# Prove: If n is an odd natural number, there are $\ s = \lceil \frac{n}{2} \rceil$ odd integers less than or equal to n

If n is an odd natural number, there are $\ s = \lceil \frac{n}{2} \rceil$ odd integers less than or equal to n

This is intuitive as the sum of n odd numbers is s squared, where s is the quantity of odd numbers in that sequence.

1 + 3 = $\ 2^2$

1 + 3 + 5 = $\ 3^2$

1 + 3 + 5 + 7 = $\ 4^2$

1 + 3 + ... + n = $\ s^2$

Since every other number is odd, it would seem that the sum of n odd natural numbers less than or equal to n would be the sum of $\ s = \lceil \frac{n}{2} \rceil$ odd natural numbers total.

However, I am puzzled by how to prove it. I have attempted a proof by induction, but I am not sure how to handle the ceiling function. It seems that it is necessary to represent $\ { \lceil \frac{n}{2} \rceil}^2$ in order to prove it.

• For an odd natural number $n=2k-1$ (where $k\geq 1$), there are $k$ odd integers $\leq n$ and we have $\lceil n/2\rceil=\lceil k-1/2\rceil=k$. Conclude. – Prasun Biswas Oct 24 '17 at 3:59

I think you're overthinking the problem. This doesn't need to have anything to do with squaring numbers; your task is just to count the odd natural numbers less than or equal to $n$. (Note: your problem text says "odd integers," but there are clearly infinitely many of those, so I assume you mean "odd natural numbers.")
As far as how to prove this, induction is fine. Whatever you do will require the definition of an odd number -- that is, you'll need to use that $n = 2k + 1$ for some integer $k$. Note that $\lceil \frac{2k + 1}{2} \rceil$ is something you can compute, which will be helpful in your proof. I'll leave the rest of the details for you to fill in but can provide more hints if you need.