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It is quite obvious that my parents know me well. For my birthday, they got me DVDs from a course on number theory. This meant I was awake far too long last night watching the first set. In the process, I learned a wicked kewl way of converting kilometers to miles using the Fibonacci series.

The Fibonacci sequence starts with 1,1 as the “seeds.” Then it follows the rule, “The next number in the Fibonacci sequence is the sum of the previous two numbers.” So the sequence begins:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55

The 2 comes from adding the previous two numbers (1 + 1). The 3 comes from adding the previous two numbers, which now are (1 + 2). The 5 follows the same rule, only now the numbers are (2 + 3). Etc.

Now it happens that the ratio between numbers in this sequence approaches the golden ratio, and at infinity it is equal to the golden ratio. The golden ratio is defined as the Greek letter phi (which doesn’t show up here, so I’ll use x instead) and can be expressed as x = 1 + 1/x. This gives us a recursive equation, one that includes itself within the definition. However, this equation can still be solved by multiplying both sides by x² = x + 1. This gives us a quadratic equation, which means we want all the terms on one side. That gives us x² – x – 1 = 0. Using the binomial theorem, we can solve this to show that x = (1 +/- sqrt(5))/2.

[In the previous equation, “sqrt(5)” stands for the square root of 5.]

Now even without a calculator, we know that the square root of 5 is just a little more than 2, since the square root of 4 is exactly two. So if we subtract the square root of 5 from 1, we will end up with a negative number as our end result. But we only want the positive version, which is x = (1 + sqrt(5))/2 so we can safely ignore the negative version.

In any case, we know that 1 + a number that is a little bigger than 2 gives us a number that is a little bigger than 3. And if we divide that number by 2, we will end up with a number a little bigger than 1.5. And the golden ratio begins 1.608…

Now it just happens that 1.6 is pretty close to the ratio between a kilometer and a mile. In other words, we have about 1.6 km per mile. So if the ratio between the numbers in the Fibonacci series in about 1.6, then we can use the Fibonacci series to convert km to miles. Let’s start easy by reprinting the Fibonacci series we had above:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55

If we want to know how many miles are in 13 km, we find 13 on the series and then look for the previous Fibonacci number, which is 8. So there are about 8 miles in 13 km. And if we measure it, we see that 13 km is roughly 8.08 miles.

But what if we want to know how to convert a number that is not part of the Fibonacci series? Well, it turns out that you can add up Fibonacci numbers to make other numbers. So suppose we want to know how many miles are in 20 km. What you do is start with the closest Fibonacci number that is smaller than the number you’re looking for. We see in this case that that would be the number 13.

So keep 13 in mind. Now if you subtract 13 from 20, you’re left with 7. Let’s find the closest Fibonacci number smaller than 7, and that’s 5. So keep 5 in mind too. Finally, if we subtract 5 from 7, we’re left with 2, which is itself a Fibonacci number. So we end with 2.

So we’ve pulled out 13, 5, and 2. And if you’ll notice, 13 + 5 + 2 = 20, our original number. So 20 is composed of those three Fibonacci numbers. What we do to convert it to miles, then, is to find the next lowest Fibonacci number for each of those composite numbers, and add them together. So the next smallest from 13 is 8, the next smallest from 5 is 3, and the next smallest from 2 is 1. 8 + 3 + 1 = 12.

So there are about 12 miles in 20 km. And by measurement, we see that 20 km is actually about 12.4 miles, so we’re pretty close.

Now you may be wondering how the above works. Well, we know that the ratio between Fibonacci numbers is the golden ratio, and the ratio between miles and km is also close to the golden ratio. This means that so long as we can express the ratio of any number as a sum of Fibonacci numbers, we can use the previous “trick” to convert them.

To demonstrate how this works, let’s use a simpler ratio: 1/2. Suppose we wanted to know what half of 20 is. Obviously, we know that 20/2 = 10. But we can express 20 as 8 + 12 too. If we then take half of each of those numbers, we get 4 +6. 4 + 6 = 10, which is identical to 20/2. Or suppose we expressed 20 as 18 + 2. Half of 18 = 9, and half of 2 = 1. 9 + 1 = 10.

Because the ratio is the same, then no matter how you compose 20, dividing each of those numbers by the same ratio before you add them together is equivalent to finding the value of the original number divided by that ratio. So the same thing happens with Fibonacci numbers, except the ratio is the golden ratio instead of a half. Therefore, if we can compose a number as the sums of Fibonacci numbers (and if I am not mistaken, I believe all natural numbers can be expressed as the sums of Fibonacci numbers), then we can use the same technique as shown above to find a conversion between km and miles.

Now I should point out that while there are many things in nature that seem to use the golden ratio, the fact that the relationship between km and miles is close to the golden ratio is actually a matter of serendipity rather than due to some natural law of the golden ratio. Apparently, the mile was defined by the English Parliament in 1592 as being 1760 yards. The survey mile was derived as being 8 furlongs (each furlong being 10 chains, each chain being 4 rods, and each rod being 25 links—yeah, you figure it out). There is no intrinsic pattern or order to any of these gradations, hence the continual demand in science to use the metric system.

A meter, on the other hand, was originally defined as the length of a pendulum with a half-period of 1 second. It was later changed to being 1/1000000 of the distance between the North Pole and the equator. In 1983, the definition was changed so that a meter is now officially the distance light travels through a vacuum in 1/299792458 seconds. This number was formed by using an updated version of the previous definition of a meter and measuring the speed of light as 299792458 meters per second; but it has now become the benchmark itself.

