1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What was your solution ? $\endgroup$ – Yves Daoust Nov 11 '14 at 19:39
  • $\begingroup$ 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. $\endgroup$ – Brenda Nov 11 '14 at 19:49
  • $\begingroup$ 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 $n>2$. $\endgroup$ – Yves Daoust Nov 11 '14 at 19:56
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Yves Daoust Nov 11 '14 at 19:49
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.