Understanding the Problem: Subset Properties

The question asks us to consider a set containing the first $2n$ natural numbers, specifically the set $S = \{1, 2, 3, \dots, 2n\}$. We are then asked to examine any subset of this set that contains exactly $(n+1)$ elements. We need to determine which of the given statements is always true for every such subset.

Let $A$ be a subset of $S$ such that $|A| = n+1$. We need to analyze the properties of the numbers within this subset $A$.

Analyzing the Options

Let's look at each option provided:

1. There exist at least two natural numbers which are prime to each other.
2. Exist at least three natural number which are prime to each other.
3. There exist no consecutive natural number.
4. There exist more than two natural numbers which are prime to each other.

The term "prime to each other" means that the numbers are coprime. Two natural numbers $a$ and $b$ are coprime if their greatest common divisor ($\gcd$) is 1, i.e., $\gcd(a, b) = 1$. For a set of three or more numbers to be "prime to each other", it usually implies that they are pairwise coprime, meaning every distinct pair of numbers within the set is coprime.

Proving Option 1 Using the Pigeonhole Principle

Let's focus on Option 1: "There exist at least two natural numbers which are prime to each other". We can prove this statement is always true using the Pigeonhole Principle.

Consider the set $S = \{1, 2, \dots, 2n\}$. We are selecting a subset $A$ of size $n+1$. Let's create pairs of consecutive integers from $S$. We can form $n$ such pairs:

• Pair 1: $\{1, 2\}$
• Pair 2: $\{3, 4\}$
• Pair 3: $\{5, 6\}$
• ...
• Pair $n$: $\{2n-1, 2n\}$

These $n$ pairs partition the set $S$. Each number from $1$ to $2n$ belongs to exactly one of these pairs.

Now, consider the subset $A$, which has $n+1$ elements. We can think of the $n$ pairs as "holes" and the $n+1$ elements of the subset $A$ as "pigeons".

According to the Pigeonhole Principle, if you have more pigeons than holes, at least one hole must contain more than one pigeon. In this case, since we have $n+1$ elements (pigeons) and $n$ pairs (holes), at least one of the pairs $\{2k-1, 2k\}$ must contain two elements from the subset $A$.

This means that for some value of $k$ (where $1 \le k \le n$), both $2k-1$ and $2k$ are elements of the subset $A$.

Now, let's consider the relationship between these two numbers, $2k-1$ and $2k$. These are consecutive natural numbers. We know that any two consecutive natural numbers are always coprime. The greatest common divisor of any two consecutive integers is always 1.

Specifically, $\gcd(2k-1, 2k) = \gcd(2k-1, 2k - (2k-1)) = \gcd(2k-1, 1) = 1$.

Therefore, the subset $A$ must contain at least one pair of consecutive numbers, and these consecutive numbers are prime to each other. This proves that there exist at least two natural numbers in the subset which are prime to each other.

Thus, Option 1 is always correct for any $(n+1)$-element subset of $\{1, 2, \dots, 2n\}$.

Analyzing Other Options

Now let's look at the other options to see if they are always correct.

Option 3: There exist no consecutive natural number.

Our proof for Option 1 clearly shows that any $(n+1)$-element subset from $\{1, 2, \dots, 2n\}$ must contain at least one pair of consecutive natural numbers. Therefore, the statement "There exist no consecutive natural number" in the subset is false. This option is incorrect.

Option 2: Exist at least three natural number which are prime to each other.

Option 4: There exist more than two natural numbers which are prime to each other.

These options imply the existence of a set of at least three numbers within the subset that are pairwise coprime. Let's check if this is always true with a counterexample.

Let's take a small value for $n$. Let $n=3$. The set is $\{1, 2, 3, 4, 5, 6\}$. The subset size is $n+1 = 4$.

Consider the subset $A = \{2, 3, 4, 6\}$. This is a valid subset of size 4 from the set $\{1, 2, 3, 4, 5, 6\}$.

Now, let's check for pairwise coprime numbers among any three elements in this subset $A$. We need to look at all possible combinations of three numbers from $A$:

• Consider $\{2, 3, 4\}$: $\gcd(2, 4) = 2$. Not pairwise coprime.
• Consider $\{2, 3, 6\}$: $\gcd(2, 6) = 2$ and $\gcd(3, 6) = 3$. Not pairwise coprime.
• Consider $\{2, 4, 6\}$: $\gcd(2, 4) = 2$, $\gcd(2, 6) = 2$, $\gcd(4, 6) = 2$. Not pairwise coprime.
• Consider $\{3, 4, 6\}$: $\gcd(3, 6) = 3$ and $\gcd(4, 6) = 2$. Not pairwise coprime.

In this specific subset $\{2, 3, 4, 6\}$, we found that no combination of three numbers is pairwise coprime. We do have pairs that are coprime, like $\gcd(2, 3)=1$ and $\gcd(3, 4)=1$, which aligns with Option 1 being true.

Since we found a subset where there are no three numbers that are pairwise prime to each other, Options 2 and 4 are not always correct.

Conclusion

Based on the analysis using the Pigeonhole Principle, any $(n+1)$-element subset of the set $\{1, 2, \dots, 2n\}$ must contain at least one pair of consecutive natural numbers, which are always prime to each other. Therefore, the only statement that is always correct is that there exist at least two natural numbers which are prime to each other.

Revision Table: Subset and Coprime Properties

Additional Information: Coprime Numbers and Pigeonhole Principle

Coprime Numbers: Two integers $a$ and $b$ are coprime or relatively prime if the only positive integer that divides both of them is 1. This is equivalent to saying that their greatest common divisor ($\gcd(a, b)$) is 1. For example, 7 and 10 are coprime because $\gcd(7, 10) = 1$. Consecutive numbers like $k$ and $k+1$ are always coprime.

Pigeonhole Principle: This principle states that if you have $N$ items to put into $M$ containers, and if $N > M$, then at least one container must contain more than one item. In our solution, the items were the $n+1$ numbers in the subset, and the containers were the $n$ pairs of consecutive integers from the original set $\{1, 2, \dots, 2n\}$.

This problem demonstrates a classic application of the Pigeonhole Principle in number theory, showing how a basic counting argument can reveal properties of elements within a set or subset.