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    Looking at the Real Numbers the Z2 Way
    By Doug Sweetser | July 9th 2012 11:15 PM | 57 comments | Print | E-mail | Track Comments
    About Doug

    Trying to be a semi-pro amateur physicist (yes I accept special relativity is right!). I _had_ my own effort to unify gravity with other forces in...

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    What I enjoy most is upsetting my own apple cart, presumptions about how things are as provided by my earliest teachers.  The image I learned about real numbers has negative numbers on the left, positive numbers on the right, and a zero sitting in the middle, with the integers marked out as even as can be:

    One of the things I have struggled to learn is about the real numbers and the group Z2 which consists of only two numbers, {-1, +1}.  It still sounds like an empty statement to discuss the group Z2 over the real numbers.  The group Z2 is a way of introducing the minus sign to a set that only has positive values, yet the real numbers already have a sign, so nothing is accomplished by the group.  Math is a game of intellectual minimalism, so the mention of the group Z2 can be deleted, and nothing is altered about one's understanding of the real numbers.

    I began to think about the positive real numbers with the group Z2 serving as a basis.  


    Here is my notation for this situation, with the u, v, and w  in bold italics serving the roles of my combination of the positive real numbers and Z2:

    v = a{-1} + b{+1}
    u = c{-1} + d{+1}
    w = e{-1} + f{+1}

    a-f >= 0

    Why didn't I just call these vectors?  The definition of a vector space says that would be illegal.  The positive real numbers are not a mathematical field, and a mathematical field is required for a vector space.  Ergo, I am not working with a vector space.  Let's see if these structures are "nice" despite that flaw.  I will call them proto-vectors because they are not vectors over a mathematical field, but they sure have lots of the properties of vectors.  Let us go through the list of properties found on the wiki page for vector space.  
    They can be added together, and multiplied by a positive scalar:

    v + u = (a+c){-1} + (b+d){+1}
    v + w = (a+e){-1} + (b+f){+1}
    u + w = (c+e){-1} + (d+f){+1}

    s v = s a{-1} + s b{+1}

    1. Associativity of addition:

    u + (v + w) = (u + v) + w = (a+c+e){-1} + (b+d+f){+1}

    2. Commutativity of addition:

    u + v = v + u = (a+c){-1} + (b+d){+1}

    3. Identity element of addition:

    0 = 0{-1} + 0 {+1}

    v + 0 = a{-1} + b{+1}

    4. Inverse elements of addition - here is where I tripped up before.  

    [method sidebar]
    My first reply to this issue in a comment was right, but since I have very little self-confidence, when challenged I will often go down other roads to see where they lead.  If it is a dead end, I go back to the start point.  Drives some readers nuts, but it is a way to explore.  It is also a way to decimate one's reputation.  My reputation is of no economic value to me, so I'd rather feel like an issue was examined completely than just take the one direct path to a truth others follow by rote.
    [end sidebar]

    Here at least is the wiki definition of this property:
    For every v in V, there exists an element -v in V called the additive inverse of v, such that v + (-v) = 0.
    Note, those were statements all about vectors.

    What I am going to do now is quote from "The Skeleton Key of Mathematics" by D. E. Littlewood, a book recommended by Barry Barrett.  Littlewood is discussing the transition from the positive integers to the signed integers.

    The integers so far have no signs, and the impossibility of subtracting say 3 from 2 is sometimes inconvenient.  This inconvenience is felt especially by a man with £2 in a purse who needs very urgently £3.  He could evade the difficulty perhaps by keeping his money in a banking account.  Then having a balance of £2 he could write a cheque for £3 and leave a negative balance of £1.

    Consider how this is possible in a banking account but not with money in a purse.  The bank associates two totals with each customer's account, the total of moneys credited and the total of moneys withdrawn.  The net balance is then regarded as the same if, for example, the credit amounts to £102 and the debit is £100, as if the credit were £52 and the debit £50.  If the debit exceeds the credit the balance is negative.

    This model is adopted in the definition of the signed integers.  Consider pairs of cardinal numbers (a, b) in which the first number corresponds to the debit, and the second to a credit.

    A definition of equality is adopted such that
    (a, b) = (c, d)
    if and only if a+d=b+c.
    Addition is defined by the rule:
    (a, b) + (c, d) = (a + c, b + d)
    and multiplication by (a, b) * (c, d) = (a d + b c, a c + b d)
    then the pair (0, a) or any equivalent pair is denoted by a, and the pair (a, 0) or any equivalent pair by -a.

    Alternatively, the signed integers could be regarded as a transition from one cardinal integer to another, e.g. the transition 100->102, or 50->52 is denoted by +2, and the transition 102->100 by -2.  The definition as a pair of cardinal integers is well adapted to this interpretation.

    The positive real numbers and the reals are a generalization of this quote.  What is interesting is seeing what these look like in the proto-vector space.  

    The real numbers are somewhat like line segments in this representation.  The start is defined with the {-1} proto-vector, and the end by the {+1} proto-vector.  These are not exactly like line segments since the notion of distance is different. 
    We are in the position to ask if every proto-vector has an inverse.

    This raises a question whether 3{-1}+3{+1} is equal to the zero vector.  The question can be answered by using equality as Littlewood defined it:

    [Note on notation and meaning... I will use three equal signs, ===, to indicate that these are equal when one is also employing equivalence classes.  An equivalence class reduces the dimensions of the space.]
    [Additional note: the === should only be used when going from the 2D proto-vector space to the 1D real number field.]
    3{-1}+3{+1} === 00{-1}+0{+1} === 0

    if and only if:

    3 + 0 = 3 + 0

    Looks good.

    5. Distributivity of scalar multiplication with respect to vector addition

    s(u + v) = su + sv = s((a+c){-1} + (b+d){+1}) = s(a+c){-1} + s(b+d){+1}

    6. Distributivity of scalar multiplication with respect to field addition

    (s + t)v = sv + tv = (s + t)a{-1} + (s+t) b {+1}= s a{-1} + s b {+1}+ t a{-1} + t b {+1}

    7. Compatibility of scalar multiplication with field multiplication

    s(v) = (st)v = s(t a{-1} + t b {+1}) = s t(a{-1} + b {+1})

    8. Identity element of scalar multiplication

    1 v = v = 1 (a{-1} + t b {+1}) = a{-1} + t b {+1}

    The proto-vectors - proto because the positive reals are not a mathematical field - are quite well behaved.  But they also have the properties of the group Z2.  Write out its multiplication table:

     x  1    -1 
     1    1  -1
     -1  -1  1

    Now we can officially and completely legally multiply two proto-vectors together.  The results are identical to what Littlefield suggested, even if written in a different notation:

    v x w = (a d + b c) {-1} + (a c + b d) {+1}

    There is a multiplication identity for the group Z2, good old +1, or 0{-1} + 1{+1}:

    v x 1 = (a 1 + b 0) {-1} + (a 0 + b 1) {+1} = a{-1} + b{+1}

    Does every element has a multiplicative inverse?  Do two simple cases first, ones where the number is like -2 or +3:

    Negative example:
    v = a{-1} + 0 {+1}
    v-1 = 1/a {-1} + 0{+1}
    v x v-1 = (a*0 + 0*1/a){-1} + (a*1/a + 0*0){+1} = 0 {-1} + 1{+1} = +1

    Positive example:
    u = 0{-1} + b{+1}
    u-1 = 0{-1} + 1/b{+1}
    u x u-1 = (0*1/b + b*0){-1} + (0 * 0 + b * 1/b){+1}

    Two down, just one more case to go: what if neither of the coefficients in the proto-vector is zero?  There's a fun puzzle.  Give it a try.

    OK, few will be willing to do so, but a little thought brings up the follow riddle: the term from the product of a non-zero proto-vector and its inverse hat goes with {-1} doesn't want to be zero.  This is where we need to Littlewood's definition of equality:

    [True if one uses equivalence classes]
    0{-1} + 1{+1} === +1
    e{-1} + (e + 1){+1} === +1

    if and only if

    0 + (e + 1) = 1 + e

    That will always be true [with equivalence classes].  Now look at the product of a proto-vector and its inverse:

    v = a{-1} + b{+1}
    v-1 = c{-1} + d{+1}
    v x v-1 = (a d + b c){-1} + (a c + b d){+1}

    An inverse can be found if the following equations hold:

    (a d + b c) = e
    (a c + b d) = e + 1

    [From CuriousReader below:A quick road to the inverse is to subtract one from the other, then:
    (a d + b c) - (a c + b d) = 1
    (c-d)(b-a) = 1
    (c-d) = 1/(b-a)
    Nice, quick and direct.]

    I did get a little lazy and ask Mathematica to solve the two equations for c and d:

    c = (a + e(a - b))/(a2 - b2)
    d = (-b + e(a - b))/(a2 - b2)

    Will c and d always be positive?  It all depends on the value of a, b, and e.  Let's assume a is greater than b.  In that case, c is always a positive number, a required thing since I am using the positive real numbers.  The value of d does depend on picking a legal value for e:

    d > 0

    if and only if:

    e >= b/(a - b)

    This idea is so new to me, I wanted to see it work with numbers.  Let a=3, b=1, and e = 1.  The value for e is OK since 1 > ½.

    c = (3 + (3-1))/(3^2 - 1^2) = ⅝
    d = (-1 + (3 - 1)) /(3^2 - 1^2) = ⅛

    (a d + b c) = e
    (a c + b d) = e + 1
    (3 * ⅛ + 1 * ⅝) = 1
    (3 * ⅝ + 1 * ⅛) = 2

    (3{-1} + 1{+1}) x (5/8{-1} + ⅛{+1}) = 1{-1} + 2{+1} === +1 

    [Note: the check on the last equality is the sum of the outies must equal the sum of the innies, and 2 does equal 2.]

    I can well imagine that some of my dear readers will be upset by the freedom to set e at any value that happens to be greater than a half.  I find that property interesting because it surprised me.  For those who want to always find the same multiplicative inverse, all one has to do is impose the rule that the value for e is exclusively the equality.  

    c = (3 + (3-1)/2)/(32 - 12) = 1/2
    d = (-1 + (3 - 1)/2) /(32 - 12) = 0

    (a d + b c) = e
    (a c + b d) = e + 1
    (3 * 0 + 1 * 1/2) = 1/2
    (3 * 1/2 + 1 * 0) = 3/2

    (3{-1} + 1{+1}) x (1/2{-1} + 0{+1}) = 1/2{-1} + 3/2{+1} === +1 

    This calculation is easier, zeroes are our friends.

    What happens if e is less than a half, what is that multiplicative inverse?  Well, it just cannot be done with the group Z2 over the positive real numbers.

    The previous section on inverses was written as I was figuring it out. It later occurred to me that finding the multiplicative inverse should be trivial because it is trivial to do with the real numbers.  If you have picked up the pattern, the inverse of this Z2 over the positive reals should take you all of two seconds:

    1/(2 {-1} + 5 {+1}) = ?

