course Mth 151 Vúá]ó¿ËOÑëß©àªÊ÷¾àÂî ¯ÿñÔ¼ææù`Í{assignment #006
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17:27:38 Query 1.1.4 first 3 children male; conclusion next male. Inductive or deductive?
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RESPONSE --> Inductive confidence assessment: 3
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17:28:58 Query 1.1.8 all men mortal, Socrates a man, therefore Socrates mortal.
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RESPONSE --> Deductive becuase it is characterized by general principles confidence assessment: 3
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17:30:44 ** this argument is deductive--the conclusions follow inescapably from the premises. 'all men' is general; 'Socrates' is specific. This goes general to specific and is therefore deductive. COMMON ERROR: because it is based on a fact, or concrete evidence. Fact isn't the key; the key is logical inevitability. The argument could be 'all men are idiots, Socrates is an man, therefore Socrates is an idiot'. The argument is every bit as logical as before. The only test for correctness of an argument is that the conclusions follow from the premises. It's irrelevant to the logic whether the premises are in fact true. **
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RESPONSE --> It is applying a specific example not repeated observation. self critique assessment: 3
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17:31:42 Query 1.1.20 1 / 3, 3 / 5, 5/7, 7/9, ... Probable next element.
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RESPONSE --> 9/12. confidence assessment: 3
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17:32:59 **The numbers 1, 3, 5, 7, 9 and 11 are odd numbers. We note that the numerators consist of the odd numbers, each in its turn. The denominator for any given fraction is the next odd number after the numerator. Since the last member listed is 7/9, with numerator 7, the next member will have numerator 9; its denominator will be the next odd number 11, and the fraction will be 9/11. There are other ways of seeing the pattern. We could see that we use every odd number in its turn, and that the numerator of one member is the denominator of the preceding member. Alternatively we might simply note that the numerator and denominator of the next member are always 2 greater than the numerator and denominator of the present member. **
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RESPONSE --> The numerators are the next odd number and the denominator is increased by 3 every time. self critique assessment: 3
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17:41:05 Query 1.1.23 This problem wasn't assigned, but you should be able to make a good attempt: 1, 8, 27, 64, ... What is the probable next element?
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RESPONSE --> The next probable number is 125. confidence assessment: 3
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17:42:56 ** This is the sequence of cubes. 1^3 = 1, 2^3 = 8, 3^3 = 27, 4^3 = 64, 5^3 = 125. The next element is 6^3 = 216. Successive differences also work: 1 8 27 64 125 .. 216 7 19 37 61 .. 91 12 18 24 .. 30 6 6 .. 6 **
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RESPONSE --> The number is mutipled by itself 3 times. self critique assessment: 2
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17:45:25 Query 1.1.36 11 * 11 = 121, 111 * 111 = 12321 1111 * 1111 = 1234321; next equation, verify.
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RESPONSE --> 11111 * 11111 = 12345432 confidence assessment: 3
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17:46:03 ** We easily verify that 11111*11111=123,454,321 **
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RESPONSE --> You follow the pattern 1, 11, 111, 1111, 11111 self critique assessment: 2
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17:57:59 Do you think this sequence would continue in this manner forever? Why or why not?
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RESPONSE --> Probably because if you keep adding a number 1 to the numbers 111111, 1111111 and mutiply them it would probably give you the numbers. confidence assessment: 2
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17:59:06 ** You could think forward to the next few products: What happens after you get 12345678987654321? Is there any reason to expect that the sequence could continue in the same manner? The middle three digits in this example are 8, 9 and 8. The logical next step would have 9, 10, 9, but now you would have 9109 in the middle and the symmetry of the number would be destroyed. There is every reason to expect that the pattern would also be destroyed. **
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RESPONSE --> I would think that you would have a zero next. self critique assessment: 2
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18:00:07 Query 1.1.46 1 + 2 + 3 + ... + 2000 by Gauss' method
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RESPONSE --> 2001 * 1000 = 2001000.00 confidence assessment: 3
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18:01:19 ** Pair up the first and last, second and second to last, etc.. You'll thus pair up 1 and 2000, 2 and 1999, 3 and 1998, etc.. Each pair of numbers totals 2001. Since there are 2000 numbers there are 1000 pairs. So the sum is 2001 * 1000 = 2,001,000 **
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RESPONSE --> You would add the first number (1) and last number (2000) and mutiple it by 1000 becuase the last number is 2000 and you would 1/2 it. So you get 2001 * 1000 self critique assessment: 3
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18:05:28 Query 1.1.57 142857 * 1, 2, 3, 4, 5, 6. What happens with 7? Give your solution to the problem as stated in the text.
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RESPONSE --> 142857 * 1 = 142857 142857 * 2 = 285714 142857 * 3 = 428571 142857 * 4 = 571428 142857 * 5 = 714285 142857 * 6 = 857142 142857 * 7 = 999999 confidence assessment: 3
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18:07:05 ** Multiplying we get 142857*1=142857 142857*2= 285714 142857*3= 428571 142857*4=571428 142857*5= 714285 142857*6=857142. Each of these results contains the same set of digits {1, 2, 4, 5, 7, 8} as the number 1428785. The digits just occur in different order in each product. We might expect that this pattern continues if we multiply by 7, but 142875*7=999999, which breaks the pattern. **
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RESPONSE --> When you mutilpy 142857 by 7 it gives you 999999 which is odd to me because with the numbers being consecutive I won't think you would get an 999999 number. self critique assessment: 2
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18:08:00 What does this problem show you about the nature of inductive reasoning?
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RESPONSE --> That you don't draw conclusions to answers unless you know for sure what the anwser is. confidence assessment: 2
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