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

You have nine holes and six pigeons, so the pigeonhole principle doesn't help because $6<9$.
Just use five holes $\{1,2\},\{3,4\},\{5,6\},\{7,8\},\{9,10\}$
