Question

Please help, mathematical proofs

Prove that 3^n > 2^n for all natural numbers n>=1
you have to put:
Base Case
Inductive hypothesis
Inductive step
Not sure how to structure it, any help appreciated!

Answer 1

the base is higher. That one way of looking at it.

Answer 2

Yeah I know, the question is just an example.. im looking for how to structure or write out the info for marks..

Answer 3

For sure \(3>2\)

Suppose \(3^n>2^n\) for some \(n\in \mathbb{N}\) (this is our inductive hypothesis). We need to show that this implies \(3^{n+1}>2^{n+1}\). To that end, we have \(3^n>2^n\implies 3(3^n)>3(2^n)\) and for sure \(3(2^n)>2(2^n)\) so that \(3(3^n)>2(2^n)\iff 3^{n+1}>2^{n+1}\) as desired.

Answer 4

In general.

Show it is true for \(n=1\) (or whatever your base case is).

Assume it is true for some \(n\).

Show that this assumption implies it is true for \(n+1\).

Done \(\checkmark\)

Answer 5

Thanks