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.
