Primes and Squares

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Nathan Merrill

03:06

@DJMcMayhem lol, this is true for me as well :)

I was excited that there were a bunch of messages, and then I realized that I couldn't read them

13 hours later…

Feeds

15:52

How to Stop a Wildfire

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3 hours later…

flawr

18:46

So let me try to explain: Usually you think of vectors as some tuples like (1,42,3.14). But this is "just" a convenient notation for finite dimensional real vector spaces.

A vector space (no matter whether finite or infinite dimensional) can be defined more abstractly: You have some set V containing the vectors, plus some field F, usually called the scalars, (like the real numbers, the rational numbers, the complex numbers, finite fields etc, basically a set where you can add, subtract, multiply and divide with all the usual laws).

Now the vectors also need to satisfy some properties: It needs to be a (commutative) group: You must be able to add and subtract vectors from eachother, there must be a "zero" a.k.a "origin" and there must be a scalar multiplication. This means that multiplying a scalar (from the field) with a vector must again result in some vector. This scalar multiplication should also satisfy some laws (that you're intuitively used to) like distributivity etc.

Keep in mind that those vectors can be anything, we did not talk about tuples yet.

So for simplicity let us now take the reals as our field. As V we can take e.g. the set of affine functions on the interval [0,1] (these are the functions that look like a straight line, that you can draw with a ruler).

We can obviously add a function by adding pointwise values: (f+g)(x) := f(x) + g(x)

We can scalar-multiply functions by multiplying them pointwise: (s*f)(x) := s*f(x)

So this makes sense to talk about as a vector space.

Now it is very inconvenient to work with functions. (In this case it is not very inconvenient, but imagine using a more general class of functions.)

Would it be nice if we could choose a few basis vectors that help us express all other vectors of our vector space?

In fact we can: Consider the function a(x) = x and b(x) = 1-x (on the interval [0,1])

Since an affine function on [0,1] is defined by a straight line, it is sufficient to know two points on that line to uniquely idenitfy it.

So any vector/function f of V can be determined by knowing its values at two points, let us take the points 0 and 1 for example

Because I choose a and b very carefully, we can now write f(x) = f(0)*a(x) + f(1)*b(x) for every f in V

And if we talk long enough about vectors in our space, it will become clear that we always multiply the first value (f(0)) with a(x) and the second value (f(1)) with b(x), so we just shorten the notation to (f(0), f(1))

So if we can choose a set of vectors, such that all other vectors can be written as a linear combination of those, and everyone knows this set (a.k.a basis) it is sufficient to just mention the coefficients.

And we do not have to care anymore what those vectors actually are, we can just treat them as those tuples.

Now we can always take V it self as our basis set, but that is kinda pointless, so we always want the minimal number of vectors in our basis.

And the number of basis vectors is by definition the number of dimensions.

Now let us look at the complex numbers as a vector space, again with the real numbers as our scalars.

If we forget the multiplication within the complex numbers, we just have an additive group, and we can also scalar-multiply them with real numbers.

It turns out (1,i) forms a basis of the complex numbers "over the reals" ("over the field F" is just another way of saying that F is our field of scalars)

Because every compelx number can be written as a*1+b*i where a,b are real numbers

we could just as well take (1+2i, 0.5) as basis

Now if we again take the complex numbers as our set of vectors, but also take the complex numbers as our field, it turns out that the tuple (1) is sufficient as basis.

Any vector in V (complex number) can be written as c*1 where c is some number in our field (complex numbers).

So the complex numbers as field over the complex numbers are just 1-dimensional.

Now consider the field Q(sqrt(2)). These are all rational numbers Q, plus all real numbers that we get by adding/subtracting/dividing/multiplying rational numbers with sqrt(2).

It turns out you can write all the numbers in Q(sqrt(2)) as a+b*sqrt(2) where a,b are rational numbers. (The most important part to see here is that 1/sqrt(2) = sqrt(2)/2)

So Q(sqrt(2)) has dimension 2 over Q

So what dimension has Q(pi) over Q? It turns out it is infinite dimensional. For example (1,pi,pi^2, pi^3, pi^4,...) is a basis. (Things get a lot trickier with infinite dimensional vector spaces)

Can you tell what dimension Q(5^(1/3)) or Q(1+sqrt(2)) has over Q?

Also since Q(sqrt(2)) is a field, we can also extend it further: Q(sqrt(2),sqrt(3)) is again a field formed in the same way, allowing all operations with sqrt(2) and sqrt(3).

@flawr A better definition maybe is that it is the smallest field (as subfield of the complex numbers) containing all rational numbers and sqrt(2).

@flawr So Q(sqrt(2),sqrt(3)) turns out to have dimension 4 over Q. A possible basis is (1,sqrt(2),sqrt(3),sqrt(6)).

But what dimension does Q(sqrt(2),sqrt(3)) have over Q(sqrt(2))?

So in number theory you often look at these so called field extensions of the rational numbers.

And you pretty much always extend them with algebraic numbers. As soon as you plug in irrational numbers like pi, things become very nasty.

And similarly as the integers are a subset (a ring: it has addition and subtraction, also multiplication but not really division) of the rational numbers, you can consider "rings of integers" of field extensions of the rational numbers.

They always have a similar relationship to their field as the integers have to the rationals.

Sorry for this wall of text, it got way longer than I anticipated, but I hope it cleared up some things. I'll stop here now, but if you want to talk about it, let me know!!

There is next to no math in my current job, so I enjoy talking about it whenever I can :)