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amistre64

  • 2 years ago

So what would be wrong with this proof by induction?

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  1. amistre64
    • 2 years ago
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    I got marked off for some reason

  2. amistre64
    • 2 years ago
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    what i wrote down on the homework was more like: \[2^n < n! ~;~n\ge4\] basis step: \[2^4 < 4!\]\[16 < 24~;true\] assume \[2^k < k! ~;~k\ge4\] prove: \[2^{k+1} < (k+1)!\]\[2*2^k < k!*(k+1)\]since: \[2 < \{5,6,7,8...\}\hspace{15em}QED\]or \[2^k < k!\frac{(k+1)}{2}\hspace{16em}QED\]

  3. TuringTest
    • 2 years ago
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    I don't see anything blatantly wrong with it... I came up with a slightly different argument

  4. amistre64
    • 2 years ago
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    It hard to determine when proofing means, "reinvent the wheel", or "use common knowledge". But i got marked off 1.5 pts on a 2pt problem

  5. myininaya
    • 2 years ago
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    What does 2<{5,6,7,8...} mean?

  6. myininaya
    • 2 years ago
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    Oh. You mean 2 is less than any element in that set.

  7. amistre64
    • 2 years ago
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    right

  8. TuringTest
    • 2 years ago
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    \[2^k<k!\implies\frac{2^k}{k!}<1\]\[2^{k+1}=2\cdot2^k<(k+1)k!\]\[2\cdot\frac{2^k}{k!}<k+1\]since \(\Large \frac{2^k}{k!}<1\) that implies that \[\Large2\cdot\frac{2^k}{k!}<2<k+1\]since we have assumed that \(k>4\) not sure if your prof would have liked that any better, I'm not so great at induction...

  9. amistre64
    • 2 years ago
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    thats a fine rendition; my prof would rather have us get it into a form similar to the assumption to compare with tho

  10. TuringTest
    • 2 years ago
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    well I guess you could proceed to multiply both sides by k! again but that seems a bit redundant

  11. myininaya
    • 2 years ago
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    Turing I like that you recalled that k+1>2 since k>=4 So \[2^{k+1}=2^k \times 2 <k! \times (k+1)=(k+1)!\]

  12. myininaya
    • 2 years ago
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    This proves the thingy

  13. TuringTest
    • 2 years ago
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    yes, much more succinct myin :)

  14. myininaya
    • 2 years ago
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    You can say that middle part was by induction when I used 2^k<k!

  15. myininaya
    • 2 years ago
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    Or turing I and turing or interchangeable I guess

  16. myininaya
    • 2 years ago
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    are*

  17. amistre64
    • 2 years ago
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    i managed to eke out a 12/10 on the homework nonetheless :)

  18. TuringTest
    • 2 years ago
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    then quit yer whining :P

  19. phi
    • 2 years ago
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    You proof starts off with what looks like an assumption that the n+1 case is true, and concludes that 2^k < k! * (k+1)/2 this is true, but you want to prove the case for 2^(k+1) In other words: assume 2^k < k! 2* 2^k < 2* k! (multiply both sides by 2 does not change the relation) 2^(k+1) < 2*k! with 2< (k+1) true for all k>1 2*k! < (k+1)*k! 2*k! < (k+1)! and we have 2^(k+1) < 2*k! < (k+1)! and conclude 2^(k+1) < (k+1)!

  20. amistre64
    • 2 years ago
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    by modifying: 2^(k+1) < (k+1)! into a form that can be compared against, 2^k < k! we can deduce the truth value. By modifying it into: 2^k < k! * (k+1)/2 Since (k+1)/2 is a positive value that increases k!; and since 2^k doesnt increase at all; then by basic mathing (or logic) skills; an increase in something that is already bigger will remain bigger.

  21. amistre64
    • 2 years ago
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    at least that has been all the examples that the teacher has done on the board for us

  22. phi
    • 2 years ago
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    I think it is the difference between if/then and if and only if? Your conclusion is definitely true, and almost obviously implies "the other direction" but it is not explicitly proving the case.

  23. phi
    • 2 years ago
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    in other words you end up with 2^k < k! * (k+1)/2 but the inductive hypothesis is 2^k< k! so proving 2^k < k! * (k+1)/2 is not very interesting.

  24. amistre64
    • 2 years ago
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    i agree that its not that interesting :) what of my first idea for the proof; that: 2 < {5,6,7,8,...} ?

  25. phi
    • 2 years ago
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    that is true. But the best way is to assume the inductive hypothesis for k, and show that it implies the truth of the k+1 case

  26. amistre64
    • 2 years ago
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    i was finally able to parse thru your proofing. I like it.

  27. amistre64
    • 2 years ago
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    im still unsure as to how mine fails; but thats more a testimony to my ignorance than anyting else :)

  28. TuringTest
    • 2 years ago
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    just out of curiosity since I still have trouble with induction at times, what do you make of my attempt @phi ? (I omitted the first step obviously)

  29. phi
    • 2 years ago
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    You are assuming 2^(k+1) < (k+1)!, manipulating this to the form 2^k < k! (k+1)/2, and so proving that if the first is true, the latter is true. This is backwards.

  30. amistre64
    • 2 years ago
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    ah, so I should try to work the 2^k < k! up to it, instead of backtracking back down

  31. phi
    • 2 years ago
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    ah, so I should try to work the 2^k < k! up to it, instead of backtracking back down yes.

  32. phi
    • 2 years ago
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    @TuringTest when you write the second statement \[2^{k+1}=2\cdot2^k<(k+1)k!\] that is what you are to show. I think I would start with your first statment, and then write down you last statement, and work upwards to the conclusion.

  33. myininaya
    • 2 years ago
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    but 2<(k+1) since k>=4 and 2<k!

  34. myininaya
    • 2 years ago
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    oops 2^k<k! *

  35. TuringTest
    • 2 years ago
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    I see what you are saying @phi, but though it may not be the most kosher way to do it I don't think that it invalidates the proof

  36. phi
    • 2 years ago
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    But the way you wrote it, you are assuming each step going down implies the truth of going back up (which it is), but the reader is left to figure that out. And it is possible that one of these steps was not reversible. In such a case you may not notice....

  37. TuringTest
    • 2 years ago
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    fair point...

  38. phi
    • 2 years ago
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    And if you write it in the order suggested, it is clear what is being assumed ( 2^k < k! and 2< k+1), and the conclusion follows from simple steps.... very convincing, and it leaves no doubts.

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