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Why 0^0 = 1

  1. Nov 7, 2003 #1
    I think 0^0 has to be examined from the level of the Z* ( http://mathworld.wolfram.com/Z-Star.html ) numbers.

    By using the empty set (with the Von Neumann Hierarchy), we can construct the set of Z* numbers {0,1,2,3,4,...}:
    Code (Text):

    [b][i]0[/i][/b] = |{ }| (notation = {})

    [b][i]1[/i][/b] = |{[b]{[/b] [b]}[/b]}| (notation = {0})
                   
    [b][i]2[/i][/b] = |{[b]{[/b] [b]}[/b],[b]{[/b]{ }[b]}[/b]}| (notation = {0,1})
     
    [b][i]3[/i][/b] = |{[b]{[/b] [b]}[/b],[b]{[/b]{ }[b]}[/b],[b]{[/b]{ },{{ }}[b]}[/b]}| (notation = {0,1,2})

    [b][i]4[/i][/b] = |{[b]{[/b] [b]}[/b],[b]{[/b]{ }[b]}[/b],[b]{[/b]{ },{{ }}[b]}[/b],[b]{[/b]{ },{{ }},{{ },{{ }}}[b]}[/b]}| (notation = {0,1,2,3})

    and so on.
     
    Let us look at this question from a structural point of view, and we shall do it by using the base value expansion method where x is some Z* number.

    For example we shall use number 26 represented by base 10 and base 3:
    Code (Text):

    Number 26 represented by base 10:

              ^0- 0123456789
                  ||||||||||
                  |_||||||||
                  |__|||||||
                  |___||||||
    Base 10 =     |____|||||
                  |_____||||
                  |______|||
                  |_______||
                  |________|
              ^1- |
                  |                1      0
                             ( 2*10 + 6*10 )
    ^0- 012345678901234567890123456
        |||||||||||||||||||||||||||
        |_|||||||||_|||||||||_|||||
        |__||||||||__||||||||__||||
        |___|||||||___|||||||___|||
        |____||||||____||||||____||
        |_____|||||_____|||||_____|
        |______||||______||||__...
        |_______|||_______|||__...
        |________||________||__...
    ^1- |    0    |    1    |  2
        |_________|         |  
        |___________________|  
        |_ ...
        |
       

    Number 26 represented by base 3:

              ^0- 012
    Base 3 =      |||
                  |_|
              ^1- |         3     2     1     0
                       ( 0*3 + 2*3 + 2*3 + 2*3 )
    ^0- 012012012012012012012012012
        |||||||||||||||||||||||||||
        |_||_||_||_||_||_||_||_||_|
    ^1- |0 |1 |2 |0 |1 |2 |0 |1 |2
        |__|  |  |__|  |  |__|  |  
        |_____|  |_____|  |_____|  
    ^2- |  0     |  1     |  2    
        |________|        |        
        |_________________|        
    ^3- |        0                
        |_ ...
        |
     
    As we can see, no matter what base value > 1 is,
    it is always reduced to 1 when power level = 0.

    Now, let us check base value 1.

    Code (Text):

             ^0- 0
    Base 1 =     |
                 |
             ^1- |
     
    From the above we can learn that when base value = 1 it still can be represented by base value expansion method.

    Now, what about base value 0 ?

    Let us examine it from a structural point of view, for example base 10:
    Code (Text):

              ^0- 0123456789
                  ||||||||||
                  |_||||||||
                  |__|||||||
                  |___||||||
    Base 10 =     |____|||||
                  |_____||||
                  |______|||
                  |_______||
                  |________|
              ^1- |
                  |    
     
    We can see that there exist two basic structural types: | AND _

    | is what we call a singleton and it can be notated as {.} which is a singleton included in some set.

    But what about _ ?

    By standard Boolean logic every object which included in some set must have a single and unique value.

    If we examine _ we shall find that the structural property of this element is out of the scope of standard Boolean logic.

    And the reason is this: _ exists between any two different singletons, therefore its property is at least and simultaneously two different states, and this property (to be simultaneously in two different states)is not in the scope of Boolean logic.

    But as we can see, without this element we can't develop our number system beyond 1.

    The "job" of _ is to be the "platform of memory" that gives us the ability to move beyond 1.

    The base value of this "platform" is 0 because there are no singletons in its scope.

    But unlike nothingness this object exists, therefore its full notation is 0^0=1(set's contenet exists).

    Now we have two basic objects: |^0=1 , _^0=1 where 1 stands for "there exist some object".

    Those two different structures are indistinguishable by their quantitative property.