    The answer is at the end.

    What do I think has been done in this exercise?  I think the group Z2 over the positive real numbers is a 2D way to represent the mathematical field of real numbers.  The two basis vectors are linearly independent, so any non-zero value in one cannot be represented by the other (the two do share zero).

    In physics, one is always talking about the differences between two values.  It is differences in energy we see, not an absolute measure of energy.  Space and time are not absolutes, so we deal with differences between the two.

    Calculus requires the properties of a mathematical field.  The group Z2 over the positive reals is like the step before having a mathematical field.  So what is calculus acquiring in this transition?  I think of zero as the location of an observer in time or in space.  As an observer in time, negative is the past, positive is the future.  The past is fixed, the future is not here yet, so only at time zero can there be change.  As an observer in space, there may be things on the left.  Whatever is on the left is spacelike separated from whatever is on the right.  The only chance for things on the left to interact with things on the right is through the observer.

    It may not be worth anything, but I am enjoying thinking of the real numbers as a 2D proto-vector space.

    Doug

    Answer:
    1/(2 {-1} + 5 {+1}) = 0 {-1} + ⅓ {+1}


    Comments

    Again, there are an infinite number of your 2-tuples that are equal to each other.

    I strongly disagree with your claim that your basis are linearly independent. By the usual definition, does the following have a non-trivial solution for "a" and "b"?
    a e0 + b e1 = 0 e0 + 0 e1
    Yes. So I would considering these linearly dependent.

    This definition is also useful, in that it shows while you have two variables, because your the second basis is linearly dependent, you only have a total of 1 degree of freedom. In contrast, what usefulness does your definition of "linearly independent" give you? It appears to only be misleading you.

    Restating some of this using one of your favorite word 'isomorphic', your "2-dimensional proto-vectors" are not isomorphic to the reals. There is not a one-to-one mapping.

    ----------------------
    To help show the silliness of some of this, consider the following "3 dimensional" reals, using a 3-tuple of positive numbers.

    I will define equals by
    (a,b,c) = (d,e,f)
    if a*c + e*f = d*f + b*c
    I define addition by:
    (a,b,c) + (d,e,f) = ( a/f+d/c, b/f+e/c, c*f )
    define multiplication by:
    (a,b,c) * (d,e,f) = (ad + be, ae + bd, c*f)

    of course this was all manufactured to make the correspondence from (a,b,c) to the real value (a-b)*c. Does this mean I have 3 dimensional proto-vectors isomorphic to the reals? No.

    We can easily concoct higher dimensional examples as well. Furthermore, because we _defined_ the operations to get the answer we want, nothing is really gained here. We didn't derive anything interesting. We only got out exactly what we put in.

    What exactly do you feel you gained here that you didn't put in by hand?

    The Stand-Up Physicist
    Welcome back Anonymous.  I had hoped the quote from Littlewood might make things clearer.  To make the story more concrete, let's imagine 5 account holders at the First Positive Real Number Bank of Cambridge, England (Littlewood taught at Trinity College).  Call them A, B, C, D, and E.
    Customer A reveals nothing about his account.  She guards her privacy which she has every right to do.

    Customer B is proud of his credit account.  He tells us he has well over £30,000 in credits.  He is ashamed of his debits, so does not tell us that bit of information.  For that reason, we do not know if his balance is positive or negative.  Does he owe to bank or does the bank owe him should he decide to cash out?


    Customers C, D, and E all belong to the equivalence class +2 whose balances are all equal to +£2.  Customer C tells us he has a credit of £102.  We know immediately how much he owes the bank, £100.  This required two bits of information, the credit and the net balance.  Customer D brags that he has a smaller amount of debt, only £50.  Now we know his exact credit balance, £52.  Customer E decides not to say a thing.  We cannot know if he has a huge credit and corresponding debit, or if he has no debt at all, thus a meager £2 of credit.


    To figure out any of the balances at the First Positive Real Number Bank requires two pieces of information, the credit and the debit.  Credits and debits do sound independent.  Customer E no longer wants to be in the equivalence class +2.  He shows up with 10 pounds, deposits it in his account, and is kicked out of the +2 club.  Customer D - a student - needs to take to a loan officer, and then doubles his debt, also falling out of the +2 club.  The mechanisms for credits and debits are different.




    Let's talk about the silliness as you say.  First of all, I will not fall into the tuple trap again for these types of discussions.  I introduced the basis {-1} and {+1} for the group Z2.  That group only has two elements.  I cannot interpret your challenge.
    Telling stories to lay an analogy over the math does not change the math. You didn't address my points, although you now bring up equivalence classes, something Littlewood alluded to but which you did not discuss in the article.

    QUESTION 1] Your "2-dimensional proto-vectors" are not isomorphic to the reals, because you do not have a one-to-one mapping. Do you understand this?

    If you want to introduce equivalence classes, then please note that your equivalence classes reduces a dimension, and you are left with one dimensional objects. These equivalence classes are isomorphic to the reals.

    And regarding linear independence --
    Definitions should be useful. The usual definition of linear independence makes it clear you only have one degree of freedom. What does yours give you? Nothing. In fact it misleads you into thinking you have two separate degrees of freedom and so therefore when you find that there are an infinite number of solutions to some equations you proclaim:
    "I find that property interesting because it surprised me."

    But it shouldn't have surprised you. That is the point. You can make up your own definitions, but in exchange for confusing the terminology in discussion, what did you gain? Nothing. In fact your definition is _less_ useful. That was the point. You can tell all the stories you want to describe your definition, but it doesn't change your definition. It is still not very usefull, and clearly it also misleads you.

    "First of all, I will not fall into the tuple trap again for these types of discussions."

    You are misunderstanding the "tuple trap". In fact in this case one could argue (and David and others did argue), writing your proto-vectors as:
    a {+1} + b {-1}
    can be somewhat misleading here. What does that "plus" mean? You are using vector-notation for your proto-vectors.
    As long as you are not changing basis, the tuple-notation is just fine as there is no confusing what basis we are talking about. Notice Littlewood's description used the tuple-notation as well. Similarly, in my example, the possible 'basis' sets were fixed when I defined how the addition and multiplication worked. Sure, I could define a new mapping from other 3-tuple labels to the current one used, but that is immaterial here as I defined how the choice I was working operates. You did not learn the real issues with your coordinate changes in quaternions (you seemed to abandon the subject) so you are misunderstanding the "tuple trap", as it is not an issue here.

    "I cannot interpret your challenge."

    Do not, I repeat do NOT start playing with this new 3-d structure. The point which seemed to have escaped you, was that playing by these rules I can concoct other examples in higher dimensions and it gives you nothing useful. The question is not about the details of these structures, but whether these structures give you something new and useful. I argue no, because we _defined_ the operations to get the answer we want. The structure didn't lead us somewhere; instead we chose the definitions purposefully to force the answer we wanted, so nothing is really gained here. We didn't derive anything interesting. We only got out exactly what we put in.

    So again I end in the question:
    QUESTION 2] What exactly do you feel you gained here?

    Or more narrowly, focussing on the practical uses:
    QUESTION 3] Your definition of equals gives you only one degree of freedom. If you collect elements into equivalence classes, you then get something isomorphic to the reals. Therefore, if you wrote a theory using these objects you gain NOTHING over just using the reals. Do you understand that? Or are you claiming you gain something here over the reals?

    -- questions labelled to help separate response for clarification of material, versus resulting response for further discussion of your ideas

    The Stand-Up Physicist
    Telling stories to lay an analogy over the math does not change the math.

    Those stories are continuations of what Littlewood wrote.  It would seem like you are ignoring his position entirely.  He is not worried silly about all those ways to write +2, so neither am I.

    QUESTION 1] Your "2-dimensional proto-vectors" are not isomorphic to the reals, because you do not have a one-to-one mapping. Do you understand this?

    The 2-dimensional proto-vectors over the positive real numbers are not isomorphic to the reals for the simple reason that they are 2-dimensional whereas the reals are one dimensional.  I did not mention isomorphisms anywhere, you did, not me.

    The usual definition of linear independence makes it clear you only have one degree of freedom.

    What is the "usual definition"?  To quote from on high:

    In linear algebra, a family of vectors is linearly independent if none of them can be written as a linear combination of finitely many other vectors in the collection.

    You appear to be unable to adapt to the convention used in the blog, but the two basis proto-vectors are {-1} and {+1}, not e0 and e1.  The 2D proto-vector 

    50{-1} + 52{+1}

    is different from:

    100{-1} + 102{+1}

    Go ahead, draw them both.  Can I change the 102 without changing the 100?  Sure can.  Sounds linearly independent to me.

    What does that "plus" mean? You are using vector-notation for your proto-vectors.

    Yup, I am using vector notation for the proto-vectors because the proto-vectors are over a set that is not a mathematical field.  I demonstrated that they had 8 properties known for vector spaces that are over mathematical fields.

    As long as you are not changing basis, the tuple-notation is just fine as there is no confusing what basis we are talking about.

    My point was I don't know what you are talking about with (a, b, c), nor with (e, d, f).  The group Z2 has two elements to it, not three.  Those higher dimensional examples can be represented with the Z2 basis vectors.

    QUESTION 2] What exactly do you feel you gained here?

    "The group Z2 over the positive reals is like the step before having a mathematical field."  It is a subtle point.

    QUESTION 3] Your definition of equals gives you only one degree of freedom. If you collect elements into equivalence classes, you then get something isomorphic to the reals. Therefore, if you wrote a theory using these objects you gain NOTHING over just using the reals. Do you understand that? Or are you claiming you gain something here over the reals?

    The two dimensional vector space has, well, two dimension.  One can collect elements into equivalence classes to get something isomorphic to the reals.  Until one does the collection, Z2 over the positive reals is different (see QUESTION 1]).

    I find the roots of ideas of interest.  A mathematical field is about as important as any idea in mathematical physics.  You do understand why I consistently used the word proto-vector?  My starting ingredients was a finite group you chose not to discuss over the set of positive real numbers which are not a mathematical field.  I am looking for insights more than material gains.

    "What is the "usual definition"?"

    The definition Anon is referring to is the same one I am referring to, and the one I already linked to in the previous article discussion.
    http://en.wikipedia.org/wiki/Linear_independence#Definition

    Yes, we understand you are using a different definition. We have repeatedly acknowledged you are trying to use a different definition. The point is that your definition is not _useful_, and you are throwing out a very important concept by insisting we don't considering the usual concept of linear independence.

    "Can I change the 102 without changing the 100? Sure can. Sounds linearly independent to me."

    That is an even WORSE definition of linearly independent than you usual one. Because even if the basis are exactly the same, I can still set two scalars multiplying them "indepdently". By this newest definition of yours, ALL basis are linearly independent. So it is a tautology and _completely_ worthless as it conveys no information what-so-ever.

