Assignment 6 Query

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course Mth 151

11/19 1

006. `Query 6

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Question: `qQuery 1.1.4 first 3 children male; conclusion next male. Inductive or deductive?

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Your solution:

Inductive because a female could be next.

confidence rating #$&*:3

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Given Solution:

`a** The argument is inductive, because it attempts to argue from a pattern. **

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Question: `qQuery 1.1.8 all men mortal, Socrates a man, therefore Socrates mortal.

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Your solution:

Deductive. Valid statement.

confidence rating #$&*:3

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Given Solution:

`a** 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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Question: `qQuery 1.1.20 1 / 3, 3 / 5, 5/7, 7/9, ... Probable next element.

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Your solution:

9/11, 11/13, 13/15

All odd numbers then just add 2 to the numerator to get the denominator.

1/3 add 2 to the numerator to have the numerator in the next number which is 3 then add 2 to the numerator in the current number to find the denominator 3+2=5 3/5

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Given Solution:

`a**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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Question: `qQuery 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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Your solution:

I could not figure this out

confidence rating #$&*:0

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Given Solution:

`a** 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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Question: `qQuery 1.1.36 11 * 11 = 121, 111 * 111 = 12321 1111 * 1111 = 1234321; next equation, verify.

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Your solution: 11111= 123454321

I’m not sure why or if there is an equation for this but it follows the previous sequences

confidence rating #$&*:3

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Given Solution:

`a** We easily verify that 11111*11111=123,454,321 **

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Question: `qDo you think this sequence would continue in this manner forever? Why or why not?

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Your solution:

Yes and no. It could go on forever but after you get to 9 the 10 could mess it up but you would have to put commas like this

1,2,3,4,5,6,7,8,9,10,9,8,7,6,5,4,3,2,1

But that would be changing the sequence that did not have commas so I will lean more toward the answer of no it can not go on forever.

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Given Solution:

`a** 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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Question: `qQuery 1.1.46 1 + 2 + 3 + ... + 2000 by Gauss' method

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Your solution:

1+2000=2001

2+1999=2001

3+1998=2001

1000 sums of 2001

Sum of all numbers 1000*2001=2,001,000

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Given Solution:

`a** 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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Question: `qQuery 1.1.55 (previously 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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Your solution:

142857*1=142857

142857*2=285714

142857*3=428571

142857*4=571528

142857*5=714285

142857*6=857142

142857*7=999999

The pattern between 1-6 is that each sum has the same numbers in it but it is just in a different order. Multiplying by 7 ruins the pattern.

confidence rating #$&*:3

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Given Solution:

`a** 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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Question: `qWhat does this problem show you about the nature of inductive reasoning?

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Your solution:

Shows that I would of thought it would have been a different mix of the numbers 142857

confidence rating #$&*:3

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Given Solution:

`a** 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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Question: `qWhat does this problem show you about the nature of inductive reasoning?

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Your solution:

Shows that I would of thought it would have been a different mix of the numbers 142857

confidence rating #$&*:3

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Given Solution:

`a** 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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