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  • Feb 6, 2007, 07:12 PM
    dao
    dao
    how do I prove by mathematics induction with this problem
    2^n>n
  • Feb 6, 2007, 09:11 PM
    worthbeads
    Try using examples. For instance, if N=5, 2^5=32 and 32>n. It would work for a negative number too. If N=-2, 2^-2=1/4 and 1/4>-2. It works for zero. If N=0 then 2^0=1 and 1>0. SO if it works for positive numbers, negative numbers, and zero it should work, right? That's the way I would do it. There is probably a better way to do it.
  • Feb 7, 2007, 05:18 AM
    galactus
    Prove


    (1): is true, since

    (2): Assume is true: .

    Now, k+1<k+k=2(k) for k>1.

    From , we see that and conclude that .

    Thus, hence, therefore, is true and QED.

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