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A growing basis?

by pondzo
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pondzo
#1
May2-14, 04:09 AM
P: 66
Hi i was just wondering if there is any concept/theory/idea (or anything really) that relates to a growing basis (primarily in R) .

What i mean by a growing basis is the following; say you start off with one element in your basis and you encounter a number/vector that cannot be built by the element, this number that cannot be built then gets incorporated into the basis and becomes an element of it and this process goes on and on picking up more building elements as it encounters numbers that cant be built from the existing elements.

For example, in R, say the number you begin with is 2, then the next (integer) number you encounter is 3, which cant be built by 2, so 3 gets including in the basis. If this process were to go on and on, you would have a basis consisting of all primes (2,3,5,7,11,13,....) since all integers can be built from the primes. I hope that makes sense.

Thank you!
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UltrafastPED
#2
May2-14, 04:20 AM
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Sounds like physics - the experimental approach to mathematics!

You will find something analogous in the creation of imaginary and complex numbers; here is a brief summary:
http://www.math.uri.edu/~merino/spri...umbers2006.pdf
pondzo
#3
May2-14, 10:48 AM
P: 66
Thank you for the links Ultra. Do you imply there is no such approach that exists to date? That's interesting you say it sounds like physics, I've heard that before!

micromass
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May2-14, 10:54 AM
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A growing basis?

Check out the Sieve of Eratosthenes: http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
pondzo
#5
May2-14, 01:53 PM
P: 66
That is very comparable to the ideas I'm thinking of, cheers for that. Another question, the set i gave as an example before, the set of primes, that isn't a basis is it? since a basis under the current definitions and axioms in linear algebra say that a basis is linearly independent, however, 2,3 and 5 are elements of the set of primes but 2+3=5 so 5 is a linear combination of other elements in the basis, and hence the set isn't linear independent.

This brings me to ask another question. How is R a subspace and how can it have 1 dimension (or am I completely wrong and R is not 1 dimensional?)? since by definition a subspace needs to be closed under scalar multiplication. In R, all numbers are a linear combination of 1, so you could say a basis for R is {(1)}, but i don't see how this basis is closed under scalar multiplication since it has, to me, no meaning to multiply (1) by a scalar quantity, c, as all scalar quantities are not defined within the basis {(1)}. Except for maybe ( 1 +/- 1 +/- 1 +/- 1 +/-...=c |c [itex]\in[/itex] R) but then if instead you used the basis {(2)} of R, there are scalar multiples, c, that cant be built by purely adding or subtracting 2 (ie 2 +/- 2 +/- 2 +/-...≠ c |c [itex]\in[/itex] R).
micromass
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May2-14, 02:00 PM
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Quote Quote by pondzo View Post
That is very comparable to the ideas I'm thinking of, cheers for that. Another question, the set i gave as an example before, the set of primes, that isn't a basis is it? since a basis under the current definitions and axioms in linear algebra say that a basis is linearly independent, however, 2,3 and 5 are elements of the set of primes but 2+3=5 so 5 is a linear combination of other elements in the basis, and hence the set isn't linear independent.
There are various notions of "basis" in mathematics. But you're right, the set of primes is not a basis in the sense of linear algebra.

This brings me to ask another question. How is R a subspace and how can it have 1 dimension (or am I completely wrong and R is not 1 dimensional?)?
The reals ##\mathbb{R}## are indeed ##1##-dimensional when you see it as a ##\mathbb{R}##-vector space.

since by definition a subspace needs to be closed under scalar multiplication. In R, all numbers are a linear combination of 1, so you could say a basis for R is {(1)},
I'm not sure what the two brackets mean. I would say that ##\{1\}## is a basis for ##\mathbb{R}##.

but i don't see how this basis is closed under scalar multiplication since it has, to me, no meaning to multiply (1) by a scalar quantity, c, as all scalar quantities are not defined within the basis {(1)}. Except for maybe ( 1 +/- 1 +/- 1 +/- 1 +/-...=c |c [itex]\in[/itex] R) but then if instead you used the basis {(2)} of R, there are scalar multiples, c, that cant be built by purely adding or subtracting 2 (ie 2 +/- 2 +/- 2 +/-...≠ c |c [itex]\in[/itex] R).
I don't see why a basis should be closed under scalar multiplication. In fact, a basis should not be closed under scalar multiplication. Since if ##\{e_1,...,e_k\}## is a basis and if ##e_1 = \alpha e_2## (for example), then the basis is linear dependent.
All we demand for a basis is that each element in ##\mathbb{R}## can be written as a unique linear combination of the basis ##\{1\}##. And indeed, any element ##x\in \mathbb{R}## can be written as ##x = x\cdot 1## and this is unique.
Of course, ##\{2\}## is a basis too and so is ##\{3\}##. In fact, if ##x## is a nonzero real, then ##\{x\}## is basis for ##\mathbb{R}##.
pondzo
#7
May3-14, 02:25 AM
P: 66
Quote Quote by micromass View Post
There are various notions of "basis" in mathematics. But you're right, the set of primes is not a basis in the sense of linear algebra.
Is it a basis in any other sense?

Quote Quote by micromass View Post
I'm not sure what the two brackets mean. I would say that ##\{1\}## is a basis for ##\mathbb{R}##.
My mistake, it was meant to be {1}

Quote Quote by micromass View Post
I don't see why a basis should be closed under scalar multiplication. In fact, a basis should not be closed under scalar multiplication. Since if ##\{e_1,...,e_k\}## is a basis and if ##e_1 = \alpha e_2## (for example), then the basis is linear dependent.
All we demand for a basis is that each element in ##\mathbb{R}## can be written as a unique linear combination of the basis ##\{1\}##. And indeed, any element ##x\in \mathbb{R}## can be written as ##x = x\cdot 1## and this is unique.
Of course, ##\{2\}## is a basis too and so is ##\{3\}##. In fact, if ##x## is a nonzero real, then ##\{x\}## is basis for ##\mathbb{R}##.
Sorry, I worded my question poorly and left it open for misinterpretation. What i mean was, say you have a basis for R, {x}, which as you say is a basis given its non zero. My problem is i do not see how {x} can span all of R. I realise that there are an infinite amount linear combinations of {x} that make up the infinite space of R (or is it Q, the rationals, since you cant make an irrational with a lin comb of {x}?). However, what does it mean to multiply a basis {x} by a scalar, c, since they are both 1 dimensional numbers and not all c can be built by x? for example say your basis is {2} then one such linear comb would be, for example, 5*{2}, but what does 5 mean to the basis {2} ( since 2 +/- 2 +/- 2 +/-....≠ 5), is it defined? does it exist? (in reference to the basis {2}).

Also, When designing a new mathematical model, is it acceptable and sometimes necessary to change the axioms that we assume in current mathematics, provided that we deduce logical and sound conclusions from these new axioms. And as long as the axioms are "sensible". I think this may relate to Godel's Incompleteness theory, but I'm not entirely sure how to apply it.

Cheers,
Michael
HallsofIvy
#8
May3-14, 06:47 AM
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Quote Quote by pondzo View Post
Also, When designing a new mathematical model, is it acceptable and sometimes necessary to change the axioms that we assume in current mathematics, provided that we deduce logical and sound conclusions from these new axioms. And as long as the axioms are "sensible". I think this may relate to Godel's Incompleteness theory, but I'm not entirely sure how to apply it.

Cheers,
Michael
You can have any axioms you want as long as they are consistent. Of course, any proof you give in that system would NOT say anything about other systems used in "current mathematics". In particular, you cannot add a new axiom in order to prove a statement that is unprovable in current mathematics.


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