1=0.999... infinities and box of chocolates..Phliosophy of Math...

Discussion in 'General Philosophy' started by Quantum Quack, Nov 2, 2013.

  1. Layman Totally Internally Reflected Valued Senior Member

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    I thought about it, and I think if you graphed it out, it wouldn't actually be able to become that line unless it was an infinite distance away from where the curve approaches that line. The reason being that it being a infinitely repeating number, you wouldn't be able to graph an infinite number of lines until it was infinitely far away from where you started. It would make me think that 0.999... = 1 at infinite distance down that line even if you assumed that 0.999... = 1.
     
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  3. Quantum Quack Life's a tease... Valued Senior Member

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    Totally true the base line of any graph would be infinitely long with an infinite number of data points on it...
    It is the sheer fact that that base line would be infinitely long and not terminate at zero that is very suggestive of this possible way of highlighting the non-terminating nature of infinity.
    Of course this doesn't alter the fact that mathematics deems 0.999...= 1. However your idea poses and interesting scenario IMO as it demonstrates as Pi does for example, an endless sequence of digits...
     
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  5. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    Absolutely correct. Division is the inverse operation of multiplication and multiplication of "M" by "n" is defined as adding M to its self n times. Algorithms have been developed to do these operations more quickly, especially when n is large integer. When n is not an integer we define the result of applying the valid n is an integer algorithm as the valid multiplication.
    For example you can not add 3 to itself 7.5 times, but the known valid algorithm works like this: 0.5x3=1.5 & 3x7=21 and 1.5 + 21 = 22.5 I. e. one can add 0.5 to itself 3 times.
    Or, alternatively you can say: I made use of a special case algorithm that applies only when you want to multiply (or divide) M by an n which is a power of the base being used. For example the notational change of value from (in base 10) 12.34 to 1234.0 is equivalent to adding 12.34 to itself 100 times. Note motion is always relative. So instead of moving the decimal point I could have moved the integers one (or two) at a time. I'll illustrate with moves of blocks of two, as less typing and plan to soon used blocks of the repeat length of a rational number's decimal equivalent.
    12.34 ==> 1200.34 ==> 1234. (where ==> means: "leads to") Or
    1/70 = 0.0142857,142857,142857,142857,.... and 1/7 = 0,142857,142857,142857,... and 100/7 = 14.2857,142857,142857,142857,142857,... etc.
    There is absolutely no reason why multiplying an infinite decimal by a power of the base can not be done this way, as I have just illustrated for three cases.

    Thus Someguy1's objection (in post 302) to my proof that 0.999... = 1 is invalid. My proof (in post 301) is just a special case of the general proof I gave, and
    the burden of proof is on those making the extraordinary claim that proof I gave fails in this one special case of the more general proof,
    even though it is valid in an infinite number of other cases that they agree to! I. e. they agree 1/3 = 0.3333... etc. but not that 9/9 = 0.999... even thought these are
    just two special cases of the general proof given in post 301.

    In closing, I note there is really only one fundamental operation in math: addition. Subtraction is the inverse of addition. The actual executions of the four operation, (+, -, x, & / ) on infinitely long strings of digits may be troublesome (or even impossible?) except when one wants to multiply or divide by a power of the notational system's base.
     
    Last edited by a moderator: Nov 26, 2013
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  7. Yazata Valued Senior Member

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    I was imagining decimal fractions in what I suppose is a constructivist manner, yes. I was imagining decimal points, followed by strings of 9's of various lengths, from one 9 to an infinite number of 9's.

    I don't understand the reference to long division.

    Or at least that's what school kids are told, when they are being taught fractions and decimals.

    The question in this thread seems to be whether there's an actual mathematical identity between 1/3 and a 0.3333... (a decimal point followed by an infinitely long string of 3's), or whether the infinitely long string of threes is a infinitly close approximation of 1/3.

    I can understand why school teachers don't get into that kind of discussion with their pupils (they probably don't understand it themselves in many cases), but I can also understand how a few of the brighter and more mathematically talented school kids might think of it themselves. (Students thinking critically and creatively about what they are taught is a GOOD thing.)

    I don't entirely understand that.

    Again, the question seems to me to be whether the decimal

    0.333333... (with an infinite number of 3's) is equal to 1/3, or whether it's an infinitely close approximation of 1/3.

