Let S be a subset of size n+1 where each element is from {1,2,..,2n}. Show that there are two distinct numbers, a and b, from S where a divides b

We are tasked with proving that for any subset \( S \) of size \( n+1 \) where each element is chosen from the set \( \{1, 2, \dots, 2n\} \), there exist two distinct elements \( a \) and \( b \) in \( S \) such that \( a \) divides \( b \).

This problem can be solved using the **Pigeonhole Principle**. Here's how we can approach it step by step:

### Step 1: Group elements by their largest odd divisor
Each element \( x \in \{1, 2, \dots, 2n\} \) can be written in the form \( x = 2^k \cdot m \), where \( m \) is an odd number and \( k \geq 0 \). The number \( m \) is called the largest odd divisor of \( x \), because \( m \) is odd and dividing \( x \) by any higher power of 2 will give a non-integer result.

For example:
- \( 4 = 2^2 \cdot 1 \) (largest odd divisor is 1)
- \( 12 = 2^2 \cdot 3 \) (largest odd divisor is 3)

### Step 2: Consider how many odd divisors exist
The odd numbers from 1 to \( 2n \) form a subset of \( \{1, 3, 5, \dots, 2n-1\} \), and their total count is exactly \( n \). Hence, the possible largest odd divisors are the odd numbers in \( \{1, 3, 5, \dots, 2n-1\} \), and there are \( n \) such numbers.

### Step 3: Apply the Pigeonhole Principle
Since the subset \( S \) has \( n+1 \) elements and there are only \( n \) possible largest odd divisors, by the **Pigeonhole Principle**, there must be at least two elements in \( S \), say \( a \) and \( b \), which share the same largest odd divisor.

### Step 4: Show one divides the other
Suppose \( a = 2^k \cdot m \) and \( b = 2^l \cdot m \), where \( m \) is their common largest odd divisor and \( k \neq l \). Without loss of generality, assume \( k < l \). Then:

\[
b = 2^l \cdot m = 2^{l-k} \cdot (2^k \cdot m) = 2^{l-k} \cdot a
\]

This shows that \( a \) divides \( b \) because \( b = 2^{l-k} \cdot a \).

### Conclusion
Thus, for any subset \( S \) of size \( n+1 \) chosen from \( \{1, 2, \dots, 2n\} \), there must exist two distinct elements \( a \) and \( b \) such that \( a \) divides \( b \).

Would you like further details on any part of this solution? Here are five related questions you might find interesting:

1. Can the Pigeonhole Principle be applied to other problems in number theory?
2. How can this proof be extended to other sets, not just \( \{1, 2, \dots, 2n\} \)?
3. What are some common applications of divisibility in combinatorics?
4. Can we generalize this problem to larger sets where the size of \( S \) is different?
5. What is the significance of the largest odd divisor in number theory?

**Tip**: When using the Pigeonhole Principle, always focus on identifying a structure that partitions elements into fewer groups than the size of your set!

Let S be a subset of size n+1 where each element is from {1,2,..,2n}. Show that there are two distinct numbers, a and b, from S where a divides b.

Explore how to prove that in any subset of size n+1 from the set {1, 2, ..., 2n}, there exist two distinct elements where one divides the other. This proof leverages the Pigeonhole Principle and concepts of divisibility in number theory.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation