1. ## 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

2. ## Induction

Hello JoeCrow
Originally Posted by JoeCrow
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.