Recall the interpretation of the rational plane: points are ordered pairs $(x, y)$ with $x, y \in \mathbb{Q}$; lines are solution sets of equations $ax + by + c = 0$ with $a, b, c \in \mathbb{Q}$ and $a, b$ not both zero; $(x, y)$ is incident to $ax + by + c = 0$ if the ordered pair satisfies the equation; and betweenness is the standard Euclidean notion. We showed (or at least stated) that this interpretation is a model of incidence geometry with betweenness, i.e.,the incidence and betweenness axioms are all satisfied. Let’s declare two segments to be congruent if they have the same Euclidean length.
Determine if this interpretation satisfies axiom $C-1$. If so, prove it. If not, find specific. examples of $A, B, A'$and $r$ such that there is not a unique $B'$satisfying $AB \equiv A'B'.$
No, it cannot.
Consider the line $x-y=0$ (i.e. $a=1, b=-1, c=0$). Consider the two points on the line : $A(0,0)$ and $B(1,1)$.
Consider now the line $r$ : $y=0$ (the $x$ axis) and on it the point $A'(0,0)$.
Then, you cannot find on $r$ a point $B'$ such that $A'B' \equiv AB$ because the lenght of this segment is $\sqrt 2$ and the point with coordinates $(\sqrt 2,0)$ has not both coordinates in $\mathbb Q$.