So these two methods, one complete arbitrary and seemingly random, and the other apparently strict measurements of distance covered through time, somehow happened to form a ratio fairly close to the golden ratio.

Since a commenter recently noted that Steve’s been writing almost all of the Triablogue posts of late, I figure I can post this one on the T-Blog even though I’m not quite sure there’s any practical apologetic use for it. On the other hand, it’s stuff that I find “wicked kewl” and therefore is interesting to me. But it’ll have a bit of math in it, so if you don’t like that, well I’m sure Steve will write something new shortly :-)

One of the questions that cosmologists have pondered is whether the universe is open or closed. An open universe would extend infinitely in all directions, whereas a closed universe would have a “boundary.” However, even a closed universe could still be infinite. If space was curved in such a manner that, just like you could always travel East on Earth and return to the point you started from, in the universe you could always pick a direction and travel long enough and you’d return to your starting point. In other words, you could travel infinitely in one direction yet always return to your starting point (this would assume space was curved in the fourth, or higher, dimension that we cannot physically see).

I have to admit that I have a strange attraction to these kinds of loops. I don’t know why, but they appear “pleasing” to me. And therefore I find it no surprise that I’ve discovered one such loop within numbers themselves. In other words, just as we could say that the universe is infinite yet closed because it loops back (assuming that theory is correct, I must add—by far this is not proven!), I say that numbers themselves are infinite and yet closed because they loop back on themselves too.

For a simple proof (simple in that it requires nothing more than algebra), consider the following.

1. The number 1 (one) is that number which has no factors other than 1.

This can be restated as:

1′. If a number n has only 1 as a factor, then n = 1.

This seems fairly straightforward to me, yet by the end of this you’ll see why it might be tempting to deny the above.

Now we need to give one other tidbit of information. I’ll explain it below (and note that because we are dealing with factors, by definition we’re only considering positive values and whole numbers, so all the numbers below are positive integers):

2. Let c be a factor of w.

3. Since c is a factor of w, the next integer greater than w that c can likewise be a factor of is w + c.

Since many people don’t like thinking with letters instead of numbers, let me give a concrete example. Let’s say that c = 7 and w = 21. 7 is a factor of 21, so (2) above is satisfied. (3) states that if 7 is a factor of 21, the next number greater than 21 that 7 could be a factor of would be 21 + 7, or 28. And this is obvious because 22, 23, 24, 25, 26, and 27 cannot have 7 as a factor. Indeed, (3) is really nothing more than restating the definition of a factor.

Now let my proof begin in earnest:

4. Let x be the product of all positive integers. That is x = 1 x 2 x 3 x 4 x … x infinity.

5. Since x is the product of all positive integers, then x has all positive integers as factors.

6. Let a be a factor of x.

7. The next number greater than a than will be a factor of x is x + a.

8. Consider x + 1.

9. Let a = 1.

10. a is a factor of x + 1 (per (7)).

11. Therefore, 1 is a factor of x + 1.

12. Let a be greater than 1.

13. a cannot be a factor of x + 1 because the next greatest number than x that a could be a factor of is x + a (per (7)), and a > 1 (per (12)).

14. Therefore, 1 is the only factor of x + 1 (per (11)).

15. Therefore, x + 1 = 1 (per (1′)).

16. If x + 1 = 1, then x = 0 (algebra).

17. But(!) x = 1 x 2 x 3 x 4 x … x infinity (per (4)).

18. Therefore, 1 x 2 x 3 x 4 x … x infinity = 0.

Now the way that I see it, there are one of two options that mathematicians can take here. Either we can simply rule that when x is 1 x 2 x 3 x 4 x … x infinity, then x + 1 is undefined (similar to the way that division by zero is undefined), or we can say that numbers themselves contain some sort of looping mechanism, wherein by the time you reach infinity (the infinity defined as the product of all positive integers), you “loop back” to zero.

You already know which way I’ll go because I like loops. :-) But there is more evidence. I think we can see the “loop back” when looking at a tangent graph. Since I don’t want to throw in Greek symbols here, assume that a is an angle: tan(a) = sin(a)/cos(a). So, whenever cos(a) = 0, tan(a) is undefined because of division by zero.

The tangent graph looks like this:

That’s with the classic orientation, where the origin (where the arms of the graph cross) is located at (0,0). You can tell that since the right-hand portion of the graph is running up toward infinity and the left-hand portion is running down toward negative infinity why there would be a sudden “jump” in the graph at pi/2 (since cos(pi/2) = 0). If the x value is just slightly less than pi/2, you have positive infinity, but if it’s just slightly more than pi/2 you have negative infinity.

Instead of assuming these things just go off to infinity, what happens if we assume that they “connect” at infinity and redraw the graph from that perspective? If I did it correctly (and since it’s late at night right now, I am subject to correction), you’d get something that looks like this:

For this graph, we’re looking at how it relates to infinity. Basically, what I did was assume that the graph “rolls over” at infinity, and made the horizontal axis the point where positive infinity and negative infinity intersect. In essence, you move the lower left to the upper right on the tangent graph and vice-versa. Naturally, the graph is horrifically distorted since it’s representing two infinities on the vertical axis—the lines would actually appear to be virtually synonymous with the vertical axis for most of the trip, with the hook out at the very end; but I think this is sufficient to at least give a faint picture. (Note: technically, the origin on this graph would still be undefined, since the origin in this view is the point where the division by zero takes place.)

In any case, note that this graph would continue in sequence, just like the tangent graph does. That means that you could print out a row of these figures. The interesting thing about them is that you can then take the top of the graph and “fold” it down so that the 0s appear on the same line (the graph would now be on a donut-shaped paper rather than a 2D screen). At this point, the line graphed would look continuous (bearing in mind that at the origin of each cross point (multiples of pi/2) the graph would still be undefined).

This would imply that the graph, represented flat on a 2D surface, takes on the characteristics of a bent 3D object. Though only two dimensions are present in the tangent graph, there is an assumed third dimension where the graph “rolls over” from positive to negative infinity. In this curved 3D representation, the graph no longer has an infinite jump from positive to negative infinity, but rather that jump is a mere point, more akin to switching from positive to negative numbers at 0.

In short, it would be a curved space of infinite length, curved in a higher dimension.

This might actually affect physics. If it is true that math on the number line itself assumes a higher dimension of curved “space” then one could question whether that means reality really is curved, or whether it means that our math will always make it appear to be curved regardless of what it really is. In other words, is the fact that the math involved in physics seems to indicate a curved universe the result of the way that the universe actually is, or is it because the only method by which we have of probing the universe on such levels is mathematically, and math itself is curved? To use an analogy, suppose you use a level and see that a board appears warped; is the board warped or is the level warped? If we define the level as being level, then the board is warped; but what if we begin to see evidence that the level itself shows a curve?

Being me can be somewhat fun, even if it is definitely somewhat weird. For example, I have the day off today because I’ll be heading up to Denver with my sister to pick up our parents from DIA since they’re back in the states for a month. Since I wasn’t working, and since I had some time to kill before heading off to Denver, I was reading a book. And then, right in the middle of a sentence, I suddenly had a thought about geometry. It’s hard for me to put into words what exactly the thought was, but after about thirty seconds I had come to a conclusion.

Namely, if you take an isosceles right triangle and get the hypotenuse, then you construct another isosceles right triangle with the base and arms equal to that hypotenuse, the hypotenuse of the second triangle will be twice the size of the base of your first triangle.

How many other people have that thought in the middle of a stinking novel?

In any case, since I’ve mentioned it here, I can go ahead and prove it to you just for fun. All you need to know is the Pythagorean theorem, which is a² + b² = c². Oh, and I suppose you should also know that in an isosceles right triangle, a = b. So for them, you can say 2a² = c².

So, if a = 1, then c = 2^1/2 (the square root is a number to the 1/2 power)

This follows from both equations, as you have:

1² + 1² = c²
1 + 1 = c²
2 = c²
2^1/2 = c

Or:

2(1²) = c²
2(1) = c²
2 = c²
2^1/2 = c

In any case, now we take 2^1/2 as the base of our second triangle. To keep the variables separate, I’ll call the second triangle 2x² = z², where x = 2^1/2. So:

2[(2^1/2)²] = z²
2(2) = z²
4 = z²
2 = z

So you can see that a = 1, z = 2. So z = 2a.

In any case, this works for all numbers, since we can show that by using variables instead of real numbers, as follows:

Take a triangle ABC where AB = BC, and take a triangle ACD, where AC = CD, then I say that AD = AB + BC.

Let AB be of the length a.
Therefore, BC is also of the length a.
Therefore, AB + BC is of the length 2a.

Also, therefore, AC is of the length (2a²)^1/2.

Which is also 2^1/2a since (xy)^1/2 = x^1/2y^1/2.

Since AC = CD, then CD = 2^1/2a

Therefore, AD² = (2^1/2a) ² + (2^1/2a) ²

AD² = 2a² + 2a²
AD² = 4a²
AD = [4a²]^1/2
AD = (4^1/2)(a²)^1/2
AD = 2a

Therefore, AD = AB + BC.
Q.E.D.

All that from reading a completely non-related novel.

Sometimes, I think it would be great if I could turn my brain off at will. You know, just stop thinking and float for a bit. This usually comes after I’ve spent the night tossing and turning trying to visualize the fourth dimension. Note: while I’m actually doing that, no way in the world do I want to turn my brain off! No, it’s only after I get up in the morning and drag my carcass off to work that the benefits of being able to turn my brain off exert themselves.

Oh, and if you’re trying to read between the lines, this did not occur last night, although it has in the past. Instead, it happened after I picked up my lunch at Subway this afternoon and started back to work. So it’s not an inconvenient time at least.

Here’s what I thought about. You remember the math stuff (I can see you roll your eyes, you know) about how zero = infinity and stuff like that? And you remember the cool nifty squares I made of numbers in various bases? Well, I think I’ve found a way to combine them.

As way of example, let me give you a base-12 square, where A = 10 and B = 11. Oh, and ou might want to look at this and this before you continue if you need a refresher as to what I am doing:

1 2 3 4 5 6 7 8 9 A B   Sum = 66
2 4 6 8 A 0 2 4 6 8 A   Sum = 60 (1 zero)
3 6 9 0 3 6 9 0 3 6 9   Sum = 54 (2 zeros)
4 8 0 4 8 0 4 8 0 4 8   Sum = 48 (3 zeros)
5 A 3 8 1 6 B 4 9 2 7   Sum = 66
6 0 6 0 6 0 6 0 6 0 6   Sum = 36 (5 zeros)
7 2 9 4 B 6 1 8 3 A 5   Sum = 66
8 4 0 8 4 0 8 4 0 8 4   Sum = 48 (3 zeros)
9 6 3 0 9 6 3 0 9 6 3   Sum = 54 (2 zeros)
A 8 6 4 2 0 A 8 6 4 2   Sum = 60 (1 zero)
B A 9 8 7 6 5 4 3 2 1   Sum = 66

Now, as before with base-10 math, if you give the zero an actual value (in this case, 6) then the sums add up to 66 per row in base-12. Oh, and if you’re wondering, the pattern seems to be this:

Take any even base, b. Then take the midpoint, as m = b/2. Then find m² + [(m – 1)m]. That will be what each row adds up to, if you assume a zero value = your m value.

