course Mth 151 ??·????·z????c????assignment #026
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07:22:43 `q001. There are six questions in this assignment. We defined an operation as follows: x * y (mod 4) = remainder when x * y is divided by 4. Find 3 * 9 (mod 4); 7 * 12 (mod 4) and 11 * 13 (mod 4).
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RESPONSE --> 3*9 (mod 4) (3*9) = 3 (mod 4) 7*12(mod 4) (7*12) = 0 (mod 4) 11*13 (mod4) (11*13) = 3 (mod 4) confidence assessment: 2
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07:23:39 3 * 9 (mod 4) is the remainder when 3 * 9 is divided by 4. Since 3 * 9 = 27 and 27 / 4 leaves remainder 3, we see that 3 * 9 (mod 4) = 3. 7 * 12 (mod 4) is the remainder when 7 * 12 is divided by 4. Since 7 * 12 = 84 and 84 / 4 leaves remainder 0, we see that 7 * 12 (mod 4) = 0. 11 * 13 (mod 4) is the remainder when 11 * 13 is divided by 4. Since 11 * 13 = 143 and 143 / 4 leaves remainder 3,we see that 11 * 13 (mod 4) = 3.
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RESPONSE --> Is the way I answered okay? I didn't just give the remainder. self critique assessment: 2
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07:33:34 `q002. Make a table for the x * y mod 4 operation, which we will call '* mod 4', operating on the set {0, 1, 2, 3}. Determine which of the properties, including commutativity, associative, identity, inverse and closure properties, are properties of this operation.
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RESPONSE --> * 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1 It has closure property, commutativity, associative, indentity and inverse confidence assessment: 2
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07:41:24 Whatever x is, 0 * x = x * 0 = 0, which when divided by 4 leaves remainder 0. Whatever x is, 1 * x = x * 1 = x, and if x is in the set {0, 1, 2, 3} we have get remainder x when dividing by 4 (e.g., 4 divides into 0, 1, 2 or 3 zero times, leaving that number as the remainder) and x mod 4 = x. From this we can see that 1 is the identity for this operation. Multiplying 0, 1, 2, and 3 by 2 we get 0, 2, 4, and 6, which when divided by 4 leave remainders 0, 2, 0 and 2, respectively. Multiplying 0, 1, 2, and 3 by 2 we get 0, 3, 6, and 9, which when divided by 4 leave remainders 0, 3, 2 and 1, respectively. The table for this operation is therefore * mod 4 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1 We note that this operation does contain identity 1, but since neither 0 nor 2 can be combined with any of the elements of the set to give us the identity, the operation on this set does not have the inverse property. We do see from the symmetry of the table about the main diagonal that it has the commutative property, which we could in any event have concluded from the fact that multiplication is commutative so that the product we get before calculating the remainder is independent of the order of the two numbers. In a similar matter we can reason that the operation is associative. The operation is also closed, since the remainder upon dividing by 4 must always be 0, 1, 2 or 3 and hence in the set {0, 1, 2, 3}.
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RESPONSE --> I put that it was indentity but I now see why it isn't. The top first row isn't identical to the heading table. self critique assessment: 2
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07:43:48 `q003. Repeat the preceding exercise for the operation x * y mod 5, defined to give the remainder when x * y is divided by 5, on the set {1, 2, 3, 4}. Determine which of the properties are exhibited by this operation.
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RESPONSE --> * 1 2 3 4 1 1 2 3 4 2 2 4 1 3 3 3 1 4 2 4 4 3 2 1 It has the closure, communitative, associative, indentity and inverse. confidence assessment: 2
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07:45:27 First we might wish to do a couple of example calculations to get familiar with the operation. For example: 2 * 3 mod 5 = 6, which when divided by 5 gives us remainder 1. 3 * 4 mod 5 = 12 which when divided by 5 gives us remainder 2. 2 * 4 mod 5 = 8 which when divided by 5 gives us remainder 3. The table is * mod 5 1 2 3 4 1 1 2 3 4 2 2 4 1 3 3 3 1 4 2 4 4 3 2 1 We immediately see that all the results are in the set {1, 2, 3, 4}, so that the operation is closed. This operation has identity 1, as we can see from the row and the column across from and beneath 1. We easily see from the table that the identity appears exactly once in each row and in each column, which assures us that the operation has the inverse property. Specifically we see that 1 * 1 mod 5 = 1 so that 1 is its own inverse, that 2 * 3 mod 5 = 1 so that 2 and 3 are inverses, and that 4 * 4 mod 5 = 1, so that 4 is its own inverse. The associativity and commutativity of the operation follow from the associative and commutative properties of multiplication on real numbers, as discussed in the preceding problem.
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RESPONSE --> Got the same answers. self critique assessment: 2
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07:53:12 `q004. The equation 3x + 7 = 9 (mod 5) has an integer solution for x = 0, 1, 2, 3 or 4. Which value of x is a solution to this equation?
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RESPONSE --> 3(0)+7 = 7 remainder 2 - not a solution 3(1)+7=10 remainder 0 - not a solution 3(2)+7=13 remainder 3 - not a solution 3(3)+7=16 remainder 1 - not a solution 3(4)+7=19 remainder 4 - not a solution I don't get any of them as a solution because none have a remainder of 9. ?????Since it is mod 5, 9 couldn't be the answer anyway because 5 would divide into 9 one time with a remainder of 4??? confidence assessment: 2
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07:59:00 3x + 7 = 9 (mod 4) means that 3x + 7 - 9 = 0 (mod 5) so 3x - 2 = 0 (mod 5). If 3x - 2 = 0 (mod 5) then when we divide 3x - 2 by 5 we should get remainder 0. So we substitute the different possible values for x into the expression 3x - 2 until we get a number of which when divided by 5 gives us remainder 0. If x = 0 then 3x - 2 = -2, and -2 (mod 5) = 3 (if you don't understand why -2 mod 5 = 3, think of the 5-hour clock in the text; but for now it should be obvious that -2 is not a multiple of 5 so that you cannot get remainder 0 when dividing -2 by 5). If x = 1 then 3x - 2= 1, and 1 (mod 5) = 1. If x = 2 then 3x - 2= 4, and 4 (mod 5) = 4. If x = 3 then 3x - 2= 7, and 7 (mod 5) = 2. If x = 4 then 3x - 2= 10, and 10 (mod 5) = 0. Thus x = 4 is a solution to the equation 3x + 7 = 9 (mod 5).
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RESPONSE --> I didn't work it right. I didn't subtract the 9 from both sides so I started out all wrong but I understand how to work it. Do you subtract the answer if it is higher than the (mod) to make the answer 0? self critique assessment: 2
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08:04:05 `q005. You see that x = 4 is a solution to the equation 3x + 7 = 9 (mod 5). One of the numbers x = 5, 6, 7, 8, 9 is also a solution. Which one is it?
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RESPONSE --> 3(x) - 2 = 0 3(5)-2=13 remainder 3 3(6)-2=16 remainder 1 3(7)-2=19 remainder 4 3(8)-2=22 remainder 2 3(9)-2=25 remainder 0 x = 9 confidence assessment: 2
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08:04:46 We recall that 3x + 7 = 9 (mod 5) is equivalent to 3x - 2 = 0 (mod 5). We evaluate 3x - 2 (mod 5) for x = 5, 6, 7, 8 and 9 and we find that the results are 3, 1, 4, 2, and 0. So x = 9 is our next solution. We might also note that the series of results 3, 1, 4, 2, 0 is the same as the series we got for x = 0, 1, 2, 3, 4. Our results therefore seem to indicate a repeating pattern in which the remainder 0 occurs every fifth number starting with 4. This is in fact what happens, and you might wish to think about why this happens. However, you should in a case remember that this is what happens. In general when we have an equation of the form A x + B = C (mod n), integer solutions happen at intervals of n. for some values of A, B and C integer solutions can also occur at shorter intervals, but they always do occur at intervals of n.
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RESPONSE --> I worked it right this time and got the correct answer. self critique assessment: 2
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08:13:41 `q006. What are the first five positive values of x which solve the equation 3x + 7 = 9 (mod 5) of the preceding problem?
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RESPONSE --> 4, 9, 14, 19, 24 confidence assessment: 2
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08:14:24 We just saw that x = 4 and x = 9 are solutions, and we saw that because we are solving an equation mod 5 solutions have to occur at intervals of 5. Thus the first five solutions are x = 4, 9, 14, 19 and 24.
