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T-Stone has written in defense of Dawkin’s idea that theists proposing Intelligent Design would need to have a God who was more complex than the universe is. Important to this discussion is the following point T-Stone raises:

A 1,000 x 1,000 pixel grid of random pixels, on the other hand, isn’t as pretty to look at as a rendering of the Mandelbrot set, but it is much more complex — maximally complex, as it turns out (which is part of why it’s not as appealing aesthetically as a fractal image!). It’s counterintuitive to people who don’t work with information theory and algorithmic complexity, but its a fact of the domain: randomness is the theoretical maximum for measured complexity. You can’t get any more complex than purely random. In a random grid of pixels, we cannot guess anything about any pixels at all. In a rendering of Sierpinski triangles, or the Mandelbrot or Julia set, as soon as we see one level of rendering, prior to any recursion, we no everything about the rest of image, and can reproduce the fractal to any depth of detail without the original program.

Unfortunately for T-Stone, if he paid attention to what he has written here he’d see that he’s soundly refuted Dawkins. After all, if maximal randomness is equivalent to maximal complexity, then it is easy for me to write a program that will generate completely random output. In other words, it is easy for me—a person who is not maximally complex—to produce a program with output that is maximally complex. Thus, if we want to play T-Stone’s game and use complexity in this sense, then Dawkin’s argument must be surrendered.

If I can make a program that is more complex than I am, then God can create a universe that is more complex than He is.

FWIW, I disagree with T-Stone’s version of information and complexity. And despite what his post would lead you to believe, the idea that “maximal randomness = maximal complexity” is not true for all information theories. And in fact, if I were to use T-Stone’s definition of complexity then I would ask him to explain not why there is so much complexity in the universe, but rather why there is so little complexity. If complexity = randomness, then it doesn’t take a rocket scientist to realize that there’s a lot of the universe that is not random, and therefore there is a lot of this universe that is not complex. Under his information theory, randomness is the default. We do not need to explain random data. We do need to explain structured and ordered data. Therefore, we do not need to explain complexity; we need to explain non-complexity.

T-Stone is just giving a sleight of hand here. It would be like a mathematician saying “a > b” and having T-Stone say, “The greater than sign is inverted with the less than sign, therefore ‘a > b’ means ‘a is less than b’.”

But as soon as he engages in his sleight of hand, we respond: “If the greater than sign is inverted with the less than sign, then ‘a > b’ is no longer true, rather ‘a < b’ is.”

Inverting the operator without inverting the operands does not refute the original expression.

The truth behind the Global Warming scam shines forth yet again!

[California] Attorney General Jerry Brown on Tuesday said he will sue to block a proposed water-bottling operation in Northern California unless its effects on global warming are evaluated.

…

Brown said the company must put its revisions into a new contract with the town of McCloud. He wants proper study of the environmental consequences of the bottling operation, saying the previous draft review had “serious deficiencies.”

He said it failed to include an examination of whether the operation will contribute to global warming through the production of plastic bottles, the operation’s electrical demands and the diesel soot and greenhouse gas emissions produced by trucks traveling to and from the plant.

“It takes massive quantities of oil to produce plastic water bottles and to ship them in diesel trucks across the United States,” Brown said in a statement. “Nestle will face swift legal challenge if it does not fully evaluate the environmental impact of diverting millions of gallons of spring water from the McCloud River into billions of plastic water bottles.”

A blind man could have seen this one coming. The Global Warming scam is just a lawyer trick. Note that once the myth of fuel emissions causing Global Warming becomes legal precedence, every company on Earth will face the same lawsuits. You want to drink a Coke? Guess what! Those are bottled elsewhere and driven by a truck to your supermarket. And then you drive to the market to buy it. (And what do you do with the plastic bottle when you’re done? You, my friend, are part of the problem too!)

But don’t worry. After you’re sued out of your house, you can’t even live in a cave in the ground–who knows what pollutants you’ll be putting into the groundwater supply? And besides, you’re a threat to the brown bears that want to sleep in the same cave.

