Download Aufbau der Geometrie aus dem Spiegelungsbegriff by Friedrich Bachmann PDF

By Friedrich Bachmann

Torischen Gruppenelemente sind und in den en wir geometrische Bezie hungen wie Inzidenz undOrthogonalitat durch gruppentheoretische Rela tionen erklaren. Die rein gruppentheoretisch formulierten Axiome, die wir wahlen, stellen einfache geometrische Aussagen flir die Punkte und Geraden der metrischen Ebenen dar. Dementsprechend kann guy beim Beweisen aus den Axiomen die Vorteile des gruppentheoretischen Kalktils ausnutzen, ohne den Leitfaden der Anschauung aus der Hand zu geben. Bemerkenswert ist, wie wenige Axiome notig sind. Die metrischen Ebenen, die mit den axiomatisch gegebenen Gruppen definiert sind, sind daher von recht allgemeiner Natur. Eine metrische Ebene braucht nicht anordenbar (erst recht nicht stetig) zu sein. In einer metrischen Ebene braucht nicht freie Beweglichkeit zu bestehen. Es gibt auch metrische Ebenen mit nur endlich vielen Punkten und Geraden. Der Begriff der metrischen Ebene enthalt keine Entscheidung tiber die Parallelenfrage, d.h. tiber die Frage nach dem Schneiden oder Nicht schneiden der Geraden. Die ebene metrische Geometrie, die wir ent wickeln, enthalt ebene euklidische, hyperbolische und elliptische Geo metrie als Spezialfalle, und wird daher, mit einem Ausdruck von J. BOLYAI, auch ebene absolute Geometrie genannt.

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5). That is, every point P on L is mapped to a point P′ on L′ , hence L itself is mapped ‘inside’ L′ . Fig. 5 So, how do we prove our subclaim that an isometry must always map collinear points to collinear points? 6). We are dealing with an isometry, therefore |A′C ′| = |AC|, |A′B′| = |AB|, and |B′C ′| = |BC|. Since A, B, C are collinear, |AC| = |AB| + |BC|. But then |A′C ′| = |AC| = |AB| + |BC| = |A ′B ′| + |B′C ′|. We are forced to conclude that A′, B′, C′ must indeed be collinear: otherwise one side of the triangle A ′ B ′ C ′ would be equal to the sum of the other two sides, violating the familiar triangle inequality.

Well, a clever observation is crucial here: it suffices to show that every isometry maps three distinct collinear points to three 4 distinct collinear points! ) that maps every three collinear points to three collinear points must also map every straight line L to (a subset of) a straight line L′ . Once this is done, preservation of distances shows easily that the image of L actually ‘fills’ L′ . Start with a straight line L and pick any two distinct points P 1 , P 2 on it. These two points are mapped by our function to distinct points P′1 , P′2 that certainly define a new line, call it L′ .

3. 12, which in particular shows how the translation vector’s coordinates are determined by the image O ′ of (0, 0): Fig. 5+y). More generally, every translation on the plane may be represented by a formula of the form T(x, y) = (a+x, b+y). Conversely, each formula of the form T(x, y) = (a+x, b+y) represents a translation defined by the vector ; sometimes we may even denote the translation itself by . Observe that the opposite of the translation defined by the vector is simply defined by the vector < − a, − b>.

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