course math 151 Porf. Smith,I sent you an e-mail requesting enrollment on Math152.
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08:43:05 `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=27/4=6.75 7*12=84/4=84 11*13=143/4=35.75 confidence assessment: 3
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08:44:14 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 --> the remainder of 3*9 [mod4] is 3, of 7*12 is 0, and 11*13 is 3 self critique assessment: 3
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08:46:53 `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 properties of operation: commutative associative identity closure confidence assessment: 3
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08:48:04 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 --> so when we multiply 1*1=1/4=.25, the result in the table is 1. identity is 1. doesn't have the inverse property because 0 nor 2 can be combine to give us identiy(1)
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09:03:36 `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 --> mod5 1 2 3 4 1 2 4 6 8 2 4 8 2 6 3 6 2 8 4 4 8 6 4 2 associative communitative inverse confidence assessment: 3
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09:13:46 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 --> the properties in this operation are commutative associative the operation is closed inverse self critique assessment: 3
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09:28:17 `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 --> 9 mode 5 is the answer, right? so the process should be: 3x+7=0 -7 -7 3x=-7 3x/3= -7/3 x= -7/3 (-7/3)/5= .4 confidence assessment: 3
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09:37:24 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 --> ok self critique assessment: 3
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09:47:44 `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 --> 1 and 9 confidence assessment: 1
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09:49:48 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 --> and the next five would be29,34,39,44,49 self critique assessment: 3
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