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Math induction

  1. Jan 26, 2010 #1
    I have to prove by induction that for n>= 4 that 2^n > n^2.

    So i start with the base case and I get 16 >= 16 which is true.
    Then I assume for k that 2^k >= k^2 for k.
    Now I have to that 2^(k+1) >= (k+1)^2

    Now going back to 2^k >= k^2 if I multiply both sides by 2 I get
    2*2^k >= 2*k^2
    2^(k+1) >= 2*k^2

    and from here I get stuck. Any help would be appreciated. Thanks.
     
  2. jcsd
  3. Jan 26, 2010 #2

    CRGreathouse

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    4k^2 >= 2k^2
     
  4. Jan 26, 2010 #3
    I'm sorry but I don't know what you mean by that.
     
  5. Jan 27, 2010 #4

    radou

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    It works for the basis of induction, n = 4. Assume it works for some k, so 2^k >= k^2. Then, for k+1, you have 2^(k+1) = 2*2^k >= 2*k^2. Now, can you show that 2*k^2 >= (k+1)^2, for n >= 4?
     
  6. Jan 27, 2010 #5
    Okay I see why you do that so I guess I have to prove 2*k^2 >= (k+1)^2 by induction.

    When n = 4 we have
    2*(4)^2 = 32
    (4+1)^2 = 25
    So 32 > 25 which is true.

    Then we assume it is true for all n right and now I have to show that
    2*(k+1)^2 >= ((k+1)+1)^2
    which is the same thing as
    2*(k+1)^2 >= (k+2)^2

    Okay now I'm stuck again.
     
  7. Jan 27, 2010 #6

    Mentallic

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    So you're trying to prove [itex]2k^2\geq (k+1)^2[/itex] for [itex]k \geq 4[/itex]
    Expanding gives: [itex]2k^2\geq k^2+2k+1[/itex]
    and now collecting like terms: [itex]k^2-2k-1\geq 0[/itex]

    Can you finish this off?
     
  8. Jan 27, 2010 #7
    No not really. I'm not even sure I'm going the right way because my professor was telling me that I will eventually have to prove another property which is n^2 >= 2n+1 for n>=3. Thanks.
     
  9. Jan 27, 2010 #8

    Mentallic

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    That's exactly where you're heading! :tongue:

    I just asked you to prove [itex]k^2-2k-1\geq 0[/itex] for [itex]k\geq 4[/itex] which is basically the same as what your professor said you'll have to do.

    Make the above inequality a perfect square.
     
  10. Jan 27, 2010 #9
    Okay so from k^2 - 2k - 1 >= 0 for k>= 4 we can rewrite it as

    (k-1)^2 >= 0 for k>= 4. Yes I'm sorry but I still don't see it.
     
  11. Jan 27, 2010 #10

    Mentallic

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    No that's not exactly right.
    [itex](k-1)^2=k^2-2k+1[/itex] and you have [itex]k^2-2k-1[/itex]

    What you're actually looking for is [itex](k-1)^2-2\geq 0[/itex]. So, for what positive k is [itex](k-1)^2\geq 2[/itex] ?
     
  12. Jan 27, 2010 #11
    It is 3 because (3-1)^2 >= 2 because 2^2 >= 2 which is 4 >= 2 but what does that prove then?
     
  13. Jan 27, 2010 #12

    Mentallic

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    Yes that's right.

    Well if we backtrack your steps, notice that we were trying to prove [itex]2^k\geq k^2[/itex] o:) Our first step was to show for a base case, and while n=1 and n=2 worked, n=3 is untrue so we started at the first case of n=4.
    Now in the 3rd step you've proven that [itex]2^{k+1}\geq (k+1)^2[/itex] by assuming [itex]2^k\geq k^2[/itex] in your 2nd step and then substituting this assumed result to obtain [itex]2k^2\geq (k+1)^2[/itex].

    That is, if [itex]2^k\geq k^2[/itex] then [itex]2^{k+1}\geq 2k^2[/itex] and surely if we're trying to prove [itex]2^{k+1}\geq (k+1)^2[/itex] then we can substitute [itex]2^{k+1}=2k^2[/itex].

    Now you've shown by logic that for [itex]k\geq 3[/itex], [itex](k-1)^2\geq 2[/itex] which, after giving a little word about how mathematical induction works, you've essentially proven the question for [itex]k\geq 4[/itex]
     
  14. Jan 27, 2010 #13
    Okay I kind of get. By showing that for k>=3 we know (k-1)^2 >= 2 which originally means 2k^2 >= (k+1)^2 for k>=4 but by proving the first inequality we have proved the previous inequality and thus by proving the previous inequality this means we have proved 2^k >= k^2 for k >= 4. Am I right?
     
  15. Jan 27, 2010 #14

    Mentallic

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    Yes, exactly! :approve:
     
  16. Jan 27, 2010 #15
    Cool, Thanks! :D
     
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