 1 show us that 1 is the identity. Since -1 * -1 = 1 and 1 * 1 = 1, both -1 and 1 are their own inverses. Thus both of the elements in the set {-1,1} have inverses and the operation has the inverse property. STUDENT COMMENT: so both of the elements in the set are inverse, I wasn't very clear about the inverse property, but I understand now that both elements in the set have to be able to multiply by itself and if the product is the same the operation has the inverse property INSTRUCTOR RESPONSE: The product of inverses is the identity. If two elements combine to give you the identity, then those elements are inverse to one another. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):3 ------------------------------------------------ Self-critique Rating:
********************************************* Question: `q005. Is the operation * of standard multiplication on the set {-1, 1} a group. Note that the operation does have the property of associativity. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: @ -1 1 -1 1-1 1 -1 1 We only need to establish associative property for this group because identity, inverse, and closure have been proven. It is associative , so it is a group. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: As seen in the preceding example the operation is closed and has the identity and inverse properties. Given that it is associative, it is therefore a group. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):3 ------------------------------------------------ Self-critique Rating:ok
********************************************* Question: `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 + has its common meaning of addition on real numbers, show that (3 + 4) + 5 = 3 + ( 4 + 5). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 7+5=9+3 12=12 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: (3 + 4) + 5 = 7 + 5 = 12. 3 + ( 4 + 5) = 3 + 9 = 12. Either way we do the calculation we get the same thing. This is a familiar property of addition. You have been using this property of basic arithmetic for years. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):3 ------------------------------------------------ Self-critique Rating:ok
********************************************* Question: `q007. Verify that for the operation @ defined on {0, 1, 2} by x @ y = remainder when x * y is doubled then divided by 3, we have 2 @ (0 @ 1) = ( 2 @ 0 ) @ 1. Verify also that (2 @ 1) @ 1 = 2 @ ( 1 @ 1). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2@0=0@1 0=0 2@1=2@1 2=2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):3 ------------------------------------------------ Self-critique Rating:ok
********************************************* Question: `q008. Does the result of the preceding exercise prove that the @ operation is associative on the set {0, 1, 2}? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are many other possibilities that would have to be verified. These are just a couple. It soes seem like it would be associative, but it only takes one example to disprove. (0@2)@1=0@(2@1) 0@1=0@2 0=0 (0@1)@2=0@(1@2) 0@2=0@2 0=0 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: Not quite. In order to prove that the operation is associative on the set we would need to prove that ( a @ b) @ c = a @ ( b @ c ) for all possible combinations all of a, b, c. We verified the property for (a, b, c) = (2, 0, 1) (that is, it was verified when a = 2, b = 0 and c = 1). We also verified the propert for (a, b, c) = (2, 1, 1) (i.e., for a = 2, b = 1 and c = 1). We would also have to verify it for every possible combination of (a, b, c). That would take a lot of work, but it could be done. The possible combinations of (a, b, c) are (0,0,0), (0,0,1), (0,0,2), (0,1,0), (0,1,1), (0,1,2), (0,2,0), (0,2,1), (0,2,2), (1,0,0), etc., etc.. There are 27 possible combinations. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):3 ------------------------------------------------ Self-critique Rating:ok
********************************************* Question: `q009. Earlier we verified the properties of closure, identity and inverse for the multiplication operation * on the set {-1, 1}. We asserted that this operation was associative, so that this set with this operation forms a group. It would fairly time-consuming to prove that * is associative on {-1, 1}. So you are asked here to just list the possible combinations of a, b, c, then verify associativity for any three of them. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (1*1)*1=1*(1*1) 1*1=1*1 1=1 (-1*-1)*-1=-1*(-1*-1) -1*-1=-1*-1 -1=-1 (-1* 1)* 1=1*(-1*1)*1 -1*1=-1*1 -1=-1 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):3 ------------------------------------------------ Self-critique Rating:ok
********************************************* Question: `q010. Let the & operation be defined as follows: x & y is the remainder obtained when x is multiplied by y, then divided by 5. The table below is for the & operation on the set {1, 2, 3, 4}. The table for this operation is shown below: & 1 2 3 4 1 1 2 3 4 2 2 4 1 3 3 3 1 4 2 4 4 3 2 1 Show whether (a & b) & c = a & (b & c) for the following combinations (a, b, c): (1, 2, 3), (2, 3, 4), (4, 3, 4). If all three of these combinations check out, you may assume that the operation is associative. Then check whether the operation is closed, whether it is commutative, and whether it has the inverse and identity properties. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I think it has all of these properties. 