    If we want to move beyond "there exist some object" (which are notated as 1^0=1 or 0^0=1) we have to associate between tham.

    The result is the Natural numbers > 1, which are combinations of at least 2*1^0 singletons connected by at least 0^0 "platform of memory", for example:
    Code (Text):

              ^0- 0   1
                  .   .  = 2*1^0
    Base 2 =      |   |
                  |___|
              ^1- | ^  
                    |
                   0^0      

     
                 {.}  {.} = 2*1^0
    Base 2 =      | {} |
                  |{__}|
                  | ^  
                    |
                   0^0      
     
    From this point of view we have at least 3 structural types of set's contents:

    {}, {.} AND {_} .


    By quantitative point of view we have:

    |{}| = 0 (set's content does not exist)

    |{_}| = 0^0 = 1 (set's contenet exists)

    |{.}| = 1^0 = 1 (set's contenet exists)





    What do you think?





    Organic
     
    Last edited: Nov 8, 2003
  2. jcsd
  3. Nov 7, 2003 #2
    this probably belongs in that other thread about 0^0.

    i don't think 0 has to be viewed from the natural numbers because 0 could mean an operative identity element in many kinds of algebraic structures. for some of those structures, it makes sense that 0^0=1 while in others, it makes most sense that 0^0 is either undefined or indeterminate.

    for cardinal numbers, we are in the following situation. natural numbers could either be considered cardinal numbers or like cardinal numbers.

    suppose x and y are two cardinal numbers.

    x^y is the cardinal number of the set of functions from a set of cardinality y to a set of cardinality x. one can show that this is well defined by proving that there is a bijection between A^B and C^D whenever there is a bijection between A and C and a bijection between B and D (here, A^B is the set of functions from B to A).

    for example, 2^3=8 because if we look at the number of functions from a three element set like {a,b,c} to a two element set like {t,f}, then there are eight total:
    1. {(a,t),(b,t),(c,t)}
    2. {(a,t),(b,t),(c,f)}
    3. {(a,t),(b,f),(c,t)}
    4. {(a,t),(b,f),(c,f)}
    5. {(a,f),(b,t),(c,t)}
    6. {(a,f),(b,t),(c,f)}
    7. {(a,f),(b,f),(c,t)}
    8. {(a,f),(b,f),(c,f)}.

    clearly, there are 8 no matter what three element set and two element set we use because we could paint all the elements different colors but this list of 8 would be essentially the same.

    then 0^0=1 because there is one function from the empty set to the empty set: the empty function {}.

    if 0 is considered a real number, things change. one way to define real numbers is with the following:
    Q^N is the set of rational sequences. let C be the subset of Q^N consisting of sequences that are cauchy:
    {a_n : n in N} is cauchy iff for every e in Q there is a p in N such that |a_n - a_m|<e whenever m and n are greater than p.
    define an equivalence relation ~ on C so that
    {a_n : n in N}~{b_n : n in N} iff for every e in Q there is a p in N such that |a_n - b_n|<e whenever n>p.
    we then have the set of equivalence classes in C. let [{a_n}] be shorthand for the equivalence class of {a_n : n in N}. in other words {b_n : n in N} is in [{a_n}] iff {b_n : n in N}~{a_n : n in N}.
    let C/~ stand for all the equivalence classes in C. R is defined to equal C/~.
    for example, the real number pi equals [{3, 3.1, 3.14, 3.141,...}] and 0 equals both [{0,0,0,...}] and [{1, 1/2, 1/3, 1/4, ...}]. the map that sends q in Q to [{q, q, q, ...}] is a natural embedding of Q into C/~ and one can view Q *as* a subset of C/~ because it is isomorphic to a a subset of C/~. it's pretty clear how one might define addition and multiplication on C/~. if @ can stand for + or x, then [{a_n}]@[{b_n}]:=[{a_n@b_n}]. it may or not be clear that @ is well defined. the question is how to define ^. let's even say that q^0=1 for all q in Q. rational exponents have a definition. can we just say that [{a_n}]^[{b_n}]:=[{a_n^b_n}]? we could say that {a_n^b_n} is not in any class in C/~, then it's not a real number. what would 0^0 be? well, to be careful, we need to independent representatives for 0. so suppose that 0=[{a_n}]=[{b_n}]. in other words, {a_n} and {b_n} are two (rational) sequences converging to 0. is it neccessarily the case that {a_n^b_n} converges to 0? this question is equivalent to asking if [{a_n^b_n}]=0. no. if a_n=2^(-n) and b_n=1/n, note that {a_n} and {b_n} converge to 0 yet {a_n^b_n} converges to 1/2. ok, so perhaps 0^0=1/2. but wait, there's more. suppose b_n=2^(-n) and a_n=1/n. then {a_n^b_n} converges to 1. therefore, the symbol 0^0 is not well defined because [{a_n^b_n}] varies for different representatives of 0. another question is is [{a_n}]^[{b_n}] well defined for the rest of the real numbers? if not, then i guess i picked a bad choice for ^. however, if it is well defined for all other real numbers (with the exception of when {a_n^b_n} is not a cauchy sequence of rational numbers) yet undefined only for 0^0, then this is a pretty convincing case that 0^0 is undefined when viewing 0 as an element of C/~=R.
     
  4. Nov 8, 2003 #3

    HallsofIvy

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    On the contrary, 00 can't be "examined from the level of the natural numbers" because 0 isn't a natural number.

    That would be a lot like examining ii "from the level of the real numbers". You can't even see it much less examine it!
     
  5. Nov 8, 2003 #4

    jcsd

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    I'm confused Halls of Ivy, ii is a real number (e-&pi;/2), but otherwise I agree with you, 0 is not an element of Z+.
     
  6. Nov 8, 2003 #5
    0 is a Natural number because:


    By using the empty set (with the Von Neumann Hierarchy), we can construct the set of all positive integers {0,1,2,3,4,...}:
    Code (Text):

    [b][i]0[/i][/b] = |{ }| (notation = {})

    [b][i]1[/i][/b] = |{[b]{[/b] [b]}[/b]}| (notation = {0})
                   
    [b][i]2[/i][/b] = |{[b]{[/b] [b]}[/b],[b]{[/b]{ }[b]}[/b]}| (notation = {0,1})
     
    [b][i]3[/i][/b] = |{[b]{[/b] [b]}[/b],[b]{[/b]{ }[b]}[/b],[b]{[/b]{ },{{ }}[b]}[/b]}| (notation = {0,1,2})

    [b][i]4[/i][/b] = |{[b]{[/b] [b]}[/b],[b]{[/b]{ }[b]}[/b],[b]{[/b]{ },{{ }}[b]}[/b],[b]{[/b]{ },{{ }},{{ },{{ }}}[b]}[/b]}| (notation = {0,1,2,3})

    and so on.
     
    Please return to my first post and give your remarsk.

    Thank you.

    Organic
     
    Last edited: Nov 8, 2003
  7. Nov 8, 2003 #6

    jcsd

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    No, 0 is not a postive integer, it has no sign. The natural numbers are defined as 1,2,3,4,....etc.
     
  8. Nov 8, 2003 #7
    OK, I change my first post to Z* ( http://mathworld.wolfram.com/Z-Star.html ) instead of Natural numbers.


    Now, please read my first post at the beginning of this tread and give your remarks.

    Thank you.


    Organic
     
  9. Nov 8, 2003 #8

    Hurkyl

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    0 is a natural number iff you use the convention that 0 is a natural number.
     
  10. Nov 8, 2003 #9
    Hi Hurkyl,

    I have changed it to Z* .

    Can you read my first post at the beginning of this thread ?

    Thank you.

    Organic
     
  11. Nov 8, 2003 #10
    some people define 0 to be a natural number and obviously some people don't.
    wikipedia: http://en.wikipedia.org/wiki/Natural_numbers
    mathworld: http://mathworld.wolfram.com/NaturalNumber.html
    can anyone explain why the author might use the word "unfortunately?"
    0^0 depends on (at least) three things: the definition of 0, the context (ie where 0 is and what structure the context has), and the definition of exponentiation. two examples are 0 being the smallest element in N with exponentiation being defined so that m^n is the size of the set of functions from a set of n elements to a set of m elements in which case 0^0=1 because there is one function from the empty set to the empty set: the empty function. the other example mentioned was 0 as a member of R with it's own special definition of exponentiation in which case 0^0 is not a well-defined instance of said definition of exponentiation and so it doesn't have a meaning using that definition. since 0 appears in a wide variety of algebraic structures, each possibly having their own version of exponentiation if any, i don't think it is the case that 0 must be viewed as a natural number. however, if it is, 0^0=1.
     
  12. Nov 9, 2003 #11
    Hi phoenixthoth,

    Please give some examples that clearly demonstrate the difference between 0 of Z*, 0 of Q and 0 of R.

    Thank you.


