If $x$ and $y\in\mathbb R$ are computable, is $x+y$ computable?

Question

I know that a real number $x\in \mathbb R$ is computable if there is a computable function $f:\mathbb N\rightarrow\mathbb Q$ such that, for all $r\in\mathbb N$, $|f(r)-x|≤2^{-r}$. I'm 99% confident that yes, $x+y$ is computable here, but am having difficulty formally proving this with confidence.

Answer 1

Here's the intuition. Suppose $f$ and $g$ approximate $x$ and $y$ to the desired accuracy, respectively. The intuitive guess at an approximating function for $x+y$ should just be the pointwise sum $$f+g: r\mapsto f(r)+g(r).$$ Keep in mind that in the "classical analysis" setting we have a commutation between sums and limits when dealing with convergent sequences, so this is a reasonable thing to do.

However, this isn't guaranteed to be a good enough approximation since the errors could add up to too large a number. That said, we do have a bound on the error of $f+g$: namely, we have that $\vert(f+g)(r)-(x+y)\vert$ is at most

$2\cdot 2^{-r}=2^{-r+1}$, by the triangle inequality (do you see how?).

Based on this, there's an easy way to modify our first guess $f+g$ to get a function which has the desired approximation property (HINT: remember that making $r$ bigger makes the approximation better):

Just set $h(r)=(f+g)(r+1)$.

Note that the "error control" ideas here closely resemble similar ideas around continuity and $\epsilon$-$\delta$ arguments. This is no coincidence: there is a tight analogy between computability and continuity, and for "naturally occurring" continuous functions the proof of continuity will in fact basically be a proof of computability.