A community for students.
Here's the question you clicked on:
 0 viewing
amistre64
 4 years ago
So what would be wrong with this proof by induction?
amistre64
 4 years ago
So what would be wrong with this proof by induction?

This Question is Closed

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

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

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

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1It 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

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

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

TuringTest
 4 years ago
Best 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...

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

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

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

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1This proves the thingy

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

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

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

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1then quit yer whining :P

phi
 4 years ago
Best ResponseYou've already chosen the best response.1You 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)!

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1by 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.

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

phi
 4 years ago
Best ResponseYou've already chosen the best response.1I 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.

phi
 4 years ago
Best ResponseYou've already chosen the best response.1in 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.

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

phi
 4 years ago
Best ResponseYou've already chosen the best response.1that 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

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

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1just 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)

phi
 4 years ago
Best ResponseYou've already chosen the best response.1You 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.

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

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

phi
 4 years ago
Best 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.

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1I 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

phi
 4 years ago
Best ResponseYou've already chosen the best response.1But 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....

phi
 4 years ago
Best ResponseYou've already chosen the best response.1And 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.