Don’t let the title of this post fool you; I’m actually only going to say a really short thing. Given my previous post, which deals with Einstein’s Relativity, I thought it would be interesting to pass on something I once read, I believe in one of Brain Green’s books (i.e. The Fabric of the Cosmos or The Elegant Universe). However, it might have also been in Genius by James Gleick. Don’t quote me on it, but it went like this:

Relativity is counter-intuitive and doesn’t make sense at first. But it obeys the rules and once you figure them out, you can train yourself to make sense of it. Quantum Mechanics, on the other hand, is not only counter-intuitive, but you can’t train yourself to make sense of it either.

And of course I could throw in the comment usually attribuited to Niels Bohr: “If you think you’ve understood Quantum Mechanics, you haven’t understood it.” (Of course I’ve heard variations on this quote too; but this is my favorite version of it.)

Since Paul C. is having difficulty understanding why causality is linked to a logical order, not a temporal order, and since others might be interested in seeing why this is the case, I decided to write another post spelling it out clearly. Before I get into the main point, we already know that temporal order is not sufficient to infer causality because that is the post hoc ergo propter hoc (after this, therefore because of this) fallacy. A simple example will suffice: the Oklahoma City bombing happened before 9/11 happened, therefore 9/11 was caused by Timothy McVeigh. This is an obvious example of the post hoc fallacy. Others are not so obvious, and we see this many times in movies about crime. For example: The victim is killed moments before the defendant leaves the premise.

So we know that temporal order is not sufficient to infer causality. In this post, I am going to take it one step further. To do so, I must talk a bit about Einstein’s theory of Relativity. In order to follow what will occur, the most important aspect to grasp is the fact that light moves at a constant velocity regardless of the framework of the observer. This is counter-intuitive, and a simple example should show why.

Suppose you are travelling in a car that is moving at 60 miles per hour. If you throw a baseball at 60 miles an hour in the same direction that you are travelling, the ball will look (from your perspective) like it is travelling 60 miles per hour. From someone on the ground, however, the ball will look like it is travelling at 120 miles per hour. That is because the observer outside your car sees the ball’s velocity as the sum of your throw (60 miles per hour) plus your vehicles velocity (60 miles per hour).

Suppose that you saw the observer on the side of the road and wanted to throw the ball back at him after you’ve already passed. Your car is still travelling at 60 miles an hour, but you give a little extra effort and throw the ball at the observer at 70 miles per hour. The observer on the side of the road will have the ball come toward him at 10 miles per hour. (The car is moving 60 miles per hour in one direction, and you throw the ball in the opposite direction (indicated by a – sign), so the result is 60 – 70 = -10 miles per hour; or 10 miles per hour in the opposite direction that the car is moving.)

This makes sense to us because we’ve seen it in action. Suppose, however, that instead of a baseball, the person in the car turns on a flashlight. Relativity states that light will appear at approximately 3 x 10⁸ m/sec for both the observer in the vehicle and the observer outside the vehicle. That is, there is no adding on the velocity of the observer to light. It moves at a constant speed through all frames of reference.

So with this in mind, let me give a slightly different version of Einstein’s train. Suppose there are two people on opposite ends of a train and these people are named Adam and Bill. At the midpoint of the train is a bomb. Adam and Bill both have buttons they can press. This will send an electrical signal that travels at the speed of light to the bomb in the middle of the train. Adam wants to blow up the train; Bill wants to keep Adam from blowing up the train. As a result, Adam’s signal will cause the bomb to detonate while Bill’s signal will keep the bomb from detonating. Furthermore, let us stipulate that Bill is at the front of the train (i.e., toward the engine) while Adam is at the back of the train (i.e., the caboose).

For ease of math, let us stipulate that light moves at exactly 3 x 10⁸ m/s. Let us also stipulate that the distance between Bill and the bomb is exactly 1,000 meters. However, due to an error when the experiment was set up, Adam is slightly closer to the bomb: he’s only 900 meters away. Let us stipulate that from the train’s framework, Adam and Bill press their buttons at the exact same time.

Now it is obvious without even doing math that because Adam is closer to the bomb and because light travels at a constant velocity that the bomb will detonate if both press their buttons at the same time. Nevertheless, let us do the math on it.

If light travels at 300,000,000 m/s, how long does it take light to go 1,000 meters? This is a simple physics problem: t = d/v. In this case, t = 1,000 / 300,000,000 or 3.3 x 10^-6 seconds.