    Pretty clearly it isn't equal to 1/3 if the string of 3's terminates at some finite length n, no matter how large n is. And pretty clearly making that terminating string longer by another 3, so that it's now (n+1) 3's long, won't make the string equal 1/3 either. That observation would seem to hold true indefinitely, by mathematical induction.

    I'm not too embarassed to write that, even if it's wrong, since Isaac Newton and Gottfried Leibniz seem to have had similar ideas. So it's probably not totally stupid.

    But today, and certainly here in this thread, it's being asserted that 0.3333... is literally identical with 1/3,

    I'm just speculating now, but it looks like it might have something to do with how infinity is conceived. If a finite string of 9's obviously doesn't equal one, but an infinite string supposedly does, then it seems to be the infinity that's responsible for closing the gap somehow. My constructivist idea of infinity, in which some constructive process (adding 3's or 9's in these examples) is continued forever with no termination, doesn't seem to provide the desired result. The gap will always remain, it will just get forever smaller. So perhaps infinity is being conceived in some philosophically different (mathematically realist?) way that (somehow) leads to different results?
     
  8. Tach Banned Banned

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    There is:

    \(0.(3)=\lim_{n \to \infty}(\frac{3}{10}+\frac{3}{10^2}+.....+\frac{3}{10^n})=\frac{3}{10}\lim_{n \to \infty}(1+\frac{1}{10}+\frac{1}{10^2}+.....+\frac{1}{10^{n-1}})=\frac{3}{10} \lim_{n \to \infty} \frac{1-\frac{1}{10^n}}{1-\frac{1}{10}}=\frac{3}{10} \frac{1}{1-\frac{1}{10}}=\frac{3}{9}=\frac{1}{3}\)

    Not asserted, proven. See above.
     
  9. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    23,198
    As Tach said and showed in post 325: "yes there is."

    However if you don't like limiting procedures, you can apply my general proof of post 301, the essences of which is:
    Note that 0.3333333 3333333 3333333 ... does have a repeat length of 7 as well any other positive integer, but as my general case results were illustrated (in green text) for repeat length of 7, I'll take the repeat length of your RD =0.333... as 7 to avoid need to make any changes in the illustration. For your case of interest a=b=c=d=e=f=g = 3 and immediately you see (from the more general green results):

    RD = 0.3333333,3333333,.... = 3,333,333 / 9,999,999 = 1/3 without any limiting procedures.

    My general procedure for finding the fraction equal to any rational decimals, as above, always gives the numerator of the results the same as the repeat sequence and the denominator with that number of 9s. E. g. when finding the rational fraction that has RD = 0.123123123123... you end up with RD = 123 / 999 but that can be reduced to 41/ 333, if you like.
     
    Last edited by a moderator: Nov 26, 2013
  10. rpenner Fully Wired Valued Senior Member

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    4,833
    Billy T, someguy1's objection seems rooted in pedantic formalism, for which a proof isn't a proof until traced to axioms. This somewhat misses the point.

    We know that \(\frac{1}{10} \sum_{k=0}^{n} \frac{a_k}{10^k} = \sum_{k=1}^{n+1} \frac{a_{k-1}}{10^k} \) and \(\sum_{k=0}^{n} \frac{a_k}{10^k} = a_0 + \sum_{k=1}^{n} \frac{a_k}{10^k} \) are true for any finite n from finite induction and the axioms of addition and multiplication of rational numbers but it does not follow that \(\frac{1}{10} \sum_{k=0}^{\infty} \frac{a_k}{10^k} = \sum_{k=1}^{\infty} \frac{a_{k-1}}{10^k} \) and \(\sum_{k=0}^{\infty} \frac{a_k}{10^k} = a_0 + \sum_{k=1}^{\infty} \frac{a_k}{10^k} \) follow in rational numbers without transfinite induction which requires some axiom relating to infinite sets of rationals. (Those axioms exist but as the the compleness axioms (or theorems) of real numbers.) In making this objection, someguy1 is pedantically correct.

    But I think someguy1 is wrong in that this thread should be an exercise of reasoning as to what our axioms should say.

