course MTH 151 lľɨ~ZҖzkڨassignment #020
......!!!!!!!!...................................
17:00:06 query 4.3.6 number following base-six 555
......!!!!!!!!...................................
RESPONSE --> There is no base six. confidence assessment: 0
.................................................
......!!!!!!!!...................................
17:02:43 ** COMMON ERROR: 556. INSTRUCTOR COMMENT: The digit 6 does not exist in base six for the same reason as there is no single digit for 10 in base 10 (digits stop at 9; in general there is no digit corresponding to the base). CORRECT SOLUTION: 555{base 6} = 5 * 6^2 + 5 * 6^1 + 5 * 6^0. If you add 1 you get 5 * 6^2 + 5 * 6^1 + 5 * 6^0 + 1, which is equal to 5 * 6^2 + 5 * 6^1 + 6 * 6^0. But 6 * 6^0 is 6^1, so now you have 5 * 6^2 + 6 * 6^1 + 0 * 6^0. But 6 * 6^1 is 6^2 so you have 6 * 6^2 + 0 * 6^1 + 0 * 6^0. But 6 * 6^2 is 6^3 so the number is 6 * 6^3 + 0 * 6^2 + 0 * 6^1 + 0 * 6^0. So the number following 555{base 6} is 1000{base 6}. The idea is that we can use a number greater than 5 in base 6, so after 555 we go to 1000, just like in base 10 we go from 999 to 1000. **
......!!!!!!!!...................................
RESPONSE --> I understand that I had to get a number higher than 555 which would be 1000. self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:16:32 query 4.3.20 34432 base five
......!!!!!!!!...................................
RESPONSE --> 3*5^4 + 4*5^3 + 4*5^2 + 3*5^1 + 2*5^0 (((3*5+4)*5+4)*5+3)*5+2=2492 confidence assessment: 2
.................................................
......!!!!!!!!...................................
17:16:40 **34432 base five means 3 * 5^4 + 4 * 5^3 + 4 * 5^2 + 3 * 5^1 + 2 * 5^0. 5^2 = 625, 5^3 = 125, 5^2 = 25, 5^1 = 5 and 5^0 = 1 so 3 * 5^4 + 4 * 5^3 + 4 * 5^2 + 3 * 5^1 + 2 * 5^0= 3 * 625 + 4 * 125 + 4 * 25 + 3 * 5 + 2 * 1 = 1875 + 500 + 100 + 15 + 2 = 2492. **
......!!!!!!!!...................................
RESPONSE --> correct self critique assessment: 3
.................................................
......!!!!!!!!...................................
17:18:45 Explain how you use the calculator shortcut to get the given number.
......!!!!!!!!...................................
RESPONSE --> I took 3*5+4*5+4*5+3*5+2 in that order to reach the total 2492
.................................................
......!!!!!!!!...................................
17:19:08 ** 2 + 5 (3 + 5(4 + 5(4 + 3 * 5))) multiplies out to the given number because, for example, the 3 * 5 at the end will be multiplied by the three 5s to its left to give 3 * 5 * 5 * 5 * 5 = 3 * 5^4, which matches the 3 * 5^4 in the sum 3 * 5^4 + 4 * 5^3 + 4 * 5^2 + 3 * 5^1 + 2 * 5^0. So we do 3 + 5, then 4 + the answer, then 5 * the answer, then 4 + the answer, then 5 * the answer, then 3 + the answer, then 5 * the answer, then 2 + the answer. **
......!!!!!!!!...................................
RESPONSE --> correct self critique assessment: 3
.................................................
......!!!!!!!!...................................
17:24:06 query 4.3.40 11028 decimal to base 4
......!!!!!!!!...................................
RESPONSE --> I don't understand how to get this answer if there is no base four? confidence assessment: 0
.................................................
......!!!!!!!!...................................
