CalvinDude.com

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 :-)

I’ve made a recording of how to make a Factor Field using VBScript. Those who’ve paid much attention to this blog will already know what I’m talking about. Otherwise, here’s the video.

The VBScript code is extremely simple:

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 = 100
MaxDepth = 100

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

wscript.Echo "Done."

In VBScript, starting a line with an apostrophe “comments out” the line. This is important because if you want to make the whole thing, you’ll have to “uncomment” (i.e., delete the apostrophe) from the two variables that control the size of the file and then either delete or comment out the lines that set the width and heights to 100 each. I should note that this script will take quite some time to run, and the bigger you make it the longer it will take. Also, there’s a huge difference between what Office 2007 and Office 2003 can do. So keep that in mind if you try to modify this. As you can see, you’re responsible for error checking (and as usual, this code is provided AS IS with no warranties, expressed or implied, and I’m not responsible for anything that happens to your computer should you decide to use this script).

In any case, the meat of the script is contained in the nested loops. I’ll explain this in broader detail so those curious can figure out what’s going on. The outer loop is fairly straightforward: it controls the columns. Since we start at column 1, then the loops starts at column 1. It then moves to whatever variable we’ve set for our last column. If you change nothing in the code, it’ll be 100 columns. If you use Excel 2003, you can go to 256 columns. If you use Excel 2007, you can go to 16384 columns.

(As a quick aside, if you’re wondering where those numbers come from, since it’s computer related you’d be correct in assuming they’re related to the powers of 2. 256 is 2⁸, which is the width of the 2003 version. The length of the 2003 version is 65,536. Which happens to be 256²; or 2¹⁶. The 2007 version is slightly different. First, the width is 16384, which is actually 128²; or 2¹⁴. The length is 1048576, which is 1024², or 2²⁰. So the width moved from 2⁸ to 2¹⁴ and the length moved from 2¹⁶ to 2²⁰. Or to put it another way, the width alone of the 2007 version is one quarter of the length of the 2003 version.)

Anyway, back to the nested loops. We see them here:

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

So the first loop just goes linearly from 1 to whatever the max width we decide for our graph. That’s pretty straightforward since we want the graph to hit every single column. But the second loop is where the interesting things happen. The second loop is in the format:

 for i = x to MaxDepth step x

This means that not only is this loop nested, but it’s also driven by the values of the outer loop. Thus, when x = 1, then i = 1 to the maximum number with step 1. But when x = 240, for example, then i = 240 to the maximum number with step 240.

Now the step feature tells the loop how many numbers to skip each time through. Thus, step 1 would be every number in sequence. Step 2 would be every other number; step 3 every third, etc.

How it goes together is that the cells in Excel are given numeric values. So we see that when x = 1, then the first time through i, i = 1 too (for i = 1 to the maximum number with step 1). Thus, cell (1,1) is painted black. Then i increases by the step value, which is 1, so i = 2. This paints cell (2,1). Etc. When the i loop is finished, x becomes 2.

So the second time through, i starts as 2, not 1! This means the first time through we start painting at cell (2,2), not at (1,2). Then the step is by 2, so the next time through i has become 2 + 2, or 4. Etc.

So that’s it. That one simple loop controls it all.

About now I know what you’re thinking. Not another VBScript program! But yes, another one :-D This one is actually setup for a future thing I’m currently working on (yeah, mixed tenses baby!). What this does is simulate the results of tossing 1,000 coins 100 times. It creates pseudorandom data (although I should point out that it’s pretty much impossible to distinguish between pseudorandom and actual random data), which is what I’ll be using it for.

In any case, here’s the code:

' coinflip.vbs

Option Explicit

Const xlFormat = -4143

Dim intHighNumber, intLowNumber, intCount, intNum, strOutput
Dim numHigh, numLow, outerLoop
Dim fso, outFileName, outFilePath, excel, dataSheet, intRow

' notify script is working....
wscript.Echo "Click OK to begin."

