course Mth 151 assignment 22 22:51:08
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RESPONSE --> the operation is closed and yes its a group. the identity is 2 (1,2) IS THE INVERSE confidence assessment: 1
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22:55:31 `q002. Which of the properties closure, identity, commutativity, inverse, does the standard addition operation + have on the set {-1, 0, 1}?
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RESPONSE --> + -1 0 1 -1 -2 -1 0 0 -1 0 1 1 0 1 2 Its not closed, 2 is not a set of -1,0, 1. the identity is 0, the inverse is -1 &1, and the operation is commutativity. confidence assessment: 3
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22:57:58 `q003. Does the operation * of standard multiplication on the set {-1, 0, 1} have the properties of closure, identity and inverse?
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RESPONSE --> * -1 0 1 -1 1 0 -1 0 0 0 0 1 -1 0 1 yes there is closure, the identity is 1, and the inverse is 1 confidence assessment: 2
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23:02:41 `q004. Does the operation * of standard multiplication on the set {-1, 1} have the properties of closure, identity and inverse?
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RESPONSE --> * -1 1 -1 1 -1 1 -1 1 yes it has closure, the indentity is 1, and -1 and 1 is the inverse. confidence assessment: 2
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23:03:48 `q005. Is the operation * of standard multiplication on the set {-1, 1} a group. Note that the operation does have the property of associativity.
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RESPONSE --> yes its a group b/c it has a identity, and the inverse is closed. confidence assessment: 1
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23:05:14 `q006. We've referred to the property of associativity, but we haven't yet defined it. Associativity essentially means that when an operation (technically a binary operation, but don't worry about that a terminology at this point) is performed on three elements of a set, for example a + b + c, it doesn't matter whether we first perform a + b then add c, calculating (a + b) + c, or group the b and c so we calculate a + (b + c). If + means addition on real numbers, show that (3 + 4) + 5 = 3 + ( 4 + 5).
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RESPONSE --> (3+4) +5= 7+5=12 3+(4+5)= 3+9=12 confidence assessment: 2
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23:11:41 `q007. Verify that for the operation @ defined on {0, 1, 2} by x @ y = remainder when x * y is double then divided by 3, we have 2 @ (0 @ 1) = ( 2 @ 0 ) @ 1. Verify also that (2 @ 1) @ 1 = 2 @ ( 1 @ 1).
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RESPONSE --> 0 @ 1 =0, 2 @ ( 0 @ 1)= 2@0=0 2@0=0, (2@0) @1= 0@1=0 2@0@1=2@0@1 2@1=1, 2@1@1=1@1=1 1@1=2, 2@1@1=2@2=2 2@1@1=2@1@1 confidence assessment: 1
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23:13:25 `q008. Does the result of the preceding exercise prove that the @ operation is associative on the set {0, 1, 2}?
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RESPONSE --> no confidence assessment: 1
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23:17:42 `q009. Earlier we verified the properties of closure, identity and inverse for the multiplication operation * on the set {-1, 1}. We asserted that this operation was associative, so that this set with this operation forms a group. It would still be too time-consuming to prove that * is associative on {-1, 1}, but list the possible combinations of a, b, c from the set and verify associativity for any three of them.
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RESPONSE --> {-1, -1, -1},{-1, -1, 1},{-1, 1,- 1},{-1, 1, 1},{1,-1, -1},{1, -1, 1},{1, 1,- 1},{1, 1, 1} confidence assessment: 3
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23:36:33 asignment 23 23:36:33 `q001. There are twelve questions in this assignment. The number 12 is evenly divisible by 1, 2, 3, 4, 6, and 12. We say that 1, 2, 3, 4, 6 and 12 are the divisors of 12. Each of these divisors can be multiplied by another to get 12. e.g., 2 * 6 = 12, 1 * 12 = 12, 3 * 4 = 12. List the numbers from 2 to 20 and list all the divisors of each.
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RESPONSE --> the divisors of 2 is 1 & 2 "" "" of 3 is 1&3 "" "" of 4 is 1,2,4 "" "" of 5 is 1,5 "" "" of 6 is 1,2,3,6 "" "" of 7 is 1,7 "" "" of 8 is 1,2,4,8 "" "" of 9 is 1,3,9 "" "" of 10 is 1,2,5,10 "" "" of 11 is 1&11 "" "" of 12 is 1,2,3,4,6,12 "" "" of 13 is 1&13 "" "" of 14 is 1,2,4,8,14 "" "" of 15 is 1,3,5,15 "" "" of 16 is 1,2,4,8,16 "" "" of 17 is 1&17 "" "" of 18 is 1,2,3,6,9,18 "" "" of 19 is 1,19 "" "" of 20 is 1,2,4,5,10,20 confidence assessment: 3
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23:40:06 `q002. Some of the numbers you listed have exactly two divisors. Which are these?
