To Infinity…and Beyond!

In researching Gödel’s Incompleteness Theorem, I stumbled upon an article that stated no one has proven a line can extend infinitely in both directions. This is shocking, if it’s true, and after a quick Google search, I couldn’t seem to find anything that contradicts the claim. So, in the spirit of intellectual adventure, I’ll offer a fun proof-esque idea here.

Consider a line segment of length \ell that is measured in some standard unit of distance/length (e.g., inches, miles, nanometers, etc.). We convert the length of \ell—whatever units and length we’ve chosen (say, 0.5298 meters)—into a fictitious unit of measurement we’ll call Hoppes (hpe) [pronounced HOP-ease]. So, now, one should consider the length of \ell to be 2 hpe such that \ell/2 = 1 hpe. We then add some fraction (of the length of) \ell to (both ends of) itself, and let’s say the fraction of \ell we’ll use, call it a, is 3\ell/4, which equals 3/2 hpe. The process by which we will add a to \ell will be governed by the following geometric series: 

s_n(a) = 1+a+a^2+a^3+\dots+a^{n-1} = (1-a^n)(1-a)^{-1}=\frac{a^n-1}{a-1}.

Let us add the terms of s_n(a) to both sides of \ell; first, we add 1 hpe to both sides (\ell=4 hpe), then 3/2 hpe (\ell=7 hpe, then 9/4 hpe (\ell=23/2 hpe)and so forth. If we keep adding to \ell units of hpe based on the series s_n(a), then we’re guaranteed a line that extends infinitely in both directions because \lim_{n\rightarrow\infty} (a^{n}-1)(a-1)^{-1} = \infty when \vert a\vert \geq 1.

Now, suppose we assume it is impossible to extend our line segment infinitely in both directions. Then s_n(a) must converge to (1-a)^{-1}, giving us a total length of 2+(1-a)^{-1} hpe for \ell, because \lim_{n\rightarrow\infty} 1-a^{n}=1, which is only possible when \vert a\vert < 1. (We cannot have a negative length, so a\in \text{R}^+_0.) But this contradicts our \vert a\vert value of 3/2 hpe above, which means the series s_n(a) is divergent. Q.E.D.

N.B. Some might raise the “problem” of an infinite number of discrete points that composes a line (segment), recalling the philosophical thorniness of Zeno’s (dichotomy) paradox; this is resolved, however, by similarly invoking the concept of limits (and is confirmed by our experience of traversing complete distances!):

\sum_{i=1}^{\infty} (1/2)^i=\frac{1}{2}\sum_{i=0}^{\infty} (1/2)^i=\frac{1}{2} s_n (\frac{1}{2})=\frac{1}{2}\Big( 1+\frac{1}{2}+(\frac{1}{2})^2+\cdots\Big)=\frac{1}{2}\Big(\frac{1}{1-\frac{1}{2}}\Big) = 1,

a single unit we can set equal to our initial line segment \ell with length 2 hpe.

Special thanks to my great friend, Tim Hoppe, for giving me permission to use his name as an abstract unit of measurement.


A Proposed Proof for the Existence of God

Assume it is impossible to prove God does not exist. Then the probability that God exists, p(\text{G}), however minuscule, is greater than zero: p(\text{G}) = ab^{-1} \in (0,1). Also assume, as many important physicists and cosmologists do, that (1) the multiverse exists and is composed of an infinite number of independent universes and (2) our current universe is but one of those infinite universes existing in the multiverse.*

If the probability of the non-existence of God, denoted p(-\text{G}), in some universe is defined as

p(-\text{G}) = (1 - ab^{-1})\in\left(0,1\right)

then as the number of universes (n) approaches infinity,

\lim_{n \rightarrow \infty} (1 - ab^{-1})^n = 0.

That is, the sequence \left(1-ab^{-1}\right)^n\to 0 as n\to\infty. Any event that can happen will ineluctably happen given enough trials. This means God must exist in at least one universe within the multiverse, and if He does, then He must exist in all universes, including our universe, because omnipresence is a necessary condition for God to exist.


* This is certainly a reasonable, if not ubiquitously held, concept that follows from the mathematics of inflationary theory. In Our Mathematical Universe, for example, Max Tegmark suggests if “inflation…made our space infinite[, t]hen there are infinitely many Level I parallel universes” (121).