NP-hard or not: partition with irrational input


Original Problem

Given a set $ N=\{a_1,…,a_{n}\}$ with $ n$ positive numbers and $ \sum_i a_i=1$ , find a subset whose sum is $ x_*$ , where $ x_*$ is a known irrational number and $ x_*\approx 0.52$ .

I proved its hardness by the following arguments.

Instance

Given a set $ N=\{a_1,…,a_{n+2}\}$ with $ n+2$ numbers where

  • $ a_1,…,a_n$ are positive and rational
  • $ \sum_{i=1}^n a_i = .02$
  • $ a_{n+1}=x_*-0.01$
  • $ a_{n+2}=0.99-x_*$

determine whether we can find a subset of $ N$ , such that the sum of the subset is $ x_*$ . .

NP-complete

  • Since 𝑥∗ is irrational, the desired subset cannot contain both of the last two numbers.

  • Since the sum of any subset not containing the n+1st element is smaller than 𝑥∗, the desired subset must contain the n+1st element.

  • The remaining question is to find a subset of the first n numbers whose sum is .01

So the original problem is NP-complete.

My problem

Someone argued that since $ x_*$ is irrational, I can’t store irrational numbers in a machine properly and my proof is not correct. How to address it?