[FOM] 715: Foundations of Geometry/3

Harvey Friedman hmflogic at gmail.com
Tue Sep 27 01:08:20 EDT 2016

Once again Dana showed me that I have still not put enough conditions
down to characterize reflection about a line. Even in 2 dimensions,
notice that

x,y goes to (-x+2y,y)

is an involution and the set of fixed points is the line y = x.

Since 0,1 goes to 2,1
0,2 goes to 4,2

it is not distance preserving. So don't pay attention to any "proofs"
in http://www.cs.nyu.edu/pipermail/fom/2016-September/020087.html

So does stubborn me just give up on this treatment of Euclidean geometry?

Not at all.I found the missing condition that characterizes
reflections about a line.

Recall that there was never an issue about how I handled pure
line/betweeness geometry with my midpoint and completeness axiom.

Now let me do the distance part correctly.


1. Take distance as usual as primitive, to be unique characterized.
2. Pick some vividly visual permutations of R^2 that "look" distance
preserving, identifying a key property of such, and just assert that
there is a permutation of R^2 with the key property that is distance
3. Give explicitly the trivial distances - distances along horizontal
or vertical or both lines.

Then we have characterized distance this way IF we can show two things.

4. That in fact there is only one permutation of R^2 satisfying those
respective key properties. And that it is distance preserving - we
need that to ensure that our characterization even has a model.
5. Given any two distinct points, one of the actual vividly visual
permutations of R^2 send one of these points to the other. So the
"distance" between any two points has in fact been determined.

It's all now very simple, and hopefully cannot be screwed up.

THEOREM 1. Let (a,b) be given. There is a unique line preserving
permutation mapping (a,b) to (a,b), (a+1,b) to (a,b+1), and (a,b+1) to
(a-1,b). It is 90 degree counterclockwise rotation.

INTUITIVE DEFINITION. Two lines are perpendicular if and only if there
is a line preserving permutation mapping (a,b) to (a,b), (a+1,b) to
(a,b+1), and (a,b+1) to (a-1,b), mapping the first line onto the
second, where (a,b) is the unique point common to both.

THEOREM 2. Let L be a line. There is a unique line preserving
involution T of R^2 where if x is in L then Tx = x, and if x is not in
L then Tx not= x and the line through Tx,x is perpendicular to L. This
involution is reflection about L.

INTUITIVE CONDITION. Let L be a line. There is a line preserving and
distance preserving involution T of R^2 where if x is in L then Tx =
x, and if x is not in L, then Tx not= x and the lines through Tx,x is
perpendicular to L.

INTUITIVE CONDITION. The distance between (a,b) and (a,b+c) is |c|.

THEOREM 3. There is exactly one distance function on pairs of points
in R^2 obeying the two Intuitive conditions above, plus the intuitive
conditions used to autonomously treat linear geometry.in


Existence is by 90 degree counterclockwise rotation with origin put at (a,b).

For uniqueness, we first assume (a,b) = (0,0), and so T(0,0) = (0,0),
T(1,0) = (0,1), and T(0,1) = (-1,0). Then T is linear and is linearly
independent at the two basis vectors. So T is uniquely determined.

For (a,b), let T(a,b) = (a,b), T(a+1,b) = (a,b+1), T(a,b+1) = (a-1,b),
T affine. Let T' also be like that. Let S(x) = T(x+(a,b))-(a,b) and
S'(x) = T'(x+(a,b))-(a,b). Then S,S' are as in the previous paragraph,
and hence S = S'. Hence T(x+(a,b)) = T'(x+(a,b)). Hence T = T'. QED


There is an automorphism of R^2 with lines and incidence which maps
any given line onto any other given line, and which preserves
perpendicularity of lines. So it suffices to prove Theorem 2 where the
given line L is the x-axis (the part before the last sentence, which
trivially follows anyway from the part before the last sentence).

So we have line preserving involution T of R^2 where if u is on the x
axis, then Tu = u, and if u is not on the x-axis then Tu not= u and
the line through Tu,u is perpendicular to the x-axis. Since T fixes
the origin, T is linear. Write T(x,y) = (ax+by,cx+dy).

The fixed points are the (x,y) such that (ax+by,cx+dy) = (x,y). Hence
(x,0) = (ax,cx), and so a = 1, c = 0.

Hence T(x,y) = (x+by,dy). Now ToT(x,y) = (x+by+bdy,(d^2)y) = (x +
by(1+d), (d^2)y) = (x,y). Hence by(1+d) = 0 and d^2 = 1.

case 1. d = 1. Then b = 0. T(x,y) = (x,y). Impossible, too many fixed points.

case 2. d = -1. T(x,y) = (x+by,-y). This has fixed points exactly the
x-axis, and is an involution, and of course line preserving (even

Now the line from (x,y) to (x+by,-y) is required to be vertical,
assuming y not= 0. I.e., x = x+by assuming y not= 0. Hence b = 0. So T
must be reflection about the x-axis as required. QED

My website is at https://u.osu.edu/friedman.8/ and my youtube site is at
This is the 715th in a series of self contained numbered
postings to FOM covering a wide range of topics in f.o.m. The list of
previous numbered postings #1-699 can be found at

700: Large Cardinals and Continuations/14  8/1/16  11:01AM
701: Extending Functions/1  8/10/16  10:02AM
702: Large Cardinals and Continuations/15  8/22/16  9:22PM
703: Large Cardinals and Continuations/16  8/26/16  12:03AM
704: Large Cardinals and Continuations/17  8/31/16  12:55AM
705: Large Cardinals and Continuations/18  8/31/16  11:47PM
706: Second Incompleteness/1  7/5/16  2:03AM
707: Second Incompleteness/2  9/8/16  3:37PM
708: Second Incompleteness/3  9/11/16  10:33PM
709: Large Cardinals and Continuations/19  9/13/16 4:17AM
710: Large Cardinals and Continuations/20  9/14/16  1:27AM
711: Large Cardinals and Continuations/21  9/18/16 10:42AM
712: PA Incompleteness/1  9/2316  1:20AM
713: Foundations of Geometry/1  9/24/16  2:09PM
714: Foundations of Geometry/2  9/25/16  10:26PM

Harvey Friedman

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