#$&* course mth151 7:15pm
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Given Solution: By the definition, 2 @ 5 is the remainder when the product 2 * 5 is doubled then divided by 3. We start with 2 * 5 = 10, then double that to get 20, then divide by three. 20 / 3 = 6 with remainder 2. So 2 @ 5 = 2. We follow the same procedure to find 3 @ 8. We get 3 * 8 = 24, then double that to get 48, then divide by three. 48 / 3 = 12 with remainder 0. So 3 @ 8 = 0. Following the same procedure to find 7 @ `2, we get 7 * 13 = 91, then double that to get 182, then divide by three. 182 / 3 = 60 with remainder 2. So 7 @ 13 = 2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. If we define the @ operation from the previous exercise just on the set {5, 6, 7} , we can use the same process as in the preceding solution to get 5 @ 5 = 2, 5 @ 6 = 0, 5 @ 7 = 1, 6 @ 5 = 0, 6 @ 6 = 0, 6 @ 7 = 0, 7 @ 5 = 1, 7 @ 6 = 0 and 7 @ 7 = 2. We can put these results in a table as follows: @ 5 6 7 5 2 0 1 6 0 0 0 7 1 0 2. Make a table for the @ operation, with the values of x and y restricted to the set {2, 3, 4}. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: @ 2 3 4 2 2 0 1 3 0 0 0 4 1 0 2 confidence rating #$&*: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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: {0, 1, 2} confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The results of the operation x @ y, which ultimately consist of the remainder when some number is divided by 3, must all be division-by-3 remainders. The only possible remainders we can have when dividing by three are 0, 1 or 2. Thus all the results of the operation x @ y are members of the set {0, 1, 2}. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. Are the results of the operation x @ y on the set {2, 3, 4} all members of the set {2, 3, 4}? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: @ 2 3 4 2 1 0 2 3 0 0 0 4 2 0 1 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* 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}? Your solution 2 @ 2 = 2 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: When operating on the set S = {0, 1, 2} the operation must as always give one of the division-by-3 remainders 0, 1 or 2. Thus the results are all in the set S. When operating on the set T = {0, 2} the possible results are 0 @ 0 = 0 @ 2 = 2 @ 0 = 0, or 2 @ 2 = 2. Since the possible results are 0 and 2, both of which are in T, all the results come from the set T on which we are operating, and the operation is closed on the set T. When operating on the set R = {1, 2} the possible results are 1 @ 1 = 2, 1 @ 2 = 2 @ 1 = 1, or 2 @ 2 = 2. Since the possible results are 1 and 2 and both are in the set R, we can say that all the results come from the set R on which we are operating, and the operation is indeed closed on the set R. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. How can we tell by looking at the table whether the operation is closed? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: come from left hand side so its closed confidence rating #$&*: 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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: yes bc its communitive. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since the first step in calculating x @ y is to multiply x * y, it would make no difference whether we multiplied x * y or y * x. So in the first step it makes no difference whether we calculate x @ y or y @ x. Since all we do after that is double our result and calculate the remainder when dividing by 3, the order of x and y won't make a difference there either. We conclude that for this operation x @ y must always equal y @ x. This property of the operation is called the commutative property. Speaking roughly, an operation is commutative if order doesn't matter. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q008. Does the operation of subtraction of whole numbers have the commutative property? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: no because its - confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q009. Is the operation of subtraction closed on the set of whole numbers? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: not closed because you getting an answer. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Whole numbers are the numbers in the set {0, 1, 2, 3, ... }. If we subtract a smaller number from a larger, we will again get a whole number. However if we subtract a smaller number from the larger, we will get a negative result, which is not a whole number. Since it is possible to subtract two numbers in the set and to get a result that is not in the set, the operation is not closed. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q010. Is the operation of addition closed and commutative on the set of whole numbers? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: to at least get 0 so closed. confidence rating #$&*: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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q011. When we multiply a number by 1, what must be our result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: it doesn't change the number confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Any number multiplied by 1 will give us the same number. Any number is unchanged if we multiply it by 1. That is 10 * 1 = 10, or -37.27 * 1 = -37.27, or 72 * 1 = 72. Multiplication by 1 does not change any number. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q012. A number which does not change any number with which it is combined using a certain operation is called the identity for the operation. As we saw in the preceding exercise, the number 1 is the identity for the operation of multiplication on real numbers. Does the operation @ (which was defined in preceding exercises by x @ y = remainder when x * y is doubled and divided by 3) have an identity on the set {0, 1, 2}? Does @ have an identity on the set {0, 2}? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 0, 1, 2 0, 1 confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q013. Does the set of whole numbers on the operation of addition have an identity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a number that doesn't chaNGE confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The identity must be a number that doesn't change the number with which it is combined. The number 0 has this property. Whenever we add to 0 is what we get. 0 doesn't change the number it is combined with under the operation of addition. For any x, 0 + x = x and x + 0 = x. Therefore we can say that 0 is the identity for addition on the set of whole numbers. ********************************************* Question: `q014. Let the & operation be defined as follows: x & y is the remainder obtained when x is multiplied by y, then divided by 5. The table below is for the & operation on the set {1, 2, 3, 4}. The table is partially filled in. & 1 2 3 4 1 1 * * * 2 * 4 * * 3 * * 4 * 4 * * * 1 We note that if 4 is combined with 4 the result is 1. • What number, if any, would be combined with 2 to obtain the result 1? • What number, if any, would be combined with 3 to obtain the result 1? • Give the complete table. • In what ways would the nature of the table change if the operation was applied to the set {1, 2, 3, 4, 5}? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: yes it can change. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q015. The operation of multiplication on the whole numbers has identity 1. Is there a whole number which when multiplied by 2 gives us the identity? Can you two different whole numbers which when multiplied gives us the identity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: yes there is prime confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course mth151 7:32 021. `query 21
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q4.5.24 a, b, c values that show that a + (b * c) not equal to (a+b) * (a+c). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (a+b) * (a+c) = (2+5) + (2+7 7 + 12 = 19 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** For example if a = 2, b = 5 and c = 7 we have a + (b + c) = 2 + (5 + 7) = 2 + 12 = 14 but (a+b) * (a+c) = (2+5) + (2+7) = 7 + 12 = 19. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q4.5.33 venn diagrams to show that union distributes over intersection YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A U (B ^ C) A U (B ^ C) confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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). ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ________________________________________ #$&*