The other answers are correct, but I find them a bit technical. Suppose you have swept (eliminated) a matrix and are looking for any solution and the matrix looks like this:
column x1 x2 x3 x4 x5 x6 | solution ----------------------------------- 1 1 1 | 3 1 | 1 1 1 | 2 2 1 1 | 1 In lineair programming, you can now set the non-pivot columns (x5,x6) to 0, and set x4 to 0.5 and find a trivial solution. In integer programming, you cannot just set the non-pivot columns to 0. The solution is harder (NP-hard) to find. Note also that the solutions are in $Q$, so this is not directly related to finite/infinite precision.