' set up file

Set fso = CreateObject("Scripting.FileSystemObject")
outFilePath = fso.GetAbsolutePathName("")
outFileName = "\coin.xls"
outFileName = outFilePath & outFileName

set excel = CreateObject("Excel.Application")
excel.DisplayAlerts = 0

If (Err.Number <> 0) Then
	On Error Goto 0
	MsgBox("Excel application not found...quiting")
	wscript.Quit
End If
On Error Goto 0

excel.Workbooks.Add
set DataSheet = excel.ActiveWorkbook.Worksheets(1)
dataSheet.Name = "Output"

dataSheet.Cells(1,1).Value = "Num 0"
dataSheet.Cells(1,2).Value = "Num 1"
dataSheet.Cells(1,3).value = "Difference"

' set up random variables
intHighNumber = 1
intLowNumber = 0

intRow = 2

For OuterLoop = 1 to 100

For intCount = 1 to 1000
	Randomize
	intNum = Int((intHighNumber - intLowNumber + 1) * Rnd + intLowNumber)
	strOutput = strOutput & intNum & " "
	if intNum = 0 then
		numLow = numLow + 1
	else
		numHigh = numHigh + 1
	End if
Next

' Output to file

	dataSheet.Cells(intRow,1).value = numLow
	dataSheet.Cells(intRow,2).value = numHigh
	dataSheet.Cells(intRow,3).Value = abs(numLow - numHigh)

intRow = intRow + 1
numLow = 0
numHigh = 0

Next
dataSheet.Cells(intRow,1).Value = "AVERAGE DIFF"
dataSheet.Cells(intRow,3).Value = "=Average(C2:C" & intRow - 1 & ")"

excel.ActiveWorkbook.SaveAs outFileName, xlFormat
excel.ActiveWorkbook.Close
excel.Quit

wscript.Echo "Done."

And here’s the data I got for one trial run (formatted with number of 0s, number of 1s—the 0s can be heads, the 1s tails or vice verse, depending on how you feel—and the difference between the number of occurrences):

Num 0	Num 1	Difference
516	484	32
520	480	40
502	498	4
482	518	36
501	499	2
487	513	26
504	496	8
509	491	18
526	474	52
501	499	2
520	480	40
498	502	4
487	513	26
530	470	60
504	496	8
449	551	102
471	529	58
531	469	62
513	487	26
558	442	116
548	452	96
541	459	82
478	522	44
489	511	22
536	464	72
467	533	66
535	465	70
472	528	56
532	468	64
515	485	30
527	473	54
498	502	4
545	455	90
505	495	10
526	474	52
454	546	92
530	470	60
447	553	106
490	510	20
525	475	50
498	502	4
512	488	24
498	502	4
502	498	4
483	517	34
485	515	30
507	493	14
484	516	32
493	507	14
505	495	10
479	521	42
486	514	28
491	509	18
500	500	0
530	470	60
499	501	2
479	521	42
518	482	36
488	512	24
493	507	14
495	505	10
487	513	26
502	498	4
497	503	6
498	502	4
500	500	0
509	491	18
500	500	0
492	508	16
499	501	2
490	510	20
523	477	46
510	490	20
488	512	24
503	497	6
525	475	50
491	509	18
498	502	4
497	503	6
492	508	16
507	493	14
494	506	12
503	497	6
526	474	52
486	514	28
502	498	4
513	487	26
551	449	102
539	461	78
509	491	18
535	465	70
492	508	16
506	494	12
482	518	36
514	486	28
507	493	14
495	505	10
546	454	92
507	493	14
503	497	6

The average difference between the “heads” and “tails” column for this data run: 32.32

BTW, you can see a nifty graph of the differences here (although it unfortunately went in reverse order):

I’ve run it a few times and it tends to be in the 20 range for the averages; but of course since it’s random data, your results will vary!

Don’t worry, I will explain more of why I am doing this later on. For now, time for me to get to work.

I freely admit that I think about weird things. And yes, I blame Bush for that. In this case, however, I’ve been thinking a bit about numbers and how abstract they are.

I remember when I first learned algebra, the most difficult part of it was the mental part of recognizing that we are moving from the “concrete” into the “abstract.” When I saw 5x + 2x written on the chalkboard, I wanted to know what x was. I didn’t care that it could stand for “anything” at all—I wanted a concrete answer.

Ironically, as I’ve thought about it lately, I think that the x actually is just as concrete as our non-algebraic math. See, when we do “regular” math, we can easily solve 5 + 2. But we never ask ourselves “What is 5 referring to? What is 2 referring to?” In point of fact, the 5 + 2 is itself abstract because these numbers do not exist in the universe at all.

