What's the 2-adic value of $\sqrt3 + \dfrac1{1+\sqrt3}$?
To start off, I have that to admit a square root, a 2-adic unit must be of the form $4^a (8b+1)$ for some $a \in \mathbb{N}$, $b \in \mathbb{Z}$.
I don't think $3$ is of that form, so I can only make this work if I extend $\Bbb Z_2$
I'm tentatively interpreting this answer as saying that I can simply add $\omega=\sqrt3$ as an additional unit without too much going wrong.
From there I'm running blind. I feel like I need to turn it into a polynomial then Hensel lift to find my answer. Total guesswork here but I have that my number will be a solution to the quotient of the polynomials $\omega^2+\omega+1$ and $\omega+1$ so if I can find the 2-adic value of those I'll be done.
I have that $\omega^2+\omega+1$ will be odd and $\omega+1$ will be even so I'll get a number in $\Bbb Q_2\setminus\Bbb Z_2$, and from there I'm even out of guesswork.
Let me try to cast some light on the issue.
First, don’t use omega here for an irrationality, ’cause half the people reading your question will immediately think of a primitive cube root of unity, $\omega^2+\omega+1=0$. If you don’t want the pain of repeately typing “$\sqrt3$”, then call this irrationality, say, $r$, for “root”.
Second, you should choose either the multiplicative valuation $|*|_2$ for which $|2|_2=\frac12$, or the additive valuation $v_2(*)$, for which $v_2(2)=1$. I prefer the additive one, partly because of typographical simplicity. You seem to have had this in mind in your question.
You know that $3$ is a $2$-adic unit, so that $v(3)=0$. Since $r^2=3$, you have $v(r)=\frac12v(3)=0$ as well. But a crucial quantity here is $1+r$, whose $\Bbb Z_2$-minimal polynomial is $X^2-2X-2$: the (field-theoretic) norm of $1+r$ is $-2$, and the rule for extending valuations is: if $\alpha$ is of degree $m$ over $\Bbb Q_p$, then $$ v(\alpha)=\frac1mv_p\bigl(\mathbf N^{\Bbb Q_p(\alpha)}_{\Bbb Q_p}(\alpha)\bigr)\,. $$ That’s the norm there in the formula, a function from the superscripted field to the subscripted field. In our case $\frac12v_2(-2)=\frac12$. Much more directly, in this case, you certainly expect that $v_2(1+r)=v_2(1-r)$, just because they’re conjugates, so that $v_2(1+r)=\frac12v_2\bigl((1+r)(1-r)\bigr)=\frac12v(-2))=\frac12$.
As to the quantity that stimulated this question, namely $$ \sqrt3+\frac1{1+\sqrt3}=r+\frac1{1+r}\,, $$ your two terms have valuation $0$ and $-\frac12$, so that the sum has valuation $-\frac12$