# Why do we round from 1/2 when converting the LP to ILP for the weighted vertex cover problem?

I understand that to approximate a solution to the weighted vertex cover, we need to relax the integer linear program to a linear program which can be solved in polynomial time, but why do we round from 1/2 when converting it back to an ILP? Why aren't the bounds say, round down to 0 from 1/3 if a given vertex has value <= 1/3 and round up to 1 if said vertex has value > 1/3?

The constraints are $$x_u + x_v \geq 1$$ for all $$uv \in E$$. To get an integral solution, if you round up only values strictly more that $$1/2$$, then when your LP solution has $$x^*_u = x^*_v = 1/2$$, the rounding gives $$x_u = x_v = 0$$ which is not valid. So we must round from some threshold value $$\alpha \leq 1/2$$.
Can we choose $$\alpha < 1/2$$? Yes, but the approximation ratio is $$1/\alpha$$ (because if $$x^*_u = \alpha$$, then $$x_u = 1$$ and the associated cost goes from $$w_u x^*_u$$ to $$w_u x_u$$, hence is multiplied by $$1 / \alpha$$). So rounding from $$1/3$$ is worse that rounding from $$1/2$$, but is nevertheless a valid approximation algorithm.