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Math Help - Mathematical Induction to Prove Inequalities

  1. #1
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    Mathematical Induction to Prove Inequalities

    I was wondering if anyone could help me out on this problem. Use mathematical induction to establish the following inequality.
    (1+1/2)^n>=1+n/2, for n belonging to N (natural numbers)
    n=1
    (1+1/2)^1>=1+1/2 true for n=1
    Assume n=k
    (1+1/2)^k>=1+k/2
    I am stuck on how to do the n=k+1. Any help is appreciated.
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  2. #2
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    Quote Originally Posted by discretec View Post
    I was wondering if anyone could help me out on this problem. Use mathematical induction to establish the following inequality.
    (1+1/2)^n>=1+n/2, for n belonging to N (natural numbers)
    n=1
    (1+1/2)^1>=1+1/2 true for n=1
    Assume n=k
    (1+1/2)^k>=1+k/2
    I am stuck on how to do the n=k+1. Any help is appreciated.

    The inductive hypothesis is that \left(1+\frac{1}{2}\right)^n\geq 1+\frac{n}{2} , so:

    \left(1+\frac{1}{2}\right)^{n+1}=\left(1+\frac{1}{  2}\right)^n\left(1+\frac{1}{2}\right)\buildrel\tex  t{ind. hyp.}\over \geq \left(1+\frac{n}{2}\right)\left(1+\frac{1}{2}\righ  t) =1+\frac{n+1}{2}+\frac{n}{4}\geq 1+\frac{n+1}{2} . QE.D.

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    The inductive hypothesis is that \left(1+\frac{1}{2}\right)^n\geq 1+\frac{n}{2} , so:

    \left(1+\frac{1}{2}\right)^{n+1}=\left(1+\frac{1}{  2}\right)^n\left(1+\frac{1}{2}\right)\buildrel\tex  t{ind. hyp.}\over \geq \left(1+\frac{n}{2}\right)\left(1+\frac{1}{2}\righ  t) =1+\frac{n+1}{2}+\frac{n}{4}\geq 1+\frac{n+1}{2} . QE.D.

    Tonio
    Can I ask what you mean by QE.D.?
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    Quote Originally Posted by discretec View Post
    Can I ask what you mean by Q.E.D.?

    Q.E.D. = Quod erat demonstrandum = what had to be demonstrated.

    According to a friend of mine, Q.E.D. = Queer and Ethereal Dork (it's tough to love mathematics...**sigh**)

    Tonio
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  5. #5
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    Quote Originally Posted by tonio View Post
    The inductive hypothesis is that \left(1+\frac{1}{2}\right)^n\geq 1+\frac{n}{2} , so:

    \left(1+\frac{1}{2}\right)^{n+1}=\left(1+\frac{1}{  2}\right)^n\left(1+\frac{1}{2}\right)\buildrel\tex  t{ind. hyp.}\over \geq \left(1+\frac{n}{2}\right)\left(1+\frac{1}{2}\righ  t) =1+\frac{n+1}{2}+\frac{n}{4}\geq 1+\frac{n+1}{2} . QE.D.

    Tonio
    Thank you for your help. I am going to stare at this one for awhile. These types of induction problems give me so much trouble. lol
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  6. #6
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    Quote Originally Posted by tonio View Post
    Q.E.D. = Quod erat demonstrandum = what had to be demonstrated.

    According to a friend of mine, Q.E.D. = Queer and Ethereal Dork (it's tough to love mathematics...**sigh**)

    Tonio
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