Numbers were invented by people who counted actual real objects. Originally it could have been something as simple as “if you give me two of your sheep, I’ll give you three of my goats.” And someone would think, “I’ve got four goats now. If I have three more goats, I’ll have seven goats” thus adding 4 goats + 3 goats = 7 goats. This could be simplified by saying 4g + 3g = 7g, which looks amazingly enough like algebra. But what we did is go one step further and just dropped the letters completely to end up with 4 + 3 = 7.

In the process, we began to treat these numbers as if they are real entities of themselves, and not tied to some actual objects. And that’s fine, because we can do some nifty things with abstract math. But one thing that we forgot was that when we taught math we needed to be clear on this.

Perhaps other places teach it better than my school did (in fact, it would be difficult to have any school teach it worse than mine did!) but I think algebra would be a lot easier to grasp by simply showing that everyone who’s done non-algebraic math in elementary school has already done more abstract math than algebra is. Algebra simply brings the abstractness to light so we can take it even further.

Recently, I read an article about solar energy. Unfortunately, I don’t remember where I read it. I think it was Reader’s Digest. In any case, the claim of the article was that the sun gives us more than enough energy every single day to supply all the energy needs of the entire United States by solar power alone, if there was just some way to harness all this energy!

Now while that sounds impressive, I happen to know a bit of physics. And so I went down a different path.

First, let’s see how much energy consumption there is in the United States. According to Wiki (so it must be true—don’t laugh), the United States used 100 qaudrillion BTUs (105 exajoules, or 29000 TWh) in 2005. We’ll go with that for fun, using the 105 exajoules because that’ll be easier to convert later on.

Now we need to know how many joules are in an exajoule. Thankfully, if you Google “How many joules are in an exajoule” a nifty formula pops up saying 1 exajoule = 1.0 x 10¹⁸ joules. So, we’re looking at 105,000,000,000,000,000,000 joules, or 1.05 x 10²⁰ joules, for an entire year.

If we divide that by 365, we get 2.88 x 10¹⁷ joules per day.

Divide it by 86,400 seconds in a day (60 seconds per minute x 60 minutes per hour x 24 hours per day), and we get: 3.32 x 10¹² joules per second.

Now for the point I am attempting to make! Y’all remember that Einstein feller, right? E = mc² and all? Well, it just so happens that we have our Energy above (3.32 x 10¹² joules) and we have the speed of light (3 x 10⁸ m/s), so now we can just plug those numbers in and solve for the mass.

Namely:

E = mc²
E/c² = m

So we can plug in the numbers. BTW, I should note that a joule is defined as 1 kg x m²/s², so we have:

(3.32 x 10¹² kg x m²/s²) / (3 x 10⁸ m/s)²

(3.32 x 10¹² kg x m²/s²) / (9 x 10¹⁶ m²/s²)

= 3.69 x 10^-5 kg (note the m²/s² cancel each other out, leaving us with the mass unit, which is what we want.

This means that less than a gram (in fact, 0.0369 grams) of atoms can provide all the energy the United States uses in a second. So let’s multiply that by our 86,400 seconds, and our 365 days and we get: 1200 kg/year. A kilogram is 2.2 pounds, so we’re talking about 2640 pounds.

In other words, just over a ton of atoms has all the energy that we need to run the entire United States for a year. Given that the Earth has a mass of approximately 5.9742 x 10²⁴ kilograms, we have enough energy just in the Earth to last for 4.98 x 10 ²⁴ years. Or: 4,980,000,000,000,000,000,000 years. That’s nearly 5 million trillion years, at current consumption rates.

The problem with this method is, of course, the same problem as with solar power. It’s one thing to say, “The energy is there.” It’s another thing completely to say, “We can harness this energy and use it.”

Our economy is currently in a sad state. It may, therefore, be somewhat of a surprise for some readers to see that I’m not at all tempted to stop the deductions from my paycheck that go to my 401(k). Several of my coworkers have already done so, and this is a tragic mistake, which I can illustrate below. Bear in mind, of course, that there are no guarantees in the stock market, and nothing you read here should constitute anything approaching professional advice. This is just my take on the situation as I see it from a mathematical point of view.

