CalvinDude.com

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.

I’ll let you bask in the wonder that is this (somewhat edited so as to not give out corporate information) memo attached to an account that I just looked up:

“All future [items] that went to [area 1] should now go to [area 2].”

I, for one, think it’s amazing that I have a co-worker who managed to hit all three tenses in one sentence.

And the (almost) final stats for the day are in. Given what I’ve said previously, I’m pretty sure you can spot which one was me:

Yeah, I know. If you take out the second place person and add everything else together in the Data Key column (which is the important one), you get 1060. So that means excluding the person in second place, I did more 26 items then everyone else combined.

Yeah. BTW, just for clarification, not ever single line that’s reading all zeros is a real person who works here (some are former people, for example). But 9 of those blank lines are people who supposedly “work” here….

Take it as ye will. I’m going home.

Well, I no longer have to ask why there are 20 people in my department, since as of 10:30 this morning there are now only 19 people in my department. Yup, one person is gone now. Furthermore, on May 8 another person is leaving. Both are people whose absence will, IMNSHO, only make things better here. So we’re about halfway there.