How long does it take light to travel 900 meters? In this case, t = 900 / 300,000,000 or 3.0 x 10^-6 seconds. This means that Adam’s signal will reach the bomb 3 x 10^-7 (0.0000003) seconds before Bill’s signal will reach the bomb.

Now suppose that there is an outside observer named Charlie. The train is moving. But because light has a constant velocity irrespective of the observer, he will see both signals travel at 3 x 10⁸ m/s just like those inside the train. Suppose that at the exact instant (from Charlie’s perspective) the bomb is in front of him, both Adam and Bill press their buttons, what does Charlie see? He sees two signals travelling at 3 x 10⁸ m/s. But he also sees the bomb travelling toward Bill’s position (Bill is at the engine) and away from Adam’s (Adam is in the caboose).

This means from Charlie’s perspective, if Adam and Bill were exactly the same distance apart and pressed their buttons at exactly the same instant, the signal from Bill’s button would reach the bomb before the signal from Adam’s button would reach the bomb. But because we know that Adam is 100 meters closer to the bomb than Bill, we ask a question: how fast must the train be moving so that from Charlie’s perspective both signals will reach the bomb at the same time?

As we calculated above, Bill’s signal will reach the bomb 3 x 10^-7 seconds after Adam’s. And we know that the difference in distance is 100 meters. So we need the train to cover 100 meters in 3 x 10^-7 seconds. However, this distance is split between Adam and Bill. That is, because the signal is moving toward Bill and away from Adam, the train needs to actually only cover 50 meters in 3 x 10^-7 seconds. This gives us 50m /0.0000003s = 1.67 x 10⁸ m/s, or just over 50% the speed of light.

So let us suppose that the train is moving at 2 x 10⁸ m/s, or 2/3s the speed of light. What will Charlie see?

He sees Adam press his button. The signal moves out at 3 x 10⁸ m/s and covers 900 meters. However, when it hits the 900 meter mark (from Charlie’s perspective) 3.0 x 10^-6 seconds later, the bomb has moved. The bomb is moving at 2 x 10⁸ m/s, and it does so for the same 3.0 x 10^-6 seconds. That means the bomb has moved 600 meters further down the track after that 3.0 x 10 ^-6 seconds. Ultimately, this means it takes Adam’s signal 9.0 x 10^-6 seconds to actually read the bomb.

At the same instant, Charlie sees Bill press the button. Bill’s signal travels out at 3 x 10⁸ m/s and the bomb has moved toward him at 2 x 10⁸ m/s too. This means that it takes only 2.0 x 10^-6 for Bill’s signal to reach the bomb. From Charlie’s perspective, Bill’s signal reaches the bomb 7.0 x 10^-6 seconds before Adam’s does.

What will the train do? Answer: it will explode. Even from Charlie’s perspective, it will still explode. Why is that? Because on the train, which is where the bomb is located, Adam’s signal reaches the bomb 3 x 10^-7 seconds before Bill’s signal does. Charlie observes Bill’s signal arriving 7.0 x 10^-6 seconds before Adam’s does, however. From Charlie’s perspective, the signal that causes the bomb to explode arrives after the signal to keep the bomb from exploding should have neutralized it.

So what caused the train to explode? Adam’s signal did. But from Charlie’s perspective, it shouldn’t have. But Charlie is still left with an exploding train, one that does not fit in a temporal causative sense. It does, however, fit logically. He knows that logically Adam’s signal must have caused the train to explode, and that Bill’s counter-signal did not neutralize the bomb.

Naturally, the train had to be going extremely fast: 2/3 the speed of light. Since we never reach those speeds on Earth, cause and effect usually follow the temporal scheme. However, it is a fallacy for us to believe that causes are temporal causes for the reasons illustrated above. The only thing that matters is whether logically they are causes. If we know that A and only A logically causes B, then even if we observe B occurring before A we know that A is the cause of B. This must be the case.

This is also why we can have logical precedence (that is, a logical before) without having a temporal before. This is commonly seen in theology when, for instance, we talk about the decrees of God. The difference between Infralapsarians and Supralapsarians boils down to the logical order of the decrees of God, not the temporal order (since all agree that temporally each decree occurred before the foundation of the world, in eternity past; that is, outside of time). There is no temporal before in causality; there is only a logical before.

