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amistre64 Group TitleBest ResponseYou've already chosen the best response.1
I got marked off for some reason
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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 \[2^k < k!\frac{(k+1)}{2}\hspace{16em}QED\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
I don't see anything blatantly wrong with it... I came up with a slightly different argument
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
What does 2<{5,6,7,8...} mean?
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Oh. You mean 2 is less than any element in that set.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
\[2^k<k!\implies\frac{2^k}{k!}<1\]\[2^{k+1}=2\cdot2^k<(k+1)k!\]\[2\cdot\frac{2^k}{k!}<k+1\]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...
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
thats a fine rendition; my prof would rather have us get it into a form similar to the assumption to compare with tho
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
well I guess you could proceed to multiply both sides by k! again but that seems a bit redundant
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Turing I like that you recalled that k+1>2 since k>=4 So \[2^{k+1}=2^k \times 2 <k! \times (k+1)=(k+1)!\]
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
This proves the thingy
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
yes, much more succinct myin :)
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
You can say that middle part was by induction when I used 2^k<k!
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Or turing I and turing or interchangeable I guess
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
i managed to eke out a 12/10 on the homework nonetheless :)
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
then quit yer whining :P
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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)!
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
at least that has been all the examples that the teacher has done on the board for us
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
i agree that its not that interesting :) what of my first idea for the proof; that: 2 < {5,6,7,8,...} ?
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
i was finally able to parse thru your proofing. I like it.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
im still unsure as to how mine fails; but thats more a testimony to my ignorance than anyting else :)
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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)
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
ah, so I should try to work the 2^k < k! up to it, instead of backtracking back down
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
ah, so I should try to work the 2^k < k! up to it, instead of backtracking back down yes.
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
@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.
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
but 2<(k+1) since k>=4 and 2<k!
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
oops 2^k<k! *
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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....
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
fair point...
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago
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