I am sorry in advance for my English. I am studying math in a different language so I've tried to translate all terms as precisely as possible, but an edit to this question would be great. To the topic. Here's the task:

Using automorphisms method prove that "Dot product is positive" is not expressible in (R^2,E) where E stands for |x-y|==|x-z|

It's all about the 2 dimensional vectors. I have struggled on this task for almost three days for now, but no solution. I have tried reflecting the axis, doing so only with some vectors, taking normal vectors. Nothing worked for me

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    $\begingroup$ The sentence "Dot product is positive" is not expressible in (R^2,E) where E stands for |x-y|==|x-z|, is still a bit sybilline. Can you elaborate on that? The dot product is in $\mathbb{R}^3$? $\endgroup$ – lcv Dec 18 '15 at 5:48
  • $\begingroup$ @lcv We have ℝ2 here as I have mentioned above. The task is to think of an automorphism which will keep the E condition OK for all the vectors for which it was OK before but there will be at least one pair of vectors for which the sign of their dot product would not remain positive $\endgroup$ – tna0y Dec 18 '15 at 9:54
  • $\begingroup$ In the "condition" you mention a variable '$z$' which made me think of $\mathbb{R}^3$. Can you explain what the set $E$ is supposed to be? $\endgroup$ – lcv Dec 18 '15 at 11:15
  • $\begingroup$ @lcv $E$ is a ternary predicate symbol - $x, y, z$ are each elements of $\mathbb{R}^2$, viewed as vectors with the usual vector subtraction, norm, and dot product. $\endgroup$ – Noah Schweber Dec 23 '15 at 20:58

The predicate $E$ asserts that $y$ and $z$ are the same distance from $x$. So in particular, any bijection from $\mathbb{R}^2$ to $\mathbb{R}^2$ which preserves distance will be an automorphism of the structure $(\mathbb{R}^2; E)$.

So it's enough to find an isometry $f$ of $\mathbb{R}^2$ such that, for some $v, w\in\mathbb{R}^2$, we have $v\cdot w>0$ but $f(v)\cdot f(w)\le 0$.

HINT: Rather than looking at two vectors at once, think about a single vector all on its own. When is $v\cdot v$ not positive? Fix $v_0$ such that $v_0\cdot v_0$ is not positive, and some other $v_1$ such that $v_1\cdot v_1$ is positive. Can you think of an isometry of $\mathbb{R}^2$ which moves $v_1$ to $v_0$?


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