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.
 A: $2n$ is an even number and $2n+1$ is odd, and 
$$3n=2n+n\geq 2n+2>2n+1$$
A: Observe that $$\;2n<2n+1<3n\;$$ Can you prove these two inequalities?
A: We start with the assumption that $n \ge 2$. Then we have
$$3n = 2n + n \ge 2n + 2.$$
So we get that $2n+1$ is an odd integer, and $2n < 2n+1 < 3n$.
A: Write numbers between $2n$ and $3n$
$$
2n<2n+1 ,2n+1 ,2n+3 ,...,2n+(n-1)<3n 
$$
if $n>2$ 
there is $n-1$ numbers between $2n$, $3n$ 
number of terms=
$$ \frac{\text{last} -\text{first}}{\text{step}}+1=\\\frac{3n -2n}{1}+1=n+1\\$$
between $2n$, $3n$ are $(n+1)-2$ terms.
A: $\color{green}{2n+1}$ is odd and $2n<\color{green}{2n+1}<3n\impliedby 0<1<n$.
A: 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.
