Question

Prove that given a real number\(x\)there exist unique numbers \(n\)and\(\varepsilon \) such that\(x = n + \varepsilon \),\(n\) is an integer, and\(0 \le \varepsilon < 1\).

Short answer

The expansion\(x = n + \varepsilon \)exists and is unique.

Step-by-step solution

Introduction

If the number is a positive integer and there are no repeated digits, it will be unique.To put it in another way, a number is considered to be unique if and only if the digits are different.

Proof using contradiction

Let n is the largest integer less than or equal to x.

Let \(\varepsilon = x - n\)then\(n \le x\), so\(\varepsilon \ge 0\) .

If\(\varepsilon \ge 1\), then\(\varepsilon - 1 \ge 0\).

\(\begin{aligned}{}\left( {\varepsilon - 1} \right) + 1 = x - n\\n + 1 = x - \left( {\varepsilon - 1} \right)\\n + 1 \le x\end{aligned}\)

As \(\varepsilon - 1 \ge 0\)it contradicts that n is the largest integer less than or equal to x.

So \(\varepsilon < 1\)and it follows that\(0 \le \varepsilon < 1\).

Uniqueness of the given expansion

An integer m and real number \(\delta \)with \(x = m + \delta \) and \(0 < \delta < 1\).

\(\begin{aligned}{}n + \varepsilon = m + \delta \\n - m = \delta - \varepsilon \end{aligned}\)

But, if \(0 \le \varepsilon ,\delta < 1\),then\(\left| {\delta - \varepsilon } \right| < 1\).

If\(\delta - \varepsilon = n - m\), then \(\delta - \varepsilon \)is an integer and the only integer with absolute value less than 1 is 0.

So,

\(\begin{aligned}{}\delta - \varepsilon = 0\\ = n - m\end{aligned}\).

Thus, \(\delta = \varepsilon \) and \(n = m\).

Therefore, the expansion\(x = n + \varepsilon \)is unique.