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

Well, in your case when $n=5$, you shouldn't make the pairs $\lbrace1,2\rbrace,\lbrace2,3\rbrace,...,\lbrace8,9\rbrace,\lbrace9,10\rbrace$, but rather $\lbrace1,2\rbrace,\lbrace3,4\rbrace,...,\lbrace7,8\rbrace,\lbrace9,10\rbrace$, as in the first case the same integer appears in two different pairs.
When we have the correct pairs established, we can directly apply the pigeonhole principle on our $n=5$ "holes" $\lbrace1,2\rbrace,\lbrace3,4\rbrace,...,\lbrace7,8\rbrace,\lbrace9,10\rbrace$, and we see that if we have $n+1=6$ "pigeons" in the "holes", at least one of our "holes" must contain two pigeons.
We then apply the same reasoning to the general case.
