Prove: For any integer $n \geq 2$, there is an odd number P such that $2n \lt P \lt 3n$ I am in high school and had this for a homework problem.  I got it wrong, but the teacher did not post the correct answer.  Any help would be appreciated.  It is about writing proofs.
Prove that for any integer n greater than or equal to 2, there is always an odd number P between 2n and 3n.

By induction.
Assume that the property is true for $n$, and prove it is true for $n+1$: there is an odd number between $2n$ and $3n$ $\implies$ there is an odd number between $2n+2$ and $3n+3$.
Indeed, take some odd number that is between $2n$ and $3n$, let $m$. Then $m+2$ is also an odd number, it is larger than $2n+2$, and smaller than $3n+2$, hence smaller than $3n+3$.
The base case is: there is an odd number between $2.2$ and $3.2$: it is 5.
