Suppose that the Elementary Continuity Principle holds for this interpretation. Let $O=(0,0),A=(3,0),P=(1,0)\text{}and\text{}R=(3,4)\text{}in\text{}Q2$

Let C be the circle in $Q}^{2$ with center O and radius OA. Note that C consists of all points (x, y) in $Q}^{2$ such that ${x}^{2}+{y}^{2}=9$.

Since $O\cdot P\cdot A,OP<OA$. Hence, P is inside C. Let $B=(\frac{9}{5},\frac{12}{5})$. Then $O\cdot B\cdot R\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}OA\stackrel{\sim}{=}OB$. This implies that $OA<OR$ and, hence, that R is outside C. Hence, by the Elementary Continuity Principle, there exists a point Q in $Q}^{2$ such that Q lies on P R and C.

Since Q lies on $PR,Q=P+t(R-P)$ for some real number t such that $0\le t\le 1$. Thus, $P=(1,0)+t(2,4)=(1+2t,4t)$. Since P is in $Q}^{2$, we conclude that t is a rational number. Since P lies on C, $OP\stackrel{\sim}{=}OA$. Hence, ${(1+2t)}^{2}+{\left(4t\right)}^{2}=9$. In other words, $5{t}^{2}+t-2=0$. Since $0\le t\le 1$, it follows that $t=\frac{-1+\sqrt{41}}{10}$ and, hence, $10t+1=\sqrt{41}$. Since t is a rational number, it follows that $10t+1$ is a rational number. In other words, $\sqrt{41}$ is a rational number. On the other hand, since 41 is not a perfect square, $\sqrt{41}$ is not a rational number. This is a contradiction. Hence, the Elementary Continuity Principle fails to hold for this interpretation