If $A\subseteq \{ 1, 2, ... 2n \}$ has $n+1$ elements, then it contains $a$, $b$ such that $a\mid b$ Let $ A $ be subset of of $ \{ 1, 2, ... 2n \} $, such that: $ Card(A) = n + 1$.
Prove that there exist at least two distinct elements $a,b$ in $ A $ such that: $ a \mid b $.

We have:
$$ \{ 1, 2, ... 2n \} =  \{ 1, 2, ... n \} \cup  \{ n+1, n+2, ... 2n \} $$
Intuitively, I know that every subset $ A $ of $ \{ 1, 2, ... 2n \} $ with $ n + 1 $ elements contains at least 2 elements $ a $ and $ b $ such: $ b = 2a $.
I don't know how to prove it mathematically. Thank you for your help. 
 A: This is is a generalized version, of a few problems on here. 
Basically, in the consecutive natural numbers, either they are odd or even. 
Up to $2n$, there are precisely $n$ odd positive natural numbers. All the even numbers, in the consecutive numbers, are two times a previous odd number. 
It follows, that if we add another value, to make A's cardinality $n+1$ the even element we add must divide by one of the odd numbers we already have put in $A$. 
A similar argument can be made starting with all the even elements. This generalizes, to things related to the pigeonhole principle. 
You can then make statements like, if you pick 6 random numbers, at least 2 of them have to share the same remainder, when divided by 5. 
Also, you can argue, that you can say that, for a much smaller set, the primes less than $2n$. Either there are $n+1$ primes less than $2n$ the end, or you can take the primes the number up to that point, and add some of their multiples. 
This is because, all numbers that aren't prime, up to the square of the next prime after 2n, will divide by one of the primes in the set.
A: The easiest way to do it is induction!


*

*$n = 2$. We have a subset of three elements of $A=\{1, 2, 3, 4\}$. Obviously, this either has 1 or both 2 and 4, so we can find two numbers, with one being a multiple of the other.

*Assume we know that statement is correct for all numbers $n\ge k\ge2$ and we want to prove it for $n=k+1$.
Our set $A=\{1,2,..., 2n+2\}$ and we have some subset $B$ of size $n+2$. If $B$ has $n+1$ numbers from the set $\{1, 2, ... , 2n\}$, then by induction assumption the statement is correct. Therefore we can see that only $n$ numbers from the set $\{1, 2, ... , 2n\}$ are in $B$, so $B$ contains both $2n+2$ and $2n+1$.
Let $C$ be a subset of $B$ of all numbers less than $2n+1$. We know it has exactly $n$ numbers, and also does not contain $n+1$ (otherwise the desired pair in $B$ would be $n+1$ and $2n+2$. But if we add number $n+1$ to C, then this set should have two numbers that are multiples of each other. Since to have this condition we added a number $n+1$, and there are no multiples of $n+1$ we have that set $C$ has a divisor of $n+1$, and hence it has a divisor of $2n+2$, which gives us two desired numbers.
