#$&* course Mth 151 7/23 11 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
.............................................
Given Solution: The table for @ on {1, 2} is @ 1 2 1 2 1 2 1 2 The table shows us that all the possible results of x @ y on the set {1, 2} are in the set, so the operations is closed. We see that 2 when combined with anything does not change that number number, so 2 is the identity for this operation on this set. We see also that 1 @ 2 = 2 @ 1 = 1 so 1 and 2 are inverses. Therefore both of the elements in the set {1, 2} have inverses in the set. Since it has already been stated that the set has the associative property, we conclude that the set is a group. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q002. Which of the properties closure, identity, commutativity, inverse, does the standard addition operation + have on the set {-1, 0, 1}? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This set has the properties of identity, commutative, and inverse confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ 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: This set has the property of identity because in the operation of multiplication -1*1 = -1 and also the inverse -1*-1=1 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The table for this operation is * -1 0 1 -1 1 0 -1 0 0 0 0 1 -1 0 1 We see that every possible result of the operation is in the set {-1, 0, 1}, and the row across from 1 and the column beneath 1 show how us that 1 is the identity. -1 * -1 = 1 so -1 is its own inverse, and 1 * 1 = 1 so 1 is its own inverse. However, anything 0 is combine with gives us 0, so 0 cannot by combined with anything to get 1 and 0 therefore has no inverse. STUDENT COMMENT I see now how the -1*-1 = 1 is an inverse. I was only thinking about the 0 and how it never gives an inverse. INSTRUCTOR RESPONSE -1 does have an inverse. However since 0 doesn't have an inverse, the operation doesn't have the inverse property, which would imply that all elements of the set have inverses. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q004. Does the operation * of standard multiplication on the set {-1, 1} have the properties of closure, identity and inverse? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This set has the property of closure because only the two numbers in the set are found, the property of identity because -1*1=-1, and closure. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): Did not know that -1 and 1 had inverses but after looking at the solution I kind of understand how they are inverses ------------------------------------------------ Self-critique Rating:2 ********************************************* 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: The operation of standard multiplication is a group on the set because it contain all of the properties of order of operation. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Has seen in the preceding example the operation is closed and has the identity and inverse properties. Given that it is sensitive, it is therefore a group. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q006. We've referred to the property of associativity, but we haven't yet defined it. Associativity essentially means that when an operation (technically a binary operation, but don't worry about that a terminology at this point) is performed on three elements of a set, for example a + b + c, it doesn't matter whether we first perform a + b then add c, calculating (a + b) + c, or group the b and c so we calculate a + (b + c). If + means addition on real numbers, show that (3 + 4) + 5 = 3 + ( 4 + 5). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (3+4)+5=7 +5=12 3+(4+5)= 3+9= 12 The associative property works confidence rating #$&*: 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, and everyone in this course has used it for years. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `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). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 0 @2=0 (2@0)@1=0 @1=0 0@ 1= 0 2@(0@1)=2 @ 0= 0 Which mean 2@(0@1) = (2@0) @ 1 2 @1 = 1 (2 @ 1) @ 1= 1 @1= 1 1 @1 = 2 2@(1@1)= 2@2=2 meaning, (2@1)@1=2 @(1@1) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 0 @ 1 = 0, so 2 @ ( 0 @ 1) = 2 @ 0 = 0. 2 @ 0 = 0 so (2 @ 0 ) @ 1 = 0 @ 1 = 0. Thus 2 @ (0 @ 1) = ( 2 @ 0 ) @ 1. (2 @ 1) = 1 so (2 @ 1) @ 1 = 1 @ 1 = 1. 1 @ 1 = 2 so 2 @ (1 @ 1) = 2 @ 2 = 2. Thus (2 @ 1) @ 1 = 2 @ ( 1 @ 1). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q008. Does the result of the preceding exercise prove that the @ operation is associative on the set {0, 1, 2}? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes the associative property worked in all examples, therefore the @ operation is associative on the set {0, 1, 2}. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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) 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I was really unsure of the question but I now see that it would take a lot more work to come to the conclusion that @ operation is associative on the set {0, 1, 2}. ------------------------------------------------ Self-critique Rating:
.............................................
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):OK ------------------------------------------------ Self-critique Rating:OK