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鲛woն鎐t Student Name: assignment #020
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ƭ✅܅L Student Name: assignment #021
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15:12:10 `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 --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.
Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. 2@5 = 2*5 = 10 *2 = 20/3 = 6 with R 2, so 2@5 = 2 3@8 = 3*8=24 * 2 = 48/3=12 R 0, 3@8=0 7@13 = 7*13=91*2=182/3 = 60 R 2 7@13 = 2 I worked out all of the set, 2@2 = 2, 2@3 = 0, 2@4=1. they all had the same answers as in your table, so we have: @ 2 3 4 2 2 0 1 3 0 0 0 4 1 0 2.................................................
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15:12:24 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 --> right
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15:14:25 `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 --> They come from working out the operation, x @ y. Using the set {2,3,4} you would work out, 2@2, 2@3, 2@4, and so on. Therefore the results are taken from the same set.
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15:15:25 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
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15:16:25 `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 --> No, because you have 0 and 1 as results, which aren't in the set {2,3,4}.
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15:16:35 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 --> ok
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15:17:48 `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 --> The operation isn't closed on any of those sets because they all have numbers or members that aren't in the original set {2,3,4}.
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15:19:06 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 --> I thought you were asking in regard to the original set, (1,2,3}
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15:20:20 `q006. How can we tell by looking at the table whether the operation is closed?
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RESPONSE --> The elements of the body of the table must be the elements that make up the set.
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15:20:47 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
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15:22:49 `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 --> If the operation is x@y as given before, when the answer is the remainder of x*y *2 / 3, then it doesn't matter how it's calcluated.
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15:23:07 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 --> ok
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15:24:02 `q008. Does the operation of subtraction of whole numbers have the commutative property?
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RESPONSE --> No, 4 - 7 = -3 but 7 - 4 = 3, two different numbers.
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15:24:14 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
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15:24:35 `q009. Is the operation of subtraction closed on the set of whole numbers?
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RESPONSE --> yes
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15:25:04 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 --> I see, I wasn't thinking about negatives.
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15:26:07 `q010. Is the operation of addition closed and commutative on the set of whole numbers?
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RESPONSE --> Yes, it is since it is impossible to add two whole numbers together and get a negative number. It also doesn't matter the order in which you add two numbers: 5+3 = 8 just as 3+5 = 8.
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15:26:18 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 --> ok
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15:26:41 `q011. When we multiply a number by 1, what must be our result?
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RESPONSE --> the number we are multiplying 1 with.
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15:26:54 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
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15:32:13 `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, 1}?
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RESPONSE --> 2 is the identity element in { 0,1,2} 1 is the identity in (0,1}.
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15:33:01 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 --> ok
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15:33:27 `q013. Does the set of whole numbers on the operation of addition have an identity?
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RESPONSE --> 0, since any number plus 0 ='s that number.
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15:33:37 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
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