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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

Hi,

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

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

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
infinity.

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
http://mathforum.org/dr.math/

```

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

Dr. Tom-

Thanks!

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.)

Now,

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
number)

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
http://mathforum.org/dr.math/

```

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

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

```

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

Sorry!

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
http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus
High School Non-Euclidean Geometry

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