    " The 2D proto-vector
    50{-1} + 52{+1}
    is different from:
    100{-1} + 102{+1}"

    In the article, you said they were equal. Now you are saying they are different.
    You can't have it both ways. If we call the two basis e0 and e1, either they are linearly independent and
    50 e0 + 52 e1 != 100 e0 + 102 e1
    OR they are linearly dependent and
    e0 + e1 = 0 e0 + 0 e1

    You cannot have it both ways. Chose something and stick with it.

    The Stand-Up Physicist
    Thanks for the link that was already in my reply.  Oh, and I quoted the link too.  so I am not using a different definition.  
    Let me demonstrate how one can "have it both ways."  They are actually two ways.
    Someone has a bank account with a credit of $102, a debit of $100, and a balance of +$2.  The credit and debit are in what I defined here as the Z2 group over the positive real number proto-vector space.  The balance which is the difference of those two is a one dimensional space that is isomorphic to the real numbers so long as one mods out by the equivalence classes.  The balance can be positive or negative.


    For the proto-vector space, you MUST include two dots on the line, not one.  The sole exception is for the additive inverse, zero.
    "so I am not using a different definition."

    YES YOU ARE!
    Did you actually read the section CuriousReader linked? That is, did you read the definition section of that article?
    Regardless of what you read there, how can you possibly claim after the discussion here that you are using the same definition CuriousReader and I are using? Please give a "good faith" effort and reading and understanding what we write.

    Let's clear the air and start over.
    I'll write everything self-contained here, to condense the information in a single place.

    Your definition:
    "a family of vectors is linearly independent if none of them can be written as a linear combination of finitely many other vectors in the collection"
    incorporating all the bits you are claiming, I'll make this precise as such:
    A vector e0 is linearly depedent with a set of proto-vectors {e1,e2,...,en} if there is a solution to:
    e0 = a1 e1 + a2 e2 + ... + an en
    _AND_ a1, a2, etc are elements of the "proto-field" you are using for your proto-vectors.

    The definition everyone else is using and is also in the link CuriousReader gave, if you read the DEFINITION section:

    "A finite subset of n vectors, v1, v2, ..., vn, from the vector space V, is linearly dependent if and only if there exists a set of n scalars, a1, a2, ..., an, not all zero, such that
    a1 v1 + a2 v2 + ... + an vn = 0 v1 + 0 v2 + ... + 0 vn (the zero vector)
    If such scalars do not exist, then the vectors are said to be linearly independent."
    This can be trivially updated to include your requirement that a1..an are elements of your "proto-field".

    And again, for clarify, everyone understands you are using a different defintion than us. And also for clarification (should be obvious, but just in case), these definitions would be equivalent if we use fields instead of "proto-fields" for our scalars.

    ---break---
    So hopefully we are now all on the same page.
    Can we agree up to here?
    ---continuing---

    Sure, you CAN choose a different way to extend the definition to your "proto-fields". The point is: Even if you want to use a different definition, there is still a clear concept described by the definition we want to use, and we don't understand your insistence on throwing this concept away. This concept is USEFUL, as opposed to the concept your definition is giving you which doesn't really give you much. A strong case can be presented (and has been) that the defintion we're presenting more appropriately extends the concept of linear independence to your "protofields".

    You concept of linear dependence of two basis is telling you more about the restriction on the "proto-field" than it is telling you anything about your basis vectors. It is not conveying useful information about the basis vectors, nor the space that can be built from them.

    I like CuriousReader's "walk in the woods" explanation. It makes the issue abundantly clear if the math didn't already. Due to your restriction on the scalars, it is possible that two basis vectors can allow reaching more locations, even if with these vectors would be proportional if using real numbers are scalars, yet still it would only be possible to reach points on a one dimensional slice on the map of the woods. Only if the two basis are linearly independent in the usual sense, would it be possible to reach points on a 2-dimensional area on the map of the woods.

    Do you understand the difference between the definitions now?
    Do you understand the usefulness of this other concept even if you refuse to call it linear dependence?

    The Stand-Up Physicist
    YES YOU ARE!

    All caps, how retro.

    Did you actually read the section…

    Read and quoted.

    I have seen zero acknowledgement on your part that you can defined what this blog is about, which has two parts: the finite group Z2 over the positive real numbers.  Those technical details are essential.

    Here is the imprecise statement:

    a1 v1 + a2 v2 + ... + an vn = 0 v1 + 0 v2 + ... + 0 vn (the zero vector)

    should be:

    a1 v1 + a2 v2 + ... + an vn === 0 v1 + 0 v2 + ... + 0 vn (the zero vector)

    The three equals signs indicate that it is true only in a situation were one mods out the equivalence classes.

    If one does not mod out the equivalence classes:

    a1 v1 + a2 v2 + ... + an vn != 0 v1 + 0 v2 + ... + 0 vn (the zero vector)

    That should be obvious from these three examples from Z2 over the positive reals which has been updated to include the indication where an equivalence class is used:

    Doug, be reasonable.
    You change details mid discussion, blame your readers, and say that it should be obvious from an image that you just now post which is different from your article.

    And it still isn't consistent or clear exactly what you are concluding.

    Note that your title "Looking at the Real Numbers the Z2 Way" implies you ARE talking about the equivalence classes of these "proto-vectors" and not the proto-vectors themselves. Note that the "=" in Littlewood's discussion is saying the same thing. He is talking about equivalence classes: "the pair (0, a) or any equivalent pair is denoted by a, and the pair (a, 0) or any equivalent pair by -a." You say you are looking at a generalization of this to the reals. So lots of things point to you wanting to use equality in the sense of the equivalence class of these proto-vectors.

    But then you make the comments like above, and also sum up your article saying: "I think the group Z2 over the positive real numbers is a 2D way to represent the mathematical field of real numbers."

    So which is it? You seem to be trying to have it both ways.
    It doesn't matter how much you refine your idea of equals, you either have a set of 2D proto-vectors that don't have additive inverses (you don't get your precious negative numbers), OR you have a 1D space of equivalence classes that are like the mathematical field of real numbers.

    Take a stance, stick to it, and learn what the results of the idea are.

    Halliday
    Anonymous:

    Can you not hold two (or more) conflicting (let alone mutually exclusive) concepts within your mind at one time?  ;)

    David

    Let me try.

    Doug, in the decimal representation:
    1.99999... = 2

    So even if we can agree they are equal, would you claim they are different numbers? Or would you claim they are not different? How are you distinguishing the concept "equal" from "not different"?

    I personally would say they are the same number; they are equal; they are not "different"; they are the same. Don't get hung up on the label. I may say "two" and you may say "dos", but they are the same number. In your case, you have a wildly redundant labeling system. You have an infinite number of labels for each number. You stated so when you stated your definition of equality.

    "The two dimensional vector space has, well, two dimension."

    Two numbers does not equal two dimensions. If I label three dimensional space with 5 basis vector (which obviously must have some linear dependency), I still only have a three-dimensional space. (<-- SERIOUS QUESTION: Can we at least agree on that? Describing a 3 dimensional space with 5-tuples doesn't turn it into a 5 dimensional space.)

    I consider this getting back to one of the "root issues" Henry pointed out -- a tendency to confuse the object for the thing you are labeling it with.

    Take your psuedo-vectors. Look at the addition and multiplication rules. Imposing only these you can have a two-dimensional system. But when you impose your definition of "equals", this choice clearly shows you have linearly dependent basis, and are describing only a one-dimensional space. You even draw it as a line!

    The Stand-Up Physicist
    The SERIOUS ANSWER.  The three dimensional space with 5 basis vectors can be easily constructed.  The basis vectors would be a set of linearly dependent vectors.   So yes, we agree on that point.

    Perhaps the stumbling point is that in my proto-vector space I really am working wiht only positive real numbers.  That is not a normal thing to do.  Neither is requiring one write two dots on the line.  Shouldn't one dot be enough?  It is not for the structure I described which has two elements to it.  We can agree Z2 has two elements I presume.  And because it is over the positive real numbers, the negatives come from Z2.

    Here is a picture:



    I certainly am used to the real line representation, how there is no issue with drawing a single dot.  I haven't thought much about those arrows at the end.  They represent a continuation, keep going until you reach infinity which ain't going to happen.

    My arrowhead mean something different.  They are directions from the origin.  Either positive or negative.  One MUST draw two points on the line, just like every account at a bank needs both debits and credits.  I could calculate the balance and then draw it on the real number line above.  It is undefined to draw one dot on the Z2 over the reals unless that one dot is zero.

    "What do I think has been done in this exercise? I think the group Z2 over the positive real numbers is a 2D way to represent the mathematical field of real numbers. The two basis vectors are linearly independent, so any non-zero value in one cannot be represented by the other (the two do share zero)."

    What do you mean by "represent"? And what do you mean by "2D"? It sounds to me you are claiming to have made an isomorphism between the first quadrant of a 2D plane, and the reals. I would not agree with that statement.

    I still object to your definition of linear independence.
    Since you like stories, imagine it this way:
    Let's say I'm in the woods, and choose two directions A and B (and I can only walk forward in each direction, not backwards).
    If I can take 7 steps in direction A and can return to where I started by taking some number of steps in direction B, would you really say those two directions are linearly independent? Can I explore a 2 dimensional space of the woods using these two directions? Or can I only reach a 1 dimensional slice on the map?

    While there is nothing preventing you from redefining terms, in this case your new definition is essentially useless. By your definition, direction A != direction B means A and B are linearly independent. So you redefined a word to state something we already have a concept for: equality. In the process you lost a word for a very useful concept, which you appear to have abandonned.

    Do you understand now why people object? Do not abandon the useful concept just because you really want to redefine "linearly independent" for some reason.

    -------------
    I feel you are getting lost in some of the math here.

    Everything is being defined so that: (a,b) maps to the real number (b-a).
    It feels like somehow somehow this gets lost and you come to conclusions that should be red flags if you remembered where you started.

    For example:
    (a d + b c) = e
    (a c + b d) = e + 1

    Why introduct e at all? It is unconstrained. Yet strangly you "derive" a constraint on it later. This should be an immediate red flag that something fishy is going on.

    Let me work though to make it more clear what is going on.
    Start with:
    (a d + b c) = e
    (a c + b d) = e + 1

    Then
    (a d + b c) - (a c + b d) = 1
    and so
    (c-d)(b-a) = 1

    Which is exactly what one would expect. This is the more natural way to write it, which could be obtained directly from the required relation to the reals.

    Given "a","b", there are of course an infinite number of solutions to "d" and "c" given the definition of equality as one should expect instead of being surprised.

    Where is the error?
    What gives the false constraint on e?
    I think the error is in sometimes using = as the usual equals, and sometimes using it as your redefined equality. There should be no restriction on e.

    (2,1)*(2,1) = (2*1+1*2,2*2+1*1)=(4,5)=(e,e+1)
    this seems to say e=4. It doesn't. e can still be any value according to your definition of equality. So it only means e=4 if you forget that you redefined what the equals sign means.

    The Stand-Up Physicist

    People are used to the power of the real number line.  I sure do understand that.  When thinking of how to program Z2 over the positive real numbers, or Q8 over the positive real numbers, the project quickly creates headaches.  For example, trig functions have no problems switching signs, it is part of their nature.  I have considered real numbers to be one dimensional since I have heard of them.  I have considered quaternions to be four dimensional since I have heard of them.  But I want multiplication to arise directly from group theory.  The group Z2 does it over the positive reals, even if it is inconvenient.

    Let's say I'm in the woods, and choose two directions A and B (and I can only walk forward in each direction, not backwards).

    I like it :-)  So I am in the woods, and give myself a simple assignment: I need to walk in the two directions, A and B, at the same time.  Damn, I failed.  I do need to make a decision, A or B.  Over the course of a few minutes, I could get around to doing both.  Space is generous like that.  The same is not true for time.  I can only go forward in time.  I can tell you stories from my past, but I can only share the stories as we both go forward in time.  This will only give a slice of a map, sorry.

    There is a minor error with the suggestion.  It is more like tossing a rock in the directions A and B from a defined origin.  Your model sounds like the origin may wander through the woods.

    You found the fast road to the equality relation by eliminating e, congrats.  In the context of the algebra for finding a multiplicative inverse, there is a constraint on e.  I calculated what it would be if e=1/10.  Then c = 2/5 and d = -1/10, which is a problem since -1/10 is not a positive real number.  It should not be a surprise to find a bit of algebra that is out of bounds for this construction.  A similar thing happens with the real numbers on the road to imaginary numbers.

    "But I want multiplication to arise directly from group theory."

    You used the multiplication of two reals in your definition of the multiplication. What is 1.5*3.14 ? Your group does not tell you this. How do you calculate something like a*b+c*d without using multiplication of the reals?

    "I like it :-) So I am in the woods, and give myself a simple assignment: I need to walk in the two directions, A and B, at the same time. Damn, I failed."

    I explicitly said A then B, not both at the same time. Are you seriously unable to understand my points? Or are you twisting the story to try to make it fit your analogies? It feels like you are purposely ignoring my points to respond with an unrelated "story".

    "This will only give a slice of a map, sorry."

    Exactly, it supports my claim, not yours, so why are you saying sorry?
    Only a one-dimensional piece is available to you when the basis are linearly dependent. You only have freedom in one dimension. Where-as if you chose the directions A and B to be linearly independent, the points that are reachable from the starting point form a 2-d region.

    "Your model sounds like the origin may wander through the woods."

    What the heck are you talking about?
    Are you purposely twisting what I'm saying? It's straining credulity to believe you are trying to read carefully and giving a good faith effort at understanding the points. Please reread.

    "In the context of the algebra for finding a multiplicative inverse, there is a constraint on e."

    Again, NO. To claim so is to change your definition of equality in the middle of the calculation.
    It is very clear if you look at the big picture that e cannot be constrained in the manner you state. So if you still are claiming e can be constrained, even after pointing out the big picture, and even after I showed the math, there is a serious disconnect here.

    It case it wasn't clear, that was me (CuriousReader), not Anon. I must have forgot to put my name there that time.

    I can't believe I'm saying this, but David could you please help us understand what "equals" means precisely enough to allow us to clear up all this confusion?

    By saying (1,2)=(2,3)=(3,4)=... I take it that Littlewood is already discussing the equivalence class. I'm not sure how to interpret it otherwise.

    Is there some sense in which Doug can say those are all equal, without bringing in the concept of equivalence classes?

    The Stand-Up Physicist
    I believe this may be the issue that causes confusion.  Equals with equivalence classes is different from equal.  Using the equivalence classes effectively reduces the number of dimensions from 2D to one D.  If I do not use equivalence classes, the two basis vectors can be viewed as linearly independent, debits are different from credits.  If I do use equivalence classes, then we are comparing bank balances, so obviously the guy with debits of 50 and credits of 52 has the same balance as the fellow with 100 and 102.

    I will modify the text of the blog to use "===" when I mean equal using equivalence classes which reduces the dimensions of the proto-vector space.
    Okay, you are getting there, but you don't seem to be understanding the complaints still.

    Note that if you do not use equivalence classes, then you no longer have additive inverses. Which is what you appeared to be trying to correct from your previous attempt. You can't have it both ways.

    Without additive inverses, you don't really have negative numbers anymore.
    The equivalence relation is what allows you to get the mapping to the reals.

    Either you have a 2-d space, or you have additive inverses.
    You can't have it both ways.

    The Stand-Up Physicist
    I looked over the 8 properties of a vector space, and only one property, the additive inverses, appears to require the equivalence class.  The only time the equals moding out the equivalence class is needed is when one tries to compare the group Z2 over the positive real proto-vector space with the 1D real numbers, so in this kind of case:
    3{-1} + 4{+1} === +1

    On the left is the proto-vector, on the right are the real numbers.  The additive inverse of:

    2{-1} + 1{+1}

    is

    1{-1} + 2{+1}


    because if I add them together I get:


    3{-1} + 3{+1}


    I can map that to the real numbers if I choose by moding out the equivalence classes:


    3{-1} + 3{+1} === 0
    Yes, I think we are all finally on the same page. Thank you.

    "I looked over the 8 properties of a vector space, and only one property, the additive inverses, appears to require the equivalence class."
    Yes, as long as every reference to scalar is replaced with your restriction to real numbers >= 0. I assume you are including that restriction.

    -----------------
    Okay, so now that everyone appears to be on the same page, let's discuss the consequences of looking at the proto-vectors themselves, instead of an equivalence classes of them.

    Given the definition of multiplication of your proto-vectors:
    (a {-1} + b {+1}) * (c {-1} + d {+1}) = ( (ad+bc) {-1} + (ac+bd) {+1} )
    There is a unique multiplicative identity:
    (a {-1} + b {+1}) * (0 {-1} + 1 {+1}) = (a {-1} + b {+1})
    But for there to be a multiplicative inverse, one would need a solution of c,d given a,b in:
    (a {-1} + b {+1}) * (c {-1} + d {+1}) = ( (ad+bc) {-1} + (ac+bd) {+1} ) = (0 {-1} + 1 {+1})
    So:
    ad + bc = 0
    ac + bd = 1

    Let's separate this into four cases, keeping in mind your restriction to reals >= 0:
    1) a and b are both zero
    No values of c and d can satisfy the second question.
    -- there is no solution
    This one is expected for the zero element.
    2) a and b are non-zero
    The only solution to the first equation is c=d=0, but this is not possible for the second equation, so:
    -- there is no solution
    3) a is zero, b is non-zero
    First equation gives c=0, d unconstrained. Second equation then gives d=1/b.
    -- So this has a unique inverse.
    4) a is non-zero, b is zero
    First equation gives d=0, c unconstrained. Second equation then gives c=1/a.
    -- So this has a unique inverse.

    So you can't do subtraction (adding with the additive inverse) with your proto-vectors, nor can you do division (multiplication by the multiplicative inverse) for all but two special 1D slices of your 2D space.

    So if you wanted to 'extend' this to your Q8 proto-vector space, you don't have a division operation there either. You are losing the cherished properties of your Quaternions. You are even losing the cherished properties of Q8, as your proto-vectors don't even form a group anymore as you are missing inverse elements.

    Let that sink in.
    You don't get the properties of the algebra you want, and you fail to even have a group despite starting with a group that you were trying to extend.

    Looking at the bigger picture: How do we judge the worth of a mathematical object for some physics problem?
    There's no quick answer of course, but avoiding the issue is like pretending every mathematical object is equally appropriate to the task.

    While many theories use complex numbers, spinors, and other interesting objects, ultimately in the lab we make measurements which result in real values. If we wanted to, we could write all our theories purely in real number valued objects ... but while possible, it would be difficult to work with and obscuring many useful properties and features of the theory. Here though, you merely defined your proto-vectors to have the proto-vector mimic the real number (a(-1) + b(+1)) after looking at the equivelence classes. You can't derive any new useful properties or relations or insights from these objects, because all the information was put in by hand to meet the expectations. And ultimately this object is _less_ convenient that the real numbers, as it is missing some basic group properties like inverses. So instead of gaining useful properties, you lose some, and the only good things you have are what you put in by hand. If you feel these protovectors have interesting properties that may be useful to you, I think you are not being honest with yourself about the situation.

    Can you honestly say otherwise?

    This would be a good time to look back at what sent you down these tangents. What were you looking for? Do you even know what you were looking for?
    (In my opinion, it looks like you got a bit frustrated with quaternions in your theories and wanted to either come up with a new binary operation for them, or a number system other than the quaternions, that you could use in your quest for a gravity theory. Since you didn't clearly define what you wanted, and had no process to try to seek this out, you seemed to forget what you were doing and got lost on tangents.)

    Did you find anything useful here to use in your quest for an alternative theory of gravity? What if anything, did you gain from this?
    (My take: You _can_ gain from this if you take the time to learn from the process to improve your process in the future. It is the difference between only gaining a single useless fact (I tried it and these objects are not helpful), and gaining a better ability to hone your approach as well as critically analyze it in the future.)

    It sounds like you are leaving, so I also want to say good luck with whatever you try next.

    Oops.
    I wrote < a , b > (without the spaces) and it caused a formatting glitch.
    Sorry about that.

    The Stand-Up Physicist
    I am somewhat surprised you did this exercise as CuriousReader showed how to find the inverse every time in a way that was simpler than what I did in the blog.  That was the:
    ad + bc = e
    ac + bd = e + 1

    then eliminate the e exercise.  This was the trick required to find one and only one multiplicative inverse.  If you insist that e must necessarily equal zero, then of course your critique stands as is.  On the hunt for multiplicative inverses, one does avoid zero, so I think your constraint is unreasonable.

    Here is another way to see that I am being rational.  Let's start with 4 little drawings all readers can accept that only use the real numbers.  Here is the real number line with no points on it:




    It is important to mark both the origin and at least one number (doesn't have to be one, but must be something).  The arrows mean "keep on trucking."  Add two points:




    The two points happen to be -3 and +2.  Now we can easily draw in their additive inverses:




    The additive inverse are the hollow dots written on the other side of zero.  And the multiplicative inverses can also be found:





    The multiplicative inverses are on the same side of zero as the initial points, drawn in a light color.  So far, I don't see how we might disagree on these figures.

    Here is how I draw the group Z2 over the group of positive real numbers:




    The difference is that the arrowheads now have a purpose: they are basis vectors.  I have a negative direction, without negative numbers.  And I can draw points, additive inverses, and multiplicative inverses:












    In this situation, I am obligated to always have two points.  The multiplicative inverse is 1/5{-1} + 0{+1}.  That was different from the two being treated separately.

