Let \(m\) be a positive integer. Show that for every real number \( x \ge 1 \), the number of multiples of \(m\) in the interval \([1, x]\) is \( \lfloor \frac{x}{m} \rfloor \); in particular, for every integer \( n \ge 1 \), the number of multiples of \(m\) among \(1, ..., n\) is \( \lfloor \frac{n}{m} \rfloor \).

First, every real number \(x\) can be formulated as the sum of an integer and an \( \epsilon \in \mathbb{R} \text{ where } 0 \le \epsilon \lt 1 \). Using the "division with remainder property" (which says for every \(a, b \in \mathbb{Z} \) with \(b > 0\), there exist unique \(q, r \in \mathbb{Z} \) such that \( a = b \cdot q + r \) and \( 0 \le r < b \)), there exist \(q,r \in \mathbb{Z} \) such that:

\[ \begin{aligned} x = q \cdot m + r + \epsilon \end{aligned} \]

where \(q \ge 0 \) and \( 0 \le r < m \)

Obviously, \(q\) is the number of multiples of \(m\) in the interval \([1, x]\). To find \(q\), we can start with dividing both sides by \(m\):

\[ \begin{aligned} \frac{x}{m} &= q + \frac{r + \epsilon}{m} \\ \lfloor \frac{x}{m} \rfloor &= \lfloor q + \frac{r + \epsilon}{m} \rfloor \\ \end{aligned} \]

Observe \( 0 \le r + \epsilon < m \), so

\[ \begin{aligned} \lfloor \frac{x}{m} \rfloor &= q \\ \end{aligned} \]

Since every integer \(n\) is also a real number, the second statement is obviously true by substituting \(x\) by \(n\) in the first statement.

\( \Box \)

# posted by rot13(Unafba Pune) @ 12:30 AM