Can every real number be uniquely represented as a sum of a rational number and an irrational number $\in [0, 1)$?

Question

I've needed to prove the transitivity of the following relation on the set of all real numbers:

$x − y$ is a rational number.

Immediately I've thought "Every real number can be uniquely represented as a sum of a rational number (let's call it $q$) and an irrational number $q' \in [0, 1)$"; from which it follows that $\forall x \in \mathbb{R}$ s.t. $x = q + q'$, $x$ is rational number iff $q' = 0$.

I am not going into the details of how to prove the transitivity property using this lemma as it is easy enough and unrelated to my question anyway. Instead, what I am asking is, how can I prove my assumption which I've realised might not be obvious enough. :)

Edit: Or, do you think it's obvious enough, and that I shouldn't worry about it?

Thanks!

Answer 1

This claim is wrong. Let's take $\frac{1}{\sqrt{2}}$. Then we may subtract two rationals $0\leq{}r≠r'\leq{}\frac{1}{\sqrt{2}}$ from $\frac{1}{\sqrt{2}}$ and the result is an irrational. Adding the difference with $r$ respectively $r'$ gives us back our irrational number. So your representation is not unique. Aside from that you can subtract rational numbers from irrational ones.

Answer 2

A rational number can not be represented as a sum of a rational number and an irrational number because the difference of two rational numbers is rational number. Of course, an irrational number can be represented as a sum of a rational number and an irrational number because an irrational number minus a rational number is again an irrational number.