    Nature is always using differences and ratios, so might as well start with two numbers at the beginning.  An event in spacetime is meaningless.  With two events, all kinds of things can be calculated.  It is so common, people forget they are dealing with differences or ratios of numbers, so they are seduced to thinking that numbers can stand by themselves.  That is my big picture view.
    Doug, I thought all of us were finally on the same page, but it looks like you have still not learned your lesson regarding the equivalence classes.

    If you DO NOT look at the equivalence classes, then you DO NOT have additive inverses or, as I showed, multiplicative inverses for the vast majority of proto-vectors. Do you understand this?

    If you use the "equals" to mean instead statements regarding the equivalence classes, then you DO have additive inverses and multiplicative inverses (except for the class (0,0) of course), but then you have a 1D space.
    You cannot have it both ways.
    I though you finally understood this in your previous message, but it is clear you are still not getting the consequences of choosing to look at the proto-vectors, vs looking equivalence classes of the proto-vectors.

    "If you insist that e must necessarily equal zero, then of course your critique stands as is."

    Look at the definition of multiplication for your protovectors. There is no freedom to set an "e" however you want. Only in the previous case where you were looking at the equivalence classes is this possible.

    You seem to be flip-flopping again.
    QUESTION: Do you want to consider the properties of the proto-vectors themselves (so you have a 2D space), or do you want to consider the equivalence classes of these proto-vectors which is isomorphic to the reals (a 1D space)?

    We can discuss both, that is not a problem, but you cannot mix and match the conclusions between them. You can't try to have it both ways. This isn't a difficult concept, and I'm worried that your logical disconnect it coming from an unwillingness to let go of a failed picture you constructed in your mind.

    In conclusion:
    If you want to look at the 2-D space of your proto-vectors, you do not have additive inverses, and the vast majority of proto-vectors do not have multiplicative inverses. There is no arguing with this. It follows directly from your definitions. If this still doesn't make sense to you, reread and think about it, for there you have some kind of logical disconnect here. (I hope, just once, we could agree on some math without having to resort to David restating it before you believe it. The truth of a mathematical conclusions is independent of who said it.)

    Regarding the "vast majority" statement, I'm not sure how to state it rigorously, but it should be conceptually obvious that the fraction of proto-vectors that don't have a multiplicative inverse is --> 1 (since only two 1D slices of a 2D space have multiplicative inverses). I'd actually be interested to hear from David on this one. For instance the "vast majority" of complex numbers are not real valued. This is conceptually obvious, but I'm not sure how to state it rigorously. David? Care to add some useful discussion for the rest of us? The conversion is getting quite stale without you.

    Damn formatting issues again. The statement is
    "fraction of proto-vectors ... is --> 1"
    With the arrow I was trying to imply some kind of limiting statement which (David help?) I don't know how to specify rigorously.

    Due to formatting issues (on my computer at least), it came out looking like I was claiming the fraction is > 1. But of course that isn't what I meant.

    Carry on.

    Halliday
    LangrangiansForBreakfast:

    The "usual" way to express what you wanted is to say that fraction of proto-vectors with inverses is order zero (often symbolized as O(0), that's Big-O of zero) [AKA measure zero].  What this means operationally is that if you were to pick a random value (from this 2D space) the probability (at least in a frequentist sense) that it would have an inverse is zero.

    There are many cases where the set of points/solutions/whatever are "order zero" [or of measure zero] in this very same way.  (For instance the set of rational numbers are a set of order zero [or measure zero] within the real numbers.  [On the other hand, the set of rational numbers are also "dense" within the real numbers.])

    David

    The Stand-Up Physicist
    If you want to look at the 2-D space of your proto-vectors, you do not have additive inverses, and the vast majority of proto-vectors do not have multiplicative inverses. There is no arguing with this.
    I hope someday you will recognize how pompous this sounds.  I have drawn the graphs of the additive inverses that will work for every member of the 2D proto-vector space.  I have drawn the graphs of the multiplicative inverses for half the cases (the other half come from b>a).  I can always do that in the 2D proto-vector space.  

    The proofs that these do play their expected roles are easier to do my reducing the dimension to 1D by moding out the equivalence classes.
    HenryB
    The proofs that these do play their expected roles are easier to do my reducing the dimension to 1D by moding out the equivalence classes. 
    You are then talking about additive inverses not in the space of the proto-vectors, but in the space of the equivalence classes of these proto-vectors for your given equivalence relation.

    So this isn't proving your proto-vectors have additive inverses.  This in no way "undoes" any of the math presented to you, and it is unclear to all the rest of us why you think so.
    I hope someday you will recognize how pompous this sounds.  
    The math says what it says.  I hope someday you recognize how pompous it sounds when you, after all previous experiences, still assume you can ignore everyone's math just because it contradicts vague notions you are carrying around in your head.  If you see the math, and don't see a mistake, you should at this point give serious consideration to the possibility that you are wrong.  It is even more frustrating that time and time again, the issue comes down to definitions.


    In this case, the issue is the definition:
    Does, or does not: a {-1} + a {+1} = 0 {-1} + 0 {+1} ?


    You have recently made it very clear the answer is no.  But you still seem to want to keep all the benefits of claiming yes.  You can't have it both ways.
    "I am somewhat surprised you did this exercise as CuriousReader showed how to find the inverse every time in a way that was simpler than what I did in the blog."

    What I wrote was when you were still defining the "=" such that:
    a {-1} + a {+1} = 0 {-1} + 0 {+1}
    You have since complained about us readers for thinking this when you "obviously" meant otherwise. But if you do not want to use equals as such, then you can't use the inverses as I wrote them. Again, you can't have it both ways. Everyone keeps repeating that phrase, but it aptly described what you seem to keep doing here.

    What LFB wrote above looks correct to me.
    If you want to look at the 2D space of your proto-vectors, you don't get additive inverses (negative numbers) or the ability to divide two arbitrary proto-vectors. So you fail to get the negative numbers you set out to find, and even fail to obtain a group despite using a group (Z2) as your starting point. You didn't get the features you wanted, and lost some of the features of the very object you were extending.

    I really don't think you have gained anything useful here.

    (David, I too would be interested in hearing if or how one can make rigorous statement regarding stuff like: the fraction of proto-vectors which don't have multiplicative inverses = 1.)

    The Stand-Up Physicist
    a {-1} + a {+1} != 0 {-1} + 0 {+1}

    Draw it, they are different.


    a {-1} + a {+1} Mod {however I should say the equivalence class} = 0


    0 {-1} + 0 {+1} Mod {however I should say the equivalence class} = 0

    The proofs require equivalence classes, taking the 2D to 1D.  The 2D things remain 2D.

    This is the rule that applies to find every additive inverse:

    The additive inverse of

    a{-1} + b{+1}

    is

    b{-1} + a{+1}


    See, all in 2D.  Do the proof in 1D by moding out the equivalence class, and those rules are the only rules that work.


    The multiplicative inverse of the same if a>b is:


    1/(a - b){-1} + 0 {+1}


    If a<b


    0{-1} + 1/(b-a) {+1}


    If a=b, there is no inverse (mod out the equivalence class, and it is zero).  Again, the proof requires using the equivalence classes in 1D, but the rule in the 2D proto-vector space is the rule in 2D proto-vector space.

    All those rules live in the 2D proto-vector space.  Each has a graphic representation that you conveniently ignore since it doesn't help your case.
    HenryB
    Doug,You are making gross logical fallicies here.

    Consider the following...
    If after doing some work you are able to prove:
     f(a) = f(b)
    Does this mean you proved:
     a=b
    Well?


    Hopefully that seems obvious to you.  But know that your errors are similar, and so seems just as obvious to everyone else.  After all this time, it is frustrating that you are unwilling or unable to confront these logical disconnects.


    Since the error has been pointed out before (multiple times now), I will try to write more equations to help you hopefully see the problem.


    First, some notation.  Explicitly carrying around these basis, when you have only used this one basis through-out the entire discussion is not needed.  So there is no worry of confusion and I will define the 2-tuple representation of your proto-vectors as:
    (a,b) = a {-1} + b {+1}

    where a and b are elements of the reals which are >= 0.

    You defined the operation "+" as:
    (a,b) + (c,d) = (a+c, b+d)
    You defined the operation "*" as:

    (a,b) * (c,d) = (ad+bc, ac+bd)


    There, that defines your 2D proto-vectors.
    Restating results from before:


    1) there are no additive inverses, except for the additive identity which is its own inverse


    Existence of unique additive identify:

    From the definition of the addition operator on these proto-vectors
    (a,b) + (c,d) = (a+c, b+d) = (a,b)
    This constrains the additive identity to the unique solution (c,d)=(0,0).

    The additive identify is its own additive inverse:
    This is true just from definitions, but can be checked explicitly.
    (0,0) + (0,0) = (0,0)


    Non-existence of additive identify for (a,b) != (0,0):
    From definitions, the additive inverse (c,d) of (a,b) must satisfy:
    (a,b) + (c,d) = (a+c, b+d) = (0,0)

    Separating into two covering options, a!=0, b!=0:
    if a!=0, then from the definition of the proto-vectors a>0 and c>=0, therefore a+c>0, and there is no solution to a+c=0.
    Same result holds for b!=0.

    2) there are no multiplicative inverses except for (a,0) or (0,a) with a>0

    For the most part copying LFB, so we have everything in one place:



    Existence of unique multiplicative identity:
    From the definition of multiplication of two proto-vectors, and definition of multiplicative identity
    (a,b) * (c,d) = (ad+bc, ac+bd) = (a,b)
    For (c,d) to be a multiplicative identity, this must hold for all (a,b). This constrains the multiplicative identity to the unique solution (c,d)=(0,1).


    Conditional existence of multiplicative inverse of (a,b):
    From the definition of multiplication of two proto-vectors and the definition of multiplicative inverse:
    (a,b) * (c,d) = (ad+bc, ac+bd) = (0,1)

    Therefore:
    [eq1]     ad+bc = 0
    [eq2]     ac+db = 0

    From the definition of the proto-vectors c>=0, d>=0, and we can separate a,b into for cases:
    case 1] a=0  and  b=0

    No values of c and d can satisfy the [eq2].
    -- there is no solution
    This one is expected for the zero element.
    case 2] a>0 and b>0
    The only solution to the first equation is c=d=0, but this is not possible for the second equation, so:
    -- there is no solution
    case 3] a=0, b>0
    First equation gives c=0, d unconstrained. Second equation then gives d=1/b.
    -- So this has a unique inverse.
    case 4] a is non-zero, b is zero
    First equation gives d=0, c unconstrained. Second equation then gives c=1/a.
    -- So this has a unique inverse. 


    side note:
    Regarding, LFB and CR's question of whether someone can rigorously say something like: the fraction of these proto-vectors that don't have an inverse is 1.  I think it would be something like: the measure of proto-vectors that have an inverse is 0.  But then you need to specify a measure, or at least properties of it.  Since you are comparing 1D slice to 2D area, I'd expect any 'reasonable' measure to do.  David will have to fill in details on this if you guys are still curious, but wikipedia can probably guide you from here.