By the way, the proof for my previous post is as follows:

The statement “The difference between two consecutive squares” can be represented as:

(n + 1)² - n²

…and the statement “is the sum of the two numbers being squared” can be represented as:

= (n + 1) + n; or = 2n + 1

So:

(n + 1)² - n² = 2n + 1

First, let’s get the square out of the first term:

(n + 1)(n + 1) - n² = 2n + 1

Now we factor:

n² + 2n + 1 - n² = 2n + 1

Cancel out the n² due to the -n² and you get:

2n + 1 = 2n + 1.

The difference between two consecutive squares is the sum of the two numbers being squared. I.e. 6² = 36. 7² = 49. Difference = 13, which is 6 + 7.

Put that in your smoke and pipe it!

This article kinda helps validate my argument that the reason people do so poorly on math is because it’s taught so poorly. Yes, I know that’s a profound and utterly non-intuitive concept and all…but if teachers are too dumb to know math how can they teach it?

Naturally, I’m somewhat biased since I don’t think math is all that difficult of a concept to grasp. Then again, at work I am apparently the only one who can tell the difference between “one” and “more than one” (as evidenced by the fact that the batches of stuff I have to run that are supposed to contain “singles” — that is, one item per transaction — often have “multies” — that is, more than one item per transaction — smuggled into them). Now I know what you’re thinking: The philosophical problem of the one and the many was solved in roughly 1996.

B.C.

Nevertheless, it remains a difficult concept for some to grasp, for reasons I cannot begin to fathom (mainly because they’re irrational, and thus not “reasons” in the first place). Anywho, all that to get back to the main point which is: while I don’t think math is all that difficult, I fully understand the fact that there are those who do. These people are dumb, and I have to work with them. Sigh.

J/k.

Actually, I do think the basics of math take a while to grasp, kinda like forming the foundations of logic. Once it “clicks” however, the rest becomes fairly simple. The problem with math (and logic) is of course that teachers make it as hard as possible to understand the mechanics of math. They want to teach you the right “method” but the right method without understanding what the method does is the wrong approach. Hence the article’s statement:

Teacher candidates know their multiplication tables, but “they don’t come to us knowing why multiplication works the way it does,” said Denise Mewborn, who heads the University of Georgia department of math and science education.

And this is the problem. This is also why I continue to rebell against the notion that students must “show their work” when doing math; because that just means that students must “follow the correct method” when the only “correct” method is “whatever method gets the right answer.” And I’m sorry, but if you have a deeper insight into how numbers work than your teacher, why should you be penalized for doing math faster, easier, and more accurately just because your teacher can’t understand what you just did?

You can ignore this post, if you wish. I know I will!

(a, a²)

y = -(x - a)² + a²

Today, as I walked down to the store to pick up groceries for the upcoming week, I thought some more on the math formula that I worked on yesterday. The end result is that I have no figured out a trick to easily multiply two “teen” numbers together.

Of course, it behooves us to define what a “teen” number is then! I’m defining a “teen” number as any number between (and including) 10 and 19. This can be expressed in the following way as well (which will benefit us shortly):

A “teen” number is any two-digit number in the format 1x, where x is a whole number between 0 and 9.

If we multiply two “teen” numbers together, then it is equivalent to the format 1x x 1y where x and y are both integers between 0 and 9, and where x may or may not equal y.

Now that we have it defined, let me give you an example of the trick first, and then I will demonstrate the reasons why the trick works after that. Suppose you wanted to find out what 17 x 15 is. Here’s how the trick works:

1. Begin with 100. This is your “baseline” and will always be the baseline for teen numbers.
2. Take off the beginning “1”s on your numbers (i.e. 17 becomes 7, 15 becomes 5. This gives us 7 & 5).
3. Add those numbers together (i.e. 7 + 5 = 12).
4. These are our “tens”, so multiply the result by 10 by sticking a 0 at the end (i.e., 12 becomes 120).
5. Add that to the baseline 100 (i.e. 100 + 120 = 220).
6. Now multiply the two numbers we got in step 2 together (i. e. 7 x 5 = 35).
7. Add that result to the number we got in 5. (220 + 35 = 255).

