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Is Infinity a Number ... in Inversive Geometry?

Date: 01/15/2010 at 10:03:38
From: rainer
Subject: Another attempt at showing that infinity is a number


This is YET ANOTHER question about infinity as a mathematical concept/

I have seen in other exchanges here at the Math Forum that the 
general consensus of mathematicians is that infinity is NOT a number, 
but rather a concept.  The quote below is typical of mathematicians' 
responses to curious laypeople like myself:

"The very sentence '1/infinity = 0' has no meaning.  Why?  Because 
'infinity' is a concept, NOT a number.  It is a concept that means 
'limitlessness.'  As such, it cannot be used with any mathematical 
operators... Would you say that 1/justice has meaning?  No, because 
'justice' is a concept."

All right then, I'm going to take another shot at it.  Are you 
ready?  Here goes:

In inversive geometry, as a point x within a unit circle of radius r 
approaches the center of that circle, its inversion point y outside 
of the circle approaches infinity.  The relation of the distances of 
the points from the circle center is:

   distance_x = r^2/distance_y

When point x reaches the origin, i.e. the center of the circle, its 
inversion point y outside of the circle reaches infinity.  We have 
here then a definite, concrete, measurable point at infinity.

Granted, algebraically, both points x and y are undefined when point 
x reaches the center according to the equation above.  The former 
involves dividing by zero, while the latter involves dividing by 

And yet ... geometrically, we cannot object to a point being at the 
origin of the Cartesian plane, i.e., the circle center.  Such a point 
is not "undefined."  And if a point at the origin is defined, 
its inversion point at infinity must be, too.

Either this must be true, or its converse must be: that a point at 
the origin is as undefined as a point at infinity.

How do you get around that one?

Date: 01/15/2010 at 11:13:11
From: Doctor Tom
Subject: Re: Another attempt at showing that infinity is a number

Hello Rainer,

If you attempt to interpret infinity as a real number and fit it in 
with the rest of the real numbers, it will work only to the extent that
you're willing to give up on some of the laws that the rest of the reals 
satisfy.  It can be used as a concept in the reals in the sense that 
you can say things like, "as x approaches infinity, 1/x approaches 
zero," and this can be made precise by talking about what it means 
when we say "x approaches infinity."  The sentence above means that 
for any tiny number you name, say E, then there is a number N 
(usually large, if E is small) such that 1/x is always smaller than E 
if x is larger than N.

In the example you mentioned, you CAN add an artificial point at 
infinity to the rest of the points on the Euclidean plane to create 
something called "inversive geometry."  The axioms of inversive 
geometry are not the same as those in Euclidean geometry, but they 
are completely consistent, and many useful things can be derived in 
this geometry.

Similarly, in projective geometry, we add not just one point to the 
plane, but an infinite number of "points at infinity," corresponding 
to "moving infinitely far in a particular direction," so there's one 
for every direction.  This sort of infinity makes sense in 
perspective drawings, where the painter (or graphics computer 
program) draws a point where the parallel railroad tracks meet.  This 
corresponds to a perspective projection that takes some points at 
infinity and maps them to finite points and vice-versa.

There are other situations where we can add points.  Called things 
like "point(s) at infinity," they intuitively correspond to things 
infinitely far away; and can be included in complete, consistent 
theories.  Still, you do have to be careful to modify the axioms of 
the original system to account for the added points. 

As an example of axiom modification, you can consider the projective 
planar geometry as I described above, with the addition of points at 
infinity in every possible straight-line direction.  When you add 
this idea, however, the parallel postulate no longer holds: there are 
no longer parallel lines, since two lines that were parallel before 
the addition of the points at infinity now meet at the infinite point 
corresponding to their common direction.

In a sense, this projective geometry is simpler than Euclidean.  In 
Euclidean geometry, you've got these axioms that are asymmetric:

1.  Every two points determine a line.
2.  Every two lines determine a point ---- EXCEPT if the lines are 
parallel, in which case ...