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RESPONSE --> You add 5 to the get the next numbers. self critique assessment: 2
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course Mth 151 ??·????·z????c????assignment #026
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07:22:43 `q001. There are six questions in this assignment. We defined an operation as follows: x * y (mod 4) = remainder when x * y is divided by 4. Find 3 * 9 (mod 4); 7 * 12 (mod 4) and 11 * 13 (mod 4).
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RESPONSE --> 3*9 (mod 4) (3*9) = 3 (mod 4) 7*12(mod 4) (7*12) = 0 (mod 4) 11*13 (mod4) (11*13) = 3 (mod 4) confidence assessment: 2
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07:23:39 3 * 9 (mod 4) is the remainder when 3 * 9 is divided by 4. Since 3 * 9 = 27 and 27 / 4 leaves remainder 3, we see that 3 * 9 (mod 4) = 3. 7 * 12 (mod 4) is the remainder when 7 * 12 is divided by 4. Since 7 * 12 = 84 and 84 / 4 leaves remainder 0, we see that 7 * 12 (mod 4) = 0. 11 * 13 (mod 4) is the remainder when 11 * 13 is divided by 4. Since 11 * 13 = 143 and 143 / 4 leaves remainder 3,we see that 11 * 13 (mod 4) = 3.
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RESPONSE --> Is the way I answered okay? I didn't just give the remainder. self critique assessment: 2
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07:33:34 `q002. Make a table for the x * y mod 4 operation, which we will call '* mod 4', operating on the set {0, 1, 2, 3}. Determine which of the properties, including commutativity, associative, identity, inverse and closure properties, are properties of this operation.
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RESPONSE --> * 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1 It has closure property, commutativity, associative, indentity and inverse confidence assessment: 2
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07:41:24 Whatever x is, 0 * x = x * 0 = 0, which when divided by 4 leaves remainder 0. Whatever x is, 1 * x = x * 1 = x, and if x is in the set {0, 1, 2, 3} we have get remainder x when dividing by 4 (e.g., 4 divides into 0, 1, 2 or 3 zero times, leaving that number as the remainder) and x mod 4 = x. From this we can see that 1 is the identity for this operation. Multiplying 0, 1, 2, and 3 by 2 we get 0, 2, 4, and 6, which when divided by 4 leave remainders 0, 2, 0 and 2, respectively. Multiplying 0, 1, 2, and 3 by 2 we get 0, 3, 6, and 9, which when divided by 4 leave remainders 0, 3, 2 and 1, respectively. The table for this operation is therefore * mod 4 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1 We note that this operation does contain identity 1, but since neither 0 nor 2 can be combined with any of the elements of the set to give us the identity, the operation on this set does not have the inverse property. We do see from the symmetry of the table about the main diagonal that it has the commutative property, which we could in any event have concluded from the fact that multiplication is commutative so that the product we get before calculating the remainder is independent of the order of the two numbers. In a similar matter we can reason that the operation is associative. The operation is also closed, since the remainder upon dividing by 4 must always be 0, 1, 2 or 3 and hence in the set {0, 1, 2, 3}.
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RESPONSE --> I put that it was indentity but I now see why it isn't. The top first row isn't identical to the heading table. self critique assessment: 2
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07:43:48 `q003. Repeat the preceding exercise for the operation x * y mod 5, defined to give the remainder when x * y is divided by 5, on the set {1, 2, 3, 4}. Determine which of the properties are exhibited by this operation.
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RESPONSE --> * 1 2 3 4 1 1 2 3 4 2 2 4 1 3 3 3 1 4 2 4 4 3 2 1 It has the closure, communitative, associative, indentity and inverse. confidence assessment: 2
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07:45:27 First we might wish to do a couple of example calculations to get familiar with the operation. For example: 2 * 3 mod 5 = 6, which when divided by 5 gives us remainder 1. 3 * 4 mod 5 = 12 which when divided by 5 gives us remainder 2. 2 * 4 mod 5 = 8 which when divided by 5 gives us remainder 3. The table is * mod 5 1 2 3 4 1 1 2 3 4 2 2 4 1 3 3 3 1 4 2 4 4 3 2 1 We immediately see that all the results are in the set {1, 2, 3, 4}, so that the operation is closed. This operation has identity 1, as we can see from the row and the column across from and beneath 1. We easily see from the table that the identity appears exactly once in each row and in each column, which assures us that the operation has the inverse property. Specifically we see that 1 * 1 mod 5 = 1 so that 1 is its own inverse, that 2 * 3 mod 5 = 1 so that 2 and 3 are inverses, and that 4 * 4 mod 5 = 1, so that 4 is its own inverse. The associativity and commutativity of the operation follow from the associative and commutative properties of multiplication on real numbers, as discussed in the preceding problem.