That’s your answer. Now at first glance, the above looks difficult. However, as you practice it, you’ll see it makes mental math quite simple. For instance, to figure out what 15 x 13 is, you only need to calculate 5 + 3 and 5 x 3, which are both simple. The rest follows simply too: since 5 + 3 = 8, then you simply add a zero to the end (equivalent to multiplying by 10) to make 80, and add that to your 100 baseline (180), and then add 5 x 3, or 15 and you’ve got 195.

Thus far, I think the method is easier than doing the multiplication the way we’re all taught to do it. However, where this method really shines is when our x and y variables add up to 10. For example, 18 x 12. 8 + 2 = 10. When you have those variables that add up to 10, you do this:

1. Begin with 200 (instead of 100!)
2. Take off the beginning “1”s on your numbers (same as step 2 in the first method)
3. Multiply those numbers together (i.e. 8 x 2 = 16)
4. Add to our baseline. (200 + 16 = 216).

And you’re done. Yup, that easy. So, 17 x 13. Start with 200, add 7 x 3 = 21. 221 is your answer. This even works if you break out of the single-digit mould. For instance, 10 x 20 = 200. You can take your x as 0 and your y as 10, and find out that 200 + 0 x 10 = 200. But of course it becomes a little more difficult to mentally think of 20 as 1A (where A = 10).

However, if you did continue with that, 21 x 9 (where you think of 21 = 1B, (B = 11), and 9 = 1-A, (-A = -1) in the same format: 9 x 21 = 189. So you have 200 + 11 x – 1, or 200 – 11 = 189. So the method still works, but at this point it becomes difficult to do it mentally.

Now on to the reason why this method works. Let’s look at the equation I used yesterday:

n² = (n – x)(n + x) + x²

The (n –x)(n + x) section of the equation is where we put in our multiplication terms. So what would happen if I wanted to see what (n + a)(n + b) would need?

n² = (n + x)(n + x) + ?

Once again, we factor:

(n + x)(n + x) = n² + nx + xn + x² Or:
(n + x)(n + x) = n² + x² + 2nx.

So the ? would have to get rid of everything except for the n² to satisfy the equation. The ? would therefore be –x² - 2xn. Therefore:

n² = (n + x)(n + x) – x² - 2xn.

But remember last time I made it relative by using (n – a) and (n + b). What would that look like here?

n² = (n + a)(n + b) + ?
(n + a)(n + b) = n² + bn + an + ab.

Therefore, we need to subtract bn, an, and ab:

n² = (n + a)(n + b) – bn – an – ab.

Now let us make these numbers “teen” numbers. In that instance, n = 10:

100 = (10 + a)(10 + b) – 10b – 10a – ab

The formula can be rearranged to show us:

(10 + a)(10 + b) = 100 + 10b + 10a + ab

And there you have the trick I showed above. You start with the baseline of 100. You take the “1” off of the number and get your x and y, which are equivalent to the a & b in the above formula. The rest follows.

The only part that would be tricky to immediately see at this point is the fact that:

10b + 10a = 10(a + b).

However, if you know how to factor, you’ll see that that is the case. And we can demonstrate it by plugging in some values too. For instance, a = 7, b = 9:

(10 x 7) + (10 x 9) = 10(7 + 9)
70 + 90 = 10 x 16
160 = 160.

This also gives us our reason for why, if the digits (a and b, or x and y) add up to 10, we start with 200 instead of 100. If:

(10 x a) + (10 x b) = 10(a + b)

Then if a + b = 10, then we are adding another 100 to the baseline.

Therefore, we can do that from the start and don’t have to worry about adding 100 + 100. We can simply say, if a + b (or x + y in the original formula) = 10, then start with 200 instead of 100.

So there you have it. The general trick to multiplying teen numbers is to start with 100, add the right-hand digits together and put a zero at the end, multiply the right hand digits together, and add all of those together. If a + b = 10, then start with 200 and add the multiple of the right-hand digits to that number.

Just to give a brief detailed explanation for my previous post, while I was at work today I began to think about something that had very little to do with work. This happens all the time, yet somehow my error rate stays fairly low… Anyway, I started with a simple supposition:

I believe that the reason most people hate math is because our schools do such a horrific job of teaching it.