    Organic
     
  13. Nov 9, 2003 #12
    the post made on 11-07-2003 06:42 PM discusses 0 in the two contexts N and R. in N, 00=1 and in R, 00 is not a well defined expression.

    let's see about Q. in Q, (p/q)m/n could be defined to mean the element j/k in Q, if it exists, such that
    jn/kn = pm/qm, where jn means basically what it does for natural numbers except if n<0 then jn can be define in terms of positive exponents by jn=1/j-n and if j<0 and n>=0, then jn can be defined as (-j)n if n is even and -(-j)n if n is odd. now in this definition, what about 00? how about rewriting this as (0/1)0/1. then (0/1)0/1 is the element j/k in Q, if it exists, such that j1/k1 = 00/10. then j/k=00, which is what we expect. with this definition of exponentiation, 00 makes as much sense in Q as it does in N. if 00=1 in N, then 00=1 in Q as well.
     
  14. Nov 9, 2003 #13
    No dear phoenixthoth,

    Please think simple, is there different definition to 0 of Z*, 0 of Q or 0 of R ?
     
  15. Nov 9, 2003 #14
    yes.

    there are many possible definitions of 0.
    for N, 0 could be defined to be the empty set.

    for Q, 0 could be defined to be 0/1. technically, from one perspective, the natural number 0 is not an element of Q but the pre-image of the element 0/1&isin;Q under an algebraic isomorphism from N to {x/1:x&isin;Z}. some people say that if f:A-->B and A and B are two algebraic structures of a certain kind and f is an injective homomorphism appropriate to that kind of structure, then A is a subset of B when, in fact, it is only isomorphic to a subset of B. some people say that N&sub;Z&sub;Q&sub;R while i say N is embedded in Z is embedded in Q is embedded in R. actually, i would say N&sub;Z&sub;Q&sub;R with the understanding that &sub; means embedded and not subset. you know how they say a topologist can't tell the difference between a coffee cup and a doughnut? well an algebraist can't tell the difference between Q and the subset of R it is isomorphic to.

    for N, Z, and Q, i think it's pretty safe to say 00=1.

    for R, 0 could be the set of all rational cauchy sequences equivalent to the sequence {0,0,0,...} where this ~-equivalence was defined in an earlier post.

    for R, the expression 00 is not well defined. in other words, the definition for xy that works when x and y are not both zero breaks down when x and y are zero. this breakdown does not occur when looking at N, Z, or Q. now if you could define an R that is a complete ordered field with the least upper bound property having Q as a "subset" and define xy so that it extends the definition for xy on rationals in a continuous way (and so that its derivative will be what it should be), and make it so that xy works for x=y=0, then that would be very nice. my suspician is that in order to make xy=1 for x=y=0, some other property will be sacrificed but i'm not sure. the reason i think that is that R is in some sense the only field with the properties it has and i doubt 00 will make sense in one creation approach to a field and not another.
     
    Last edited: Nov 9, 2003
  16. Nov 10, 2003 #15
    Experiment with exponential functions.

    Graph exponential functions with different bases. First start (type into box) with 4^x, 3^x, 2^z, clicking "graph it" button for each one. Then graph 1^x (exponential function goes flat at b=1). Finally, try bases like (0.1)^x, (0.01)^x, (0.001)^x, etc. Observe what is happening to b^x as b->0 (x at negative, zero and positive values). Also, observe what the graph of x^0 shows when x=0.

    EDIT- additional

    The simple radical functions x^(0.5), x^(0.4), x^(0.3), ..., on down to x^(0.1), x^(0.01), x^(0.001), ... are interesting to graph. They all pass through points (0,0) and (1.1), Going relatively flat as x->0, then diving down to value 0.

    These two function types, exponential and radical, illustrate the two extreme attractors, (0,0) and (0,1), as the base for one function and the exponent for the other gets reduced through positive values to 0.

    --->
    Function Grapher
    http://people.hofstra.edu/staff/steven_r_costenoble/Graf/Graf.html

    Here's another grapher, with settable parameter A. Type in A^x for the function.

    --->
    EZ Graph
    http://id.mind.net/~zona/ezGraph/ezGraph.html
     
    Last edited: Nov 11, 2003
  17. Nov 12, 2003 #16
    I think that way to answer what is 0^0 must deal with two basic questions:


    1) What is 0?

    2) What is ^0?


    My answers are:

    1) Zero means: no singletons.

    2) ^0 means: the simplest level of any object's existence.

    Therefore 0^0 means the existence an object with no singleton(s)
    in its scope = the continuum itself without any number {___}.

    Only this object has the power of the continuum,
    and no infinitely many singletons have the power of the continuum.
     
    Last edited: Nov 13, 2003
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