Which seems almost ironic, because continuous linear optimization almost certainly doesn't exist really because real numbers can only be approximated, and so we're always doing discrete linear optimization at some level.
If all the numbers that appear in your constraints are rational (p/q with finite p and q), then any solution is also a rational number (with finite nominator and finite denominator).
(Well, any finite solution. Your solution could also be unbounded, then you might have infinities in there.)
