09-240:HW4: Difference between revisions
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|'''Just for Fun.''' |
|'''Just for Fun (1).''' |
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* Take a large integer and write it in base 10. Cut away the "singles" digit, double it and subtract the result from the remaining digits. Repeat the process until the number you have left is small. Prove that the number you started from is divisible by 7 iff the resulting number is divisible by 7. Thus the example on the right shows that 86415 is divisible by 7 as 0 is divisible by 7. |
* Take a large integer and write it in base 10. Cut away the "singles" digit, double it and subtract the result from the remaining digits. Repeat the process until the number you have left is small. Prove that the number you started from is divisible by 7 iff the resulting number is divisible by 7. Thus the example on the right shows that 86415 is divisible by 7 as 0 is divisible by 7. |
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* Find a similar criterion for divisibility by 17 and for all other divisibilities and indivisibilities. |
* Find a similar criterion for divisibility by 17 and for all other divisibilities and indivisibilities. |
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'''Just for Fun (2).''' Is there a problem with the following inductive proof that all horses are of the same color? |
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We assert that in all sets with precisely <math>n</math> horses, all horses are of the same color. For <math>n=1</math>, this is obvious: it is clear that in a set with just one horse, all horses are of the same color. Now assume our assertion is true for all sets with <math>n-1</math> horses, and let us be given a set with <math>n</math> horses in it. By the inductive assumption, the first <math>n-1</math> of those are of the same color and also the last <math>n-1</math> of those. Hence they are all of the same color as illustrated below: |
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{{Equation*|<math>(H,[H,\ldots,H),H]</math>}} |
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(The horses surrounded by round brackets <math>(\cdots)</math> are all of the same color. The horses surrounded by square brackets <math>[\cdots]</math> are all of the same color. Therefore the first and the last horses have the same color as the ones in the middle group, and hence all horses are of the same color.) |
Latest revision as of 16:02, 7 October 2009
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- Read sections 1.5 through 1.7 in our textbook. Remember that reading math isn't like reading a novel! If you read a novel and miss a few details most likely you'll still understand the novel. But if you miss a few details in a math text, often you'll miss everything that follows. So reading math takes reading and rereading and rerereading and a lot of thought about what you've read.
- Solve problems 3, 8, 9, 10, and 11 on pages 41-42, but submit only your solutions of problems 8, 9, and 11.
- Solve problems 1, 2, 4, 5, 9, 12, 13, and 16 on page 53-56, but submit only your solutions of problems 4, 5, 9, and 12.
- This assignment is due on Thursday October 15 at the tutorials.
Just for Fun (1).
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Just for Fun (2). Is there a problem with the following inductive proof that all horses are of the same color?
We assert that in all sets with precisely horses, all horses are of the same color. For , this is obvious: it is clear that in a set with just one horse, all horses are of the same color. Now assume our assertion is true for all sets with horses, and let us be given a set with horses in it. By the inductive assumption, the first of those are of the same color and also the last of those. Hence they are all of the same color as illustrated below:
(The horses surrounded by round brackets are all of the same color. The horses surrounded by square brackets are all of the same color. Therefore the first and the last horses have the same color as the ones in the middle group, and hence all horses are of the same color.)