Solutions

a) We have this inequality:

$$\lfloor x\rfloor \le x < \lfloor x\rfloor +1 \le x+1<y.$$

Hence, $\lfloor x+1\rfloor$ is the integer $k$.

b) By Theorem 3, we can find $n\in\mathbb{N}$ such that

$$\frac{1}{n}<y-x.$$

This implies that $ny-nx>1$, and by a), there exists an integer $m$ such that

$$nx<m<ny.$$

Hence,

$$x<r=\frac{m}{n}<y.$$

c) We know that $\sqrt{2}$ is an irrational number, so is $\frac{\sqrt{2}}{2}$. Also, $0<\frac{\sqrt{2}}{2}<1$. Hence,

$$\begin{align*} 0<\frac{\sqrt{2}}{2}<1 & \implies 0<\frac{\sqrt{2}}{2}(s-r)<s-r \\ & \implies r<\frac{\sqrt{2}}{2}(s-r)+r<s. \end{align*}$$

Thus,

$$\frac{\sqrt{2}}{2}(s-r)+r$$

is the irrational number we're looking for. We proved in Exercise 2.12 that such a number is irrational, given the operations on $\sqrt{2}$.

d) By b), there exists a rational number $r$ such that $x<r<y$. By applying b) again, there exists a rational number $s$ such that $x<r<s<y$. By applying c), there exists an irrational number $z$ such that $x<r<z<s<y$. Hence, the irrational number $z$ is between $x$ and $y$.