course Mth 151 This is the real text assignment, promise! ~Ҕ{ⵁ~ҋL_assignment #006
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00:54:32 Query 1.1.4 first 3 children male; conclusion next male. Inductive or deductive?
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RESPONSE --> This wasn't on the homework as a multiple, but I believe this would be inductive reasoning, because it follows a pattern. There is no fact to prove that the next child will be male. confidence assessment: 2
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00:54:38 ** The argument is inductive, because it attempts to argue from a pattern. **
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RESPONSE --> OK self critique assessment: 3
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00:55:20 Query 1.1.8 all men mortal, Socrates a man, therefore Socrates mortal.
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RESPONSE --> This isn't a multiple of the homework either, but, since all men are mortal, and that is a fact, this is deductive reasoning. confidence assessment: 2
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00:55:26 ** 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 --> OK self critique assessment: 3
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00:56:31 Query 1.1.20 1 / 3, 3 / 5, 5/7, ... Probable next element.
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RESPONSE --> 7/9. The numerator and denominator increase by odd numbers. So the next element is sure to be 7/9. confidence assessment: 3
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00:56:38 **The numbers 1, 3, 5, 7 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 5/7, with numerator 5, the next member will have numerator 7; its denominator will be the next odd number 9, and the fraction will be 7/9. 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 --> OK self critique assessment: 3
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00:58:46 Query 1.1.23 1, 8, 27, 64, ... Probable next element.
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RESPONSE --> This wasn't on my homework either, and I am unsure of the answer. confidence assessment: 0
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00:58:55 ** 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 --> self critique assessment: 3
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01:01:21 Query 1.1.36 11 * 11 = 121, 111 * 111 = 12321 1111 * 1111 = 1234321; next equation, verify.
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RESPONSE --> 11111*11111= 12,354,321 The numbers added to the end of each sum are descending. But this can't be verified because the calculated sum is 123,454,321 confidence assessment: 2
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01:02:16 ** We easily verify that 11111*11111=123,454,321 **
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RESPONSE --> I went about solving the problem the wrong way. I was using inductive reasoning to find the next pattern, and THEN verify it. But missed the sum. self critique assessment: 2
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01:02:44 Do you think this sequence would continue in this manner forever? Why or why not?
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RESPONSE --> I think it would continue, because as the pattern increases, the number continues to be consistent. confidence assessment: 2
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01:02:51 ** 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 --> OK self critique assessment: 2
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01:04:04 Query 1.1.46 1 + 2 + 3 + ... + 2000 by Gauss' method
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RESPONSE --> This question was not a multiple of 3 or 5 either, but I can take 2001 and multiply it by 1000 to find the sum, which is 2001000. confidence assessment: 2
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01:04:28 ** 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 --> OK self critique assessment:
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01:05:53 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 --> 999,999. The solution to the problem is as follows 1=142,957 2= 285,714 3= 428,571 4= 571,428 5= 714,285 6= 857,142 confidence assessment: 2
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01:06:11 ** 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 --> OK self critique assessment: 3
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01:06:28 What does this problem show you about the nature of inductive reasoning?
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RESPONSE --> That inductive reasoning does not always guarantee a concrete answer. confidence assessment: 2
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01:06:50 ** Inductive reasoning would have led us to expect that the pattern continues for multiplication by 7. Inductive reasoning is often correct it is not reliable. Apparent patterns can be broken. **
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RESPONSE --> I understood that the patterns can be broken, which is why I said it was unreliable. self critique assessment: 2
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