course math 151 é]Ó›¸ÔÀyƒÉÓîí½»îO|Á›â¹º’˜assignment #021
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09:04:54 By the definition, 2 @ 5 is the remainder when the product 2 * 5 is doubled then divided by 3. We start with 2 * 5 = 10, then double that to get 20, then divide by three. 20 / 3 = 6 with remainder 2. So 2 @ 5 = 2. We follow the same procedure to find 3 @ 8. We get 3 * 8 = 24, then double that to get 48, then divide by three. 48 / 3 = 12 with remainder 0. So 3 @ 8 = 0. Following the same procedure to find 7 @ `2, we get 7 * 13 = 91, then double that to get 182, then divide by three. 182 / 3 = 60 with remainder 2. So 7 @ 13 = 2.
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RESPONSE --> so we multiply, double the product then divide by three self critique assessment: 3
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09:19:06 `q002. If we define the @ operation from the previous exercise just on the set {5, 6, 7} , we can use the same process as in the preceding solution to get 5 @ 5 = 2, 5 @ 6 = 0, 5 @ 7 = 1, 6 @ 5 = 0, 6 @ 6 = 0, 6 @ 7 = 0, 7 @ 5 = 1, 7 @ 6 = 0 and 7 @ 7 = 2. We can put these results in a table as follows: @ 5 6 7 5 2 0 1 6 0 0 0 7 1 0 2. Make a table for the @ operation restricting x and y to the set {2, 3, 4}.
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RESPONSE --> {2,3,4} 2@2=2 2@3=4 2@4=5 3@2=4 3@3=6 3@4=8 4@2=5 4@3=8 4@4=10 table: @ 2 3 3 2 2 4 5 3 4 6 8 4 5 8 10 confidence assessment: 3
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09:28:29 Using the same process as in the solution to the preceding problem we find that 2 @ 2 = 2, 2 @ 3 = 0, 2 @ 4 = 1, 3 @ 2 = 3 @ 3 = 3 @ 4 = 0, 4 @ 2 = 1, 4 @ 3 = 0 and 4 @ 4 = 2. The table is therefore @ 2 3 4 2 2 0 1 3 0 0 0 4 1 0 2.
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RESPONSE --> I'm confused now! this is what I did: 2*2=4 double =8/3= 2 2*3=6 double =12/3=4 2*4=8 double=16/3=5 3*2=6 double=12/3=4 3*3=9 double=18/3=6 3*4=12 double=24/3=8 4*2=8 double= 16/3 = 5 4*3=12 double= 24/3 = 8 4*4 = 16 double = 32/3 = 10 self critique assessment: 3
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09:35:41 `q003. All the x and y values for the table in the preceding problem came from the set {2, 3, 4}. From what set are the results x @ y taken?
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RESPONSE --> from associating all the numbers in set confidence assessment: 2
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09:47:50 The results of the operation x @ y, which ultimately consist of the remainder when some number is divided by 3, must all be division-by-3 remainders. The only possible remainders we can have when dividing by three are 0, 1 or 2. Thus all the results of the operation x @ y are members of the set {0, 1, 2}.
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RESPONSE --> ok self critique assessment: 3
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09:48:43 `q004. Are the results of the operation x @ y on the set {2, 3, 4} all members of the set {2, 3, 4}?
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RESPONSE --> yes, they are confidence assessment: 3
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10:45:10 `q005. Since the operation x @ y on the set {2, 3, 4} can result in at least some numbers which are not members of the set, we say that the @operation is not closed on the set {2, 3, 4}. Is the @ operation closed on the set S = {0, 1, 2}? Is the @ operation closed on the set T = {0, 2}? Is the @ operation closed on the set R = {1, 2}?
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RESPONSE --> S= not closed T= closed R= not closed confidence assessment: 2
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10:48:39 When operating on the set S = {0, 1, 2} the operation must as always give one of the division-by-3 remainders 0, 1 or 2. Thus the results are all in the set S. When operating on the set T = {0, 2} the possible results are 0 @ 0 = 0 @ 2 = 2 @ 0 = 0, or 2 @ 2 = 2. Since the possible results are 0 and 2, both of which are in T, all the results come from the set T on which we are operating, and the operation is closed on the set T. When operating on the set R = {1, 2} the possible results are 1 @ 1 = 2, 1 @ 2 = 2 @ 1 = 1, or 2 @ 2 = 2. Since the possible results are 1 and 2 and both are in the set R, we can say that all the results come from the set R on which we are operating, and the operation is indeed closed on the set R.
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RESPONSE --> when the result are found in the set the operation is closed self critique assessment: 3
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10:50:04 `q006. How can we tell by looking at the table whether the operation is closed?
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RESPONSE --> because the answer on the body of table are all the elements of the set confidence assessment: 3
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10:51:47 If all the numbers in the table come from the far left-hand column of the table--e.g., the column underneath the @ in the tables given above, which lists all the members of the set being operated on --then all the results of the operation are in that set and the operation is therefore closed.
