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  1. #1 integers 
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    The set of integers may come into existence in the following way
    1. the number 1 comes into existence independently
    2. prime numbers are formed
    3. prime numbers form other positive integers
    4. the equation of symmetry forms negative integers
    5. before each negative integer comes into existence an equation sets the point of reference to 0


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  3. #2  
    Forum Professor serpicojr's Avatar
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    I'm not sure exactly what you want to accomplish here--are you trying to come up with a formal theory of the integers? In any case, if I were to describe the cosmogony of the integers, additive structure would come before multiplicative structure. Prime numbers would not be described until after all integers (or all positive integers) entered the picture.

    Why addition first? The integers reflect counting, and counting is an additive process (you're adding one at each step, say). And once you have the additive structure, multiplication is natural and easy to describe (you're trying to count the number of objects in n groups of m things, and you realize you're just adding n to itself m times). It is also important to note that counting and addition carry a natural notion of order--whenever you add one, you get something bigger.

    Going the other way seems quite tricky... You start off with an infinite number of building blocks and combine them in every which way. I can't think of a natural question that this answers--can you? Even beginning with the notion of the infinite seems to be putting the cart before the horse, as I feel that the infinitude of the integers is the origin of the idea of infinity. Addition does not seem to arise as a natural extension of this mutiplicative structure--we can assume that if we add two of these building blocks, we get a product of a bunch of them, but how do we know what that product is, or how can we stipulate what it should be? (If you have a good answer for this, get together with me and we can solve the ABC conjecture.) If we had a notion of order, we might have a leg up for defining addition, but there does not seem to be a natural notion of such. You could put them in some arbitrary (unnatural, noncanonical) line, but then how would you know how to compare products? How would you know 2*2 > 3?


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  4. #3  
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    Umm, I don't think you can define, or derive, the integers without 6 unprovable axioms. Here are mine (without too much pondering).

    Assume the existence of:

    the unit, 1

    concatenation, 1|1 as many times as we choose

    closure; 1|1 as many times as we chooses is integer

    total order; 1|1 > 1

    equality relation; 1 = 1

    identity integer 0; 1|0 = 0|1 = 1.

    One is now free to make arbitrary definitions. Define concatenation as +, then 1|1 = 1 + 1 > 1,

    Define 1 + 1 = 2, then 2 > 1.
    By the closure axiom, there is some 1 + 1' = 0. Define 1' as -1, likewise define 1 + (-2) = 1' = -1. And so on for all n in Z. I think that recovers the integers.
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  5. #4 integers 
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    let us discuss my first statement.i have mentioned that the integer 1 comes into existence independently. this is because we cannot obtain the integer 1 by adding any two integers(of course you can do 0+1=1).but again here we have to assume that 1 and 0 exist independently. that is my point. that is why i have stated in my first statement that the integer 1 comes into existence independently.that clears my stand on my first statement.
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  6. #5 Re: integers 
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    Quote Originally Posted by parag
    let us discuss my first statement.i have mentioned that the integer 1 comes into existence independently.
    You did, and I assumed that by "comes into existence independently", you meant it is an axiom that is unprovable from any thing else in the theory. Was I wrong to make that assumption?
    but again here we have to assume that 1 and 0 exist independently.
    Once again I am assuming that by "exist independently", you mean they are part of the integer axioms? If so, then we agree. They are elements in the "axiom set"
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  7. #6  
    Forum Professor serpicojr's Avatar
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    Yeah, Guitarist raises a good question about what you mean by "exist independently". I reiterate my related question: what are you trying to do? Are you trying to come up with a formal system describing the integers? Are you trying to describe the thought process of coming up with the integers? What do you mean by "coming into existence"?[/i]
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  8. #7 integers 
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    i agree with the guitarist.
    now my second statement. prime numbers are formed. do we agree that it is the prime numbers which are responsible for creation of other composite numbers by the addition process?
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  9. #8  
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    No i don't, I'm afraid. Here's why.

    Lets call our set together with its axiom set as Z. First, by my axioms Z can be generated by repeated applications of the operation + to the unit 1 (this entitles Z to a special name, which I'll tell you in a mo). Define a subset P of Z as follows:

    elements in P have the property that there is no element in Z, other than -1 and -p, for some p, for which repeated applications of the operation + yield the identity, 0.

    We have thus used our axioms to define P, so the existence of P cannot possibly be axiomatic. (Hmm.. maybe this is what you mean by "formed"? Dunno)

    Anyway, our axioms say that Z has a closed operation, an identity and an inverse. In the grown-up world this is sufficient to define it as a Group.

    The fact that this group can be generated by repeated applications of the operation to a single element in Z, here 1, is sufficient to define it as a cyclic group.
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  10. #9  
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    you stil haven't answered my q-estion.
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  11. #10  
    Forum Professor serpicojr's Avatar
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    I agree with Guitarist, and that was the intent of my original reply. Defining the integers in terms of addition is natural, and multiplicative structure falls out naturally. Defining them in terms of prime numbers is not natural, and additive structure does not fall out. In hindsight, we do know that the prime numbers multiplicatively generate the integers, but this should not suggest that we use this fact as the logical basis of the integers.
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  12. #11  
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    Actually, this is a lot of fun, thanks parag and serpicojr. Let's see......

