Is the set of all non-computable numbers closed under addition?

My instinct tells me no. I would like to imagine, since most real numbers are uncomputable, there exist two real uncomputable numbers $$\alpha,\beta$$ such that $$\alpha+\beta=1$$ but I certainly don't know how to show it.

• The computable numbers are a field, therefore an subgroup of $(\Bbb R,+)$. The complement of a subgroup is never non-empty and addition-closed. Commented Sep 5 at 1:11
• @SassatelliGiulio: Is it or isn't it addition-closed? Your sentence can mean either.
– pts
Commented Sep 5 at 13:12
• @pts Whichever is true is mine Commented Sep 5 at 14:23
• @SassatelliGiulio: pts's question was legitimate! I would have said: "The complement of a subgroup can never be both non-empty and addition-closed." Commented Sep 5 at 16:59

The set of non-computable numbers is not closed under addition. Take some non-computable number $$a$$. Then $$1-a$$ and $$1+a$$ are also non-computable, but their sum is 2, a computable number.
• The dumber answer is just a - a. Commented Sep 5 at 9:16