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Math Help - Inductive Proof help!

  1. #1
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    Inductive Proof help!

    I am having some trouble figure out the steps to prove these two problems using induction.


    In the two questions F(n) is the product of the first n odd positive integers. And G(n) is the product of the first n even positive integer. I'm suppose to show the four steps of the inductive proof and justify my changes of inequalities by each rule.

    1. Use induction to show that F(n)G(n) = (2n)! n >=1

    2. Use induction to show that F(n) < G(n) for all n>=1
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  2. #2
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    Induction

    Hello JoeCrow
    Quote Originally Posted by JoeCrow View Post
    I am having some trouble figure out the steps to prove these two problems using induction.


    In the two questions F(n) is the product of the first n odd positive integers. And G(n) is the product of the first n even positive integer. I'm suppose to show the four steps of the inductive proof and justify my changes of inequalities by each rule.

    1. Use induction to show that F(n)G(n) = (2n)! n >=1

    2. Use induction to show that F(n) < G(n) for all n>=1
    1. The (n+1)^{th} odd positive integer is (2n+1) and the (n+1)^{th} even integer is 2n+2.

    So F(n+1) = F(n)\times(2n+1) and G(n+1) = G(n) \times (2n+2)

    So if P(n) is the propositional function F(n)G(n) = (2n)! :

    P(n) \Rightarrow F(n+1)G(n+1) = (2n)!(2n+1)(2n+2) = (2n+2)! = (2(n+1))! \Rightarrow P(n+1)

    P(1) is F(1)G(1) = 2, which is true. So P(n) is true for all n \ge 1


    2. Let P(n) be F(n) < G(n)

    Then P(n) \Rightarrow F(n)(2n+1)< G(n)(2n+1), since (2n+1) > 0

    \Rightarrow F(n)(2n+1)<G(n)(2n+1)+G(n), since G(n) > 0

    \Rightarrow F(n)(2n+1)<G(n)(2n+2)

    \Rightarrow F(n+1)<G(n+1)

    \Rightarrow P(n+1)

    P(1) is ... etc.

    Grandad
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