    We have a set with axioms which we called Z. Let's now introduce a partition on Z, say, A, B (this means, in the language of set theory that A ∩ B = ∅ and that A ∪ B = Z)

    Hmm, do I need more axioms for this? not sure, let's plod on.

    Define A as follows: for all a<sub>i</sub> in A, a<sub>i</sub> + a<sub>i</sub> is in A.

    Define B as: for all b<sub>j</sub> in B, b<sub>j</sub> + b<sub>j</sub> is in A and for all a<sub>i</sub> in A and all b<sub>j</sub> in B, a<sub>i</sub> + b<sub>j</sub> is in B.

    Call A as the "evens" and B as the odds". Note that by this construction the identity 0 is even. And as A ∩ B = ∅, the identity cannot also be in B. Note also that by these definitions, the operation + is not closed in B. Then by our definition of a group, A is a group, B is not. But as every element in A is also an element in Z, we'll call A a subgroup of Z, but B is not a subgroup of Z!

    Something slightly weird pops into to mind that someone might have a pop at:

    Let a be any even integer, b any odd. Let's stretch notational convention to breaking point and say that +<sup>a</sup> is application of the operation an even number of times, and +<sup>b</sup> denotes addition an odd number of time (all a in A, all b in B)

    Then for all b in B, b +<sup>b</sup> b is in A, and for all b in B, b +<sup>a</sup> b is in B. Likewise, or rather, unlike-wise, a +<sup>a</sup>a is in A and a + <sup>b</sup>a is in A.

    Weird, if you think about it too much
    Anyone?
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  13. #12  
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    These are the axioms for the integers .

    1. If a, b are integers, a+b is an integer.

    2. If a, b are integers, a+b = b+a.

    3. If a, b, c are integers, (a+b)+c = a+(b+c).

    4. There is an integer 0 such that a+0 = a for any integer a.

    5. For any integer a, there is an integer -a such that a+(-a) = 0.

    (1-5 is an abelian group.)

    6. If a, b are integers, ab is an integer.

    7. If a, b are integers, ab = ba.

    8. If a, b, c are integers, (ab)c = a(bc).

    9. There is an integer 1 0 such that a1 = a for any integer a.

    10. If a, b, c are integers, a(b+c) = ab+ac.

    (1-10 is a commutative ring with unity.)

    11. If a and b are integers, either a b or b a.

    12. If a and b are integers and a b and b a, then a = b.

    13. If a, b, c are integers and a b and b c, then a c.

    (11-13 is a totally ordered set. (Note that 11 also implies reflexivity.))

    14. If a, b, c are integers and a b, then a+c b+c.

    15. If a, b are integers and 0 a and 0 b, then 0 ab.

    (1-15 is an ordered ring. They also imply that is an integral domain: If a, b are integers and ab = 0, then either a = 0 or b = 0.)

    Let .

    16. Every nonempty subset of has a least element. That is, if such that for all .

    1-16 together imply that is an well-ordered ring. This uniquely characterizes the integers because any two well-ordered rings are isomorphic.
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  14. #13  
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    Quote Originally Posted by Guitarist

    Then for all b in B, b +<sup>b</sup> b is in A, and for all b in B, b +<sup>a</sup> b is in B. Likewise, or rather, unlike-wise, a +<sup>a</sup>a is in A and a + <sup>b</sup>a is in A.
    I would have thought that the sum of an odd number of odd numbers would be an odd number. So either the first, or the second, of the two propositions here should be wrong. No?
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  15. #14  
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    No. It is quite true that this thread is so old I have forgotten exactly where I was going with my post, but I stand by it.

    The single sum of any pair of odd integers is even, the triple sum of any 4 odd integers is even and so on.

    It seems you misunderstood slightly - we are supposed to be counting the number of operations, that is, the number of "+" signs, not the number of integers entering into the operation.

    So that: 3 + 5 = 8 (even). 3 + 5 + 7 + 9= 24 (even) and so on for any odd integer.
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  16. #15  
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    Quote Originally Posted by Guitarist
    No. It is quite true that this thread is so old I have forgotten exactly where I was going with my post, but I stand by it.

    The single sum of any pair of odd integers is even, the triple sum of any 4 odd integers is even and so on.

    It seems you misunderstood slightly - we are supposed to be counting the number of operations, that is, the number of "+" signs, not the number of integers entering into the operation.

    So that: 3 + 5 = 8 (even). 3 + 5 + 7 + 9= 24 (even) and so on for any odd integer.
    It is an incredibly boring but instructive exercise to start with the following axioms

    1. 1 is a natural number
    2. For each natural number x ther exists exacly one natural number, x', called the successor to x.
    3. x' is never equal to 1.
    4. If x'=y' then x=y
    5. Let M be a seet of natural numbers such that
    I) 1 belongs to M
    II) If x belongs to M then x' also belongs to M
    Then M contains all natural numbers.

    and from these axioms derive all the usual properties of the natural numbers, the integers, the rational numbers and the complex numbers.

    You can see this done in Foundations of Analysis by Landau or, I think, in a Schaums's outline on real analysis.

    In any case those five axioms, called the Peano Axioms, are all that is needed, if one accepts basic naive set theory.
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