So we show this in base 14. In base 14, b = 14, so m = 7. 7² = 49, while [(7 – 1)(7)] = 6 x 7 = 42. 42 + 49 = 91. We can test it by seeing that 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 +13 = 91 (this would be your first row if you constructed a box like the above, although it would look like 1 2 3 4 5 6 7 8 9 A B C D).

In any case, you can convert

m² + [(m-1)m] = 2m² – m

which is easier to use on a calculator.

In any case, look at what would happen if we included a zero:

1 2 3 4 5 6 7 8 9 A B 0  Sum = 66 (1 zero)
2 4 6 8 A 0 2 4 6 8 A 0  Sum = 60 (2 zeros)
3 6 9 0 3 6 9 0 3 6 9 0  Sum = 54 (3 zeros)
4 8 0 4 8 0 4 8 0 4 8 0  Sum = 48 (4 zeros)
5 A 3 8 1 6 B 4 9 2 7 0  Sum = 66 (1 zero)
6 0 6 0 6 0 6 0 6 0 6 0  Sum = 36 (6 zeros)
7 2 9 4 B 6 1 8 3 A 5 0  Sum = 66 (1 zero)
8 4 0 8 4 0 8 4 0 8 4 0  Sum = 48 (4 zeros)
9 6 3 0 9 6 3 0 9 6 3 0  Sum = 54 (3 zeros)
A 8 6 4 2 0 A 8 6 4 2 0  Sum = 60 (2 zeros)
B A 9 8 7 6 5 4 3 2 1 0  Sum = 66 (1 zero)
0 0 0 0 0 0 0 0 0 0 0 0  Sum = 0  (12 zeros)

Now, let’s add the value of 0 = 6 to the equation above:

1 2 3 4 5 6 7 8 9 A B 0  Sum = 72 (1 zero)
2 4 6 8 A 0 2 4 6 8 A 0  Sum = 72 (2 zeros)
3 6 9 0 3 6 9 0 3 6 9 0  Sum = 72 (3 zeros)
4 8 0 4 8 0 4 8 0 4 8 0  Sum = 72 (4 zeros)
5 A 3 8 1 6 B 4 9 2 7 0  Sum = 72 (1 zero)
6 0 6 0 6 0 6 0 6 0 6 0  Sum = 72 (6 zeros)
7 2 9 4 B 6 1 8 3 A 5 0  Sum = 72 (1 zero)
8 4 0 8 4 0 8 4 0 8 4 0  Sum = 72 (4 zeros)
9 6 3 0 9 6 3 0 9 6 3 0  Sum = 72 (3 zeros)
A 8 6 4 2 0 A 8 6 4 2 0  Sum = 72 (2 zeros)
B A 9 8 7 6 5 4 3 2 1 0  Sum = 72 (1 zero)
0 0 0 0 0 0 0 0 0 0 0 0  Sum = 72 (12 zeros)

And what is the relationship of m = 6 to the result of 72?

2m²
2(6² = 2(36) = 72.

In some way, it appears that because of the way the pattern works, the zero (which is where the pattern repeats) seems to take on the value of the midpoint of the pattern. In other words, zero doesn’t act like “nothing” in these patterns. Zero acts as if it has value, but the value comes not because of the digit itself, but because of the patterns that are formed.

Now here’s the thing. You can extend this out to infinity, at least as long as you’ve got an even infinity for your pattern. Of course, we’d quickly run out of digits to express our pattern, so we couldn’t make a physical box like we’ve done above. But if you have an infinite number base that’s even (when I have time later I may see how this works in odd bases too, but at the moment my lunch break is almost over), then the zero value in the patterns would seem to take on 1/2 infinity as its value.

Of course, this begs the question: does it really take on this value? Obviously in terms of pure math, we’d say, “of course not.” On the other hand, looking at the patterns above, it seems so elegant to get the rows all to the same value that it feels like there’s something to this notion.

In any case, I think it’s safe to agree that zero isn’t what we usually think it is.

As I came to work this morning, something dawned on me about the Factor Field. Long story short, the value I assigned for *! would actually be 0, since I had already established that *! = 0; and therefore, *! can’t be the place where infinity switches its sign.

But that’s okay, because I think I figured out where that switch would be anyway. To demonstrate this, I have to start with the fact that if *! = 0, as I showed in this post, then we are faced with what seems to be a contradiction. Numbers get larger and larger, yet at some point they have to switch to become smaller and smaller in order to get us back down to zero. And, in fact, this happens if you think about infinity switching signs.

So I thought to myself, Self, is there another place on the graph where we can extrapolate from something we see to form an analogy of what would happen at infinity? And I responded, Why, yes, Self, there is!

That’s because basically what we are looking for are the properties when x =1, 2, 3, 4, 5…infinity are all valid. So we can look at a place where some of these properties are valid and deduce what would happen if they were all valid. So I picked the graph at 420.