    If we say \({a_0.a_1a_2a_3a_4\dots}_{\tiny 10} = \sum_{k=0}^{\infty} \frac{a_k}{10^k} = A\) is a mapping of bounded infinite sequences of digits into the [real] numbers in the most reasonable way possible, then the partitioning of the sum on the left and the decimal shift are justified by everyones intuitive concepts of what is reasonable and the actual properties of infinite sequences. The existence of the completeness axiom is our secret weapon to be sure, but even if our opponents have never heard of Cauchy-completeness or Dedekind-completeness, the strength of the claim (9.999... / 10 ) = 0.999... is in it's reasonableness. Reasonableness implies the intuitive notion that A is linear in \(a_n\) so 9.999... is the largest number you can create without filling in the next decimal place. 0.999... shares that distinction thus ( 9.999... / 10 ) = 0.999... = 9.999... − 9. And having argued the central algebraic claim from reasonableness, algebra is used to close the argument, for if (A / 10) = A − 9, it follows that A = 10.
    Against this argument, which I concede is not in formal mathematics, even if our justification for it is, we invite reasonable alternatives.

    The assumption 9.999... ≠ 10 invites the question what is the number 10 − 9.999... -- a question that no one has answered in the way that the number is either not zero or has the formal properties of any non-zero number. For example asserting (10 − 9.999...) = ( 1 − 0.999...) > 0 (which is just one possible proposal) requires that 10 × 0.999... = 9.999... − 9 × ( 1 − 0.999... ) and that rule is consistent with there being a "last digit" when by the assumption of the repeating decimal there can be no last digit.

    A second argument, even more basic, is that 0.999... = 1 in the reals and the reals are useful. Thus Prof. Gowers makes 0.999... = 1 an axiom of his system of equating the reals with infinite decimals. And countering this argument from utility requires a demonstration that the alternative is at least as useful.

    That this is not the level that those that assert 1 > 0.999... are using in their posts is not really our fault, I submit. We have not seen proof that 1 > 0.999... , we have not seen an argument from utility or reasonableness. We have not seen a non-useful or non-reasonable but concrete proposal on what the properties of ( 1 − 0.999...) as a number should be. If the last digit proposal is formalized, the value of infinite decimals is reduced to rational numbers and infinitesimals which are insufficient to make √2 or π a representable number but this formalism and its consequences has not been endorsed by the side that embraces 1 > 0.999...

    We are trying to meet them half-way and they aren't even looking in our direction.
     
  11. Yazata Valued Senior Member

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    5,701
    I don't believe that anyone has disputed that 1/3 is the limit of 0.3333... as the number of 3's goes to infinity.

    The question is whether 0.3333... ever precisely equals its limit, or whether all it can ever be is an infinitely close approximation of that limit.

    Tach's mathematical hieroglyphs include this move, over towards the right-hand side:

    \(lim({1} - 1/{10^n})= 1\) as n goes to infinity.

    Which is fine. But the question in this thread seems to be more long the lines of whether

    \({1} - 1/{10^n}\) ever exactly equals 1 at any point, when n has finally become sufficiently large. I don't see how it can. (I'm not claiming that it can't, I mean precisely what I say.)

    I can easily see why Newton and Leibniz resorted to dropping infinitesimals from their physical calculations. Since they are infinitely small they are of no conceivable import in real world applications.

    The more philosophical problem seems to me to arise in justifying disregarding the infinitesimal entirely and simply equating the value of the function with the value of a limit that, to my eye at least, it appears to be approaching aymptotically but never reaches.
     
  12. Motor Daddy Valued Senior Member

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    5,425
    \((\frac{1}{\2}=.5)\) and \((.5*2=1.0)\)
    \((\frac{1}{\4}=.25)\) and \((.25*4=1.0)\)
    \((\frac{1}{\5}=.2)\) and \((.2*5=1.0)\)

    So it appears to me, Tach, that according to your concept of the above, zero times infinity should equal one according to you.

    \((\frac{1}{\infty}=0)\) and \((0*\infty=1)\)

    Is that what you are saying, Tach, that since \(\frac{1}{\infty}=0\) then it stands to reason that \((0*\infty=1)\)???

    Dude, get your meds adjusted, you're not thinking straight!!
     
  13. Tach Banned Banned

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    The point that I made earlier to you is that 0.(9) IS the limit when n goes to infinity. The point that I just made to you is that 0.(3) IS the limit when n goes to infinity. 0.(p) is a shorthand for limits. So, you have some very serious misconceptions about calculus, they can be resolved by taking an intro class. BTW, "Quantum Quack" filled hundreds of posts with the same type of misconception.
     
  14. Tach Banned Banned

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    5,265
    Good, you pass 3-rd grade arithmetic.

    Nope, basic calculus says that \(0* \infty=undetermined\)


    Yep

    Nope, you fail. Again.