    --------------------------------------------------------------------------------------------


    Alright, that should all be unquestionable math.
    The 2D proto-vectors were defined, and properties were derived from the definition.
    Doug, for some reason you think you can do some manipulations to avoid these conclusions.  Why?  Why do you think these above math is not conclusive?


    The results follow from the definitions.  This is done.


    So, moving onto a new space.
    You appear to want to also discuss an equivalence relation for these proto-vectors, and look at the properties of this new space: the set of the equivalence classes of these proto-vectors.



    Definition of your equivalence relation:
    (a,b) ~ (c,d) iff  b-a = d-c

    1) We can label the equivalence classes with a single real value (the space is one-dimensional):

    By construction:
    [z] = { (a,b) is an element in the space of Doug proto-vectors | b-a=z }
    Note that since a>=0, b>=0, z can have any real value.

    2) Addition in this space can easily be defined to be compatible with the proto-vector addition:
    With the definition
    (a,b) + (c,d) = (a+c, b+d) 
    then existence of a compatible addition can be shown by explicit construction
    [b-a] + [d-c] = [b - a + d - c] = [(b+d)-(a+c)] 
    thus a compatible definition of addition in this space is:
    [a] + [b] = [a+b]

    3) This space has additive identity and inverses
    existence of additive identity:
    From definitions, the additive identity must satisfy for all a:
    [a] + [b] = [a+b] = [a]

    there is a unique solution with b=0.

    existence of additive inverse:
    From definitions, the additive inverse [b] of [a] must satisfy:
    [a] + [b] = [a+b] = [0]

    there is a unique solution with b=-a.

    4) Multiplication in this space can easily be defined to be compatible with the proto-vector multiplication:
    With the definition
    (a,b) * (c,d) = (ad+bc, ac+bd)
    then existence of a compatible multiplication can be shown by explicit construction
    [b-a] * [d-c] = [(b-a)*(d-c)] = [(ac+bd) - (ad+bc)] 

    thus a compatible definition of multiplication in this space is:
    [a] * [b] = [ab]



    5) This space has multiplicative identity and inverses for all but [0]

    From the definitions, the multiplicative identity must satisfy for all a:

    [a]*[b] = [ab] = [a]
    there is a unique solution with b=1.


    existence of multiplicative inverse:
    From definitions, the additive inverse [b] of [a] must satisfy:

    [a] * [b] = [ba] = [1]
    if a=0, there is no solution for b
    if a!=0, there is a unique solution with b=1/a

    6) This space is isomorphic to the real numbers
    The elements are already labelled above with real numbers, and the addition and multiplication operators are compatible.

    -------------------------------------------------------------------------------


    So you have two different things you are considering.
    SpacePV = The space of Doug "proto-vectors", which as defined above

    SpaceECPV = The space of equivalence classes of Doug proto-vectors for a given equivalence relation, which was defined above

    Summary:
    SpacePV -- a two dimensional space with binary operations "+" and "*" defined, but there are no additive or multiplicative inverses for any elements except some special cases.

    SpaceECPV -- a one dimensional space with binary operations "+" and "*" defined.  It is isomorphic to the reals.


    Because these are two different spaces, which are not isomorphic, you cannot freely swap between the two.  That is your error.  You start in SpacePV, but want an additive inverse, so you jump to SpaceECPV and conclude there is one in this space, but you want the space to be 2D, and so you jump back to SpacePV and claim you have a 2D space with additive inverses.



    My summary of the situation:

    Note that to do all this, one is using addition, additive inverses, multiplication, and multiplicative inverses of the reals to define things.  Doug, you didn't somehow derive negative numbers this way.  Littlewood does show how you can fairly generically take something with only positive numbers and extend it to the negatives though.  THAT is the interesting part of what Littlewood did.  And sure enough, you showed his idea is generic enough to work for reals instead of integers.  Good for you, enjoy that.  However you seem to have twisted this in your head to concluded you gained something with these objects, which to do so is completely forgetting what just happened.  You were trying to come up with a different "number-like" object that you could play with.  You came up with such an object (your "proto-vectors").  But they didn't have many of the properties of numbers you wanted.  So you used Littlewood's process ... but just got back to the reals.  So your "new" numbers are ... the reals.

    You failed to fix the problems of your "new numbers" and still have something distinct.  You just got back to the reals.  So you have two choices:
    1] Use elements of SpacePV, and put up with their deficiencies
    or
    2] Use elements of SpaceECPV, knowing full well that this is completely equivalent to using the reals (you cannot claim these objects gain you any ability over using the reals)

    In short:
    You can't have it both ways.





    HenryB
    To put it lightly, I am not a fan of the comment-form on this site.  It is very painful to enter equations (unlike say physicsforums.com and other sites), using the equation web-app often also pastes equations randomly instead of where the cursor is, and even simple things like copy-paste can place things randomly.  But one of the most annoying "features", due to how confusing it is to me that it can't get this correct, is the weird adding of line breaks all over the place.  I can even press edit, delete a line again, save it, and the extra lines are back again.  Sometimes I'll get anal and go into the "raw" mode and try to fix up all the unclosed tags or random spans it places in there.  This time I don't feel like wasting anymore time.


    Hank,
    Can you please look into getting a better web-app editor thing?
    If you need me to send official bug reports I can do that.
    Hank
    I use the LaTeX editor in the toolbar all of the time, you have the same toolbar I do:



    I don't get weird line breaks and I use Opera, certainly the least known of the browsers in mass use.

    Anyway, what would you suggest for a better editor?  We're the only independent science site of any size, so no corporate parent or taxpayer money, so we use open source tools because those are free.  Has someone made a better one that is free and can be incorporated into a site like this?


    HenryB
    Thanks for the reply.

    If I'm the only one having trouble with the software, it must be something with my setup.  That alone is useful information, because instead of waiting for improvements, I can just try other setups that may improve the experience.  (I currently use Chrome, which I've really liked, until now I guess.)

    To other users:
    Am I really the only one that has the line break issue?  It happens to me all the time, and if I switch into "plain editor" I can see that the software is sometimes inserting empty or unclosed spans, sometimes using <br> and sometimes using <br />.  Deleting any unclosed spans and removing some <br> by hand and putting in some <br /> by hand usually fixes the problem.  I don't understand how this can be a browser dependent issue.  This sounds like an actually bug.
    Testing to see what triggers the bug
    Note: cursor was HERE when I tried to insert the Latex.

    Edit: underline
    bold
       asdf
          there
             should be
                 no extra
                     line breaks                      from here to edit
                               test 
                                    ... test

    Pressing enter after "breaks" above will move "from here to edit" to the next line in the rich text editor, but switching to the plain text makes it disappear.  No matter how many times I try.  The only resort is to fixing it in plain-editor, and submitting.


    HenryB
    Now trying in IE9

    This is a test, lines breaks after the X
    alsdfj lasjdflkj lsjkd flja sdflkjasld as df   X
    asd fl jlaksjd fasdfsa dflaskjdfasdf  asdf asd f  X
    asldkfjlasdkjf ladskjf ladskjf aldskjf alsdjkfa sdf   X
    asdfasdf   X
      X
    asdf X
    asdf    X
    asdfasdfasdasdfasdfasdfasdf    X
    asdfasdf X
    end X test

    inserting latex HERE

    asdf

    Wait, what? Switching to plain-text and back made all the spaces disappear (the X's above should line up)
    insert latex two

    underline, bold dd m s,
    asdf
    asdf
    aasdf
    staggerred lines
     line 1
    line 2
    line 3
        line 4
       line 5
    line 6
        line 7
    line 8
       line 9
    done (now two enters)

    df


    asdf
    asdf
    EDIT: Hmm... staggerred lines were supposed to look like a staircase, after submit most spaces were deleted somehow.  Strange. IE9 is much better with line breaks and Latex positioning, but not with spaces.
    Also in plain-text editor, I now see no <br> tags, only <br /> tags. Weird. So the code is quite browser dependent.
    HenryB
    arg. I see typos, but the edit button is gone.
    Only big one I've noticed so far:
    [eq2]     ac+db = 1


    knowing me, there are likely others. Point them out if it could cause problems.
    Henry wrote:
    --------
    Consider the following...
    If after doing some work you are able to prove:
    f(a) = f(b)
    Does this mean you proved:
    a=b
    Well?
    -------------

    Let me make that even more explicit to make sure I understand the connection.
    And being more explicit may also be helpful for Doug.

    Let's consider the function f which maps a "proto-vector" to a real number. I'll define it to match the equivalence class labeling Henry used:
    define f( (a,b) ) = b-a

    Now, does the existence of a solution to:
    f( (a,b) ) + f( (c,d) ) = 0
    Mean you have proven there is an additive inverse
    (a,b) + (c,d) = (0,0)
    ?

    No, of course it does not.

    Henry,
    I like it when you come by and summarize.
    So know that it is appreciated by at least one student.
    Thanks!

    David,
    Is there another way to discuss the "amount" of an infinite set other than what Henry suggested with measures?
    To say that the faction of proto-vectors that have multiplicative inverses is 0 feels like it needs some limit since there are elements that have multiplicative inverses, but I don't see where the limit comes in.

    Halliday
    CuriousReader:

    Measures are the most general.  Of course, this is because of how general measures are.  The truth of the matter is that even finite cases, or simple things like integrals all actually use measures, it's just that the measures seem so natural that we don't even notice that that's what we have.  It's once we try to handle things like Dirac's Delta "function" (actually, a distribution or "generalized function"), or a "function" over the real numbers that yields a value of zero (0) for any rational number argument, but a value of one (1) for all other real number arguments, where we find we actually need to take a closer look and define what we mean by a measure.

    In a post, above, I used a "pick a random value" sort of approach.  Of course, what one means by picking such a "random" value actually does need some clarification:  In the above case, one may use a "uniform" probability distribution (though it can be difficult to properly define a "uniform" probability distribution on infinite sets); however, one may even use many non-uniform probability distributions that will still yield the same result (as long as the probability distribution does not "pathologically" select for the special cases).

    In a rather realistic way, the nature of the probability distribution for such "random" picking of elements is a form of measure.  Explicitly, the measure is the probability of finding a point within any chosen region (with a probability of one for the region being the entire region of interest, such as the region of a two dimensional real plane spanned by Doug's "proto-vectors").

    However, measures need not be so normalizable—they can yield positive infinity for the measure of the entire region of interest.  So measures are more general still.

    David

    The Stand-Up Physicist
    My hope is that none of you, CuriousReader, LagrangiansForBreakfast, or Henry, have jobs in finance.  Debits and credits are different, and those that are cavalier about the difference recreate finance scandals.


