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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

I came up with this idea last night while brushing my teeth!

Let K denote the field of rationals.

Let U := { u | u = a[sub]1[/sub][sup]r1[/sup]a[sub]2[/sub][sup]r2[/sup]
a[sub]n[/sub][sup]rn[/sup]} where {a[sub]1[/sub]
a[sub]n[/sub]} are prime, and {r[sub]1[/sub]
r[sub]n[/sub]} are in K

For all u,v in U and k in K:

let u+v := uv,

let ku := u[sup]k[/sup],

let 0 in U := 1

Then we have U is a vector space over K.

**Proof:**

*Associativity: *

(u+v)+w = (uv)w = u(vw) = u+(v+w)

*Zero vector existence and properties:*

0+u = (1)u = u = u(1) = u+0

*Negative vector existence and properties:*

For every vector u = a[sub]1[/sub][sup]r1[/sup]a[sub]2[/sub][sup]r2[/sup]
a[sub]n[/sub][sup]rn[/sup], there exists (-u) = a[sub]1[/sub][sup]-r1[/sup]a[sub]2[/sub][sup]-r2[/sup]
a[sub]n[/sub][sup]-rn[/sup] such that:

u+(-u) = (-u)+u = (u)(u[sup]-1[/sup]) = 1 = 0

*Commutativity:*

u+v = uv = vu = v+u

*Distributivity of scalers:*

k(u+v) = k(uv) = (uv)[sup]k[/sup] = u[sup]k[/sup]v[sup]k[/sup]= (u[sup]k[/sup])+(v[sup]k[/sup]) = ku + kv

*Distributivity of vectors:*

u(k1+k2)= u[sup](k1+k2)[/sup]=u[sup]k1[/sup]u[sup]k2[/sup]= u[sup]k1[/sup]+u[sup]k2[/sup]=k1u + k2u

*Scaler Associativity:*

(ab)u = u[sup](ab)[/sup]=u[sup](ba)[/sup]= (u[sup]b[/sup])[sup]a[/sup]=a(u[sup]b[/sup])=a(bu)

*Unit Scaler:*

(1)u = u[sup]1[/sup]=u

A logarithm is just a misspelled algorithm.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

I find this especially cool because it links linear algebra to number theory.

For instance! The fact that every number has a unique factorization follows from the fact that the primes form a basis of this vector space, and every element of a vector space can be expressed as a unique combination of its basis vectors.

Moreover, the scope of this goes beyond whole numbers because it includes numbers of the form (prime)^(rational).

So if I'm not mistaken, we can also use this to show that sqrt(2) =2^(1/2) is irrational. Because any fractional representation would be an attempt to create two distinct descriptions of sqrt(2) in terms of the basis vectors.

(edit) I take that back. I'm not sure you can prove the linear independence or primes without the FTA. Still I think the vector space notion is cool!

*Last edited by mikau (2010-01-27 11:39:59)*

A logarithm is just a misspelled algorithm.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Let U := { u | u = a1r1a2r2 anrn} where {a1 an} are prime, and {r1 rn} are in K

I assume you mean prime integers. If that's the case, every integer z is in U. Further, because the exponents can be -1, 1/z is also in U. The result of this is that U = K.

This reminds me of something a professor of mine said. Not exactly the same, but similar: "It's very easy to come up with invariants in mathematics. It's much harder to come up with invariants that aren't always zero."

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

this is a hasty response, but doesn't U include sqrt(2) = 2^(1/2), whereas K being rationals does not include sqrt(2)?

A logarithm is just a misspelled algorithm.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

no i'm allowing powers to be rational (right?)

so I think that doesn't include transcendental numbers.

*Last edited by mikau (2010-01-27 14:53:55)*

A logarithm is just a misspelled algorithm.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

This is correct (why I deleted my last post). You certainly have a subset of algebraic numbers, what I'm now trying to determine is if it's an interesting subset.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

what exactly are you talking about when you say "invariants". I know the notion of invarients in programming/comp sci but thats about it.

A logarithm is just a misspelled algorithm.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

An invariant is anything that is "fact" that holds in the "same" things. In other words, it doesn't change (remains invariant) between two things that are the "same". For example, if two topological spaces are homeomorphic, then their fundamental group is the same. Another rather simple invariant is that if two finite groups are isomorphic, then they have the same size.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

i unfortunately haven't studied topology or group theory. I take mostly computer science courses these days. *sigh* i miss math!

A logarithm is just a misspelled algorithm.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

This is not quite a useful invariant, but it may illustrate the purpose that invariants serve.

Let A and B be two finite sets of integers. Then each set has an invariant: odd or even, if the sum of their numbers is either odd or even. As with all invariants, this can tell you when two sets are not the same: a set which adds up to an even number can never be the same as a set which adds up to an odd number. Of course it can not tell you when two sets are the same as 2 + 8 = 6 + 4.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

that makes sense.

So what kind of useful invarient are we looking for in the context of this topic?

A logarithm is just a misspelled algorithm.

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