1 is the identity and is in every row in the table, so this indicates the inverse property as well. The diaganol symmetry is an indication of commutativety, and it is closed because the far left column contains the same elemants as the results. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q011. Multiplication of real numbers is associative. Determine whether multiplication is a group on the set of positive rational numbers. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1 is the identity of multiplication. It is associative as already stated in the question,it is closed, when multipying two rational positive numbers you will get a positive rational number. Integers on multiplication lack the inverse property so this is not a group.#$&* course Mth 151 11/20/2013, 1:40 p.m. If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: In order to be a group the operation has to be closed, associative, the identity must be present, and each element in the set must have an inverse in the set. So the four questions that need to be answered are: Is the operation closed? Is the operation associative? Is the identity present? Is the inverse property satisfied? 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):3 ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q002. Which of the properties closure, identity, commutativity, inverse, does the standard addition operation + have on the set {-1, 0, 1}? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: identity, 0, addition identity is 0 inverse, 0 is it's own inverse, and 1 and-1 are inverses. associative and commutative, the order in which the addition has does not matter. it is not closed, 2,-2 are not include in the set It is not a group. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The table for this operation would be + -1 0 1 -1 -2 -1 0 0 -1 0 1 1 0 1 2 .The table shows that the operation is not closed, since the results 2 and -2 are not in the set {-1, 0, 1}. The number 0 is the identity, as we can see by looking at the row across from 0 and the column below 0. In other words, any number combined with 0 gives us that number (said another way, 0 doesn't change the number it's combined with). We see that -1 + 1 = 0 so that -1 and 1 are inverses, and that 0 + 0 = 0 so 0 is its own inverse. The operation therefore has the inverse property. The order of addition doesn't affect the result, which as we see makes the table symmetric about a diagonal line from upper left to lower right (copy the table, sketch the line and see how everything above and to the right of the line is 'reflected' in the corresponding below and to the left of the line. We therefore conclude that the operation is commutative. This operation has some important properties, but since it is not closed on this set it is not an interesting operation on this set. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):3 ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q003. Does the operation * of standard multiplication on the set {-1, 0, 1} have the properties of closure, identity and inverse? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: *-1 0 1 -1 1 0-1 0 0 0 0 1-1 0 1 it has all but the inverse property. It does have an identity of 1, but 1 does not appear in the third row, which means nothing multiplied by 0 that equals 1. The far left row contains all the members, so it has closure. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The table for this operation is * -1 0 1 -1 1 0 -1 0 0 0 0 1 -1 0 1 The numbers 0, -1 and 1 all occur as results of the operation, so every possible result of the operation is a member of the original set {-1, 0, 1}. The row across from 1 and the column beneath 1 show how us that 1 is the identity. We now check to see if the inverse property holds: -1 * -1 = 1 so -1 is its own inverse, and 1 * 1 = 1 so 1 is its own inverse. However, anything 0 is combined with gives us 0, so there is nothing that can be combined with 0 to get our identity 1. 0 therefore has no inverse, so we cannot say that every element of the set has an inverse in the set. The operation therefore does not have the inverse property. STUDENT COMMENT I see now how the -1*-1 = 1 is an inverse. I was only thinking about the 0 and how it never gives an inverse. INSTRUCTOR RESPONSE -1 does have an inverse. However since 0 doesn't have an inverse, the operation doesn't have the inverse property, which would imply that all elements of the set have inverses. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):3 ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q004. Does the operation * of standard multiplication on the set {-1, 1} have the properties of closure, identity and inverse? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: * -1 1 -1 1-1 1-1 1 Yes, the set {-1,1} has closure, identity , and inverse properties. When removing 0 from the set it made the inverse property possible. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The table for this operation is easily written: * -1 1 -1 1 -1 1 -1 1 All the results come from the set {-1,1} so the operation is closed. The row across from and column beneath 1 show us that 1 is the identity. Since -1 * -1 = 1 and 1 * 1 = 1, both -1 and 1 are their own inverses. Thus both of the elements in the set {-1,1} have inverses and the operation has the inverse property. STUDENT COMMENT: so both of the elements in the set are inverse, I wasn't very clear about the inverse property, but I understand now that both elements in the set have to be able to multiply by itself and if the product is the same the operation has the inverse property INSTRUCTOR RESPONSE: The product of inverses is the identity. If two elements combine to give you the identity, then those elements are inverse to one another. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):3 ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. Is the operation * of standard multiplication on the set {-1, 1} a group. Note that the operation does have the property of associativity. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: @ -1 1 -1 1-1 1 -1 1 We only need to establish associative property for this group because identity, inverse, and closure have been proven. It is associative , so it is a group. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: As seen in the preceding example the operation is closed and has the identity and inverse properties. Given that it is associative, it is therefore a group. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):3 ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `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 + has its common meaning of addition on real numbers, show that (3 + 4) + 5 = 3 + ( 4 + 5). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 7+5=9+3 12=12 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: (3 + 4) + 5 = 7 + 5 = 12. 3 + ( 4 + 5) = 3 + 9 = 12. Either way we do the calculation we get the same thing. This is a familiar property of addition. You have been using this property of basic arithmetic for years. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):3 ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q007. Verify that for the operation @ defined on {0, 1, 2} by x @ y = remainder when x * y is doubled then divided by 3, we have 2 @ (0 @ 1) = ( 2 @ 0 ) @ 1. Verify also that (2 @ 1) @ 1 = 2 @ ( 1 @ 1). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2@0=0@1 0=0 2@1=2@1 2=2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):3 ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q008. Does the result of the preceding exercise prove that the @ operation is associative on the set {0, 1, 2}? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are many other possibilities that would have to be verified. These are just a couple. It soes seem like it would be associative, but it only takes one example to disprove. (0@2)@1=0@(2@1) 0@1=0@2 0=0 (0@1)@2=0@(1@2) 0@2=0@2 0=0 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Not quite. In order to prove that the operation is associative on the set we would need to prove that ( a @ b) @ c = a @ ( b @ c ) for all possible combinations all of a, b, c. We verified the property for (a, b, c) = (2, 0, 1) (that is, it was verified when a = 2, b = 0 and c = 1). We also verified the propert for (a, b, c) = (2, 1, 1) (i.e., for a = 2, b = 1 and c = 1). We would also have to verify it for every possible combination of (a, b, c). That would take a lot of work, but it could be done. The possible combinations of (a, b, c) are (0,0,0), (0,0,1), (0,0,2), (0,1,0), (0,1,1), (0,1,2), (0,2,0), (0,2,1), (0,2,2), (1,0,0), etc., etc.. There are 27 possible combinations. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):3 ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q009. Earlier we verified the properties of closure, identity and inverse for the multiplication operation * on the set {-1, 1}. We asserted that this operation was associative, so that this set with this operation forms a group. It would fairly time-consuming to prove that * is associative on {-1, 1}. So you are asked here to just list the possible combinations of a, b, c, then verify associativity for any three of them. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (1*1)*1=1*(1*1) 1*1=1*1 1=1 (-1*-1)*-1=-1*(-1*-1) -1*-1=-1*-1 -1=-1 (-1* 1)* 1=1*(-1*1)*1 -1*1=-1*1 -1=-1 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):3 ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q010. Let the & operation be defined as follows: x & y is the remainder obtained when x is multiplied by y, then divided by 5. The table below is for the & operation on the set {1, 2, 3, 4}. The table for this operation is shown below: & 1 2 3 4 1 1 2 3 4 2 2 4 1 3 3 3 1 4 2 4 4 3 2 1 Show whether (a & b) & c = a & (b & c) for the following combinations (a, b, c): (1, 2, 3), (2, 3, 4), (4, 3, 4). If all three of these combinations check out, you may assume that the operation is associative. Then check whether the operation is closed, whether it is commutative, and whether it has the inverse and identity properties. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I think it has all of these properties. 1 is the identity and is in every row in the table, so this indicates the inverse property as well. The diaganol symmetry is an indication of commutativety, and it is closed because the far left column contains the same elemants as the results. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q011. Multiplication of real numbers is associative. Determine whether multiplication is a group on the set of positive rational numbers. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1 is the identity of multiplication. It is associative as already stated in the question,it is closed, when multipying two rational positive numbers you will get a positive rational number. Integers on multiplication lack the inverse property so this is not a group.