Problem 12 - Discrete Mathematics with Applications

Question

Let \(n \in \mathbb{N} .\) Suppose \(n\) elements are selected from the set \(\\{1,2, \ldots, 2 n\\}\) Find a pair of integers in which one is not a factor of another integer. Use the pigeonhole principle to prove the following.

Step-by-Step Solution

Answer

In both possible scenarios of choosing \(n\) elements from the set \(\{1, 2, ... , 2n\}\), we can find a pair of integers such that one integer is not a factor of another integer. This is either by selecting all odd elements resulting in a pair \((2k+1, 2m+1)\) with odd products, or by selecting at least one even element and one odd element to form an odd-even pair \((2k+1, 2m)\) with even products.

Step 1: Split the Set into Odd and Even Elements

Let's split the given \(2n\) elements into two sets: a set of all odd elements, and the set of all even elements. Each element in the set can be either odd or even. For example, for \(n = 4\), we have the set \(\{1, 2, 3, 4, 5, 6, 7, 8\}\) and the division into odd and even sub-groups looks like: Odd: \(\{1, 3, 5, 7\}\) Even: \(\{2, 4, 6, 8\}\)

Step 2: Count Odd and Even Elements

Since there will be \(n\) odd numbers in the range \(\{1, 2, ... , 2n\}\) and \(n\) even numbers in the range. For example, when \(n = 4\), we have: Odd: \(\{1, 3, 5, 7\}\) (4 elements - equal to \(n\)) Even: \(\{2, 4, 6, 8\}\) (4 elements - equal to \(n\))

Step 3: Select n Elements from the Set

As per the given exercise, we must choose \(n\) elements from this set of \(2n\) elements. Regardless of which specific elements are chosen, we have two possibilities when it comes to the number of odd elements and even elements picked: 1. All \(n\) chosen elements are odd. 2. There is at least 1 even element among the \(n\) chosen elements.

Step 4: Finding Integer Pair with Conditions

For both of these scenarios, let's try to find a pair of integers in which one is not a factor of another integer: 1. If all \(n\) chosen elements are odd, then there must be at least one pair \((2k+1, 2m+1)\) existing among the chosen elements. Since both of these elements are odd, the product \((2k+1)*(2m+1) = 4km + 2k + 2m + 1\) will be an odd number. Thus, it's clear that one integer cannot be a factor of another integer in the pair. 2. If there is at least 1 even element among the \(n\) chosen elements, then we know there must exist at least one odd element in the chosen set as well – otherwise, we'd be choosing more than \(n\) even numbers from a set of only \(n\) even numbers. In any odd-even pair \((2k+1, 2m)\), the product \((2k+1)*2m = 4km + 2m\) will always be an even number; therefore, one cannot be a factor of the other in the pair.

Step 5: Conclusion

In both possible scenarios of choosing \(n\) elements from the set \(\{1, 2, ... , 2n\}\), we have found a pair of integers such that one integer is not a factor of another integer.

Key Concepts

Odd and Even Numbers

In mathematics, understanding the difference between odd and even numbers is fundamental. Even numbers are integers that are divisible by 2 without leaving any remainder. Examples include:

• 2
• 4
• 6

Odd numbers are integers that leave a remainder of 1 when divided by 2. Examples of odd numbers are:

• 1
• 3
• 5

Knowing whether a number is odd or even is important for many mathematical operations and problem solving. In the exercise, we separated a set of numbers up to 2n into odd and even categories. This distinction plays a key role in ensuring that when we select numbers, some properties, such as divisibility, are taken into account.

Factor Pairs

A factor pair consists of two numbers that, when multiplied together, result in a given number. For instance, the factor pairs of 12 include:

• (1, 12)
• (2, 6)
• (3, 4)

Understanding factor pairs is crucial when solving mathematical problems that involve divisibility. In our problem, a pair of integers is required such that one isn't a factor of the other. This understanding helps us identify conditions where meeting such criteria is possible. By carefully selecting sets of numbers, such as only involving odd numbers, it's clear that they can't be exact multiplies and thus unlikely to contain direct factors of each other.

Integer Selection

Integer selection refers to the process of choosing specific numbers from a given set. This concept is especially vital in problems involving constraints, such as the one in the given exercise. Here, we pick n integers from a set of 2n numbers. When making these choices, several factors influence the process:

• Balancing the number of odd and even integers selected.
• Avoiding selection configurations where one integer is a multiple of another.

The exercise explores different selection scenarios that ensure the selected pairs adhere to the stated condition – where one isn't a factor of the other.

Sets and Subsets

A set is a collection of unique objects or numbers. In mathematics, when we talk about subsets, we refer to selecting a smaller set that contains only elements from a larger set. Subsets are composed by selecting some or all of the elements from a given set. In the exercise, the original set defined as {1, 2, …, 2n} is reduced into subsets focusing on the odd and even numbers. Understanding how to divide a set into effective subsets is critical for solving complex problems since it allows pinpointing where certain conditions can or cannot be applied. By strategically forming subsets, the conditions required for establishing factor pairs or ensuring elements aren't factors of each other become more manageable.