Problem 1409

### Part (a)
We need to prove that for any $n \in M$, $\{\sqrt{n}\} \geq 0.011$.

1. Consider $n \in M$. Since $n$ is not a perfect square, we can write $n = a^2 + b$ where $0 < b < 2a$ and $a$ is an integer.
2. The fractional part of $\sqrt{n}$ is given by:
$$\{\sqrt{n}\} = \sqrt{n} - \lfloor \sqrt{n} \rfloor = \sqrt{a^2 + b} - a$$
3. We need to show that:
$$\sqrt{a^2 + b} - a \geq 0.011$$
4. Using the approximation for small $b$:
$$\sqrt{a^2 + b} \approx a + \frac{b}{2a}$$
Therefore:
$$\sqrt{a^2 + b} - a \approx \frac{b}{2a}$$
5. Since $0 < b < 2a$, we have:
$$0 < \frac{b}{2a} < 1$$
6. To ensure $\frac{b}{2a} \geq 0.011$, we need:
$$b \geq 0.022a$$
7. Since $b$ is an integer and $0 < b < 2a$, there will always be some $b$ such that $b \geq 0.022a$. Hence:
$$\sqrt{a^2 + b} - a \geq 0.011$$
8. Therefore, for any $n \in M$, $\{\sqrt{n}\} \geq 0.011$.

### Part (b)
We need to prove that there exists a number $n \in M$ such that $\{\sqrt{n}\} < 0.0115$.

1. Consider $n = 44^2 + 1 = 1937$. Since $1937$ is not a perfect square, $n \in M$.
2. The fractional part of $\sqrt{1937}$ is given by:
$$\{\sqrt{1937}\} = \sqrt{1937} - \lfloor \sqrt{1937} \rfloor = \sqrt{1937} - 44$$
3. Using the approximation:
$$\sqrt{1937} \approx 44 + \frac{1}{2 \cdot 44} = 44 + \frac{1}{88}$$
4. Therefore:
$$\{\sqrt{1937}\} \approx \frac{1}{88}$$
5. We need to show that:
$$\frac{1}{88} < 0.0115$$
6. Calculating:
$$88 \cdot 0.0115 = 1.012$$
Since $1.012 > 1$, we have:
$$\frac{1}{88} < 0.0115$$
7. Therefore, there exists $n = 1937 \in M$ such that $\{\sqrt{n}\} < 0.0115$.