Discretely simple

The Count — Sun, 25 Oct 2009 14:00:43 +0000

I’ve said it before, and I’ll say it again – I like math. Now, don’t get me wrong; I don’t run around solving Fermat’s Last Theorem for fun on Sunday nights, or anything like that. Perhaps it would be more clear to say I enjoy math, for many reasons, not the least of which is, when you prove something in math, it’s true every time!

Since merely saying, “I can’t find a contradictory example” is never accepted as “it’s always true”, mathematical proofs are often filled with a multitude of complex concepts and references to other also very complex proofs. It’s very refreshing, I think, to find a few that are so simple and elegant, they make you wonder if there aren’t very simple proofs for those other theorems, that we just haven’t discovered yet. Here I present two of my favorite math proofs. I hope you’ll enjoy them as much as I do.

It’s well known that the sum of the integers from 1 to N is N*(N+1)/2, and this can easily be verified for any number you care to choose. However, it’s the proof for this statement that’s makes it so I can say that it’s true for ALL positive integers. To find the sum for an unknown N, see the image below.

First, write the numbers from 1 to N across the top, then write the same numbers underneath those, only in reverse order. This gives us N columns of numbers which we can now sum individually; as it happens, each column totals N+1. The sum of all these N columns must be N*(N+1) and, since we used all the numbers 1 to N twice each in the rows, we need only divide that sum by 2 to find the solution to our problem… N*(N+1)/2. This elegant little exercise proves that the sum will always work out to that formula. How cool is that?

The second proof is a little more esoteric. While it may not be entirely as useful in your life as the previous proof (if you found that one so), it’s another example of just how simple and elegant some math proofs can be (though most aren’t). Look at the image on the right. The blue shape is a semicircle where A-B is a diameter. What’s not entirely intuitive and/or noticeable from the picture is that all 3 of the angles (marked 1, 2, and 3) are 90-degree or right angles. In fact, the angle formed by the segments between any point on the semicircle and the 2 diametric points must be a right angle. Useful information? Maybe not, but let’s look at how we know it’s always true.

In the picture on the left, we take a look at a single arbitrary example of the angle. By treating it in a completely generic fashion, what we do will be applicable to all possible angles. Note that an additional segment has been drawn from the point D to the center of the diameter (and of the entire circle, were it shown) at point C. The angle we’re trying to prove is a right angle is the sum of the two angles a and ß. Now, since AC and CD are both radii of the semicircle, they must have the same length, and their corresponding angles in the smaller triangle ACD must be equivalent (we’ll call that a). The same thing goes for the BCD triangle, only the angle there is probably different than the ACD pair, so we’ll call it ß. Now, since ABD is a triangle too, its angles must total 180 degrees, yet from our previous statements, it must also total 2*a + 2*ß. Basic algebra will reveal that (a+ß) must equal 90 degrees. Simple and elegant, just as I promised.

Well, these are just two of my favorite math proofs. While neither is exactly mind-shattering, I hope they show that it is possible to prove something mathematically without obscure references to complex transformational theorems and multitudinous graphs and charts. The next time you see a numeric pattern or shortcut, make a try at proving it. You may just come up with a short method that’s eluded us all.

The Count

Discrete Ideas » geometry

Discretely simple