course Mth 151 Üá¥àÑWʨ֋뫵€ˆ¼Jð¹UÊÔ•ýßÝÚ„h×ÿassignment #021
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21:08:29 `q001. If we define the operation @ on two numbers x and y by x @ y = remainder when the product x * y is multiplied by 2 then divided by 3, then find the following: 2 @ 5, 3 @ 8, 7 @ 13.
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RESPONSE --> 2@5 = 6.666666667 3@8 = 16 7@13= 60.6666666667 confidence assessment: 2
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21:09:29 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 --> Ok I see now self critique assessment: 2
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21:20:00 `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 0 1 3 0 0 0 4 1 0 2 confidence assessment: 2
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21:20:10 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 --> self critique assessment: 2
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21:20:59 `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 each set? confidence assessment: 2
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21:21:27 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: 2
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21:22:00 `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 confidence assessment: 2
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21:22:17 The possible results of the operation, whose table is @ 2 3 4 2 1 0 2 3 0 0 0 4 2 0 1 , are seen from the table to be 0, 1 and 2. Of these possible results, only 2 is a member of the set {2,3,4}. So it is not the case that all the results come from the set {2, 3, 4}.
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RESPONSE --> self critique assessment: 2
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21:23:19 `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 --> no, no, yes confidence assessment: 2
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21:23:25 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 --> self critique assessment:
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21:24:16 `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 --> no it doesn't confidence assessment: 2
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21:24:28 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 --> self critique assessment: 3
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21:25:53 `q008. Does the operation of subtraction of whole numbers have the commutative property?
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RESPONSE --> no it does not confidence assessment: 2
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21:26:20 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 --> self critique assessment: 0
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21:26:40 `q009. Is the operation of subtraction closed on the set of whole numbers?
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RESPONSE --> yes confidence assessment: 2
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21:27:05 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 --> self critique assessment:
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21:29:13 `q010. Is the operation of addition closed and commutative on the set of whole numbers?
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RESPONSE --> yes confidence assessment: 2
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21:29:25 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 --> self critique assessment: 0
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21:29:45 `q011. When we multiply a number by 1, what must be our result?
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RESPONSE --> the number you multiplied by confidence assessment: 2
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21:29:56 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 --> self critique assessment: 2
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21:31:11 `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 --> yes and yes confidence assessment: 0
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21:31:34 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 --> self critique assessment: 2
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21:31:58 `q013. Does the set of whole numbers on the operation of addition have an identity?
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RESPONSE --> yes confidence assessment: 0
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21:40:53 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 --> self critique assessment:
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course Mth 151 ¦FÜÆ҉麰ÝätÒÚ}‘اÁ”ÕN¤¿Üassignment #021
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13:07:24 4.4.6 star operation [ [1, 3, 5, 7], [3, 1, 7, 5], [5, 7, 1, 3], [7, 5, 3, 1]]
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RESPONSE --> I'm sorry but I'm not getting what a star operationis. confidence assessment: 0
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13:08:30 ** Using * to represent the operation the table is * 1 3 5 7 1 1 3 5 7 3 3 1 7 5 5 5 7 1 3 7 7 5 3 1 the operation is closed, since all the results of the operation are from the original set {1,3,5,7} the operation has an identity, which is 1, because when combined with any number 1 doesn't change that number. We can see this in the table because the row corresponding to 1 just repeats the numbers 1,3,5,7, as does the column beneath 1. The operation is commutative--order doesn't matter because the table is symmetric about the main diagonal.. the operation has the inverse property because every number can be combined with another number to get the identity 1: 1 * 1 = 1 so 1 is its own inverse; 3 * 3 = 1 so 3 is its own inverse; 5 * 5 = 1 so 5 is its own inverse; 7 * 7 = 1 so 7 is its own inverse. This property can be seen from the table because the identity 1 appears exactly once in every row. the operation appears associative, which means that any a, b, c we have (a * b ) * c = a * ( b * c). We would have to check this for every possible combination of a, b, c but, for example, we have (1 *3) *5=3*5=7 and 1*(3*5)=1*7=7, so at least for a = 1, b = 3 and c = 5 the associative property seems to hold. **
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RESPONSE --> ok the * operation, I'm sorry now I know what you meant self critique assessment: 0
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13:12:34 4.4.24 a, b, c values that show that a + (b * c) not equal to (a+b) * (a+c).
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RESPONSE --> I'm not for sure what you are asking for here, in my book 4.4.24 the question is about a military clock, 1400+500 would be 1900 (mod 24) confidence assessment: 0
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13:12:49 ** For example if a = 2, b = 5 and c = 7 we have a + (b + c) = 2 + (5 + 7) = 2 + 12 = 14 but (a+b) * (a+c) = (2+5) + (2+7) = 7 + 12 = 19. **
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RESPONSE --> self critique assessment:
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13:13:18 4.4.33 venn diagrams to show that union distributes over intersection
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RESPONSE --> confidence assessment:
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13:13:41 ** For A U (B ^ C) we would shade all of A in addition to the part of B that overlaps C, while for (A U B) ^ (A U C) we would first shade all of A and B, then all of A and C, and our set would be described by the overlap between these two shadings. We would thus have all of A, plus the overlap between B and C. Thus the result would be the same as for A U (B ^ C). **
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RESPONSE --> where is this coming from? self critique assessment: 0
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¸¦ÍåÑTó^ßÓv½z¦ÚŽ‚zÑ…Ÿ assignment #022 022. `query 22 College Algebra 11-10-2008
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14:35:23 4.5.9 {-1,0,1} group on multiplication?
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RESPONSE --> * -1 0 1 -1 1 0 -1 0 0 0 0 1 -1 0 1 confidence assessment: 0
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14:51:06 ** There are four criteria for the group: closure, identity, inverse property, and associativity. The lack of any one of these properties means that the set and operation do not form a group. The set is closed on multiplication. The identity is the element that when multiplied by other elements does not change them. The identity for this operation is 1, since 1 * -1 = -1, 1 * 0 = 0 and 1 * 1 = 1. Inverses are pairs of elements that give you 1 when you multiply them. For example -1 * -1 = 1 so -1 is its own inverse. 1 * 1 = 1 so 1 is also its own inverse. However, 0 does not have an inverse because there is nothing you can multiply by 0 to get 1. Since there is an element without an inverse this is not a group. **
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RESPONSE --> self critique assessment: 0
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