On a line segment (that is, a line bounded by two endpoints, one point on each end of the line, the line connecting the two endpoints – OK, we understand they are END points!), there are signed distances. Call them A,B,C,and D. They are signed and so, lacking a ‘negative’ or ‘minus’ sign, they are all positive. We are looking at a line, looking at positive distances on the line, so they are ‘laid out’ on the number line to the right of zero. Points to the left of zero on the number line – that is, points corresponding to negative numbers – are not included.

The Theorem says that the product of the distance, whatever it may be, of segment AB and segment CD added to the product of segment AC and segment DB added to the product of segment AD and BC, all together add up to the grand total of 0. Zip. Nada.

I’m looking at the operations expressed in that equation and I’m thinking, ‘How can you multiply a piece of a line times a piece of a line?’ That’s because when I see something mathematical that is not familiar to me, I revert back to fifth grade. In this case, I think: I can add two lines together, because that’s like laying two sticks end to end, but no way, no how, can I multiply two sticks together! Sigh. Can’t be done.

The older me kicks in and says, ‘You’re not multiplying two sticks, and the number line is not one long endless stick for kindling!’ It is measurable but not wood. So discard the wood idea and keep the measurement idea.

What sticks and the number line do have in common, for one thing, is distance, which is just numbers. As Wikipedia says, ‘distance is a numerical description of how far apart objects are,’ so if the points on a number line can be called, loosely, objects, they have distance.

Now, distance can be measured, pure and simple, using inches, centimeters, pounds, or whatever makes a measure… even just a simple, nondescript, unnamed unit such as the difference between 1 and 2.

No matter. When they are used in the equation, they add up to zero.

Euler reasoned that they add up to zero because:

(b-a)(d-c)+(c-a)(b-d)+(d-a)(c-b)=0

Which is also true!

In this equation, the explanation and justification for the first equation, there are minus signs galore. It really is a case of a positive distance being subtracted from another positive distance.

Anyone who’s taken a basic algebra class will get right to work and multiply out those pairs of polynomials. Then add the sums. Then see the result, after crossing out pairs that are identical except for their number signs. The end result will be zero.

Sure, it will be messy.

Using FOIL to multiply the first pair of polynomials, (b-a)(d-c), you get:

(b)(d) First

(b)(-c) Outer

(-a)(d) Inner

(-a)(-c) Last

Keeping those products (which are pairs, not expressions of one variable subtracting another) and remembering that a minus times a minus produces a plus-signed product, your result is:

[(b)(d)]+[(b)(-c)]+[(-a)(d)]+[(a)(c)]

I won’t give the other two sums using FOIL because I’m lazy. It all starts looking a little bit tidier when you realize that [(b)(-c)] is -bc, so just write it that way! Then, each variable pair looks cleaner and nicer and ready to be strung out on this bracelet of math Chicklets:

bd-bc-ad+ac+cb-dc-ab+ad+dc-db-ac+ab=0

Scratch out bd and -db, etc., until you have nothing left. 0!

How did Euler see that? The man was one of the greatest mathematicians in Western history, some might call The Greatest. Somewhere in the vast Beyond The West, there may be an unpublished master who matches Euler for genius. Don’t know where to begin to think about that.

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I, for example, still fall into a deep pit when I approach certain areas of the multiplication table. I need to stay away from 7 x 8, for example, but this is not always possible.

It’s my own secret insecurity (until I just posted my confession). Anyone young enough to wear pants falling off their backside may not be as troubled by weaknesses in sums or the ‘times tables.’ After all, there’s a calculator for that. There’s an app for that.

I am working on a problem right now that I’m attempting with only a pen, paper, and a calculator for the addition operations. It is the first problem posed on the website, Project Euler. net.

The problem asks what is the sum of all numbers divisible by 3 and 5, up to but not including 1,000.

There’s a perhaps-apocryphal story about the schoolboy Gauss that places him in a math class. Stop me if you’ve heard this one before:

The teacher told the class to add all the integers from 1 to 100. Gauss took less than a minute to compute the sum and then put his pencil down. The teacher was, I suppose, making ‘busy work’ for the class while he worked on other things, so he noticed Gauss’ gesture of finality at once. He looked at Gauss’s answer – correct! Gauss had paired all the number from 1 to 100 with corresponding numbers 100 to 1. Each pair yielded the same sum: 101. There were 50 such sums. Multiplying 50 times 101 gave the product: 5050.

Done and done!

In real life, I can’t stand people who do things like that. They sit back in their chair while the rest of the class grinds away at sums. They beam with confidence – oh, how I wish I could give the young Gauss a smack right now!

But I digress. I’m working on the Project Euler problem and imagine this: I have my shoes and socks off just in case I need those toes! A Gauss I will never be. Nor an Euler (pronounced ‘oiler’) unless I’m making spanakopita!

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