Show that if n+1 integers are choosen from set $\{1,2,3,...,2n\}$ ,then there are always two which differ by 1 Considering n=5 i have 
$\{1,2,3,...,10\}$ .Making pairs such as $\{1,2\}$ ,$\{2,3\}$ ... total of $9$ pairs  which are my holes and $6$ numbers are to be choosen which are pigeons .So one hole must have two pigeons ,i am done .Is this correct ?
In general if we have $2n$ numbers and then we have $2n-1$ holes and $n+1$ pigeons ,But for $n=3$ onwards it is valid ? There  is something wrong i think
Thanks

Suppose not, then numbers differ by at least two. Order the numbers $a_1<a_2<\dots <a_n$. Then $a_1\geq 1$. We prove $a_i\geq 1+(i-1)2$ by induction
Base step $i=1$ is clear (we have to prove $a_1\geq 1+(1-1)2+1=1$)
Inductive step. We have $a_{i}\geq 1+(i-1)2$ and $a_{i+1}\geq a_i+2$ so $a_{i+1}\geq 1+(i-1)2+2=1+((i+1)-1)2$. Taking $i=n+1$ we get $a_n\geq 1+(n+1)2=2n+1$, a contradiction since all number must be in set $\{1,2,3\dots 2n\}$
