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The headline on the paper today screams out: “This Is Worse than Our Worst Case Scenario”

A little paranoia much, eh?

I think it proves that whoever came up with the worst case scenario has a distinct lack of imagination. Because, seriously, I can come up with far worse cases than this…without even thinking about it.

Sheesh.

On the other hand, we have the Washington Post asking pertinent questions of President Um. Like: “What’s your take on A-Rod?”

Perhaps if the press would stop genuflecting before the Anointed Light-bearer they’d see that so far he’s nominated at least three tax cheats, and two others who have been investigated for pay-to-play schemes. You know, things that are almost as important as A-Rod taking ‘roids.

As I walked home today from the bus stop, my mind mulled over the math I’ve been doing lately. I figured that since I had presented the concepts of multiplying two numbers that are one apart from each other [i.e., n (n + 1)] I had the mathematical skills to actually prove this. So why not just do that?

So here goes. The proof for n(n + 1) = (n – 1)(n + 2) + 2 (e.g., similar to 6 x 7 = (5 x 8) + 2)

n(n + 1) = (n – 1)(n + 2) + 2
n² + n = n² + 2n – n – 2 + 2 via factoring
n² + n = n² + n -2 + 2 cancelling
n = n

And indeed seeing this gets me my general rule, for as you can see above the numeric part of the factorization of (n – 1)(n + 2) has to cancel out the +2 at the end of the equation. So we have the general rule that:

n(n + 1) = (n – a)[n + (a + 1)] + a(a + 1)
n² + n = n² + n(a + 1) – an – a(a + 1) + a(a + 1)
n = n(a + 1) – an the n²’s cancel, as does –a(a +1) + a(a+ 1)
n = na + n – an factor
n = n since an = na, then na – an cancels itself out

It might be easier to see how the math works if we use substitution:
b = a + 1
n(n + 1) = (n – a)(n + b) + ab
n² + n = n² + nb –an –ab + ab
n = nb – an – ab + ab
n = n(a + 1) – an substitution of (a + 1) for b, and note –ab + ab cancels
n = na + n – an
n = n

And finally, the opening formula can be rewritten to:

n(n + 1) = (n – a)[n+ (a + 1)] + a² + a

So let’s test it with n = 13 and a = 7

13(13 + 1) = (13 – 7)[13 + (7 + 1)] + 7² + 7
13 x 14 = (6)[13 + 8] + 49 + 7
13 x 14 = (6 x 21) + 56
13 x 14 = 126 + 56
13 x 14 = 182

So, yup, it works. Of course as I originally noted, this is going to be in most cases a less efficient manner of doing multiplication. But where it will work is if you’re multiplying low numbers and you can’t remember your multiplication table. So if you can’t remember what 7 x 8 is, you know that it’ll be whatever 6 x 9 is plus 2, or (and here’s where it’s easy!) 5 x 10 + 6. Anything beyond that gets way complicated way fast. :-)

I have just come up with a rather inefficient, yet wickedly cool, way to multiply number together. It’s based the following pattern I discovered (although “discovered” isn’t the right word):

20 x 20 = 400
21 x 19 = 399 (1)
22 x 18 = 396 (3)
23 x 17 = 391 (5)
24 x 16 = 384 (7)
25 x 15 = 375 (9)
26 x 14 = 364 (11)
27 x 13 = 351 (13)
28 x 12 = 336 (15)
29 x 11 = 319 (17)
30 x 10 = 300 (19)
31 x 9 = 279 (21)
32 x 8 = 256 (23)
33 x 7 = 231 (25)
34 x 6 = 204 (27)
35 x 5 = 175 (29)
36 x 4 = 144 (31)
37 x 3 = 111 (33)
38 x 2 = 76 (35)
39 x 1 = 39 (37)
40 x 0 = 0 (39)
41 x -1 = -41 (41)
42 x -2 = -84 (43)

And:

21 x 20 = 420
22 x 19 = 418 (2)
23 x 18 = 414 (4)
24 x 17 = 408 (6)
25 x 16 = 400 (8)
26 x 15 = 390 (10)
27 x 14 = 378 (12)
28 x 13 = 364 (14)
29 x 12 = 348 (16)
30 x 11 = 330 (18)
31 x 10 = 310 (20)
32 x 9 = 288 (22)
33 x 8 = 264 (24)
34 x 7 = 238 (26)
35 x 6 = 210 (28)
36 x 5 = 180 (30)
37 x 4 = 148 (32)
38 x 3 = 114 (34)
39 x 2 = 78 (36)
40 x 1 = 40 (38)
41 x 0 = 0 (40)
42 x – 1 = -42 (42)
43 x – 2 = -86 (44)

To explain what’s there, you have the product of various numbers in a sequence. The number on the left side of the product always increments by 1 while the number of the right side of the equation always decrements by 1. The numbers in parentheses are the differences between the current line’s product and the previous line’s product.

