Lesson 1: On Rigor

It was a very long time before I saw my first "rigorous proof". Geometry in 8th grade purported to be a proof-based class, what with lists of theorems and lemmas, but it felt completely inorganic. There was no deep result made more meaningful by beautiful argument, no cleverness.

We literally listed theorems and lemmas on one side of the page, with definitions on the other.

In college, however, you very quickly run in to the type of people who swing entirely the other direction. They desire perfect rigor--infuriating pedantry that similarly detracts from what, in my opinion, makes mathematics so interesting. Does it matter if every line of a proof is written in flawless logical notation, upside-down As and backwards capital Es abound? Or is it more powerful to understand the proof itself, and to eschew formality in favor of clarity?

I would say that whatever path leads you to understand why a given statement is true, or why a proof actually is a proof is more useful--more powerful. I find that pedantry in notation obfuscates the truth of things, but I've known people that wouldn't have it any other way--they would literally fill chalkboards with excruciatingly detailed proofs of honestly very simple concepts.

Much of this is besides the point, though, because you probably don't even have a good idea of what perfect mathematical rigor even is. But don't worry, mathematicians didn't for a couple thousand years, either.

Lesson One: A Formula You Probably Memorized
A Preface to Infinite Series

Throwing aside rigor for the moment, it shouldn't be difficult to convince yourself there's an infinite number of colored triangles along the band in the image to the left.

There's an infinite number of purple-shaded triangles, and an infinite number of yellowish-orange-dirt colored triangles. I don't really know what to call either of those colors.

Regardless, the square's area is clearly a². It follows from the formula for an area of a triangle (1/2 * base * height) that the bigger, white triangle has an area of (1/2) a², and the smaller "big" white triangle has an area of (1/4) a².

This leaves a total area of (1/4) a² for the entire strip of infinitely many colored triangles.

Now consider the largest pair of colored triangles. One has an area of (1/2)*(1/2)*a*(1/2)*a, or (1/8)a², and the other has an area of (1/2)*(1/2)*a*(1/4)*a, or (1/16)a². The goofy looking quadrilateral formed by the union of two largest colored triangles then has an area of (3/16)a².

Notice now that each triangle of the same color has roughly the same shape--their dimensions are just scaled down. Again, throwing aside rigor, it should not be difficult to convince yourself that successively smaller triangles differ in area by a factor of 1/4. If you're having difficulty seeing this, write out the areas for the first handful of triangles of either type (which isn't terribly rigorous), or simply notice that a triangle of a given size can be constructed by 4 triangles of the next smallest size, like so:

We can now construct an expression for the total area of any given pair of colored triangles.

It is simply (3/16)a² * (1/4)ⁿ, where n=0 would correspond to the first pair of triangles, and so on.

If we wish to represent the area of that entire strip, it follows that we simply sum every single pair of colored triangles.

But we already know what this area is! We've constructed our strip such that it only takes of (1/4)a² of area of our square.

This must mean that the sum

(3/16)a² * (1/4)⁰ + (3/16)a² * (1/4)¹ + (3/16)a² * (1/4)² + (3/16)a² * (1/4)³ + ... must equal (1/4)a².

If you're clever, you'll notice the quantity (3/16)a² is present in every term, so it can be factored out, leaving us with:

(3/16)a² *[1 + (1/4)¹ + (1/4)² + (1/4)³ + ... ] = (1/4)a²

Which, by a little rearrangement, implies that:

[ 1 + (1/4)¹ + (1/4)² + (1/4)³ + ... ] = (1/4)a² * (16/3) * (1/a²) = (4/3)

Now, the mathematician must ask, what have we actually accomplished? All we did was cut a square into a bunch of pieces, and subsequently show that the sum of these pieces was in fact the total area of the square, and then rearranged everything a little bit. Is this really surprising at all?

How about we take it a little bit further.

Consider, again, the infinite series:

[ 1 + (1/4)¹ + (1/4)² + (1/4)³ + ... ]

We've already demonstrated what it equals, but let's play with it a little bit. We know each successive term differs only by a factor of (1/4), just by inspection. Then, it follows that:

[ 1 + (1/4)¹ + (1/4)² + (1/4)³ + ... ] - (1/4) * [ 1 + (1/4)¹ + (1/4)² + (1/4)³ + ... ] =

[ 1 + (1/4)¹ + (1/4)² + (1/4)³ + ... ] - [ (1/4)¹ + (1/4)² + (1/4)³ + (1/4)⁴ + ... ] = 1

Well that's all well and good. But isn't this little relation true if we replace 4 with any natural number?

Let's see:

[ 1 + (1/m)¹ + (1/m)² + (1/m)³ + ... ] - (1/m) * [ 1 + (1/m)¹ + (1/m)² + (1/m)³ + ... ] = 1

This appears to be alright, as long as m behaves properly, but well get to that in a second. Our sum appears twice on the left hand side, so let's factor it:

[ 1 + (1/m)¹ + (1/m)² + (1/m)³ + ... ] * (1 - (1/m)) = 1

And, by division:

[ 1 + (1/m)¹ + (1/m)² + (1/m)³ + ... ] = 1 / (1 - (1/m))

Which might be more familiar as

[ 1 + (1/m)¹ + (1/m)² + (1/m)³ + ... ] = 1 / (1 - r)

Where r is just that common factor by which successive terms are related. This is just the formula for a geometric series, which we appear to have derived from an intuitive geometric argument, along with some algebraic hand-waving. What isn't clear from our argument is under what conditions this formula fails.

For instance, it doesn't really make sense that this formula would work if (1/m) where greater than or equal to one, because our infinite series would grow infinitely large, whereas our formula predicts it would it would either be a particularly infuriating indeterminate, 1/0, or it would be negative.

It's also unclear if this formula would still hold for negative numbers greater than -1 (it does), and if it would fail for negative numbers less than -1 (again, it does). Does it work for complex numbers as well? Interestingly enough, you might have heard the phrase "Radius of Convergence" with regards to when infinite series like our example above are convergent. This phrase doesn't make much sense when you're talking about Real numbers, because you can only really go two directions, more positive, or more negative. On the complex plane, however, the Radius of Convergence is actually a radius, centered around 0, and in this case, means that all complex numbers of modulus or "length" less than 1 satisfy our nifty little formula.

But, again, we haven't really proved why our formula works in the region that it does. It's just intuitively clear that it would work in that region, and not outside of it. Unfortunately, the machinery required to rigorously prove all of this would require several more pages of theorem proving, so we'll just put that off...for a while.

More interesting, to me at least, is that such a profound result can be constructed so simply. You don't need a command of any math higher than basic algebra. And if I have any single objective in writing about mathematics, it's to prove (there's that word again) that there is no topic in mathematics, physics, or anything really, that can't be explained in an intuitive and simple way.