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Proof for a simple inequality

  1. Mar 4, 2010 #1
    1. The problem statement, all variables and given/known data

    Prove that:
    [tex]2^n \geq n + 5[/tex] for all integer n > 5

    2. Relevant equations
    n/a

    3. The attempt at a solution
    This is what I did, just want to see if it's right.
    Using mathematical induction:

    Basis step:
    When n = 3,
    [tex]2^3 \geq 3 + 5[/tex]
    Since 8 = 8, base case is proven

    Inductive step:

    Assume [tex]2^k \geq k + 5[/tex] for all integers k greater or equal to n.

    We want to prove that [tex]2^{k+1} \geq k + 1 + 5[/tex]

    Using our assumption, [tex]2^k \geq k + 5[/tex], I know that:

    [tex]2(2^k) \geq 2(k + 5)[/tex]

    =[tex]2^{k+1} \geq 2k + 10[/tex]

    From here, if I can prove 2k+10 is greater or equal to k+6, then it will prove
    [tex]2^{k+1} \geq k + 6[/tex]

    So:

    [tex]2k+10 \geq k + 6[/tex]

    [tex] 2k \geq k - 4[/tex]

    [tex] k \geq -4[/tex]

    Thus is proven because k was said to be greater or equal to n which is at least 3. So the inequality is true.

    Since inductive step is proven, [tex]2^3 \geq 3 + 5[/tex] must be true.

    QED


    Does this proof stand? I haven't had much practice with inequalities... so I'm very unsure about this proof.

    Thanks.
     
    Last edited: Mar 4, 2010
  2. jcsd
  3. Mar 4, 2010 #2
    Your reasoning is fine, but you should clean up your proof a bit. If you were told to prove the inequality for n > 5, and it evidently holds for n greater than or equal to 3, then the first thing you should do is state that this latter claim is the claim you are going to prove (even though it is a slightly "stronger" or better claim) by induction.

    Under this claim, your base case is fine. For your inductive step, you should stick to one letter. If you are going to use k instead of n, which is fine since you are already given an n in the problem statement, then you actually need to say that the statement holds for k. This means that the inductive step is

    Assume [itex]2^k \geq k + 5 [/itex] for all integers [itex]k \geq 3[/itex]

    (It doesn't matter whether you use k or n or some other letter to denote the natural number that you will increment, but referring to multiple letters in your proof might cause slight confusion)

    Besides this, there is nothing wrong with the proof, except at the very end, you should state that the statement you proved is true, not just its base case (this may be a typo).
     
  4. Mar 5, 2010 #3
    Yeah, that last bit was a typo. I'll post future proofs on here to see if I'm improving. Thanks for the tips, appreciate it.
     
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