Irrationality and $\sqrt 2$

The word “irrational” is an element of second level, or semi-technical, vocabulary we all pick up in school.

First level vocabulary – everyday words Third level vocabulary – technical terms specific to the discipline Second level vocabulary – every day words that take on specific meaning in the context of a discipline.

Second level vocabulary is usually the toughest to acquire because words you already know now take on a different meaning. That new meaning is difficult to pick up until you are familiar enough with its context of use. One example would be how the meaning of the word “dribble” changes once you know we’re talking about basketball, but if you’ve never seen a game played, this meaning will be hard to come by.

At some point, some math teacher asked us to memorize a definition for “irrational”, but likely didn’t really go into why or why it mattered. In everyday life, irrational means absurd or crazy. And most people tend to bring a bit of this connotation to the math world. And why not? Irrational numbers, and most people remember at least part of this into adulthood, are those that are a bit of a pain in the ass. If you want to write them out, it’s not fun.

But we need to go back a step in the etymology if we want to really understand “irrational” as it pertains to numbers. One thing that irrational means in both contexts is “not rational”. It’s just that the word “rational” has a technical meaning that we obfuscate.

The root of “rational” is “ratio”, what we today see as interchangeable with the notion of fractions. The ratio 1:2 is the same to us as the fraction $\frac{1}{2}$.¹ Then, an irrational number is one that is not a fraction. Let’s say your number is $x$. It is irrational if there are no two whole numbers $a$ and $b$ so that \[x=\frac{a}{b}.\] An irrational number is thus one that is not equal to any fraction. i.e. It is impossible to write it as a fraction.

This is already more thought than most school-based work with irrationals gets, except for a mnemonic fact to remember: irrational numbers are those whose decimal expansions go on forever without repeating.

So Many Questions

If we stop for a minute, we know enough to ask a bunch of questions.

How would you ever know it was impossible to write a number as a fraction? Maybe you just haven’t found the right one?
Do irrational numbers exist? If you can’t write out their decimals, or give a pattern to make their decimals, how do we know they exist?
What’s the connection between being a fraction and decimal representation?
On what basis do we agree that numbers of various kinds exist? And what do we mean by exist?

Now some of these questions have subtle, contingent answers, or maybe none at all, but surprisingly for those on the outside, math as a discipline has a very simple answer to question 1. Questions like these are the bread and butter of mathematical methods and epistemology (what math knowledge can be). Here’s the basic gist.

$\sqrt 2$ Is Irrational – Our First Proof

This game is called “proof by contradiction”, and it lets us decide that the very idea of a certain number being a fraction is absurd, sidestepping the need to check every possible fraction in existence. Let’s say that we want to prove that √2 is irrational.

The game starts by pretending the opposite is true. We say, “okay smarty pants, let’s just say that there is a fraction of whole numbers $a$ and $b$ so that $\frac {a}{b}=\sqrt 2$.”

Rather than going into the details right away, the game plays out by forming a chain of logically airtight deductions from this initial assumption. We hope to end up claiming something to be true that cannot possibly be true, like $0=1$. If $\frac {a}{b}=\sqrt 2$ then this other thing “A” is true, and if “A” is true, then certainly “B” holds, … , and so this other impossible thing has to be true.

Where does this get us? Well, since the impossible thing, the conclusion at the end of this chain, cannot possibly be true, then the only possibility is that our original assumption, in this case that $\frac {a}{b}=\sqrt 2$ is true. So no matter which $a$ and $b$ we choose, that equation never holds. That’s proof by contradiction. Sometimes its called reductio ad absurdum.

And the details? Let’s see.

Being Careful – Equivalent Fractions

Now, to be careful, we have to take a step back. Often, in these proofs, we have to do some work to make sure something stupid doesn’t get in our way. In this case, the idea of “equivalent” fractions could get in our way if we’re not careful. In most situations, you and I don’t consider $\frac{1}{2}$ and $\frac{2}{4}$ to be different; you can cancel a two in the top and bottom of the latter to see their equivalence. Since we’re going to do some manipulation of $a$ and $b$, we should make sure the letters aren’t hiding something important, some critical detail for our proof.

There’s a lot of ways to be careful. One could be assuming that the fraction is in lowest terms, that we’ve done all the cancelling we can. This is a fine assumption to make as long as we believe that fractions work the way we were taught. If the fraction wasn’t in lowest terms, we’d just cancel a bit until we got one that was.

But maybe you start, now that we’re questioning irrational numbers, to question more. So let’s take a simpler assumption. Let’s assume that $a$ and $b$ aren’t both even. If they are, we can cancel a 2 out of both and get smaller numbers that give an equivalent fraction. If these two smaller numbers are not both even, we’ll use those instead of $a$ and $b$. If they are both even, we cancel again. We can’t cancel forever because each of these numbers, being finite, can only have so many twos in them.

Does that assumption seem watertight enough? That we can assume $a$ and $b$ are not both even?

Good. I knew it would.

Okay so if we’re assuming not both even, what are the other possibilities?

$a$ is even and $b$ is odd
$a$ is odd, and $b$ is even
Both $a$ and $b$ are odd

The way we finish the proof is to look at each of these three cases independently. In each case, we will show that our original assumption (that $\frac {a}{b}=\sqrt 2$) and the assumption defining the case (the parity of $a$ and $b$) lead to a contradiction (an absurdly untrue thing).

