In Euclidean geometry, I frequently use concepts related to invariance under scaling. For example, I know that if two squares have different side lengths, the ratio of their side lengths is the square root of the ratio of their areas.
I became curious about the justification of this principle recently when I was asked to explain pi, the ratio of circumference to diameter of a circle, to a young student. It occurred to me half way through the explanation that I do not know a reason why pi exists; I'm not sure why all circles have the same ratio.
I haven't studied non-Euclidean geometry, but I presume this scale-invariance is a special feature of Euclidean geometry. On the surface of a sphere, for example, the sum of the angles of a triangle is related to the ratio of the area of the triangle to the surface area of the sphere, whereas Euclidean triangles have the same sum of angles regardless of size.
I know that Euclidean geometry is distinguished from other geometries by the parallel postulate, the statement that for any line and any point not on the line, there exists a unique line parallel to the original and through the point.
My question is, "How can I go from this parallel postulate and other relevant features of geometry to understanding why Euclidean geometry has no innate length scale involved?"
Best Answer
Similarity is a special feature of Euclidean geometry. It follows from the Euclidean parallel postulate:
As a matter of fact, the following claim, called Wallis's postulate, is equivalent to the parallel postulate:
There is a list of 15 claims all logically equivalent to the Euclidean parallel postulate on the Wikipedia Parallel Postulate article. One of which, not surprisingly, is the Pythagorean theorem.
Perhaps the following will provide some insight.
Assume
the proof of which does not require the parallel postulate but is rather involved. Nonetheless, with it, we can show:
Proof: We can construct a unique ray $\overrightarrow{DP}$ so that $\angle EDP \cong \angle A$. We can similarly construct a unique ray $\overrightarrow{EQ}$ on the same side of $\overleftrightarrow{DE}$ so that $\angle DEQ \cong \angle B$. Since the angle measures of the constructed angles, $\angle EDP$ and $\angle DEQ$, sum to less than $180$, by Euclid's fifth postulate (equivalent to Euclidean parallel postulate), the two rays must intersect at some point $F$ forming a triangle $\triangle DEF$. By the angle-sum theorem (also logically equivalent to the Euclidean parallel postulate), $\angle C \cong \angle F$. Now by the AAA similarity theorem, the two triangles are similar. $\square$
Then given any scale factor $r$, We can construct a segment $\overline {DE}$ so that $$r=\frac{DE}{AB}.$$
From this, Wallis's postulate follows straight forwardly.
Conversely, let us assume Wallis's postulate and from it derive the Euclidean parallel postulate.
Proof of EPP from Wallis's Postulate: Suppose we have a line $\ell$ and a point $P$ not on the line. We can drop a perpendicular from $P$ to line $\ell$ and let the perpendicular intersect $\ell$ at point $F$. Then construct a new line $m$ at $P$ at $90$ degrees to our perpendicular. This new line will be parallel $\ell$. If it were not, $m$ and $\ell$ would intersect forming a triangle which would violate the exterior angle theorem.
It remains to show the line $m$ is unique. Suppose there were a different line $n$ also through $P$. We must show $n$ is not parallel, that is, line $n$ intersects $\ell$.
If $n$ is our perpendicular $\overleftrightarrow{PF}$ obviously it intersects $\ell$. So if $n$ is not $\overleftrightarrow{PF}$, lets choose a point $Q$ of $n$ between lines $\ell$ and $m$ but not on $\overleftrightarrow{PF}$. Let $R$ be the point of intersection of a perpendicular dropped from $Q$ onto line $\overleftrightarrow{PF}$. Let $r=\frac{PF}{PR}$. Wallis's theorem assures us that there is somewhere a triangle $\triangle ABC$ that is similar to $\triangle PQR$ with scale factor $r$. We can construct a new triangle $\triangle PSF$ on the segment $\overline {PF}$ congruent to $\triangle ABC$ by placing $S$ on $\ell$ at distance $BC$ from $F$ and on the same side of $\overleftrightarrow{PF}$ as $Q$. This triangle is congruent to $\triangle ABC$ by the side-angle-side congruence theorem and so is similar to $\triangle PQR$. So the angles $\angle FPS$ and $\angle RPQ$ are the same angle. Hence the lines $n$, $\overleftrightarrow{PQ}$, and $\overleftrightarrow{PS}$ are the same line, and $n$ does indeed intersect $\ell$. $\square$