17:24:17 ** 4^0 = 1 4^1 = 4 4^2 = 16 4^3 = 64 4^4 = 256 4^5 = 1024 4^6 = 4096 (4*7 = 16386, which is larger than the given 11028) So to build up 11028 we need 2 * 4^6 = 8192, leaving 2836. 2 * 4^5 = 2048, leaving 788. 3 * 4^4 = 768, leaving 20. 0 * 4^3, because we need only 20, which is less than 64. 1 * 4^2 = 16, leaving 4. 1 * 4^1 = 4, leaving 0.0 * 4^1. Thus our number is 2230110 base 4.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:27:41 query 4.3.51 DC in base 16 to binary
......!!!!!!!!...................................
RESPONSE --> 11011100two D=1101 C=1100 confidence assessment: 0
.................................................
......!!!!!!!!...................................
17:27:58 ** C stands for decimal 12, which in binary is 1100. D stands for decimal 13, which in binary is 1101. Since our base 16 is the fourth power of 2, our number can be written 11011100, where the 1101 in the first four places stands for D and the 1100 in the last four places stands for C. Note that this method works only when one base is a power of the other.**
......!!!!!!!!...................................
RESPONSE --> correct self critique assessment: 3
.................................................
......!!!!!!!!...................................
17:29:57 Is a base-9 number always even, always odd, or sometimes even and sometimes odd if the number ends in an even number?
......!!!!!!!!...................................
RESPONSE --> They can be both odd and even. confidence assessment: 1
.................................................
......!!!!!!!!...................................
17:30:23 ** You can investigate this question by trying a variety of examples. For example, 82nine = 8 * 9^1 + 2 * 9^0 = 72 + 2 = 74 in decimal. This is even because you are adding two even multiples of the odd numbers 9^1 = 9 and 9^0 = 1. You have to use an even multiple of 9^0 = 1 because the number ends with an even number. But you don't have to use an even multiple of 9^1 = 9. So we try something like 72nine = 7 * 9^1 + 2 * 9^0 = 63 + 2 = 65 in decimal and we see that the result is odd. The key is that in base nine, the powers of nine are always odd numbers. So when converting you get from a base-9 number with two even digits a number which is even * odd + even = even, while from a base-9 number with an odd digit followed by an even digit you get odd * odd + even = odd; many other possible combinations show the same thing but this is one of the simpler examples that shows why we get odd for some base-9 numbers, and even for others. For a couple of specifics, 70 in base 9 is 63 in base ten, and is odd. However 770 in base 9 is even. **
......!!!!!!!!...................................
RESPONSE --> correct self critique assessment: 3
.................................................
Y͒Ƽaզ assignment #021 021. `query 21 College Algebra 04-07-2009
......!!!!!!!!...................................
17:36:11 4.4.6 star operation [ [1, 3, 5, 7], [3, 1, 7, 5], [5, 7, 1, 3], [7, 5, 3, 1]]
......!!!!!!!!...................................
RESPONSE --> Closure uses the same set of numbers {1,3,5,7} *1357 11357 33175 55713 77531 confidence assessment: 1
.................................................
......!!!!!!!!...................................
17:36:26 ** 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. **
......!!!!!!!!...................................
RESPONSE --> correct self critique assessment: 3
.................................................
......!!!!!!!!...................................
17:41:08 4.4.24 a, b, c values that show that a + (b * c) not equal to (a+b) * (a+c).
......!!!!!!!!...................................
RESPONSE --> a=1 b=2 c=3 1 + (2*3)=7 (1+2) *(1+3)=3*4=12 so a + (b*c) does not = (a+b)*(a+c) confidence assessment: 1
.................................................
......!!!!!!!!...................................
17:41:17 ** 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. **
......!!!!!!!!...................................
RESPONSE --> correct self critique assessment: 3
.................................................
......!!!!!!!!...................................
17:44:32 4.4.33 venn diagrams to show that union distributes over intersection
......!!!!!!!!...................................
RESPONSE --> AU(B^C) A would be shaded and where B & C overlap. confidence assessment: 1
.................................................
......!!!!!!!!...................................
17:44:39 ** 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). **
......!!!!!!!!...................................
RESPONSE --> correct self critique assessment: 3
.................................................