If $\frac {a}{b}=\sqrt 2$, then it is also true that $(\frac {a}{b})(\frac {a}{b})=\sqrt 2 \times \sqrt 2$. This is the same as saying

\[\frac {a^2}{b^2} = 2.\]

And since no one likes fractions, we can multiply both sides of this equation by $b^2$, and get something really useful.

\[a^2=2b^2.\]

So we will actually use this latter equation instead of the one we first assumed. We’ll do this in all three cases, so that’s why I’m mentioning it here, ahead of time.

Case 1 – $a$ is even and $b$ is odd

The key in this case, along with the other two, is to look at the left and right sides of the equations and ask about their parity. But this is the hard case, so we need to do some extra work first.

$a^2$, with $a$ even, is very even. It is at least divisible by 4.² The machinery of algebra helps us make something of this realization, using symbols to make the fact stand out.

Since $a$ is even, we could write $a=2\times k$ for some other whole number $k$. Then $a^2 = (2k)^2=4k^2$. This says—in symbols—that $a^2$ is at least divisible by 4.

But what about the left side of $a^2=2b^2$? With our substitution, $2k$ for $a$, we can write

\[4k^2=2b^2.\]

And we can go ahead and scratch the cancelling itch, and divide both sides by 2. We get

\[2k^2=b^2.\]

Now we are ready to find a contradiction. The left hand side of this equation, $2k^2$ is even. The 2 is right there. But the right side, $b^2$ is odd since $b$ is.³ There is no whole number that is both even and odd.

Case 1 is impossible.

Case 2 and 3

Here’s the thing. Since it’s our first proof, I did the hard one for you. But you have to get used to doing some work. So I’m not going to work cases 2 and 3 to their contradictions. You have to. I will say that they yield their contradictions without as much massaging. Just look at the left and right sides of the equation $a^2=2b^2$ in each case.

You can’t learn math without doing it. It’s not about memorizing details but about training a sort of internal VR. By going through the motions, these symbols stop being substations and become their own material reality, something you can directly visualize in your brain.

So go for it!

When you’ve shown that neither of these two cases is possible either, we have together finally arrived at the conclusion that there is no fraction so that $\frac {a}{b}=\sqrt 2$. $\sqrt 2$ is irrational!

What Else Is Irrational?

So if this has been successful, we have now answered question 1 above. But notice, we haven’t had anything to do with the others. We didn’t even mention decimal representations, and we just took the existence of $\sqrt 2$ for granted. Who is to say we haven’t just twiddled our thumbs in an ivory tower, counting angels on the head of a pin.

We will look into these questions too, but not today.

Other roads to glory

So we have one reason to believe that $\sqrt 2$ is irrational. For many people that would be enough. But not for math. We have lots more questions, even about the issue we’ve settled.

Is that enough?
What other knowledge does this argument take for granted?
Is there a simpler or shorter way to show this?
Can this line of reasoning be generalized? i.e. can we use it to show that $\sqrt 3$ or other similar numbers are irrational too?
Is this proof beautiful?

And why do we look the gift horse in the mouth so? One reason could be fault tolerance of the resulting system of knowledge. Knowing that there are many analytic paths to this same fact is reassuring in a sense that there’s not some hidden mistake in our work. But seriously, it is most often that last questions, the one of aesthetics that drives mathematicians forward from the point of bare sufficiency.

In the case of $\sqrt 2$ check out this short article. It’s from a course blog, an effort of math writing not too dissimilar from what we’re after. So you also may want to read it with that meta context in mind. The post shows many proofs and their origins. I for instance am more convinced by the much simpler proof that makes use of unique prime factorization, the theorem that each number breaks up into a unique set of multiples of prime divisors.

I’ll wait here, go ahead and read that. Better than that, make sure you actually work through at least one of the new proofs.

Done? Good.

Did you find a proof you liked better? On what grounds did it strike the mark? If you feel it was easier to understand, try to figure out the features and adjectives like short, simple, etc. that help get to the bottom of it.

In my own case I said I like the prime factorization based proof. Why? And if I like that one, why not show it to you first?

Well I like it because we all tend to take prime factorization for granted, and that’s a good thing. It’s a wonderful thing to know about numbers. It does make a lot of the other things we’d want to know about numbers work and it certainly generalizes from the case of $\sqrt 2$ to any other non-square number’s irrationality in a trivial way.

So why not use this one? Well, it’s a primitivism of sorts, that and the fact that this is the proof everyone is shown first, likely because it was the first one we have historically. What I mean by primitive is that our original proof doesn’t take a lot for granted when it comes to what numbers are and how they work. You need to know or figure out a bit about even and odd numbers and their arithmetic, and that’s it. This proof is not built high atop a pile of what could turn out to be cards.

On the other hand, it asks you to follow a rather serpentine argument. You might get lost in it, or get to the end not knowing where you’ve been. There’s nothing worse than a proof that seems logically valid but which makes you none the wiser because you don’t see why it’s right.

Now you can go back to the article’s proofs and make some more explicit decisions about how each of them succeeds and fails.

This coincidence is a surprisingly recent one. It’s a bit subtle, but the Greeks who started us off down this road did not consider ratios to be fractions or even numbers, really. Just as many math classes today use vectors or matrices as examples of mathematical objects that are not numbers, until the full–on adoption of decimal fractions somewhere between 1600 and 1850, ratios were seen to be another type of beast, not necessarily commensurable with numbers directly. ↩
In the context of number theory, the word “divisible” means what we might otherwise call “strictly” or “evenly divisible”. More second level vocabulary for us! ↩
Why? Can you prove this? ↩