Dfrnhjy

Okay, the title is a bit of a clickbait, I just want to know what properties of valuations extend to $mathbb R$...

Denote an extension of the 2-adic valuation from $mathbb Q$ to $mathbb R$ by $nu$.
Suppose $nu(x)=nu(y)=0$.

Is it true that $nu(x+y)ne 0$?

What about $nu(x^2+y^2)le 1$?

I'm interested in knowing both whether these are true for every extension, as well as knowing whether there is some extension for which they are true (for every $x$ and $y$).

No. The important thing to know is that, if $K subseteq L$ is a field extension and $v: K to mathbbR$ is a valuation, then $v$ can be extended to $L$. So I can answer all of your questions by working in some easy to handle subfield of $mathbbR$. I'll work in $K = mathbbQ(sqrt5)$ for the first question and in $K = mathbbQ(sqrt3)$ for the second.

The ring of integers in $mathbbQ[sqrt5]$ is $mathbbZ[tau]$ where $tau = tfrac1+sqrt52$, with minimal polynomial $tau^2=tau+1$. Note that $mathcalO_K/(2 mathcalO_K)$ is the field $mathbbF_4$ with two elements. Your first statement is true in $mathbbQ$ only because $mathbbZ/(2 mathbbZ)$ has two elements.

Specifically, both $1$ and $tau$ are in $mathcalO_K$ but not $2 mathcalO_K$, so $v(1) = v(tau) = 0$, but $1+tau$ is also not in $2 mathcalO_K$ so $v(1+tau)=0$ as well.

Similarly, the ring of integers in $mathbbQ(sqrt3)$ is $mathbbZ[sqrt3]$ and the prime $2$ is ramified, with $2 = (1+sqrt3)^2 (2-sqrt3)$ (note that $2-sqrt3$ is a unit). We have $v(1) = v(sqrt3) = 0$, but $v(1+sqrt3^2) = 2$. In this case, the result is true in $mathbbQ$ because $2$ is unramified.