I have a question about Terence Tao's lecture notes, Proposition 5:
Let $x$ be a rational number. Then there exists a unique integer $n$ such that $n \leq x < n+1$.
Here is my proof (existence only):
Proof: If $x = 0$ let $n = 0$. If $x > 0$, then $x = \frac{a}{b}$ such that $a, b$ are positive integers. By Euclidean algorithm we know that there exists $p, r$ such that $p>0$ and $0 \leq r < b$ and $a = pb + r$. Then we have $\frac{a}{b} = p + \frac{r}{b}$, which means $p \leq \frac{a}{b} < p+1$. Let $n = p$ we get the inequality.
If $x < 0$, $x$ can be represented as $-\frac{a}{b}$ and $a, b$ are positive integers. Since we have already proved this case, we have $-(p+1)< -\frac{a}{b} \leq -p$. Let $n' = -(p+1)$ we get only $n' < x \leq n'+1$, not precisely the same thing. Now we need to show the two inequalities $n' < x \leq n'+1 $ and $n \leq x < n+1$ are equivalent.
Since $x$ is given, the interval $(n', n'+1]$ is fixed. If $x \in (n', n'+1)$, let $n = n'$ we have $n \leq x < n+1$. If $x = n'+1$ let $n = n'+1$ we have $n \leq x < n+1$. In either case we have $n \leq x < n+1$, we finished the proof.
Could you help me to improve this proof? It seems not neatly enough. Thanks!
The following is an approach very similar to yours, but perhaps a little sleeker. Take a positive rational $r=\frac{a}{b}$, where $a$ and $b$ are positive integers. There are integers $\ge r$, for example $a$. Let $n$ be the smallest positive integer which is $\ge r$. Continue.