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amistre64 Group Title

So what would be wrong with this proof by induction?

  • one year ago
  • one year ago

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  1. amistre64 Group Title
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    I got marked off for some reason

    • one year ago
  2. amistre64 Group Title
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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 \[2^k < k!\frac{(k+1)}{2}\hspace{16em}QED\]

    • one year ago
  3. TuringTest Group Title
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    I don't see anything blatantly wrong with it... I came up with a slightly different argument

    • one year ago
  4. amistre64 Group Title
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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

    • one year ago
  5. myininaya Group Title
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    What does 2<{5,6,7,8...} mean?

    • one year ago
  6. myininaya Group Title
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    Oh. You mean 2 is less than any element in that set.

    • one year ago
  7. amistre64 Group Title
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    right

    • one year ago
  8. TuringTest Group Title
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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...

    • one year ago
  9. amistre64 Group Title
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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

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

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

    • one year ago
  12. myininaya Group Title
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    This proves the thingy

    • one year ago
  13. TuringTest Group Title
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    yes, much more succinct myin :)

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

    • one year ago
  15. myininaya Group Title
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    Or turing I and turing or interchangeable I guess

    • one year ago
  16. myininaya Group Title
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    are*

    • one year ago
  17. amistre64 Group Title
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    i managed to eke out a 12/10 on the homework nonetheless :)

    • one year ago
  18. TuringTest Group Title
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    then quit yer whining :P

    • one year ago
  19. phi Group Title
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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)!

    • one year ago
  20. amistre64 Group Title
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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.

    • one year ago
  21. amistre64 Group Title
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    at least that has been all the examples that the teacher has done on the board for us

    • one year ago
  22. phi Group Title
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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.

    • one year ago
  23. phi Group Title
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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.

    • one year ago
  24. amistre64 Group Title
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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,...} ?

    • one year ago
  25. phi Group Title
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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

    • one year ago
  26. amistre64 Group Title
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    i was finally able to parse thru your proofing. I like it.

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

    • one year ago
  28. TuringTest Group Title
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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)

    • one year ago
  29. phi Group Title
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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.

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

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

    • one year ago
  32. phi Group Title
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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.

    • one year ago
  33. myininaya Group Title
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    but 2<(k+1) since k>=4 and 2<k!

    • one year ago
  34. myininaya Group Title
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    oops 2^k<k! *

    • one year ago
  35. TuringTest Group Title
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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

    • one year ago
  36. phi Group Title
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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....

    • one year ago
  37. TuringTest Group Title
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    fair point...

    • one year ago
  38. phi Group Title
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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.

    • one year ago
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