If $0<r^2<p\in\mathbb{Q}$, there exists positive integer $k$ such that $(r+1/k)^2<p$.

Question

Let $r,p\in\mathbb{Q}$ such that $0<r$ and $0<p$. Assume $r^2<p$. Then, there exists $k\in\mathbb{N}\setminus\{0\}$ for which $$\left(r+\frac{1}{k}\right)^2<p.$$

I am stuck on this problem. I know that for each $0<x\in\mathbb{Q}$ and $y\in\mathbb{Q}$, there exists $n\in\mathbb{N}\setminus\{0\}$ such that $y<nx$. I don't know how to use this property to prove the above. Can someone help me start the proof?

Answer 1

$$(r+1/k)^2 = r^2 + \frac{2r}{k} + \frac{1}{k^2} \le r^2 + \frac{2r+1}{k}.$$

Choose $k$ large enough so that $\frac{2r+1}{k} < p - r^2$.

Answer 2

Let $x = p - r^2$, then $x>0$. We know that there exists a natural $k$ such that $k(p-r^2) > 2r+1$, hence $p > r^2 + 2r/k + 1/k \ge r^2+2r/k+1/k^2$ (since $k\ge1)$, or $p>(r+1/k)^2$

Answer 3

Here is a more analytical way to prove that.

The function $x \mapsto x^2$ is continuous at $r$. Because $p-r^2>0$, there exists $\eta > 0$ such that for all $y \in [r-\eta, r+\eta]$, $|y^2-r^2|<p-r^2$.

Take $k$ such that $\frac{1}{k} < \eta$ (such a $k$ exists because $\frac{1}{k}$ tends to $0$ as $k$ tends to $+\infty$). For such a $k$, $r+\frac{1}{k} \in [r-\eta, r+\eta]$, so $$\left( r + \frac{1}{k}\right)^2 - r^2 < p-r^2$$ so $$\left( r + \frac{1}{k}\right)^2 < p$$