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Is this induction proof valid?

  1. Oct 25, 2011 #1
    1. The problem statement, all variables and given/known data

    My induction proof was really easy and simple, and people were having problems and doing substitutions and other complicated things. Normally I'd take pride in making a simpler proof but, there are really brilliant people in this class and I'm just starting to get the grasp of proofs, so it's not comforting.

    Prove that for all real numbers:

    [itex](1+\frac{1}{2})^{n} \geq 1 + \frac{n}{2}[/itex]

    2. Relevant equations



    3. The attempt at a solution

    Let S be the set of natural numbers such that:

    [itex](1+\frac{1}{2})^{n} \geq 1 + \frac{n}{2}[/itex]

    Let n = 1

    [itex](1+\frac{1}{2})^{1} = 1.5[/itex]

    [itex]1 + \frac{1}{2} = 1.5[/itex]

    Therefore, 1 E S

    Assume this is true for n = k

    [itex](1+\frac{1}{2})^{k} \geq 1 + \frac{k}{2}[/itex]

    Check if this implies truth for n = k + 1
    Left side:
    [itex](1+\frac{1}{2})^{k+1}[/itex]

    [itex](1+\frac{1}{2})^{k}(1+\frac{1}{2})[/itex]

    Right side:

    [itex]1 + \frac{k+1}{2}[/itex]

    [itex]1 + \frac{k}{2} + \frac{1}{2}[/itex]


    [itex](1+\frac{1}{2})^{k}(1+\frac{1}{2}) \geq 1 + \frac{k}{2} + \frac{1}{2}[/itex]

    Is a true statement because our initial assumption asserts that

    [itex](1+\frac{1}{2})^{k} \geq 1 + \frac{k}{2}[/itex]

    Here, the greater term is multiplied by 1.5, and the lower term has .5 added to it. A number multiplied by 1.5 is always greater than the same number added to .5, except in the case of 1, in which case they are equal. Since the first term is greater than the second, and is operated on a way such that a greater result is yielded even if the two were equal,

    [itex](1+\frac{1}{2})^{k}(1+\frac{1}{2}) \geq 1 + \frac{k}{2} + \frac{1}{2}[/itex]

    Must be true.


    Therefore S is inductive.

    Therefore S = Natural Numbers.
     
  2. jcsd
  3. Oct 25, 2011 #2

    Mark44

    Staff: Mentor

    I think you are hand-waving here. Start with the left side and show that it is >= 1 + (k + 1)/2.
     
  4. Oct 25, 2011 #3

    Dick

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    The flaw is this. "A number multiplied by 1.5 is always greater than the same number added to .5, except in the case of 1, in which case they are equal." You don't have the "same number" on both sides. Just write (1+1/2)^k*(1+1/2) as (1+1/2)^k+(1+1/2)^k/2 and try that again.
     
  5. Oct 25, 2011 #4
    Why does it matter if they aren't the same number? The left is greater than the right by induction hypothesis.
     
  6. Oct 25, 2011 #5

    Dick

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    The problem is in your exposition. The language is confusing. I think you should go back to taking pride in making a simpler proof. It's not that hard. You want to show (1+1/2)^k*(1+1/2)=(1+1/2)^k+(1+1/2)^k/2>=1+k/2+1/2.
     
  7. Oct 25, 2011 #6
    What I am saying is basically, if I assume that x is greater than y, then x + 2 is greater than y + 1.

    Of course, +1 and +2 are examples of operations.
     
  8. Oct 25, 2011 #7

    Dick

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    Sure, now why don't you try to clear up the original proof?
     
  9. Oct 25, 2011 #8
    Ok, I'll take a shot. So you're telling me that I should translate the jist of it into math?
     
  10. Oct 25, 2011 #9

    Dick

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    Yes, show (1+1/2)^k*(1+1/2)=(1+1/2)^k+(1+1/2)^k/2>=1+k/2+1/2. I've been repeating that for a while. It's pretty direct.
     
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