I have personal experience to back this up, although that might just make me biased. (Based on others’ testimony, however, I feel my personal experience is hardly unique.) When I was in elementary school, I loved math. I used to do extra credit problems in the 2nd Grade. I’d get a packet of 100 addition problems and do them all to turn in the next day. I remember how much fun that was. By the time I finished 6th Grade, I had done all the prelim work to prepare for Pre-Algebra for 7th Grade.

But at that time, I moved to a different school district. Instead of getting Pre-Algebra, I got put in remedial math. Why? Because everyone took it in 7th Grade at the new school. 8th Grade…same thing. Remedial math yet again. I was totally bored.

To make up for it, they tossed us all into Algebra I in 9th Grade. It was a sink or swim moment. Some people, my sister for example, couldn’t swim under those conditions. As a result, my sister hates math to this day. She’s probably the smartest person I know who doesn’t do Algebra. I was fortunate enough to love math enough before this happened that, by the time the school made me hate math I was still functional enough to pull myself out. If you look at my grades, you’ll see that I got a C in Algebra I. In Algebra II I had a B. Then, for Geometry and Trig, I was back up to As. I had caught on…but it was far more work than it needed to be. And I only caught on because I happen to love math despite how much they tried to make me hate it.

Separating from my personal experience, one of the biggest problems with math education on the whole is the fact that teachers require students to show their work. This is just a euphemism for: “You have to do it the way that we tell you to do it.” Granted, some of it is to protect against cheating…but if the teacher is halfway decent he or she will already know whether a student is actually learning or not.

But because we require everyone to approach math the same way, people fail. Some of the smartest people in the world can do math equations without being able to show how they know the answer. The answer is right, but they cannot do the “work” to prove it. That’s because they did not solve it in the “correct” manner. As a result, they do not get credit for their answers. The solution is unimportant; following the method is key.

This is a travesty. In math, there are multiple roads to the same answer, and if one road works easier for you…use it!

But enough sermonizing. The above is sufficient for you to get my point of view. As I sat in front of the scanner at work, I thought about mental math tricks. I’ve talked to a couple of coworkers and asked them how they would solve a problem such as: 17 + 23 = ?

Most said the same thing. “You start with 7 + 3 = 10, carry the one…”

I’m thinking, I’m too lazy to carry the freaking 1! Sure, that method will get you the answer, but I look at that and think this:

17 is pretty close to 20. In fact, it’s 3 away. 23 is also close to 20. It’s 3 away as well. The 17 is 3 under, the 23 is 3 over; the 3s cancel each other out. This is the same thing as 20 + 20, which is obviously 40.

And there, I’ve solved the problem without needing to “carry” anything at all. And I do this for everything. What is 19 + 19? Well, add 1 to 19 and you get 20. You have to do that twice, so you end up with 2 left over. 20 + 20 = 40, and 40 – 2 = 38. Problem solved. None of this 9 + 9 = 18, carry the 1 crap at all.

As a result of this, when I see the number 7, for example, I don’t see 7. I see “2 above 5 and 3 below 10.” 5s and 10s are easy to add; I seek to convert everything to them. 6 is 1 above 5; 3 is 2 below 5 (and 3 above 0). Etc.

That works well enough for addition, but I wondered today if there was a way that I could take that similar method and figure out the way it works with multiplication. I thought, if I were to add 19 + 21, I would go 19 + 1 = 20; 21 – 1 = 20; the 1s cancel each other out, and the result is 20 + 20 = 40. But what if I were to multiply instead of add? How would I solve 19 x 21 in a similar manner?

And that’s how I discovered my formula. See, I know 20 x 20 = 400. That’s pretty easy. So what does 19 x 21 equal? 399.

399 = 400 – 1.

I found that to be very interesting. Especially when I then tried 29 x 31, knowing that 30 x 30 = 900. 29 x 31 = 899.

899 = 900 – 1.

Aha! A pattern emerges! I tried it for a few more variables and it continued to work. Without using a calculator, I could instantly know that 49 x 51 = 2500 – 1, or 2499. I then tested it with numbers that weren’t next to multiples of 10. I started easy:

7 x 5 = 35. Well, 6 x 6 = 36, and 36 – 1 = 35! The pattern continued. I tested a few more numbers and found it worked through. (BTW: I should note at this point that you can obviously see that I do math as a scientist. I “test” numbers and make hypothesis and then experiment with them, etc. This is not a rigorous proof…but I felt quite confident in this process given the fact that it’s worked so well—and I should also point out, once I get to the formula, that I can prove it after all.)

