Dtjytyk

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 four 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.

You can also quite easily demonstrate it with an appeal to extremes.

Take $(a,b,c) in (2 mathbbN+1)^3, ~ c=a+b$

This defines what I will call statement * : c is an odd number. By definition, we can say :

$exists (a_0, b_0, c_0) in (2mathbbN)^3, ~ a=a_0+1, ~ b=b_0+1, ~ c=c_0+1$

and therefore $a_0+b_0+2 = c_0+1$

Since $a_0$ and $b_0$ are both even, we can say :

$exists (a_0', b_0')in mathbbN, ~ a_0 = 2a_0', ~ b_0=2b_0'$

and

$$
(a_0+b_0) = 2(a_0'+b_0')
$$
$$
Rightarrow a_0+b_0 in 2mathbbN
$$
$$
Rightarrow (a_0+b_0+2 = c_0+1) in 2mathbbN
$$
$$
Rightarrow c in 2mathbbN
$$
which we will call statement ** : c is an even number.

At this point, I think there is no need to say that * and ** are contradicting each other, which is the point of the demonstration since this can only happen if the hypothesis (statement *) is wrong.