FINAL ROUND

a.) To prove the required statement it suffices to find the number $n\in M$ such that $\{\sqrt{n}\}={\min }_{k\in M}\{\sqrt{k}\}$. Any number $n\in M$ can be uniquely presented as $$n=k^2+r,(1)$$ where $1\le r\le 2k$, and $k=[\sqrt{n}]$ ($\cdot$ is the whole part of a number). Then $$\{\sqrt{n}\}=\sqrt{k^2+r}-k=\frac{r}{\sqrt{k^2+r}+k}.$$ Since the function $y=\sqrt{x}$ is increasing on any segment $[l^2,(l+1{)}^2]$, $l\in \mathbb{N}$, and its values belong to $[l,l+1]$, we see that $\{\sqrt{x}\}$ has its minimal value for positive integers $x\in (l^2,(l+1{)}^2)$ when $x=l^2+1$. So the minimal value of $\{\sqrt{n}\}$ one can look for among $n$ such that $r=1$ in representation (1), i.e., in other words, one must find the minimum of the function $$\{\sqrt{n}\}=\frac{1}{\sqrt{k^2+1}+k},(2)$$ where $n\in M$, $k=[\sqrt{n}]$. From (2) it follows that $\{\sqrt{n}\}$, $n\in M$, takes the minimal value when $k$ takes the maximal possible value. Since $44^2=1936<2015<2025=45^2$, $k$ takes its maximal value when $n=44\in M$. Hence for any $n\in M$ we have $$\{\sqrt{n}\}\ge \{\sqrt{44^2+1}\}=\frac{1}{\sqrt{44^2+1}+44}>\frac{1}{89}>\mathrm{0.011.}$$

b.) We have $$\{\sqrt{44^2+1}\}=\frac{1}{\sqrt{44^2+1}+44}<\frac{1}{88}<\mathrm{0.0115.}$$