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RESPONSE --> 2 with the divisors 1 and 2 3 with the divisors 1 and 3 5 with the divisors 1 and 5 7 with the divisors 1 and 7 11 with the divisors 1 and 11 13 with the divisors 1 and 13 17 with the divisors 1 and 17 19 with the divisors 1 and 19 confidence assessment: 3
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23:41:58 `q003. These numbers with exactly two divisors are called prime numbers. List the prime numbers between 21 and 40.
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RESPONSE --> 23,29,31,37 confidence assessment: 2
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23:43:52 `q004. The primes through 40 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 and 37. Twin primes are consecutive odd numbers which are both prime. Are there any twin primes in the set of primes through 40?
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RESPONSE --> 17 and 19 are consecutive odd # and both prime, the same is true for 5 and 7, and 11and 13. confidence assessment: 2
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23:44:10 `q006. We can prove that 89 is prime as follows: 89 is odd and is hence not divisible by 2. If we divide 89 by 3 we get a remainder of 2 so 89 is not divisible by 3. Since 89 is not divisible by 2 it cannot be divisible by 4 (if we could divide it evenly by 4 then, since 2 goes into 4 we could divide evenly by 2; but as we just saw we can't do that). Since 89 doesn't end in 0 or 5 it isn't divisible by 5. If we divide 89 by 7 we get a remainder of 5 so 89 is divisible by 7. Since 89 isn't divisible by 2 it isn't divisible by 8, and since it isn't this will by 3 it isn't divisible by 9. At this point it might seem like we have a long way to go--lots more numbers to before get to 89. However it's not as bad as it might seem. For example once we get past 44 we're more than halfway to 89 so the result of any division would be less than 2, so there's no way it could be a whole number. So nothing greater than 44 is a candidate. We can in fact to even better than that. If we went even as far as dividing by 10, the quotient must be less than 10: since 10 * 10 is greater than 89, it follows that 89 / 10 must be less than 10. So if 10 or, more to the point, anything greater than 10 was going to divide 89 evenly, in the result would be one of the numbers we have already unsuccessfully tried. It follows that after trying without success to divide 89 by all the numbers through 9, which we didn't really have to try anyway because 9 is divisible by 3 which we already checked, we are sure that 89 has to be prime. What is the largest number you would have to divide by to see whether 119 is prime?
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RESPONSE --> confidence assessment: 0
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23:44:58 `q007. Precisely what numbers would we have to try in order to determine whether 119 is prime?
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RESPONSE --> 119 divisible by 2, by 3, by 5, by 7. confidence assessment: 2
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23:46:37 `q008. Is 119 prime?
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RESPONSE --> yes, 119 is divisbile by 7 confidence assessment: 3
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23:53:07 `q009. We can 'break down' the number 54 into the product 2 * 27, which can further be broken down to give us 2 * 3 * 9, which can be broken down one more step to give us 2 * 3 * 3 * 3. We could have broken down 54 into different way as 6 * 9, which could have been broken down into 6 * 3 * 3, which can be broken down one more step to give us 2 * 3 * 3 * 3. No matter how we break 54 down into factors, the process ends with a single factor 2 and 3 repetitions of the factor 3. Break down each of the following in this manner, until it is not possible to break it down any further: 63, 36, and 58.
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RESPONSE --> 63 63 9*7 21*3 3*3*7 3*7*3 36 36 2*18 9*4 2*2*3*3 3*3*2*2 58 58 2*9 confidence assessment: 2
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23:54:34 `q010. The results that we obtained in the preceding exercise are called the 'prime factorizations' of the given numbers. We have broken the numbers down until we are left with just prime numbers. Why is it that this process always ends with prime numbers?
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RESPONSE --> the number must be broken down. the process will stop when all the numbers are prime confidence assessment: 2
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23:55:44 `q011. Find the prime factorization of 819, then list all the factors of 819.
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RESPONSE --> prime factors are 3,7,13 819 3* 273 3*3*91 3*3*7*13 confidence assessment: 3
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23:57:22 `q012. List all the factors of 168.
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RESPONSE --> prime factor of 168 is 2,3,7 168 2*84 2*2*42 2*2*2*21 2*2*2*3*7 confidence assessment: 3
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23:57:47 12-13-2007 23:57:47 `q012. List all the factors of 168.
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NOTES ------->
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