That’s because 420 has factors 1, 2, 3, 4, 5, 6 and 7 (it also has lots more, but this make a nice “spike” in the graph for us to look at). Here’s what it looks like in part:

Now here’s the important thing to examine: the radial arms. The radial arms match the spike that shoots out from the center in terms of length. They also go at a 45 degree angle (a slope of 1/1 and -1/1) from the spike at y = 420.

Why is that important? If we ignore everything after x = 7, the graph looks exactly the same at y = 420 as it does at y = 0:

So the value of y at the maximum point will be equivalent to the length of the radial arms, and the length of the spike—so it’s the infinity formed by adding 1 + 1 an infinite number of times.

This means that the value of *! needs an adjustment. While it is true the *! = 0 if we define *! as 1 x 2 x 3 x 4 … x infinity, that’s not the only infinite value that equals zero!. Consider for a minute something about even numbers. If 4 is a factor of a number, then 2 must also be a factor of that number. If 16 is a factor of a number, then 2, 4, and 8 are all factors too!

Suppose for a moment that we’re looking for the graph of a number that has 2, 4, 6 and 8 as factors. If we have 8 as a factor, then we must already have 2 and 4 as factors. That means that we only need to multiply 6 x 8 to get a number that has 2, 4, 6, and 8 as factors. 6 x 8 = 48, which does indeed have 2, 4, 6, and 8 as factors. Now it is still true that if we multiplied 2 x 4 x 6 x 8 we’d also have a number that contains 2, 4, 6, and 8 as factors (that number would be 384). But 384 is a lot larger than 48! If the key that we were searching for was the first time we got 2, 4, 6, and 8 as factors, it would happen at 48 and we’d never get to 384.

So that means that we’re not actually going to use the infinity of 1 x 2 x 3 x 4 … in our problem. We can throw out 2 no matter what, because any even number greater than 2 will automatically have 2 as a factor. That alone halves the size of the infinity we’re dealing with. In reality, the number we’re dealing with is simply the number that is made by multiplying the highest values of numbers close to infinity that are composed of all the other factors below infinity. That number is, of course, impossible to write down. But let’s call it m, for maximum. m = 0, for the same reasons that *! = 0. Namely, m has all the factors of all the numbers, so the graph of m + 1 looks exactly like the graph of 0 + 1, so m + 1 = 1, and therefore m = 0.

However, m < *! in terms of the size of the infinity, since 1 x 2 x 3…etc. is going to be way larger than simply multiplying the highest numbers that contain all the lesser factors together! Furthermore, it would place us in the position where *! falls outside the range of the Factor Field. I think this is just Gödel making his appearance, so I’m prepared to dismiss it—but of course that may not be the rigorous mathematical thing to do!

Now if the distance from 0 to m is one oscillation, then the sign switch of infinity will occur at a distance of m/2. That’s because you go out m/2 distance, then you return m/2 distance. Let’s call m/2 by the variable M.

Let’s try to visualize what’s going on. Imagine the graph that goes up from y = 0 somehow meeting the graph that goes down from y = 0. At the point they meet, infinity switches signs. The numbers are getting larger and larger as the graph goes up; then it hits the merge point (at the sum of all whole numbers) and instantly we go from +infinity to –infinity—exactly as what occurs on the tangent graph. Then the points fall back down (or rather, still increase but now from their negative state) to the value of zero, where they switch signs and go back up toward infinity.

Now for why I think this M point is where the sign of infinity switches. Again, think of the graph at y =0. We move up in the positive values to reach infinity at y = M. But at y = M + 1, the graph has inverted. If M – 1 is positive infinity – 1, then M + 1 is negative infinity + 1. This is exactly what you’d expect, because M + 1 looks exactly the same as the line you’d get at negative infinity when the line reaches the negative value of M – 1.

I maintain that this is exactly what happens at the tangent graph, and the reason that we cannot actually graph it when cos x = 0 is because M is simultaneously positive and negative, just as zero is simultaneously positive and negative.

Just some quick notes (more for me to keep track of things to think about, rather than me trying to prove anything here). I think that given my previous two posts, -*! corresponds to the positive number line while +*! corresponds to the negative number line. That is, since zero is a reflecting point, and *! Is functionally equivalent to 0, then *! is also a reflecting point. However, if you try to picture this mentally, in order to keep positive integers from being cancelled out by the graph created at *!, you’d have to invert the signs in one of them.

Secondly, where functions are undefined (such as the tangent function wherever cos x = 0), I think that’s where the sign switches at the infinity mark. In other words, just as hitting zero on a number line switches our sign, so hitting *! switches the sign. This is sort of like saying the function actually is continuous, but in a different dimension that we cannot draw on the 2D grid. In other words, if we flip it and instead graph it from the x-axis crossing y at *!, then we’d see the line go through the point at *! and we’d be referring to the positive and negative zero, rather than the positive and negative infinity, and we’d be saying that the function is undefined at 1/n as n-> infinity (instead of how it is now, when it’s as n -> 0).

In other words, I think that the only reason these functions are considered undefined there is that we cannot visualize the two infinities meeting. Except that we can once we realize that it would look exactly like the zero line. We still couldn’t draw it, but we could at least visualize what it ought to look like, at least as far as we can mentally think of any infinite.