    Henry did a proof showing there are no additive inverse for a space he called SpacePV.  Note this was his invention, not mine.  I made a different statement about what an additive inverse meant than he did (using === to try and convey the point, but failing).  There is nothing wrong with the proof there is no additive inverse for SpacePV, and only limited multiplicative inverses per se other than it is not relevant to the discussion at hand.

    What I claimed was the additive inverse of:

    v = a{-1} + b{+1}

    is

    -v = b{-1} + a{+1}

    This concrete claim was ignored.  Why?  Does it fit with the starting thesis that there are "gross logical fallacies"?  My formula is specific to the proto-vector space defined in the article.  Is there any way to justify that it is an additive inverse?  Let a=50 and b=52, with {-1} representing debits, while {+1} is for credits.  According to my proposal for the additive inverse:

    v = 50 {-1} + 52{+1}

    -v = 52 {-1} + 50{+1}

    v + (-v) = 102{-1} + 102{+1}

    This certainly is not the zero proto-vector, which is:

    0 = 0{-1} + 0{+1}

    They do "look different" if one were to graph them.  So they are distinct points in the proto-vector space.  And we all know what that relationship is, do we not?  In the banking world, what we would say is that they both have the same balance of zero.  It is the statement about the balances that is the proof that my proposed definition of an additive inverse is sensible.

    The balances are one dimensional, based on a calculation from the 2D proto-vector space.  In the banking world, one does not keep as their gold standard the credit value, the debit value, and the balance.  What would happen if there was a power outage as the balance was being calculated or written so that the recorded balance was different from the debits minus the credits?  That would be a bad situation.  What they must be meticulous about are the recorded debits and credits, calculating the balances after that on the fly.

    The balance does have every single property found for the real numbers.  There are positive numbers, there are negative numbers, the inverses are as to be expected.  Thanks for pointing that out - except that was in the bloody title of the blog.  Where did I claim to make an utterly new number?  I thought I was hoping to find a new perspective on a powerful tool.

    Note also that debits and credits have the properties of positive real numbers.  I may owe $1000 to the bank, but not I have a credit of -$1000.

    You did get close to the point of the article, but missed the mark which essential was forced after your opening paragraph.  I think the reason lies in a question I didn't have the energy to answer from CuriousReader:
    What is 1.5 * 3.14?
    That is the trivial group over the positive real numbers.  That provide no information about what to do when one multiplies a positive number times a negative, nor a negative times a negative.  That is what the group Z2 does.  That assumption about how multiplication should work comes from that group.  Since you failed to mention the group, I failed to communicate the issue.  So it goes.
    "My hope is that none of you, CuriousReader, LagrangiansForBreakfast, or Henry, have jobs in finance. Debits and credits are different, and those that are cavalier about the difference recreate finance scandals."

    I hate your analogies, because you seem to imply you can ignore the concrete math because of a hand-wave. Look, we all agree:
    1] SpacePV is not the same as SpaceECPV
    2] we can calculate which equivalence class a particular proto-vector belongs to (as is clear from Henry's statements how the equivalence class is defined)

    All you keep doing is keep stating these things in different ways, and then ignore the conclusions.
    You even note when comparing (0,0) and (102,102) that "They do look different if one were to graph them. So they are distinct points in the proto-vector space." But then you conclude this doesn't matter because they have the same balances. That is your fallacy, exactly like Henry and Curious reader suggested. You are claiming that just because:
    f(a) + f(b) = 0
    you can conlude that b is the additive inverse of a, and are therefore claiming
    a + b = 0

    That is wrong. It is an obvious (gross) logical fallicy. You even admit (102,102) and (0,0) are distinct points in your proto-vector space. That is the end. Stop. To then claim their "balances" are the same is now looking at equality of their equivalence classes in SpaceECPV. Or alternatively, if you do not wish to invoke an entire other space, at the very least you are only saying there is a function that takes a protovector and gives you a real number the "balance", and for this function you have f( (102,102) ) = f( (0,0) ). Does that change ANYTHING about the existence of additive inverses in SpacePV? No. That is already set from the definition of the space.

    "There is nothing wrong with the proof there is no additive inverse for SpacePV, and only limited multiplicative inverses per se other than it is not relevant to the discussion at hand."

    It is incredibly relevant.
    The only difference I see with his space is that he didn't define multiplying by a scalar, as that was not relevant to the discussion. Adding that in, and you have your Proto-Vectors. This doesn't affect his derivations, so the results hold.

    Are you claiming the space of Proto-Vectors you are discussing is different than the one he defined?
    If so, be specific like he is doing, and define the space YOU are talking about.

    Instead, you continue to mix and match and say:
    "I made a different statement about what an additive inverse meant than he did (using === to try and convey the point, but failing)."

    EXACTLY. Your statement is about the equivalence classes, which are isomorphic to the reals. When you invoke the equivalence classes to claim there is an additive inverse, you are making a statement about SpaceECPV, not a statement about SpacePV.

    What are you not getting here?
    If you feel we are making conclusions about a space other than the one you are working with, then give the math specifying how your space is different.

    "Where did I claim to make an utterly new number?"

    What do you call the proto-vectors? You were tying to come up with a new 8 dimensional number system that had some relation to quaternions. You focused on the 2D case first and claim you made your proto-vectors that can add and multiply and divide.

    The 2D proto-vectors are not isomorphic to the reals. We agree on that.
    So if they really can add, subtract (add along with existence of additive inverses), multiply, and divide (multiply along with existence of multiplicative inverses), then I would consider them a new number system.

    " I think the reason lies in a question I didn't have the energy to answer from CuriousReader:
    What is 1.5 * 3.14?
    That is the trivial group over the positive real numbers."

    You completely missed the point. Consider the set of real numbers. Or even just the set of positive reals if you want to restrict to that. Multiplication is a binary operation, taking two reals and mapping it to a third. Does "the trivial group over the positive real numbers" define multiplication for you there? No. It can't tell you which number 1.5 * 3.14 maps to. It doesn't define the multiplication operation.

    This comes back to a common issue Henry pointed out. You seem to confuse mathematical structures with what they can be represented by. The group Z_2 can be represented by the set of real numbers {-1,+1}, if we also state the group operation on this set is represented by multiplcation in the reals. Instead, you just seem to think Z_2 is the set {-1,+1} and that therefore the mathematical structure, the group Z_2, is what then define multiplication of those elements. That is nonsense. Why not claim the group operation of Z_2 defines addition of those elements? You don't claim that, because you already know what multiplication and addition is for the integers.

    Do you understand the objection now?
    None of your discussion of groups can tell me what 1.5*3.14 is. You are NOT deriving details of multiplication of the reals.

    The Stand-Up Physicist
    What do you call the proto-vectors? 
    Proto-vectors.


    The trivial group provides the rule that one positive number times another positive number generates a third positive number.  Your quote deleted the critical part, that Z2 is the reason a positive number times a negative is negative, or a negative times a negative is positive.  Positive times positive being positive is the trivial group.  This is all about the signs.  The same sign issue happens for the integer multiplication.
    "What do you call the proto-vectors?
    Proto-vectors."

    Then stop claiming they have additive inverses and can divide. Stop trying to claim these give you a 2-D version of the real numbers.

    "Your quote deleted the critical part, that Z2 is the reason a positive number times a negative is negative, or a negative times a negative is positive."

    Reread what CuriousReader said. You are making the same mistake still. I don't know how to make it any more clear that CuriousReader as he really distilled the issue.

    Please reread carefully.
    The properties of integers allow the set {+1,-1} to be a representation of Z2 if integer multiplication is used to represent the group operation. You cannot invert this and try to claim Z2 enforces some structure on the integers. Z2 exists independently of the integers. If one wishes to write down a representation, there is no requirement that one uses integers.

    As Henry pointed out several articles ago, this type of error is a common logical mistake you commit.
    Please reread and think about CR's statements.

    Halliday
    Doug:

    Do you not recognize that as soon as you begin talking "balances" you are talking "equivalence classes"?

    That's what your "detractors" are trying to get you to recognize:  When you are talking your "proto-vectors" (sans equivalent classes), vs. when you have imposed the equivalence classes and no longer have your "proto-vectors" because you are now working with your equivalence classes.

    If I have some time this weekend I'll try to create a 2D image that I think brings things together:  An image of your 2D "proto-vector" space (mapped to R2), with some representative equivalence classes pointed out.

    David

    Halliday
    OK.  Here's the promised image:

    Doug's 2D "proto-vector" space

    (Please let me know if any of you cannot see this SVG image.  I can post a PNG version.)

    Doug's 2D "proto-vector" space occupies the entire upper right quadrant of the infinite 2D plane, R2, so the space extends without limit to the right and up from this image.  You'll notice that the equivalence classes form diagonal bands through Doug's 2D "proto-vector" space, and the real numbers identified with each equivalence class correspond with the location along the intersection with the horizontal axis (extending the negative equivalence classes as necessary).

    David

    The Stand-Up Physicist
    Nice.  I had no problem seeing the graphic using Chrome on a Mac.
    Wow!
    Henry's overview of the Doug Proto-Vectors is much easier to follow than the actual article. That information should definitely be folded into the article to fix it up, as currently the article's conclusions are wrong.

    Doug,
    When you say something like
    (a,a) === (0,0)
    You are saying, for the given equivalence relation
    EquivalenceClassOf( (a,a) ) = EquivalenceClassOf( (0,0) )

    So while a statement with an "===" shows your 1D space (equivalent to the reals), have an additive inverse, it does not change anything about the equalities in your 2D space and the non-existence of additive inverses.

    Maybe the confusion is coming from wrapping our heads around the idea of equivalence relations and equivalence classes. How about we consider a much simplier case for us to discuss, for example a -finite- set, an equivalence relation on this set, and its equivalence classes.

    Consider the set A: {0,1,2,3,4,5,6,7,8,9,17,91}
    We can define an equivalence relation:
    for elements a,b in A: a ~ b iff ((a is even and b is even) or (a is odd and b is odd))
    One can check that this satisfies the definition of an equivalence relation.

    The equivalence classes are
    Class_even: {0,2,4,6,8}
    Class_odd: {1,3,5,7,9,17,91}

    So using your notation, and the defined equivalence relation, I can say
    0 === 2
    What does this mean? It means
    EquivalenceClassOf( 0 ) = EquivalenceClassOf( 2 )
    Both 0 and 2 are elements of Class_even.

    Notice that 0 is not equal to the equivalence class Class_even, nor is 2 equal to Class_even.
    This may seem obvious, but it must be said:
    Set of equivalence classes = {Class_even, Class_odd} != Set A

    Just like
    SpacePV != SpaceECPV

    Do you understand better now?
    It doesn't matter if SpaceECPV has an additive inverse. This doesn't change the additive properties of SpaceEC.