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RESPONSE --> ok self critique assessment: 3
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10:51:52 `q007. When calculating x @ y for two numbers x and y, does it make a difference whether we calculate x @ y or y @ x?
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RESPONSE --> confidence assessment:
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10:54:44 Since the first step in calculating x @ y is to multiply x * y, it would make no difference whether we multiplied x * y or y * x. So in the first step it makes no difference whether we calculate x @ y or y @ x. Since all we do after that is double our result and calculate the remainder when dividing by 3, the order of x and y won't make a difference there either. So we conclude that for this operation x @ y must always equal y @ x. This property of the operation is called the commutative property, meaning roughly that order doesn't matter.
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RESPONSE --> commutative property allows us to change the order of the elements and still get the correct answer self critique assessment: 3
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10:57:36 `q008. Does the operation of subtraction of whole numbers have the commutative property?
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RESPONSE --> no because it would give us negative numbers in some cases confidence assessment: 3
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10:58:02 Subtraction of whole numbers does not have the commutative property, because it is not true for most whole numbers x and y that x - y = y - x. For example, 5 - 3 = 2 while 3 - 5 = -2. So for subtraction order usually does matter, and the operation is not commutative.
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RESPONSE --> ok self critique assessment: 3
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10:59:29 `q009. Is the operation of subtraction closed on the set of whole numbers?
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RESPONSE --> yes it is closed confidence assessment: 2
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11:01:13 Whole numbers are the numbers in the set {0, 1, 2, 3, ... }. If we subtract a smaller number from a larger, we will again get a whole number. However if we subtract a smaller number from the larger, we will get a negative result, which is not a whole number. Since it is possible to subtract two numbers in the set and to get a result that is not in the set, the operation is not closed.
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RESPONSE --> because we could get a negative number by substracting a smaller form a larger number we could get a different result not found in the set. not closed self critique assessment: 3
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11:01:36 `q010. Is the operation of addition closed and commutative on the set of whole numbers?
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RESPONSE --> yes it is closed confidence assessment: 2
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11:03:15 When we add two numbers x and y it doesn't matter which one we add to which--it doesn't matter whether we do x + y or y + x--so order doesn't matter and we can say that the operation is commutative. And if we add two whole numbers, which must both be at least 0, we get a whole number which is at least 0. So the operation is also closed.
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RESPONSE --> because it doesn't matter in what order we add, we will get a number which will be in the set self critique assessment: 3
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11:04:07 `q011. When we multiply a number by 1, what must be our result?
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RESPONSE --> it will be the same number 1 was multiplied by confidence assessment: 3
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11:04:21 Any number multiplied by 1 will give us the same number. Any number is unchanged if we multiply it by 1. That is 10 * 1 = 10, or -37.27 * 1 = -37.27, or 72 * 1 = 72. Multiplication by 1 does not change any number.
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RESPONSE --> ok self critique assessment: 3
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11:07:57 `q012. A number which does not change any number with which it is combined using a certain operation is called the identity for the operation. As we saw in the preceding exercise, the number 1 is the identity for the operation of multiplication on real numbers. Does the operation @ (which was defined in preceding exercises by x @ y = remainder when x * y is doubled and divided by 3) have an identity on the set {0, 1, 2}? Does @ have an identity on the set {0, 2}?
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RESPONSE --> on set {0,1,2} is 0 on set {0,2} is 0 also confidence assessment: 3
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11:11:37 The table for @ on {0, 1, 2) is @ 0 2 1 0 0 0 0 1 0 2 1 2 0 1 2. We see from the row across from 2 that 2 * 0 = 0, 2 * 1 = 1 and 2 * 2 = 2. We also see from the column beneath 2 that 0 * 2 = 0, 1 * 2 = 1 and 2 * 2 = 2. Thus, no matter how we combine 2 with other numbers in the set {0, 1, 2}, we don't change those other numbers. That is, for any x in the set, 2 @ x = x @ 2 = x. Therefore @ does indeed have identity 2 on the set {0, 1, 2}. On the set {0, 1} the number 2 isn't included. Since 0 * 1 = 0, not 1, the number 0 can't be the identity. Since 1 * 1 = 2, not 1, the number 1 can't be the identity. Both 0 and 1 at least sometimes change the number they operate with, and the identity can't do this.
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RESPONSE --> so the first set identity number is 2 and the second set does not have an identity number because 0,1 can changet the number they operate self critique assessment: 3
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11:12:09 `q013. Does the set of whole numbers on the operation of addition have an identity?
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RESPONSE --> yes it can confidence assessment: 3
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11:12:58 The identity must be a number that doesn't change the number with which it is combined. The number 0 has this property. Whenever we add to 0 is what we get. 0 doesn't change the number it is combined with under the operation of addition. For any x, 0 + x = x and x + 0 = x. Therefore we can say that 0 is the identity for addition on the set of whole numbers.
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RESPONSE --> ok self critique assessment: 3