There exists a perfect square between $n$ and $2n$

Question

I need help proving the following mathematical statement:

Prove that for every $n$ there is a $k$ such that $n \leq k^2 \leq 2n$ where $n,k\in \mathbb{N}$.

Could someone get me started or give me some advice on proving this statement?

Answer 1

Complete the base case of $n=1$ first. Then suppose that the inequality holds for all $n \in \{1,2\ldots, p\}$. We at least know $p \leq k^2 \leq 2p$ for some $k$ and some $p$. If $k$ is such that $p+1 \leq k^2 \leq 2(p+1)$ you're done. So suppose the case that $p+1>k^2$. This means $p = k^2$, which leads to the inequality $$p+1<(k+1)^2 = k^2+2k+1 \leq 2k^2+1 < 2k^2+2 = 2p+2 = 2(p+1)$$

The inequality above only works if $k \geq 2$. Can you prove this must be true for $n>1$?