    The Stand-Up Physicist
    A friend sent me a 24 page paper, "When is one thing equal to some other thing?" by Barry Mazur.  It is a light intro to category theory, the shift from proofs to mapping out relationships.  So let me describe what I tried but failed to do.  The real numbers are essential to doing mathematical physics.  I will pluck out five members of the real numbers:
    (-2, -1, 0, +1, +2)

    What I wondered about was how to construct the real numbers from smaller parts.  That would be:

    (0, +1, +2) - the positive reals
    {-1, +1}  - the finite group Z2

    There should be a nice way to do this it seems to me.  I would not be surprised if it was the kind of thing one would get assigned for grad students in a class on category theory ("...and this is the functor for an additive inverse...").

    I am not wandering around worried about "SpacePV != SpaceECPV".  I puzzle about the relationships between -1, 0, and +1.  That is all set with the real numbers, there is nothing to ponder, it is a given.  The flame out of this blog is mildly ironic.  A blog or two  ago I tried to argue that time was a real number and not a vector, but a scalar.  People said that real numbers were vectors, they can be positive or negative.  So I took those two directions seriously.  So it goes.
    "A blog or two ago I tried to argue that time was a real number and not a vector, but a scalar. People said that real numbers were vectors, they can be positive or negative. So I took those two directions seriously. So it goes."

    After all this time, it appears you still did not take the time to learn what the issues were, nor what your mis-steps are.

    Listen, if you learn only one "factoid" from all this, I hope it would be:
    Understand that _EVERY_ basis vector in a four-vector has a direction in the 4D spacetime. If the +/- helped you see part of this, great. But you seem to think that is ALL that gives it direction. You are still very much missing the point. Consider the time basis vector for one inertial coordinate system. Think about another time basis vector for a relatively moving inertial coordinate system. The time basis vectors of these two different coordinate systems need not point in the same direction. It is more than just an issue of plus and minus. It really is a vector. This shouldn't be a surprise since it is right there in the name basis vector.

    This error again seem to have the root in the issue of you confusing a part of an object for a representation. Yes, in the four-tuple representation, each "component" is represented by a real number. This representation doesn't make the basis vectors explicit. David encouraged you to make them explicit, but you still considering a time vector as just a real number. Even after all this time, it is not clear you even understand what a four-vector is. There are some really basic misunderstanding you have that are causing gigantic logical disconnects that you refuse to confront for some reason.

    "I am not wandering around worried about "SpacePV != SpaceECPV"."

    You should be, because your lack of caring about details is what is letting you continue to carry around wrong ideas, and wrong conclusions. Why do you even investigate an idea if you are going to just draw the conclusion you want instead of the conclusion the math gives?

    For instance, you conclude your article:
    "I am enjoying thinking of the real numbers as a 2D proto-vector space"

    Do you still not understand your mistake here?
    There was a lot of confusion on mingling ideas, but eventually it was sorted out, and then Henry even made is starkly clear. There is no excuse to still avoiding these issues and deciding you'll go on concluding whatever and not worry about a math fact that was clearly shown to you.

    Halliday
    Doug:

    I feel for you.

    It looks to me like you and your "detractors" are talking past one-another, in at least one way.

    It looks like you may be one of those individuals that can keep mutually exclusive (or, at least, incompatible) ideas within your mind at once.  (I can do the same.)

    However, I suspect most people don't understand such an ability.

    Unfortunately, to the extent my suspicion is correct, you need to recognize when you are doing this, and "sort it out" when communicating to others (even to others than have a similar ability, since, otherwise, things will "blow up" combinatorially).

    David

    The Stand-Up Physicist
    This certainly is what is going on here.  Physics is a training in both approaches.  Proposed answers are wrong (like my unified field theory GEM not conserving angular momentum).  Particles are waves (have continuity over a region of spacetime) and are discrete (have sharp boundaries in spacetime).

    Is there any chance I am fumbling around with category theory?  Take the hunt for the infamous additive inverse of a{-1}+b{+1}.  It's a unicorn, and people have done the proof that it does not exist.  I didn't see anything wrong in the proof, so the team playing the algebraic hand wins.  That is the what equality in math is all about.

    I abandon the thread to the other team since I do not think they went ahead a read an article I recommended, "When is one thing equal to some other thing?" by Barry Mazur.  I get why they (probably) didn't bother, they had an algebraic proof, that is enough.  I have enough experience in this group to know I cannot discuss the content of that paper with the kind of precision to not trip on my shoelaces.  Let me quote two paragraphs that I think are quite relevant to the unicorn hunt:

    <cut and pasted from the PDF>

    8 Equality versus isomorphism

    The major concept that replaces equality in the context of categories is isomorphism. An isomorphism f : A → B between two objects A, B of the category C is a morphism in the category C that can be “undone,” in the sense that there is another morphism g : B → A playing the role of the inverse of f ; that is, the composition gf : A → A is the identity morphism 1A and the composition f g : B → B is the identity morphism 1B . The essential lesson taught by the categorical viewpoint is that it is usually either quixotic, or irrelevant, to ask if a certain object X in a category C is equal to an object Y . The query that is usually pertinent is to ask for a specific isomorphism from X to Y .

    Note the insistence, though, on a specific isomorphism; although it may be useful to be merely assured of the existence of isomorphisms between X and Y , we are often in a much better position if we can pinpoint a specific isomorphism f : X → Y characterized by an explicitly formulated property, or list of properties.

    <end quote>
    In the main blog, I did go through a list of eight properties of the 2D proto-vectors.  I did not provide a rule from going from the real numbers to the 2D proto-vectors.  Here is one specific isomorphism:

    if R>0, then 0{-1} + R{+1}
    if R<0, then R{-1}+0{+1}

    That is not the only specific isomorphism one could write.  An alternative that works is the +3 isomorphism:

    if R>0, then 3{-1} + (R+3){+1}
    if R<0, then (R+3){-1}+3{+1}



    Isn't that kind of wiggle room bad?  To quote the paper again:<begin quote>
    A uniquely specified isomorphism from some object X to an object Y characterized by a list of explicitly formulated properties—this list being sometimes, the truth be told, only implicitly understood—is usually dubbed a “canonical isomorphism.” The “canonicality” here depends, of course, on the list. It is this brand of equivalence, then, that in category theory replaces equality: we wish to determine objects, as people say, “up to canonical isomorphism.”

    <end quote>

    What I did was to provide a uniquely specified isomorphism for the additive inverse of the 2D proto-vector to real number additive inverse:

    The additive inverse of the 2D proto-vector a{-1}+b{+1} is b{-1}+a{+1}, while the additive inverse for the real number a is -a, "up to canonical isomorphism."

    It is irrelevant to ask if the 2D vector space is equal to the real numbers.  One can construct a specific isomorphism up to canonical isomorphism.  That is good enough for me.

    Halliday
    Doug:

    Are the typographical errors in your quotes from the article your typos, or are they also found in the referenced article? [I see that the answer is that most, if not all, typos were from your transcription.  ;)] 

    I'm not so sure you are using category theory correctly.  I haven't delved into category theory much, so I'm not certain, but I think you will find that the Real numbers, and your "proto-vector" space are not within the same category, for starters.  For another, I think "up to canonical isomorphism" doesn't apply well to the Real numbers (except as a triviality), though there may be a way for it to apply to your "proto-vector" space (though I'm not certain of that).

    You are correct in the generality that equal, same, equivalent, and isomorphic are not all the same thing, and one needs to be careful, and as explicit as reasonably possible and necessary when dealing with such matters.

    There most definitely is not an isomorphism ("a morphism ... that can be 'undone' ") between your "proto-vector" space (sans equivalence classes) and the Real numbers.  However, once you have imposed the equivalence classes (which is no longer your "proto-vector" space), there is a quite natural (even "canonical") isomorphism.

    When I was referring to "keep[ing] mutually exclusive (or, at least, incompatible) ideas within your mind at once", I was thinking in terms of the concepts of your "proto-vector" space (sans equivalence classes) vs. once you have imposed the equivalence classes (which is no longer your "proto-vector" space):  These two concepts are "mutually exclusive (or, at least, incompatible) ideas".

    David

    The Stand-Up Physicist
    Typos are all mine.  The quotes were edited, using cutting and pasting from the original PDF.  I forgot that was possible these days.

    I am quite certain I am not using category theory right (what is a functor anyway? "A mapping between categories,"  - could be related to this issue, but I will never know).  I read John Baez pontificate about category theory in many of This Week's Finds in Mathematical Physics.  Years of reading that stuff, and nothing stuck.  I know there are many people who are dismissive of the subject, and others like Baez that think it is a deep idea.  I found it interesting as I batted proto-vectors around in my own mind that I might need category theory to deal with "the awkwardness of equality" to quote the title of the first chapter of the paper.
    Should I ever find a user-friendly category theory forum, I will ask a question along these lines:

    Start with two copies of the set  and the group Z2. Discuss the relationship between this structure and the real numbers.  What is similar and different between the two?

    I would include your diagonal image, along with my dual arrows


    I want to put the arrowheads to work instead of them being ornamentation.
    Halliday
    Doug:

    The arrows at the ends of the line segments have meaning for both of your linear "dual arrow" diagrams.  In fact, my "diagonal image" could usefully use similar arrows at the top of the "y" and the right of the "x" axes.  (Instead, I pointed out the infinite/unbounded aspects in prose.)

    By the way, if one transforms my "diagonal image" such as to "skew" the diagram in order to increase the angle between the "axes", one should be able to find a "limit" that is like unto your "preferred" "dual arrow" representation (the one with the two arrowheads at -1 and +1).  The actual "limit" would no longer have a true 2D character, but such is doable for all angles just short of 180 degrees.

    Anyway, just a thought.

    David

    Halliday
    Doug:

    Here is a transformed* version of my diagram, above.

    The two black diagonals are all the points ("proto-vectors") that are proportional to the two Z2 basis vectors:  {1} (down and to the right), and {-1} (down and to the left).  They were the two axes on the original diagram.

    The Real axis, at the top of the figure, is set at equal angles to the two Z2 basis vectors, so it is perpendicular to the equivalence classes.

    Now, if we change the angle between the two Z2 basis vectors, as mapped into the two dimensional R2 plane, approaching 180 degrees, we get the following figure (actually at 179.9 degrees, so the basis vectors are still linearly independent even in the usual vector space sense):

    David

    *  The original diagram is, essentially, rotated 45 degrees counter-clockwise, and then reflected vertically (as if in a horizontal mirror).

    The Stand-Up Physicist
    A most excellent figure.  I don't want to reed too much into it ;-)  One thing I wonder about is having nearly flat spacetime, but making sure the nearly is not ignored.  One of the things we learn early is that gravity is a force that pushes things so a baseball travels in a parabolic arc.  In my upcoming software project, I will resist the urge to have a simple acceleration parameter.  Instead, I will do it the "right" way, having the basis vectors change depending on where they are in a gravitational field.  That is going to be tricky to implement in code, but I will be able to tell when it is done right because it will look right.

    The subtle difference in the arrowheads I was thinking of was the fact that I can get from zero to {+1} and from zero to {-1}, but I never will reach to positive or negative infinity.