Some questions from "Mathematical Analysis" by Apostol. While solving the questions from Apostol's "Mathematical Analysis - 2nd Edition", I came across these questions:-


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*Given a real number $x$ and and integer $N > 1$, prove that there exists integers $h$ and $k$ with $0 < k \le N$ such that $\left| kx - h \right| < \dfrac{1}{N}$.

*If $x$ is irrational, prove that there are infinitely many rational numbers of the form $\dfrac{h}{k}$ with $k > 0$ such that $\left| x - \dfrac{h}{k} \right| < \dfrac{1}{k^2}$.

*Let $x$ be a positive rational number of the form $$x = \sum\limits_{k = 1}^{r} \dfrac{a_k}{k!}$$ where each $a_k$ is a non - negative integer with $a_k \le k - 1$ for $k \ge 2$ and $a_n > 0$. Let $\left[ x \right]$ denote the greatest integer in $x$. Prove that: $a_1 = \left[ x \right]$, $a_k = \left[ k! x \right] - k\left[ \left( k - 1 \right)! x \right]$ for $k = 2, 3, \dots, n$ and that $n$ is the smallest integer such that $n!x$ is an integer. Conversely, prove that every positive rational number can be expressed in this form in one and only one way.
My attempts in solving these questions:-


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*I was thinking of using the density of rational and irrationals in $\mathbb{R}$ to prove this, but was not really successful in doing so! So, I looked at the hint given in the book, which says: Consider $N + 1$ numbers $tx - \left[ tx \right]$ for $t = 0, 1, 2, \dots, N$ and then show that some pair differs by atmost $\dfrac{1}{N}$. The reason the use of density of rationals and irrationals was failing was that use of Archemedian Property to chose the $N$ was not being proper as $\dfrac{1}{N}$ is greater than some quantity.

*This solution clearly depends on the first one! If I assume that there are only finitely many rationals satisfying the given conditions, I could come to a contradiction using the above result but again, intuitively, all of it is fine! When I try to write it down, everything gets messed up!

*In this, I could solve the first part. That is proving $a_1 = \left[ x \right]$. I did this by the fact that $a_k \le \left( k - 1 \right)$ and hence $\sum\limits_{k = 2}^{r} \dfrac{a_k}{k!} < 1$. Therefore, only the first term will be an integer and it will denote the greatest integer of $x$. However, for proving the next part and conversely I am not at all getting any insights!
Help will be appreciated!
 A: Part 1
The $N+1$ numbers $tx - [tx]$ lie in the $N$ intervals $\left[\frac{k}{N}, \frac{k+1}{N} \right)$ for $k = 0,1,\ldots ,N-1$. Applying the pigeonhole principle, there must be two numbers $t_1x - \lfloor t_1 x \rfloor$ and $t_2x - \lfloor t_2 x \rfloor$ in the same interval. We can assume WLOG that $0 \leqslant t_1 < t_2 \leqslant N$. Since the length of the interval is $1/N$, we have 
$$\left|(t_2-t_1)x - (\lfloor t_2 x \rfloor -  \lfloor t_1 x \rfloor) \right|= \left|t_2x - \lfloor t_2 x \rfloor - (t_1x - \lfloor t_1 x \rfloor) \right|< \frac1{N}.$$
Taking $k = t_2-t_1$ and $h = \lfloor t_2 x \rfloor -  \lfloor t_1 x \rfloor$ we are finished.
Part 2 
From Part 1, there are integers $k$ and $h$ such that $|kx - h| < \frac1{N}$ and $\left|x - \frac{h}{k} \right| < \frac{1}{kN}$ with $k \leqslant N$.  If there are only finitely many $(h_i,k_i)$ such that $\left|x - \frac{h_i}{k_i} \right| < \frac{1}{k_iN}< \frac1{N}$ then consider $\alpha = \min_i \left|x - \frac{h_i}{k_i} \right|$. Since $x$ is irrational, we have $\alpha > 0$.  
Again from Part 1, we can  find $(h,k)$ such that $\left|x - \frac{h}{k} \right| < \alpha$. Hence, $(h,k)$ cannot be one of the pairs $(h_i,k_i)$ and there must be infinitely many such pairs.
Part 3
You have already shown that $a_1 = \lfloor x \rfloor$.  Use induction to prove that $a_k = \lfloor k!x \rfloor - k\lfloor (k-1)! x \rfloor$.
For the first step, note that 
$$2x - 2a_1 - a_2 = 2 \sum_{k=3}^r \frac{a_k}{k!} \\ \leqslant  2 \sum_{k=3}^r \frac{k-1}{k!} \\ =    2 \sum_{k=3}^r \frac{1}{(k-1)!} - 2 \sum_{k=3}^r \frac{1}{k!} \\ = 1 - \frac{2}{r!}$$
Since the RHS is less than $1$ we have 
$$\lfloor 2x \rfloor = 2a_1 + a_2 \implies a_2 = \lfloor 2x \rfloor  - 2a_1 =\lfloor 2x \rfloor - 2 \lfloor x \rfloor.$$
Try to finish.  
