#$&* course MTH 151 10:34, 4/8/14 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: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I’m finally feeling like I’m beginning to understand how the remainders work. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `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, with the values of x and y restricted to the set {2, 3, 4}. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2 @ 2 = 2 * 2 = 4 * 2 = 8 / 3 = 2 remainder 2, because 3 * 2 = 6. 2 @ 2 = 2. 2 @ 3 = 2 * 3 = 6 * 2 = 12 / 3 = 4 remainder 0. 2 @ 3 = 0. 2 @ 4 = 2 * 4 = 8 * 2 = 16 / 3 = 5 remainder 1, because 3 * 5 = 15. 2 @ 4 = 1. 3 @ 3 = 3 * 3 = 9 * 2 = 18 / 3 = 6 remainder 0. 3 @ 3 = 0. 3 @ 4 = 3 * 4 = 12 * 2 = 24 / 3 = 8 remainder 0. 3 @ 4 = 0. 4 @ 4 = 4 * 4 = 16 * 2 = 32 / 3 = 10 remainder 2, because 10 * 3= 30. 4 @ 4 = 2. Values: 2 3 4 Results (below) 2 2 0 1 3 0 0 0 4 1 0 2 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Still not quite sure that my table is correct. ------------------------------------------------ Self-critique Rating: 3
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Given Solution: 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}. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q004. Are the results of the operation x @ y on the set {2, 3, 4} all members of the set {2, 3, 4}? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No, because there are different numbers for both. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The possible results of the operation, whose table is @ 2 3 4 2 1 0 2 3 0 0 0 4 2 0 1 are seen from the table to be 0, 1 and 2. Of these possible results, only 2 is a member of the set {2,3,4}. So it is not the case that all the results come from the set {2, 3, 4}. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q005. Since the operation x @ y on the set {2, 3, 4} can have results which are not members of the set, we say that the @operation is not closed on the set {2, 3, 4}. • Is the @ operation closed on the set S = {0, 1, 2}? • Is the @ operation closed on the set T = {0, 2}? • Is the @ operation closed on the set R = {1, 2}? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: All three sets would work, because the results from {2 3 4} all include the numbers 0, 1, and 2. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q006. How can we tell by looking at the table whether the operation is closed? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If the numbers in the table all match the results of the set. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If all the numbers in the table come from the far left-hand column of the table--e.g., the column underneath the @ in the tables given above, which lists all the members of the set being operated on --then all the results of the operation are in that set and the operation is therefore closed. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q007. When calculating x @ y for two numbers x and y, does it make a difference whether we calculate x @ y or y @ x? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It would not, because either way, the sum would still be the same. For example, 2 * 3 and 3 * 2 both have the same answer, even if they are arranged differently. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. We conclude that for this operation x @ y must always equal y @ x. This property of the operation is called the commutative property. Speaking roughly, an operation is commutative if order doesn't matter. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q008. Does the operation of subtraction of whole numbers have the commutative property? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No, because the order of the numbers can make a difference in the sum. For example, 5 - 1 = 4, but 1 - 5 = -4. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Subtraction of whole numbers does not have the commutative property, because it is not true for most whole numbers x and y that x - y = y - x. For example, 5 - 3 = 2 while 3 - 5 = -2. So for subtraction order usually does matter, and the operation is not commutative. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q009. Is the operation of subtraction closed on the set of whole numbers? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It would depend on what the whole numbers were, and as long as they could be subtracted without getting a negative number. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q010. Is the operation of addition closed and commutative on the set of whole numbers? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I would assume that it would be closed, as long as all whole numbers are used. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: When we add two numbers x and y it doesn't matter which one we add to which--it doesn't matter whether we do x + y or y + x--so order doesn't matter and we can say that the operation is commutative. And if we add two whole numbers, which must both be at least 0, we get a whole number which is at least 0. So the operation is also closed. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q011. When we multiply a number by 1, what must be our result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When any number is multiplied by 1, the answer is the same as the number being multiplied. For example, 3 * 1 = 3, 4 * 1 = 4, and so on. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `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}? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I’m still not understanding identities completely, but to me it seems like both sets would have identities of 0 and 2. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The table for @ on {0, 1, 2) is @ 0 2 1 0 0 0 0 1 0 2 1 2 0 1 2. We see from the row across from 2 that 2 * 0 = 0, 2 * 1 = 1 and 2 * 2 = 2. We also see from the column beneath 2 that 0 * 2 = 0, 1 * 2 = 1 and 2 * 2 = 2. Thus, no matter how we combine 2 with other numbers in the set {0, 1, 2}, we don't change those other numbers. That is, for any x in the set, 2 @ x = x @ 2 = x. Therefore @ does indeed have identity 2 on the set {0, 1, 2}. On the set {0, 1} the number 2 isn't included. Since 0 * 1 = 0, not 1, the number 0 can't be the identity. Since 1 * 1 = 2, not 1, the number 1 can't be the identity. Both 0 and 1 at least sometimes change the number they operate with, and the identity can't do this. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q013. Does the set of whole numbers on the operation of addition have an identity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It would depend on which whole numbers are being added. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. ********************************************* Question: `q014. 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 is partially filled in. & 1 2 3 4 1 1 * * * 2 * 4 * * 3 * * 4 * 4 * * * 1 We note that if 4 is combined with 4 the result is 1. • What number, if any, would be combined with 2 to obtain the result 1? • What number, if any, would be combined with 3 to obtain the result 1? • Give the complete table. • In what ways would the nature of the table change if the operation was applied to the set {1, 2, 3, 4, 5}? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3 could be combined with 2 to result in 1, 2 can be combined with 3 to result in 1. & 1 2 3 4 1 1 * * 2 4 1 * 3 1 4 4 * * 1
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Given Solution: 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. ********************************************* Question: `q014. 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 is partially filled in. & 1 2 3 4 1 1 * * * 2 * 4 * * 3 * * 4 * 4 * * * 1 We note that if 4 is combined with 4 the result is 1. • What number, if any, would be combined with 2 to obtain the result 1? • What number, if any, would be combined with 3 to obtain the result 1? • Give the complete table. • In what ways would the nature of the table change if the operation was applied to the set {1, 2, 3, 4, 5}? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3 could be combined with 2 to result in 1, 2 can be combined with 3 to result in 1. & 1 2 3 4 1 1 * * 2 4 1 * 3 1 4 4 * * 1