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Odd proof?

  1. Dec 18, 2006 #1
    In my maths textbook it asks to prove 2^n<=n! for all n>=4

    I could prove it no problems using induction but could show 2^n<n! without the equality inequality.

    My question is why would the textbook ask for a weaker condition? Is it a misprint?

    If I can show < then it automatically implies <= holds as well dosen't it?
  2. jcsd
  3. Dec 18, 2006 #2


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    Yes, if x is strictly less than y, then x is certainly less than or equal to y. With regards to your former question, it might just be a typo- either way, you can solve the problem using induction.
  4. Dec 18, 2006 #3


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    It can't be a missprint. See what happens with n=4.


    PS. Apparently, it can be a missprint.
    Last edited: Dec 18, 2006
  5. Dec 18, 2006 #4


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    2^4=16 , 4!= 24, so 2^4<4!
  6. Dec 18, 2006 #5
    I was only asking the relationship between 2^n and n!

    Looking at the graphs for 2^n and n! it seems that no where is 2^n=n!

    n>=4 is definitely correct.
  7. Dec 18, 2006 #6
    Look at what are the factors of 2^n and the factors of n!...
  8. Dec 19, 2006 #7


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    There are two "=" signs in question. The first one is definitely a misprint but n>=4 is correct.
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