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Mathematical Induction

  1. Jan 20, 2008 #1
    I am trying to understand induction and not having much luck. Here is my problem as I understand it

    Problem:
    n > or equal to 4, 2^n < n!

    Step 1) Prove it works for n=1 no, n=2 no, n=3 no, n=4 yes 16 < 24

    step 2) assume it works for n=k 2^k < k!

    Step 3) prove it works for n= (k+1)
    2^(k+1) < (k+1)!

    So now I don't know what to do. I am unsure what I am proving. I have successfully figured other problems but not with exponents, inequalities, and factorials.

    Can you give me a push in the right direction?
     
  2. jcsd
  3. Jan 20, 2008 #2
    Start with 2^(k+1) this is equal to 2*2^k do you understand this step?
    Then you know 2^k<k!, this was your induction hypothesis.
    So then 2^(k+1)<2*k! and you want that to be less than (k+1)! So how can you get (k+1)! form k!? And remember that k>=4.
     
  4. Jan 20, 2008 #3
    I don't understand how 2^(k+1) is = to 2*2^k
     
  5. Jan 20, 2008 #4
    How does it become 2*2^k
     
  6. Jan 20, 2008 #5
    That's basic arithmetic - you're probably just not recognizing it when it's typed this way. How about this:
    [tex]2^{k+1}[/tex]
    Do you know to re-express that in terms of 2 raised to single powers? I'm guessing you do, if you're doing a problem of this level.
     
  7. Jan 20, 2008 #6
    2*2^k means
    [tex]2\ast2^{k}[/tex] - did you get that?
     
  8. Jan 20, 2008 #7
    OKay Now I see it. Now you asked How I get (k+1)! form k! I have to admit I have no idea.
     
  9. Jan 20, 2008 #8
    I think you must be making this harder than it is ... just think of the definition of exponentiation.
    [tex]2^{k+1}=2\ast2^{k}[/tex]
    What does the left side of this expression mean, anyway? In words?
     
  10. Jan 20, 2008 #9
    Left side means 2^k * 2^1 Got it Sorry


    In the last part are you asking me how to get (k+1)! and k! to equal each other?
     
  11. Jan 20, 2008 #10
    How are (k+1)! and k! related? What is the definition of k!?
     
  12. Jan 20, 2008 #11
    k! would be k * k+1 * k+2 * k+3 etc. and (k+1)! would be (K+1) * (k+2) * (k+3) etc
     
  13. Jan 20, 2008 #12
    No, then 1! would be 1*2*3*4*5*6*7*8*9*10*11*12*13*14... which is greater than 1... so obviously that is not the defnition.
     
  14. Jan 20, 2008 #13
    k! = k(k-1)...2 * 1

    I am really searching here I have been out of school for 35 years. I don't think they taught this stuff back then.
     
  15. Jan 20, 2008 #14
    ngluth ... where did you think k * k+1 * k+2 * k+3 etc. was going to end?

    n! is defined to be 1 * 2 * 3 * ... * n-1 * n for any positive integer n.
     
  16. Jan 20, 2008 #15
    BTW ... they certainly did teach this stuff 35 years ago .... that's when I first learned it!! Luckily for me, I've continued to use it in the mean time. I've forgotten much else, though, so I sympathize.
     
  17. Jan 20, 2008 #16
    Thank you for all your time and patience. Now I can finish the rest of the assignment. This was the hardest one and I just could not understand what they were asking. Thanks
     
  18. Jan 20, 2008 #17
    I'm sending this again. Don't think it went the first time. Thanks for all your time and patience. I can now finish the assignment. This was the hardest problem. I just could not understand what was going on with it. Thanks. This is an awesome service you are providing. Thanks
     
  19. Jan 20, 2008 #18
    No problem. Induction is actually quite difficult to grasp at first - many people seem to be unable ever to get it. I've had problems trying to teach it to people who were never able to accept that it was logically sound - they just kept insisting that I'd proved only one case. Oh, well.

    Good luck with the rest of your studies - it sounds like you're turning some cranks that might have gathered a little rust over the years!
     
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