In other words, if we start with no difference between two numbers, increment one side by one and decrement the other side by one, you get a difference in the products of 1. This is true no matter what number you pick. While the above was based on 20 x 20, it’s also true of 6 x 6 – (6 x 6 = 36; 7 x 5 = 35, a difference of 1) or 92 x 92 – (92 x 92 = 8464; 93 x 91 = 8463).

On the other hand, if we start with a difference of 1 and do the incrementing and decrementing, you get a difference in the products of 2. Again, this is true of 6 x 5 – (6 x 5 = 30; 7 x 4 = 28) and 92 x 91 – (92 x 91 = 8372; 93 x 90 = 8370) etc.

With that in mind, you can effectively multiply any two numbers together based on the above sequences. It’s easy enough to demonstrate by picking two random numbers. Let’s pick 75 x 53.

First we need to see which of the above patterns this follows. 75 – 53 = 22, which is an even number (in fact, any time you have two numbers that are both even or both odd, the difference will be even so you don’t even have to actually do the subtraction to know in advance which pattern you’ll use). This means it’s going to follow the first pattern where both numbers are identical. So we get to that identical number by subtracting 11 from 75 and adding 11 to 53, yielding 64 x 64. (For anyone confused, the 11 is derived from 22 divided by 2, since we are affecting both numbers.)

64 x 64 = 4096. This gives us our baseline.

Since we are dealing with both numbers being the same, we know that the difference will be the sum of 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 (in other words, we are adding all of the first 11 odd numbers—11 again because that’s how many steps away our two numbers are from the center). Adding those together gets 121.

Therefore, 75 x 53 will equal 4096 – 121, or 3975, which you can check with a calculator if you want.

Let’s pick two more random numbers: 17 x 74.

Again, we look first for the difference: 74 – 17 = 57 and 57 / 2 = 28.5. This means there are 28 and a half steps between the two numbers. If you’re concerned about that half step there, don’t be: we are just going to ignore it. Subtract 28 from 74 and add 28 to 17 and you get 46 & 45 respectively.

46 x 45 = 2070

Now add the first 28 even numbers together: 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24 + 26 + 28 + 30 + 32 + 34 + 36 + 38 + 40 + 42 + 44 + 46 + 48 + 50 + 52 + 54 + 56 = 812

2070 – 812 = 1258

Which is exactly what 17 x 74 is.

Now as I said this is a very inefficient way to do multiplication. It involves far more math with all the addition, and you’re already multiplying numbers together anyway! However, it illustrates one of the fundamental patterns in mathematics beautifully.

1. Let all k_x = x such that k₁ = 1, k₂ = 2, etc. and where x is always a positive whole number.
2. Let n = k₁k₂k₃ . . . k_infinity
3. Let f be a factor of any whole number, w.
4. If f is a factor of w, then the next largest whole number after w that f can also be a factor of is w + f. (e.g., if 2 is a factor of 16, then the next largest number that 2 can be a factor of is 16 + 2).
5. For all kx where x > 1, k_x is not a factor of n + 1 (implications of step 4).
6. n + 1 is a factor of n + 1 (law of identity).
7. n + 1 has only the factors of 1 and n + 1 (steps 5 and 6 above).
8. Therefore, n + 1 is a prime number (definition of a prime number).

Apparently, the infinity sign doesn’t show up in WordPress… :-(

There is a massive search underway to find the largest prime. So far, we’ve got one with over 9 million digits.

But consider the following number:

n = (1 x 2 x 3 x 4 … x [infinity]).

n would have factors of every single number, and thus is obviously not prime (in fact, it’s as far away from prime as you can get).

But what about n + 1?

n + 1 doesn’t have any of those factors except for 1, and it’s easy to prove.

Consider the xth factor of n. If x = 2, and x is a factor of n (which it is by the above definition), then the next number that x can be a factor of is n + 2. This is because of the nature of even numbers. They occur every other number.

When x is 3, the next number after n that can have x as a factor is n + 3. Only every 3rd number is divisible by 3.

In short, for every x, the next number that can possibly have x as a factor is n + x.

Thus, for all x > 1, it is impossible for x to be a factor of n + 1.

Or to try to show it more rigorously:
1. Let all k_x = x such that k₁ = 1, k₂ = 2, etc. and where x is always a positive whole number.
2. n = k₁k₂k₃ . . . k_[infinity].
3. For all k_x where x > 1, k_x is not a factor of n + 1 since k_x is a factor of n.
4. n + 1 is a factor of n + 1 by the law of identity.
5. n + 1 has only the factors of 1 and n + 1.
6. Therefore, n + 1 is a prime number.
Thus, we have an infinitely long prime number. And [infinity] > 9 million+.

If any atheists are upset by that, I’m vewwy vewwy sawwwy. Maybe this will help.

Remember…unicorns and puppy dogs.

Image shamelessly stolen from James White.