I'm self-learning real analysis, and I am trying to understand a part in the proof for the following theorem in Mathematical Analysis by Apostol:
Let $x\geq 0$. Then for every integer $n \geq 1$, there exists a finite decimal $r_{n} = a_{0}.a_{1}a_{2}\cdots a_{n}$ such that $r_{n} \leq x < r_{n} + \frac{1}{10^{n}}$.
Here is the given proof:
let S be the set of all nonnegative integers that are less than or equal to $x$. Since $0 \in S$, $S$ is not empty. Since $x$ is an upper bound of $S$, $S$ has a supremum $a_0=\sup(S)$. Since $a_0\le x$, $a_0 \in S$, and so $a_0$ is a nonnegative integer. Then $a_0=\lfloor x \rfloor$ is the greatest integer in $x$. Then $a_0\le x < a_0 + 1$.
Let $a_1=\lfloor 10x-10a_0\rfloor$ be the greatest integer in $10x-10a_0$. Then $0\le 10x-10a_0=\color{red}{10(x-a_0)<10}$. Then $0\le a_1 \le 9$. Since $a_1=\lfloor 10x-10a_0\rfloor$, then $a_1\le 10x-10a_0<a_1+1$. Then $a_1$ is the largest integer that satisfies $a_1+10a_0\le10x<a_1+1+10a_0$ or $a_0+a_1/10\le x<a_0+a_1/10+1/10$.
In general, choosing $a_1$, $a_2$, …, $a_{n-1}$ with $0\le a_i\le9$, let $a_n$ be the largest integer that satisfies $a_0 + a_1/10 + \cdots + a_n/10^n \le x < a_0 + a_1/10 + \cdots + a_n/10^n + 1/10^n$. Then $0\le a_n \le 9$. Then $r_n \le x < r_n + 1/10^n$ with $r_n=a_0.a_1a_2\cdots a_n$. $\blacksquare$
The part I don't understand is in red. Why is $10(x-a_{0})<10$? I thought $x-a_{0}$ is nonnegative since $a_{0}\leq x$. Is it because $x$ is not an integer?
Best Answer
Since $a_0$ is defined to be $a_0 = \left[x\right]$ in the proof, this means by the definition of "greatest integer" that $a_0 \leq x \lt a_0 + 1$, as also stated in the proof. Using the second part and multiplying both sides of the inequality by $10$ gives that $10x \lt 10a_0 + 10$. This can be rearranged to $10x - 10a_0 \lt 10$, which factors to $10\left(x - a_0\right) \lt 10$.