Last night through this morning, I’ve thought of some more implications of what would happen if we viewed infinity in the same manner as we view zero. First off, we can ask a simple question: what happens to the number line at zero? I’ll give an ASCII representation here:

< ------------------->
     -2 -1  0  1  2

So we see that as we move left to right, the numbers are negative. Ignoring the negative sign, the numbers appear to be getting smaller (they are in fact getting larger because of the negative sign, but think about the concept for a moment). Then it hits zero. At that point, the sign of the numbers change from negative to positive, and the numbers printed begin to get larger again.

The result is that we can view the negative numbers as an inversion of the positive number, and the inversion point is at 0. This is more clearly seen in the graph I showed in the previous post, which I’ll reproduce here:

If we ignore the “name” that we give to the numbers and just look at the picture, we see that the zero line gives us a “reflection surface” such that what is above the line is reflected by what is below the line. Above the line goes in one direction; below, in the opposite.

If we maintain the equivalency of zero and infinity, then when we hit the infinity line, we will exactly recreate the graph you see above, with only the “names” of the numbers different. Instead of seeing “1” and “-1” we would see infinity -1 and infinity + 1.

Now for the important part. I maintain that at that point, the sign of the numbers (in relation to infinity) changes, so we can say that the sign of infinity changes. In other words, if we say all numbers > 0 are positive and all number < 0 are negative, then we can also say that all numbers > the infinity line are positive infinity, and all numbers < the infinity line are negative infinity.

(Quick side note: when I say infinity above, I am specifically referring to the infinity that is formed by 1 x 2 x 3 x 4 x 5 … etc. Since there are different infinities, then it is critical that this be kept in mind. To help do so, and because 1 x 2 x 3 … etc. are factorials, then I will refer to this specific infinity as “infinity factorial.” Since WordPress won’t let me paste in the infinity symbol, I’m going to use the * symbol as the infinity symbol. Therefore, *! = infinity factorial. Hope that’s not too confusing.)

Back to the point now. As I was saying, if we say there is a sign switch at the *! line, then we’d have positive *! and negative *!, or +*! and -*!. The immediate question is, is there any evidence that such a thing exists? That is, is there evidence that there’s a single point where positive and negative infinity switch at all? And the answer is…yes.

Consider the tangent curve. You can see a picture of it here (look at the graph called y = tan x). The important feature is that there are “discontinuities.” What’s a discontinuity? It’s the part of the graph where the line suddenly jumps from positive infinity to negative infinity. Since tan = sin/cos, then this happens anytime that cos x = 0. This is due to the fact that you can’t divide by zero. I would argue that, given what I’ve stated above, division by zero = *!, and that’s why you get the jump between positive and negative infinity on the graph.

If this is true, then we have a startling relationship. It is loosely stated that 1/infinity = 0 (in reality, to be more precise, you have to talk about the limits, and say that 1/n = 0 as n -> infinity). What I’m proposing is that 1/0 = *!. Algebraically, this makes sense:

If a/b = c, then this is equivalent to:

a = cb

b = a/c

Now in the above, let a = 1, b = *!, and c = 0. We see that if 1/*! = 0, then it’s also true that 1/0 = *!.

Yes, I realize that division by zero is something that you ought to rebel against because it can be abused to prove many things that are contradictory. For instance, you can prove that 2 = 7 in the following manner:

a = 1
b = a
2(a – b) = 7(a – b)
Divide both sides by (a – b) to remove the common factor:
.: 2 = 7

This doesn’t work because (a – b) = 0, and this is division by zero which is officially “undefined” and not allowed.

But note that if my theory is correct, the math still works! I argue that any number (x) would solve the equation x/0 = *!, so 0/0 = *!. Therefore, what you have is 2*! = 7*! and I would say that this does not violate any rules of math, any more than 2 x 0 = 7 x 0 violates rules of math. Of course, just because i would say that doesn’t make it true…so let’s examine it a bit further.

My claim would be that x*! = y*!, no matter what x and y are. So this is equivalent to saying x*! = *!. Is there a way to prove this? Well, we could try it this way (note: it still keeps the denominator as zero, but we don’t actually do any division by zero in the following so I think it doesn’t violate any rules, but I may be wrong on that count).

*! = a/0, where a can be any number.
x*! = xa/0
.: y*! = ya/0

Now the claim is x*! = y*! so:

xa/0 = ya/0

We can multiply both sides by 0/1, which is the typical way to try to remove the denominator of a fraction:

(0/1)(xa/0) = (0/1)(ya/0)
[0(xa)]/(1 x 0) = [0(ya)]/(1 x 0)
0/0 = 0/0

Since the final line is an obvious truism, insofar as both sides have the same symbols, then it follows that x*! = y*!, and therefore it would not be a violation to say 2*! = 7*!.

The net result is that I think we may have ways to divide by zero now, although it only replaces saying “don’t divide by zero” with “divide by zero but now you have to think about a whole bunch of infinity properties.”

As I’ve done some more looking at the Factor Field, I’ve discovered an interesting phenomenon. Of course, when I say “I discovered” this I don’t think it means that I’m the first person to think this, but only that I’ve come up with it on my own. Of course, that doesn’t mean I’m not the first person to think of it, but I’ll be the first to admit that I’m not a professional mathematician and only dabble in it for fun, so I’m pretty much clueless as to where the vast majority of theorists are on this subject!

In any case, I’ve come to the conclusion that zero and infinity are functionally equivalent. This isn’t just due to the fact that you can’t divide numbers by either zero or infinity (e.g., 1/0 is “undefined”), but because when you look at the Factor Field you can SEE that these things are graphed IDENTICALLY.

