The Partition Problem supposes that we have a set $S=\{s_{i} \in \mathbb{R^{+}}\}_{i=1}^{n}$, is there a subset $T$ such that $\sum_{s \in T} s = \sum_{s \notin T}s$?
I have read a lot of proofs using the Subset Sum problem: Given an integer set $S$ and an integer t, does there exist a subset $U$ of $S$ such that $\sum_{s \in U}s = t$? My problem is that most of them does not consider that the partition problem does not require the inputs to be integers, and a lot of them transform the input $S$ of the Subset Sum problem to $S\cup \{ \text{some elements} \}$ in the Partition Problem. If the input to the partition problem is non-integer, or, worse, irrational, I do not think this method works.
I have tried to reduce the partition problem to subset sum problem, by transforming the input $S$(which could be irrational numbers) to $S^{\prime} = \{ 2\lceil s_{i} \rceil \}$ and $t = \sum_{i} \lceil s_{i} \rceil$. Besides, I have also tried a few other similar ways including considering the floor function. However, whichever I tried, I could not prove that partition problem has a solution on $S$ if and only if the subset sum problem has a solution on $(S^{\prime},t)$. I do not find any contradiction if the subset sum problem has a solution on $(S^{\prime},t)$ and the partition problem does not have a solution on $S$, and vice versa. It sort of makes sense to me because having the result in integer subset sum problem could hardly have any implications on the non-integer partition problem.
Would anyone have any hints on how to prove the non-integer version of the Partition Problem is NP-complete by reducing it to the Subset Sum Problem?