I have this mathematician friend who used to make up crazy stuff all the time, just to see how much of it I would believe. He eventually went away to grad school at Texas, because there were “tons of badass topologists there.” I wondered why, if they’re so badass, they aren’t done yet.

He once told me that there’s a fundamental distinction between the irrational numbers and . is a solution to the equation

,

but , so he claimed, is not the solution to any equation

with all the integers. (Restricting to integers is no stronger than restricting to rationals. If all the were rationals, you could multiply the whole equation by the a common denominator of all the and get a new equation using integer coefficients, so you might as well stipulate that are integers.)

Turns out he was (as best I can tell) giving me the truth that time. At least, if it’s a joke, a lot of people are in on it.

There’s a nice thing about math, which is that, in principle, no one’s tricky jokes should work. If they say such-and-such is true, you can check it for yourself, just by thinking about it. But you can only do that if you’re sufficiently clever. I do not understand the proof that is not algebraic. (Okay, to be honest, I have never even *tried* to understand any such proofs. All I’ve done is seen that such-and-such a book says, “it’s hard, here’s a reference if you’re interested”. Well, I’m interested, but not that interested.)

I can check some simpler things, though. For example, I’ve seen it stated that the algebraic numbers form a field, which implies (among other things), that if you take two of them and add them, you get another algebraic number. Curiously, the books I was looking at (and maybe they’re the wrong ones) gave no proof *and* no reference on this. But hey, it’s math. Just do it.

Let and be algebraic numbers, so that

,

for some set of integer coefficients and two integers and .

We’d like to know if is also algebraic.

Looking at the equations that have and as roots, they look very similar to the equations defining linear dependence of a set of vectors. In the language of linear algebra, those equations say about and :

The set is linearly dependent.

Likewise for . I have to admit this is not strictly accurate. Linear dependence is a concept between vectors, but the are not vectors because vectors must be defined over some field. The are only defined over a ring (the integers). But recall that defining algebraic numbers as solutions to polynomials with integer coefficients is no different than letting the polynomials have rational coefficients. The rationals form a proper field, so let’s let all our coefficients be rationals. (Maybe those rationals all happen to be integers.) Then the various can be used to define a vector space – whatever is the smallest space they span. (To wit: the space of all their linear combinations and nothing else.) Call that space . In , the vectors are linearly dependent. The dimension of is less than because we defined to be the smallest space spanned by the , and there are of those, and they’re dependent. Also, in a different space (call it ) spanned by the vectors , the are linearly dependent. (Your mom gave me a set of linearly dependent ‘s last night.)

Consider totally new space: the smallest space spanned by the first powers of . (By this I mean the smallest space spanned by set . Call this space .

If you expand out all those things like and whatever, you get a bunch of things like that come from the binomial coefficients. But say that, for example, . (Recall is larger than the dimension of .) It’s true that

,

but is can be written as a sum of lower powers of . Replacing it with that linear combination

(explicitly, ),

we see that we don’t need at all, and a sum of lower powers will suffice.

Indeed,

.

Replace all those ‘s by sums of smaller powers. Then group up all like terms, and for any powers of that remain greater than 4, do the same thing over again and again. Eventually, you’ll get rid of all powers of greater than 4.

So to write out all the binomial terms in , we don’t need any powers bigger than or . Therefore, is a subspace of the tensor product . (That just means it is spanned by the set .)

has dimension less than , but we have a list of vectors in it. was arbitrary, so we take and it follows that the vectors are linearly dependent because there are more of them than the dimension of the space in which they reside. We can then write

for some set of rationals , proving that is an algebraic number.

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