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Thread: 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 $\displaystyle (n+1)^{th}$ odd positive integer is $\displaystyle (2n+1)$ and the $\displaystyle (n+1)^{th}$ even integer is $\displaystyle 2n+2$.

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

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

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

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


    2. Let $\displaystyle P(n)$ be $\displaystyle F(n) < G(n)$

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

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

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

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

    $\displaystyle \Rightarrow P(n+1)$

    $\displaystyle P(1)$ is ... etc.

    Grandad
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