    Your basic error is that \(\frac{1}{\infty} =0\) does not imply \(0 * \infty=1\). Math is tough.
     
  15. Motor Daddy Valued Senior Member

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    So 1/2=.5 and .5*2=1, but the same concept of multiplying .5*2 and equaling 1 doesn't apply to \(\frac{1}{\infty} =0\) where \(0 * \infty=1\)? Why is that? Did you mistakenly say math is tough when you meant to say your math is BS?
     
  16. Motor Daddy Valued Senior Member

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    So then calculus must think \(undetermined* \infty=0\), right???
     
  17. Tach Banned Banned

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    Nope \(undetermined* \infty=undetermined\). Math is really tough. The fact that you'll never learn it should start sinking in.
     
  18. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    Thanks for your comments. You are much better versed in formal mathematics than I (a physicist) ever was but I do have a question:

    If rational decimal, RD = 0.ab.... where a&b are integers, perhaps equal as in 0.3333... or not as in RD =0.121212... is it less well established from axioms that 10RD =a.babab.... than that terminating decimal, TD =0.ab00000... has 10TD = a.b0000...? Or are they equally well built on the axioms?

    I only said I "multiplied" to more clearly explain my general procedure but as note earlier post, my moving of the decimal points I think is well defined and FUNDAMENTAL in the meaning of the notational system used.

    I.e. I believe if 0.abc = 0.def then a.bc = d.ef and both =10x0.abc or 10x 0.def just from what the base 10 notational system means. (First place to left of decimal point tells how many units of one, second to the left tells how many units of ten, etc. and etc. for the right side of the decimal point)

    Do you agree?
    Also do you agree that multiplication is only defined for multiplication by integers as multiplying M by n is defined as adding M to itself n times. The multiplying M by some non-integer is defined from an algorithm known to be valid for multiplying by an integer or IFF at least one of M & n is and integer with help of the commutative law (I think that is the name) which for ordinary math states: AxB = BxA.*

    E.g. 3x7.5 is not computed directly from the definitions of Mxn but makes use of this commutative law first to get 7.5 x 3 as you can add 7.5 up 3 times but not add 3 up with itself 7.5 times but 3.4 x7.5 can not be computed directly from the definition of multiplying. We can only assume (or define) the result via some algorithm known to be valid by it producing the same results as the products the definition can be applied to.

    I would claim there is an exception built into the notation system for multiplying by the base of the notation system but that would be useful mainly (only?) when the base is NOT an integer. - Possibly a never use based, but conceptually possible, I think.

    * Typically, or at least often, not true if both A & B are operators instead of numbers as anyone who has done some quantum calculations knows. If I recall correctly after nearly 60 years, the uncertainty principle exists ONLY because some pair of operators do NOT commute. Most people think it exist because one measurement effects the other but that is false and does not explain with it applies only to specific pairs of measurements. I.e. energy & when or momentum and position pairs can not be both precisely known, but position and energy can be, at least in principle.
     
    Last edited by a moderator: Nov 26, 2013
  19. Motor Daddy Valued Senior Member

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    Why are the rules different for infinity? Why does 1/2=.5, and .5*2=1.0, but when it comes to infinity you change the rules of being able to check my work? Why can I check my work successfully when it comes to other numbers, but when it comes to infinity you change the rules? How can 1/infinity equal 0? How do you make a whole disappear by dividing it into a greater quantity of parts? Don't you just get a greater quantity of smaller parts? How do you make it disappear without a trace (0)???
     
  20. Tach Banned Banned

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    It is a conspiration to make life tough for the ones like you.

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  21. Motor Daddy Valued Senior Member

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    Wrong answer. The rules are different because your math is BS, so in order to create the illusion of correctness, math makes up some BS rule that .999...=1.

    What makes life tough for people like me is people's math like yours.

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  22. Tach Banned Banned

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    Your posts remind me of a story: a guy escapes from a lunatic asylum , steals a car and enters a freeway driving against the traffic. He turns on the radio and a police report comes on:

    "Attention, there is an escapee from the lunatic asylum driving against the flow of traffic"

    The guy hits the radio and mutters "They are all lunatics, they are all driving against the traffic".
     
  23. Motor Daddy Valued Senior Member

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    I see the problem. The problem is that the lunatic was living in Einstein's world where Einstein had him believing that there is no truth, that everyone is entitled to their own BS.

    When the world changes to my world things will straighten out, but until then you all are the lunatics! LOL
     

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