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RESPONSE --> Got the same answers. self critique assessment: 2
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07:53:12 `q004. The equation 3x + 7 = 9 (mod 5) has an integer solution for x = 0, 1, 2, 3 or 4. Which value of x is a solution to this equation?
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RESPONSE --> 3(0)+7 = 7 remainder 2 - not a solution 3(1)+7=10 remainder 0 - not a solution 3(2)+7=13 remainder 3 - not a solution 3(3)+7=16 remainder 1 - not a solution 3(4)+7=19 remainder 4 - not a solution I don't get any of them as a solution because none have a remainder of 9. ?????Since it is mod 5, 9 couldn't be the answer anyway because 5 would divide into 9 one time with a remainder of 4??? confidence assessment: 2
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07:59:00 3x + 7 = 9 (mod 4) means that 3x + 7 - 9 = 0 (mod 5) so 3x - 2 = 0 (mod 5). If 3x - 2 = 0 (mod 5) then when we divide 3x - 2 by 5 we should get remainder 0. So we substitute the different possible values for x into the expression 3x - 2 until we get a number of which when divided by 5 gives us remainder 0. If x = 0 then 3x - 2 = -2, and -2 (mod 5) = 3 (if you don't understand why -2 mod 5 = 3, think of the 5-hour clock in the text; but for now it should be obvious that -2 is not a multiple of 5 so that you cannot get remainder 0 when dividing -2 by 5). If x = 1 then 3x - 2= 1, and 1 (mod 5) = 1. If x = 2 then 3x - 2= 4, and 4 (mod 5) = 4. If x = 3 then 3x - 2= 7, and 7 (mod 5) = 2. If x = 4 then 3x - 2= 10, and 10 (mod 5) = 0. Thus x = 4 is a solution to the equation 3x + 7 = 9 (mod 5).
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RESPONSE --> I didn't work it right. I didn't subtract the 9 from both sides so I started out all wrong but I understand how to work it. Do you subtract the answer if it is higher than the (mod) to make the answer 0? self critique assessment: 2
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08:04:05 `q005. You see that x = 4 is a solution to the equation 3x + 7 = 9 (mod 5). One of the numbers x = 5, 6, 7, 8, 9 is also a solution. Which one is it?
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RESPONSE --> 3(x) - 2 = 0 3(5)-2=13 remainder 3 3(6)-2=16 remainder 1 3(7)-2=19 remainder 4 3(8)-2=22 remainder 2 3(9)-2=25 remainder 0 x = 9 confidence assessment: 2
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08:04:46 We recall that 3x + 7 = 9 (mod 5) is equivalent to 3x - 2 = 0 (mod 5). We evaluate 3x - 2 (mod 5) for x = 5, 6, 7, 8 and 9 and we find that the results are 3, 1, 4, 2, and 0. So x = 9 is our next solution. We might also note that the series of results 3, 1, 4, 2, 0 is the same as the series we got for x = 0, 1, 2, 3, 4. Our results therefore seem to indicate a repeating pattern in which the remainder 0 occurs every fifth number starting with 4. This is in fact what happens, and you might wish to think about why this happens. However, you should in a case remember that this is what happens. In general when we have an equation of the form A x + B = C (mod n), integer solutions happen at intervals of n. for some values of A, B and C integer solutions can also occur at shorter intervals, but they always do occur at intervals of n.
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RESPONSE --> I worked it right this time and got the correct answer. self critique assessment: 2
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08:13:41 `q006. What are the first five positive values of x which solve the equation 3x + 7 = 9 (mod 5) of the preceding problem?
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RESPONSE --> 4, 9, 14, 19, 24 confidence assessment: 2
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08:14:24 We just saw that x = 4 and x = 9 are solutions, and we saw that because we are solving an equation mod 5 solutions have to occur at intervals of 5. Thus the first five solutions are x = 4, 9, 14, 19 and 24.
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RESPONSE --> You add 5 to the get the next numbers. self critique assessment: 2
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