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 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 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 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 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/ |
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