Suppose that you buy $100 worth of shares from a diversified fund every month. For the sake of ease of math, let’s suppose that the stock begins where $1 buys one share. Obviously, if you bought $100 of shares, you’d have 100 shares.

After the first month, the value of the shares drop to $0.75 per share. $100 now buys 133 1/3 shares. You can see that because:

1 share = $0.75, or s = 3/4(d)

So d = 4/3(s)

100d = 100(4/3)(s)
100d = (400/3)(s)
100d = (133 1/3)(s)

Now let’s put that together. After two months, you’ll have spent $200 (because you’re always spending $100 per month). You now have 100 shares for the first month, and 133 1/3 shares for the second month. That gives you 233 1/3 shares total. Which, at $0.75 per share, is worth $175.00. This means that so far your investment has lost $25 dollars.

The next month, the value drops to $0.50. Now you can buy 200 shares for $100. You’ve spent $300 to get 433 1/3 shares valued at $0.50 per share, which means you’ve spent $300 to get $216.67.

Now I know what you’re thinking. This is insane! You’re losing money! And so far that’s true. But what if the next month, the price goes back up to $0.75 per share? This means that once again you buy 133 1/3 shares. Now you’ve spent $400, and you’ve got 566 2/3 shares.

566 2/3 shares at 3/4 dollar per share is how much money?

(566 2/3) x 3/4 = ?

1700/3 x 3/4 = ?

5100/12 = …… 425.

That’s right. You’ve spent $400, but your shares are worth $425. You’ve made $25. And guess what? Your share values are still under what they were when you started investing! In fact, if you buy $100 the next month and it returns to the $1.00 per share value, you’ve got 666 2/3 shares valued at $1 per share, so you’ve got $666.67 while only spending $500. You’ve made $166.67.

That’s why I’m not panicking. Each month, money is being taken from my paycheck to buy more shares, shares that are currently very, very cheap. The market does NOT have to return to the value it was last August when I started losing money in order for me to make lots of money in my 401(k). Because I’m getting cheaper shares, I have more of them, so even a little increase will help me a lot.

Now obviously it would be best if you could avoid buying stock the first few months when it goes down, and then buy it only when the market goes back up. However, the real world doesn’t work that simplistically. You cannot tell what the market will do in advance—if you could, everyone would be rich. Instead, what you can do is know that the market will always recover at some point. You can bank on that. The only way you’ll lose value forever is if your shares never recover at all (which is why you should diversify lest you have all your investments in one company that does go belly up and you lose everything).

We’ve lost roughly ten years worth of stock market value these past few months. It may drop a little more even yet; however I personally think that there’s already overselling and if President Um would just leave the economy alone it’ll recover shortly. I also do not think it’ll take a full ten years to get back to where we were last August. But even if it did, the fact that I’m buying cheap stock right now means I don’t need the market to regain all that value before I start making more money than I’ve invested.

Naturally that doesn’t help folks like my parents who are approaching retirement age, nor does it help my grandparents who are retired and are not buying new stock right now. Their value will remain tied exactly to the stock market value because they are not buying cheaper shares, like I am. So for folks nearing retirement, it may be wise to pull out of the 401(k) before you lose too much. But for anyone who’s at least 10 years out, I don’t think you have much cause for panic at this point.

It’s been floating around the blogosphere, so I didn’t invent this. However, I will show the math on it for you all :-)

If you gave away $1,000,000 every day (that’s one million dollars, for those counting), how long would it take you to give away the $800,000,000,000 (eight hundred billion) “stimulus” package that is headed to conference for reconciliation even as I write this?

That’s pretty simple. Just looking at it, you can cut out some of the zero’s already. 800,000,000,000/1,000,000 = 800,000

So it would take 800,000 days. And how many years is that? I’m glad you asked! If you divide 800,000 by 365.25 (the 0.25 is to account for leap year, which takes place every four years, or 1/4 = 0.25 of years), you get: 2190.28…

In other words, 2,190 years.

In other other words, if you gave away $1,000,000 a day since the birth of Christ, you will not have given away $800,000,000,000 by now.

In other other other words, assuming a birth date of around 6 B.C., it would take another 175 years before we would hit the magic $800,000,000,000 mark had we started giving away $1,000,000 every day upon Christ’s birth.