That last part is important to emphasize. These are identical graphs. Which means that if we stipulate that if method A = Graph 1, and method B = Graph 1, then method A = method B, then we have to say that zero = infinity.

So let me show you what I mean. Here’s a cross section of the Factor Field. I’m getting this by using the following VBScript code, which you can run at home if you have Windows and Microsoft Excel:

set objExcel = CreateObject("Excel.Application")
Set objWorkbook = objExcel.Workbooks.Add()
Set objWorksheet = objWorkbook.Worksheets(1)
objExcel.Visible = True

' Use if you want to fill up entire spreadsheet for 2007 format
'MaxWidth = 16384
'MaxDepth = 1048576	

' Use if you want to fill up entire spreadsheet for 2003 format
'MaxWidth = 256
'MaxDepth = 65536   

' Use for a practical demonstration
MaxWidth = 201
MaxDepth = 201

a = 1

for x = (MaxDepth-1) to 1 step -1
	objExcel.Cells(a,1).Value = x
	a = a + 1
next

for x = 1 to (MaxWidth - 1)
	for i = (MaxDepth-x) to 1 step -x
		objExcel.Cells(i,x+1).Interior.ColorIndex = 1
	next
next

a = MaxDepth
for x = 2 to (MaxWidth)
	objExcel.Cells(a,x).Value = (x-1)
next

for x = MaxDepth to (MaxDepth * 2)
	for i = MaxDepth + x to 1 step x
		objExcel.Cells(i,x+1).Interior.ColorIndex = 1
	next
next

a = -1

for x = MaxDepth + 1 to ((2 * MaxDepth) - 1)
	objExcel.Cells(x,1).Value = a
	a = a - 1
next

for x = 1 to (MaxWidth - 1)
	for i = (MaxDepth + x) to ((MaxDepth * 2) - 1) step x
		objExcel.Cells(i,x+1).Interior.ColorIndex = 1
	next
next

wscript.Echo "Done."

Now I should note that this code is fairly “hacked together” so it’s not elegant, but it does the job. And what it does is make this graph (if you scroll to row 201 of the Excel spreadsheet):

Now you can see what I’ve done to modify the Factor Field is to extend the pattern into the negative numbers. But with one important thing missing. There are no cells filled in on the y = 0 line (which holds the number for the x-axis of the graph in the picture). But clearly the pattern continues, because if we look at it with the y = 0 line filled in, it fits the patterns established in every column:

Clearly, if we just look at the pattern itself, the graph wants the y = 0 line to be filled in its entirety. But this causes an immediate problem, because what the Factor Field represents is factors of numbers. That is, when you look at y = 1, you see only x = 1 as a valid answer. That’s because 1 has only 1 as a factor. When you look at y = 2, you see both x = 1 and x = 2, because 2 and 1 are both factors of 2, etc. That’s why when you see y = 9, for instance, you see x = 1, x = 3, and x = 9, because 1, 3, and 9 are the only factors of 9.

But when you look at y = 0, you get x = 1, 2, 3, 4…to infinity.

Because this graph functions to show us factors, that means that you can take the furthest out x value and it will be equivalent to the y value. Or, to state it the other way, y is equal to the highest x value. So if the highest x value is 9, then y = 9. Therefore, if the highest x value is infinity, then the graph holds for when y = infinity.

Infinity = zero on this graph.

But it’s not just that. I’ve been thinking of the mini “proof” I gave about an infinitely long prime number, and it turns out that the “proof” exists only because I stipulated that a number must have itself as a factor by virtue of the identity axiom. But let’s jettison that for a moment and ask ourselves what the proof really requires. Let me restate it here, modified slightly:

1. Let n = 1 x 2 x 3 x 4 … x infinity.

2. Let w be a factor of some number, c.

3. If w is a factor of c, then the first number greater than c that w can also be a factor of is c + w. (e.g., 7 is a factor of 14; the next number greater than 14 that 7 can also be a factor of is 14 + 7, or 21.)

4. Since n is the product of every positive whole number, every positive whole number is a factor of n.

5. n + 1 can have only the factor of 1 (via 3). (Reasoning: if 2 is a factor of n + 1, then 2 cannot be a factor of n; since 2 is a factor of n, the next possible number after n that 2 can be a factor of is n + 2. This holds true for all numbers, therefore the only possible number that can be a factor of n + 1 is 1.)

6. Since n + 1 has only the factor of 1, then n + 1 MUST BE the number 1.

7. If n + 1 = 1, then n = 0.

8. Conclusion: 1 x 2 x 3 x 4 … x infinity = 0.

Pretty amazing, huh?

As often happens when I think of math, I keep thinking of math. In this case, I’ve come up with a few more observations about the square that I described earlier, namely this one:

1 2 3 4 5 6 7 8 9
2 4 6 8 0 2 4 6 8
3 6 9 2 5 8 1 4 7
4 8 2 6 0 4 8 2 6
5 0 5 0 5 0 5 0 5
6 2 8 4 0 6 2 8 4
7 4 1 8 5 2 9 6 3
8 6 4 2 0 8 6 4 2
9 8 7 6 5 4 3 2 1

First, there’s an obvious reason as to why 1/9, 2/8, 3/7, etc. are reciprocal columns/rows. Each are the same distance away from the 0 position. One is above, one is below, but both columns are the same actual distance. So the patterns should be similar, and in fact the only difference is the inversion to describe the one that is above and the one that is below.

