Question

Prove or disprove: Any subset \(X \subseteq\\{1,2,3, \ldots, 2 n\\}\) with \(|X|>n\) contains two (unequal) elements \(a, b \in X\) for which \(a \mid b\) or \(b \mid a\).

Short answer

The proof for the statement is correct. If a subset \(X\) of the set \(\{1,2,3,...,2n\}\) has more than \(n\) elements, the set \(X\) will always contain two unequal elements \(a, b\) such that \(a | b\) or \(b | a\).

Step-by-step solution

Assume the Contrary

Assume for the sake of contradiction that there exists a set \(X\) such that \(|X|>n\), and none of its elements \(a, b\) (where \(a\neq b\)) satisfy \(a \mid b\) or \(b \mid a\).

Look at Divisibility by 2

Elements of set \(\{1,2,3,\ldots,2n\}\) fall into two categories: they are either divisible by 2 (even) or not divisible by 2 (odd). From step 1, if \(X\) does not contain elements such that one element divides another, then the subset of elements in \(X\) that are even cannot have 2 as an element. That means, the maximum number of elements in 'even' category that \(X\) can have is \(n-1\), which are \(\{4, 6, 8, ... 2n\}\).

Look at Odd Elements

Since odd numbers are not divisible by any even numbers (other than 1) and none of the odd numbers divide each other, we can have at most \(n\) odd elements that can be elements of \(X\), which are \(\{1,3,5,...,2n-1\}\).

Maximum Cardinality of X

So, the maximum possible number of elements in \(X\) is \(n-1\) even elements + \(n\) odd elements, which is equal to \(2n-1\).

Contradiction

But we are given that \(|X|>n\), so if we have \(2n-1\), \(|X|\) can be equal to \(2n\) elements which contradicts the assumption we made in step 1. So, at least two elements in a set of more than \(n\) elements divisible will divide each other or be divided by the other.

Key concepts

Set Theory and its Role in Mathematics

Set theory is the branch of mathematical logic that studies sets, which are collections of objects. In mathematics, a set can be defined simply as a collection of distinct elements that can be thought of as a whole. These elements can be numbers, letters, symbols, or even other sets. The theory of sets serves as the foundation for various fields of mathematics and is pivotal in the formulation and understanding of relations between different mathematical entities.

In the context of the given exercise, we are examining a subset of natural numbers from 1 to 2n. The concept of cardinality, which is the number of elements in a set, is critical as we investigate the properties of this subset. Set theory allows us to formalize ideas like subsets and cardinality, providing the language necessary for posing and solving problems related to divisibility within a collection of numbers.

Understanding Divisibility in Mathematics

The concept of divisibility is integral to number theory, a branch of pure mathematics that deals with the properties and relationships of numbers, particularly integers. Divisibility refers to the ability to divide one integer by another and obtain an integer result, that is without remainder. For instance, we say that an integer 'a' is divisible by another integer 'b' if there exists an integer 'k' such that \(a = bk\).

Within a set of numbers from 1 to 2n, every even number is divisible by 2, while odds are not. However, when creating a subset with particular divisibility constraints, the properties of these elements take on key importance. The solution to our exercise suggests that it isn't possible to avoid having divisible pairs above a certain set size, which emerges from the basic principles of divisibility. This understanding is essential when grappling with more complex questions in number theory.

Mathematical Contradiction and Proof Techniques

A mathematical contradiction occurs when an assumption leads logically to a situation that defies the initial premise or established facts, thereby proving the assumption to be false. This method is known as proof by contradiction, which is a powerful tool in mathematical arguments. To perform such a proof, one assumes the opposite of what they are trying to prove and, through logical reasoning, derives an absurdity or a contradiction, hence showing the original statement must be true.

In our exercise, the assumption that a set with more than 'n' elements does not have divisible pairs leads to an impossibility when considering the maximum possible unique elements. By identifying the contradiction — that the maximum number of non-divisible pairs is less than or equal to 'n' for both even and odd elements — we deduce that given the condition \(\left| X \right| > n\), there must indeed be at least one pair of elements where one is divisible by the other. This elegant technique not only solves our specific problem but also underscores the fundamental importance of mathematical contradiction in validating the truth across various domains of mathematics.