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To solve this problem, we need to prove that for each pair of real numbers \( x < y \), there exist integers \( m \in \mathbb{Z} \) and \( n \in \mathbb{N} \) such that:

\[
x < \frac{m}{5^n} < y.
\]

### **Proof Strategy:**
We will use the Archimedean property of real numbers and the density of rational numbers to find such integers \( m \) and \( n \).

#### **Step 1: Using the Archimedean Property**
The Archimedean property states that for any real numbers \( a \) and \( b \), with \( a < b \), there exists a positive integer \( k \) such that:

\[
a < \frac{1}{k} < b.
\]

This suggests that for any \( x < y \), there will be a sufficiently large natural number \( n \) such that \( 5^n \) is large enough to divide the interval \( (x, y) \) into sufficiently fine partitions. In particular, there will be an integer \( m \in \mathbb{Z} \) such that:

\[
x < \frac{m}{5^n} < y.
\]

#### **Step 2: Choosing \( n \)**
We want to find an integer \( n \in \mathbb{N} \) such that the interval \( (x, y) \) contains a rational number of the form \( \frac{m}{5^n} \). To do this, we pick \( n \) large enough so that the distance between \( x \) and \( y \) is bigger than the "spacing" between consecutive multiples of \( \frac{1}{5^n} \). That is, we choose \( n \) such that:

\[
y - x > \frac{1}{5^n}.
\]

This guarantees that there exists at least one integer \( m \) such that:

\[
x < \frac{m}{5^n} < y.
\]

#### **Step 3: Finding \( m \)**
For a given \( n \), multiply both \( x \) and \( y \) by \( 5^n \):

\[
5^n x < 5^n \frac{m}{5^n} = m < 5^n y.
\]

Since \( 5^n x \) and \( 5^n y \) are real numbers, there will exist an integer \( m \) between these two values because the integers are dense within the real line. Thus, we can always find an \( m \in \mathbb{Z} \) such that:

\[
x < \frac{m}{5^n} < y.
\]

### **Conclusion:**
For any pair of real numbers \( x < y \), we can always find integers \( m \in \mathbb{Z} \) and \( n \in \mathbb{N} \) such that:

\[
x < \frac{m}{5^n} < y,
\]

as required.

---

Do you need further details or clarification on any part of the proof?

Here are 5 related questions you might find helpful:
1. What is the Archimedean property, and how does it apply to this proof?
2. How can we use the density of rational numbers to find rational numbers between any two real numbers?
3. Why does choosing \( n \) large enough guarantee the existence of \( m \)?
4. Can this result be generalized to any base other than 5? If so, how?
5. How can this theorem be applied in real-world contexts, such as numerical approximation?

**Tip:** Understanding the Archimedean property is crucial when working with real numbers and their relation to integers.

Prove that for each pair of real numbers x < y, there exist integers m ∈ Z and n ∈ N such that x < m / 5^n < y.

This proof shows that for any real numbers x < y, there exist integers m and n such that x < m / 5^n < y. It uses the Archimedean property and the density of rational numbers, making it ideal for advanced high school or undergraduate students studying real analysis.

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