After this, I thought it might be fun to add up the various values in the square. I’ll just do the rows here:

1 2 3 4 5 6 7 8 9   T = 45
2 4 6 8 0 2 4 6 8   T = 40
3 6 9 2 5 8 1 4 7   T = 45
4 8 2 6 0 4 8 2 6   T = 40
5 0 5 0 5 0 5 0 5   T = 25
6 2 8 4 0 6 2 8 4   T = 40
7 4 1 8 5 2 9 6 3   T = 45
8 6 4 2 0 8 6 4 2   T = 40
9 8 7 6 5 4 3 2 1   T = 45

Furthermore, the long diagonals both add up to 45 as well.

Now the odd values that don’t add up to 45 seem somewhat out of place. And combining what I discovered earlier with the above chart, you can easily see why. For when we look at columns 1/9, 2/8, 3/7 etc., we see that the column pairs add up to 10 in each case. But when we get to the fifth column, it pairs to nothing.

So what happens if we treat the 0 in the above as if it held a value identical to 5? We’d get:

1 2 3 4 5 6 7 8 9   T = 45
2 4 6 8 0 2 4 6 8   T = 40 + 1 Zero character = 45
3 6 9 2 5 8 1 4 7   T = 45
4 8 2 6 0 4 8 2 6   T = 40 + 1 Zero character = 45
5 0 5 0 5 0 5 0 5   T = 25 + 4 Zero characters = 25 + 20 = 45
6 2 8 4 0 6 2 8 4   T = 40 + 1 Zero character = 45
7 4 1 8 5 2 9 6 3   T = 45
8 6 4 2 0 8 6 4 2   T = 40 + 1 Zero character = 45
9 8 7 6 5 4 3 2 1   T = 45

That’s right! If we treat the zero as if it were a five, the columns add up to 45 in each case. That these two numbers go hand in hand becomes more obvious when we look at the pattern of their placement in the square. If we replace the every character that is NOT a 5 or a 0 with an *, we get the following:

* * * * 5 * * * *
* * * * 0 * * * *
* * * * 5 * * * *
* * * * 0 * * * *
5 0 5 0 5 0 5 0 5
* * * * 0 * * * *
* * * * 5 * * * *
* * * * 0 * * * *
* * * * 5 * * * *

Obviously, only a 5 or a 0 appears on the cross line that creates this square. Let’s look at two more patterns:

1 * * * * * * * 9    * * 3 * * * 7 * *
* * * * * * * * *    * * * * * * * * *
* * 9 * * * 1 * *    3 * * * * * * * 7
* * * * * * * * *    * * * * * * * * *
* * * * * * * * *    * * * * * * * * *
* * * * * * * * *    * * * * * * * * *
* * 1 * * * 9 * *    7 * * * * * * * 3
* * * * * * * * *    * * * * * * * * *
9 * * * * * * * 1    * * 7 * * * 3 * *

As you can see, they’re sprinkled symmetrically throughout (which is to be expected).

In any case, as I said before, you can make of this what you will! :-)

Here’s something nifty. Take a multiplication table and look at just the last digit on the right (i.e., 7 x 5 = 35, so you’d look at the 5, etc.). Let’s compare the various numbers side-by-side from 1 – 9:

1 2 3 4 5 6 7 8 9
2 4 6 8 0 2 4 6 8
3 6 9 2   8 1 4 7
4 8 2 6   4 8 2 6
5 0 5 0   0 5 0 5
6   8       2   4
7   1       9   3
8   4       6   2
9   7       3   1

Astute readers will already see some cool patterns in the above. The most obvious is the fact that the 9’s column is the inverse of the 1’s column. But when you look further and ignore the zeros (for the moment), you see that the 8’s column is the inverse of the 2’s column, the 7’s is the inverse of the 3’s, and the 6’s are the inverse of the 4’s.

And now comes the fun part. What happens if we fill in the blanks on the rest of the square we’ve got by padding it with the repetitions of the previous rows? We get this:

1 2 3 4 5 6 7 8 9
2 4 6 8 0 2 4 6 8
3 6 9 2 5 8 1 4 7
4 8 2 6 0 4 8 2 6
5 0 5 0 5 0 5 0 5
6 2 8 4 0 6 2 8 4
7 4 1 8 5 2 9 6 3
8 6 4 2 0 8 6 4 2
9 8 7 6 5 4 3 2 1

The 1’s column is still the reciprocal of the 9’s column, etc. But look at what else happens! The 1’s row is equal to the 1’s column; and that means that the 1’s row is also the reciprocal of the 9’s row! This carries through with the rest of the rows too. Finally, looks at some diagonals. For instance, take the diagonal at 1.

The diagonal at 1 is: 1 4 9 6 5 6 9 4 1, a sequence that is a palindrome (it reads the same forwards as it does backwards). This is true for all the diagonals.

This is also true if we consider other even bases. Let’s look at base-6 for instance. Here we get:

1 2 3 4 5
2 4 0 2 4
3 0 3 0 3
4 2 0 4 2
5 4 3 2 1

Note, too, how the mid-point oscillates between 0 and the midpoint, just as in base 10 (i.e., in base 10 it’s 5 0 5 0 5 0… in base 6 it’s 3 0 3 0 3 0…).

So what happens with and odd base, like base 7? Let’s find out!

1 2 3 4 5 6
2 4 6 1 3 5
3 6 2 5 1 4
4 1 5 2 6 3
5 3 1 6 4 2
6 5 4 3 2 1

Once again, the pattern holds, only this time there’s no central column that oscillates between 0 and the midpoint of the base.

I’ll let you do what you will with this :-)