In projective geometry, things are totally simple and symmetric:

1.  Every two points determine a line.
2.  Every two lines determine a point.

Nice, huh?

I hope this helps!

- Doctor Tom, The Math Forum 

Date: 01/18/2010 at 10:13:43
From: rainer
Subject: Thank you (Another attempt at showing that infinity is a 

Dr. Tom-


Very interesting about the superior consistency of inversive geometry 
over Euclidean.  Very nice, indeed.  I was not aware that inversive 
geometry did not take place in Euclidean space.  Although probably 
Euclidean space is inversive geometry (or "hyperbolic"?) space at a 
"local" scale.  I can draw lines that appear to be straight and 
parallel at a local level, but if I extend them several orders of 
magnitude beyond my sheet of paper, they will connect (I think).  At 
the very least they will curve, that much I am sure of.

While I've got your attention, I am trying to resolve a related issue
about the unit circle and its inversion outside the circle.

If we think of the point approaching the center of the circle as 
spaceship x, and call the one approaching infinity spaceship y, what
is spaceship y's acceleration relative to the velocity of spaceship x?

My reasoning: If we set the time interval over which the velocity is 
measured equal to the radius of the unit circle, and substitute in V 
(velocity) and T (time) for Distance, the Ts cancel and we get the 
same formula we started with for the distance of the two points from 
the center of the circle -- only this time, in terms of velocity:

   V_xV_y = 1 instead of D_xD_y = r^2

(r is the "unit radius," so we can safely substitute 1 for it.)


   V_y = 1/V_x

Acceleration equals the derivative of velocity, right?  So just take 
the derivative:

   d(V_y)/d(V_x) = -2/V_x^3

                 = acceleration of spaceship y relative to
                   velocity of spaceship x.  (Right?)

I find it interesting that if we look at the inverse of this 
(spaceship x's deceleration relative to spaceship y), we have a rate 
of deceleration by which spaceship x never stops moving towards its 
destination at the center of the circle, and yet never reaches it 
either -- a "Zeno's paradox" rate of deceleration.

Thanks again.

Date: 01/18/2010 at 10:35:15
From: Doctor Tom
Subject: Re: Thank you (Another attempt at showing that infinity is a 

Hello Ranier,

To work out accelerations, et cetera, begin by first writing down the 
position of the point moving toward the center of the circle in terms 
of time.  For example, if the point moves along the x-axis from 1 to 
0 taking total time 1 unit, then its position is:

   x(t) = 1 - t.
   y(t) = 0,

where the circle is the unit circle centered at the origin.

The position of its inversion is thus:

   X(t) = 1/(1 - t)^2
   Y(t) = 0

The y-coordinates don't enter into it, since they will obviously both 
be zero always, but you can just take first and second derivatives of 
X(t) to find the velocity and acceleration of the inverted point at 
any time t.

The nice thing about setting it up this way is that, if you want, you 
could investigate how the acceleration of the inverted point changes 
with different rates of approach of the first point to the center of 
the circle.

- Doctor Tom, The Math Forum 

Date: 01/19/2010 at 10:11:22
From: rainer
Subject: Thank you (Another attempt at showing that infinity is a 

Dr. Tom-

Great.  But for spaceship y's position, surely you mean

   1/(1 - t), not
   1/(1 - t)^2?
The inversion equation is y = r^2/x.  We are assuming r = 1.

1/(1 - t)^2, as the first derivative of y's position, would give us 
y's velocity instead, right?

Date: 01/19/2010 at 10:33:58
From: Doctor Tom
Subject: Re: Thank you (Another attempt at showing that infinity is a 


You're right.  The position would be 1/(1 - t), the velocity would be 
its derivative, 1/(1 - t)^2, the acceleration would be the second 
derivative, 2/(1 - t)^3, et cetera.

- Doctor Tom, The Math Forum 
Associated Topics:
High School Calculus
High School Non-Euclidean Geometry

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