Anyway, at this point I then asked the next question. So far I’ve only been subtracting by 1. What happens if I subtract by 2? Let’s start with the ol’ standby, 20.

18 x 22 = 396. 396 = 400 – 4.

Let’s test the next level up!

28 x 32 = 896. 896 = 900 – 4 (!)

Again, the pattern is consistent. If you’re spaced 1 away from a number, then you subtract 1; if you’re 2 away, you subtract 4! But this could still fit several patterns. It was time to test one more:

17 x 23 = 391. 391 = 400 – 9
27 x 33 = 891. 891 = 900 – 9

The pattern was still there! If you were 3 away from a given number, then you subtract 9. This was obviously a sequence of squares: 1, 4, 9 from 1 x 1 = 1, 2 x 2 = 4, 3 x 3 = 9.

With this in mind, I got the formula:

(n – x)(n + x) = n² - x²

And from that, it was a simple switch:

n² = (n – x)(n + x) + x²

It’s easy enough to prove certain aspects of it. For instance, it’s easy to prove that this formula is true whenever n = x. Since the (n – x)(n + x) term exists, if n = x, then that multiplication would be 0. We then simply have n² = x², which we can take the square root of each side and get back to n = x, which is what we started with. So the equation is proven true under that condition.

Likewise, if x = 0, we see that n² = (n + 0) (n – 0) + 0; or n² = n x n, which is what n² means.

And it even works if n = 0. Under that condition, we have 0 = (0 – x)(0 + x) + x², which is 0 = -x² + x²; or x² – x² = 0, which is true.

I was about to write: “However, since I am publik skewled I lack the ability to prove this for all numbers”…but then I figured out a way to do so when I looked at the above n = 0 part. So, instead I will amend that to: Despite being publik skewled, I can prove this for all numbers:

n² = (n – x)(n + x) + x²
n² = n² + nx – xn – x² + x² [Factoring]
n² = n² + nx – nx – x² + x² [Cancel out like terms]
n² = n²

Thus, I’ve proven it true for all numbers after all!

And because of that relationship, we can immediately generalize further. Instead of limiting ourselves to numbers the same distance from a certain square, we can subtract by a and add by b. Therefore:

n² = (n – a)(n + b) + ?

Let’s figure out what the question mark would be. Obviously, it needs to be something that would cancel out the rest of the factorization:

(n – a)(n + b) = n² + nb – an – ab

Therefore, the ? needs to be –nb + an + ab

Therefore, for any number combination now:

n² = (n – a)(n + b) + an + ab – nb.

Test with n = 10, a = 3, b = 7

100 = 7 x 17 + 30 + 21 – 70
100 = 7 x 17 – 19
119 = 7 x 17

And that can be checked with a calculator to see that it is, indeed, correct.

And just for fun, let’s have negative numbers! n = 10, a = -3, b = 4

100 = 13 x 14 – 30 – 12 – 40
100 = 13 x 14 – 82
182 = 13 x 14

Which is again proven by the calculator.

Now this brings us immediately to an easier way to do math. If you are multiplying two numbers together and you can find the square of a number that is smaller than both the numbers, you can use the following method. It works perfectly well with 10² and multiples of 10, so why not use them?

Suppose we want to find what 16 x 19 is. Well, since I know that 10 x 10 = 100, and I know 6 x 10, 9 x 10, and 6 x 9 because those are all done by rote. 60 + 90 + 54 = 204. Since both 16 and 19 are greater than 10, we add the result to 100. We therefore have 304, which is the correct answer.

Of course, we could use 20 instead, but in this case because the numbers are both lower than the square we’re looking at, we have to do more subtraction. For instance, 16 is 4 less than 20, and 19 is 1 less than 20. We’re looking for 4 x 20 and 1 x 20 and 4 x 1. But what to do with them? Since both numbers are lower than the square, we subtract the numbers that are multiplied by the squared number (i.e. 4 x 20 and 1 x 20) from the squared number, but we add back the product of the other terms. Thus, 400 – (4 x 20) – (1 x 20) + 4 x 1, or 400 – 100 + 4, or 304.

Obviously, the first method is much easier to do in our heads.