$1\otimes \arccos \frac{1}{3}$ is not zero in  $\mathbb{R}{\otimes}_{\mathbb{Q}}(\mathbb{R}/\mathbb{Q})$ How to prove that the element
$1\otimes \arccos\frac{1}{3}\in\mathbb{R}{\otimes}_{\mathbb{Q}}(\mathbb{R}/\mathbb{Q})$ isn't equal to zero?
I know why
$$\arccos\frac{1}{3}\neq \frac{m}{n}\pi,$$
where $m\in\mathbb{Z}$ and $n\in\mathbb{N}$.
So, am I right that sufficiently state on $\mathbb{R}{\otimes}_{\mathbb{Q}}(\mathbb{R}/\mathbb{Q})$ (on decomposable elements) equivalence relation
$$x\otimes y \sim x\otimes z \Leftrightarrow (y-z)\in\mathbb{Q}?$$
Thanks.
 A: The hypothesis that $\arccos\frac{1}{3}\neq \frac{m}{n}\pi$ is irrelevant. What counts is that $\arccos\frac{1}{3}\notin \mathbb Q$.
Choose a basis $(r_i)_{i\in I}$ of $\mathbb R$ over $\mathbb Q$ with $0\in I$ and $x_0=1$.
Every $\xi \in \mathbb{R}{\otimes}_{\mathbb{Q}}(\mathbb{R}/\mathbb{Q})$ can be written uniquely as $\xi=\Sigma x_i\otimes \bar y_i$ and in particular $$1\otimes \bar y_i=0 \iff \bar y_i=0 \in \mathbb R/\mathbb{Q}\iff   y_i\in \mathbb Q$$
which proves that $1\otimes \overline {\arccos\frac{1}{3}}\neq 0$ 
A: WARNING!!! As explained in Georges's comment, the argument below is completely wrong. After having read this comment I deleted the answer, but Jason asked me very kindly what precisely was incorrect. So I decided to undelete it, hoping that it will serve as an example of what one should not do. (I hope the warning is conspicuous enough. If it isn't, please let me know.)
Hint. The tensor $1\otimes a\in\mathbb R\otimes_\mathbb Q\mathbb R/\mathbb Q$ is nonzero if and only if $a$ is irrational. 
Proof: The natural $\mathbb Q$-bilinear map $f$ from $\mathbb R\times\mathbb R/\mathbb Q$ to $\mathbb R/\mathbb Q$ sends $(1,a)$ to $a$.
(The map $f$ is defined, with obvious notation, by $f(x,\overline y)=\overline{xy}$.)
Edit. More generally, the argument shows this. Let $A$ be a subring of $B$. Then $1\otimes\overline b$ is zero in $B\otimes_AB/A$ if an only if $b$ is in $A$. (We assume that $B$ is commutative.)
