# 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.

• What was your solution ? – Yves Daoust Nov 11 '14 at 19:39
• I was subtracting, 3n - 2n = n, and saying that n-1 is the number of integers between the two. I was trying to prove that if there were at least 2 consecutive integers between them, then one had to be odd since integers are odd then even. I see now that that will not work. – Brenda Nov 11 '14 at 19:49
• Your solution is very close to be correct: indeed in any sequence of at least $2$ integers there is an odd number. But as the sequence length is $n-1$, the argument can't be used for $n=2$ (it works for all $n>2$). To fix your solution, just split in two cases: $n=2$ and $n>2$. – Yves Daoust Nov 11 '14 at 19:56

$2n$ is an even number and $2n+1$ is odd, and $$3n=2n+n\geq 2n+2>2n+1$$

Observe that $$\;2n<2n+1<3n\;$$ Can you prove these two inequalities?

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 $2n < 2n+1 < 3n$.

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.

• This solution is said to be "non-constructive" as it does not tell exactly which is the odd number (no formula given), it just guarantees that there is one. – Yves Daoust Nov 11 '14 at 19:49

$\color{green}{2n+1}$ is odd and $2n<\color{green}{2n+1}<3n\impliedby 0<1<n$.

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.