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 $(n+1)^{th}$ 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), since $G(n) > 0$

$\Rightarrow F(n)(2n+1)

$\Rightarrow F(n+1)

$\Rightarrow P(n+1)$

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