I'm planning to teach high-school geometry. As usual, this will be by building from axioms. (The axioms used are AFAICT particular to the book I've been assigned, but they're some combination of Hilbert's, SMSG's, and God knows what.) I'm considering demonstrating that geometry's axioms need not have their usual model by presenting an alternative model of at least a few basic axioms. Can anyone recommend such a model? I'd need it to be accessible to high schoolers (so, for example, not this).
[Math] Alternative model of Euclidean geometry
educationeuclidean-geometrygeometrysoft-question
Related Solutions
The eleven postulates are sufficient to prove 3.11.
Lemma 1 A line and a point not on it, two different lines in a plane, or two parallel lines define a plane.
Two points on a line and a point not on it define a plane by #7. If two lines are different there's a point on the second that's not on the first (by #6), so by the first part they define a plane. By definition two parallel lines are different lines in a plane so define it by the second part.
Lemma 2 If $a,b,t$ are different coplanar lines and $a$ is parallel to $b$ and $t$ is not parallel to $a$ then $t$ is a transversal of $a$ and $b$.
By definition $t$ intersects $a$ so call the point of intersection $A$ defining an angle $\angle at\ne 0$ (by #3). Let $S$ be a point on $b$ then $SA$ defines a line $s$ (by #6) which is a transversal of $a$ and $b$ (by definition). Then $s$ cuts off angles $\angle sb=\angle sa$ (by #10) and $\angle st\ne \angle sa$ (by #4 because they are coincident), so $t$ is not parallel to $b$ by $\angle st\ne \angle sb$ and #10, and is a transversal (by definition).
Proposition If $a,b,c$ are different lines with $a$ parallel to $b$ and $b$ parallel to $c$ then $a$ is parallel to $c$.
If the lines are coplanar then let $t$ be a line intersecting $b$, then applying Lemma 2 twice it is a common traversal of $a,b,c$. By #10 $\angle ta=\angle tb=\angle tc$ and by #11 $a$ is parallel to $c$.
If the lines are not coplanar, then let $C$ be a point on $c$. By Lemma 1 $a$ and $b$ are in a plane $\pi_1$, $b$ and $c$ are in a different plane $\pi_2$, and $a$ and $C$ are in a plane $\pi_3$. By #9 $\pi_2$ and $\pi_3$ intersect in a line $l$ that contains $C$.
$l$ cannot intersect $b$ in any point $B$, otherwise $a$ and $B$ are in both $\pi_1$ and $\pi_3$, so $\pi_1\equiv\pi_3$ by Lemma 1, $b\equiv \pi_1\cap\pi_2\equiv\pi_3\cap\pi_2\equiv l$ which would require $C$ to be on $b$, contradicting that $b$ and $c$ are parallel. So $l$ does not intersect $b$ but it intersects $c$ at $C$. Since $b,c,l$ are coplanar in $\pi_2$, by Lemma 2 they cannot all be different, so $l\equiv c$.
$l$ cannot intersect $a$ in any point $A$, otherwise $b$ and $A$ are in both $\pi_1$ and $\pi_2$, so $\pi_1\equiv\pi_2$ by Lemma 1, contradicting that $a,b,c$ are not coplanar. Since $l\equiv c$ and $a$ are both in $\pi_3$ and do not intersect it follows that $a$ is parallel to $c$.
Recall the following formulation of the scalar product of two vectors $v, w$: $$ \langle v , w\rangle = |v| \cdot |w| \cdot \cos(\varphi) $$ where $\varphi$ is the angle between $v$ and $w$. In your situation the vector $P-Q$ points in the same (or opposite) direction as $N$, because the line $PQ$ is perpendicular to $\mathit{m}$. Thus you either get $\cos(\varphi)=1$ $$ \langle P-Q , N\rangle N = (|P-Q|\cdot |N|) \cdot N = |P-Q| \cdot N = P-Q $$ if $N$ and $P-Q$ point in the same direction, or $\cos(\varphi)=-1$ $$ \langle P-Q , N\rangle N = (|P-Q|\cdot |N| \cdot (-1) )\cdot N = (-1) \cdot|P-Q| \cdot N = (-1) (Q-P) = P-Q $$ if $N$ and $P-Q$ point in opposite directions.
Best Answer
The rational plane $\mathbb{Q}^2$ is a model for Euclid's five axioms, and I would think (hope?) that it is accessible to high-schoolers. Many common geometric constructions don't work as expected for it; for example, here's an excerpt from Explanation and Proof in Mathematics, p.66: