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Topic: 10adic numbers (Read 12007 times) 

Icarus
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10adic numbers
« on: Sep 9^{th}, 2003, 5:09pm » 
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There are a couple of ways of defining the 10adics [smiley=bba.gif][sub1][sub0]. Conceptually, the easiest is to just "flip" traditional decimal notation: A 10adic number has a decimal representation that can continue indefinitely to the left, but must terminate to the right. For example, instead of the infamous 0.999..., in the 10adics you have ...999.0 = ...999. Addition, subtraction, multiplication, and division are performed exactly as they are for real numbers. Carries that "go on to infinity" are lost. E.g. ...999.3 + 1 = 0.3. For the most part, these operations obey the same rules as they do for real numbers. For instance, addition and multiplication are commutative and associative, and multiplication still distributes over addition. On the other hand, there is no ordering ("a < b" is generally undefined for 10adics). Thus, while irrational 10adics may appear to be "infinitely big", the concept is without meaning for them. The 10adics do not require a "unary minus": ...999 + 1 = 0, so 1 = ...999. Some challenges, in increasing difficulty: (1) For any 10adic number x, describe how to calculate x. (2) Find 1/7 as a 10adic number. More generally, for integer N, describe how to calculate 1/N. (3) Show that the 10adics possess zerodivisors (i.e. nonzero numbers whose product is zero). (4) Show that the 2adics (same construction, but with base 2) do not possess zerodivisors. (The 2adics form a field.) For what values of N do the Nadics possess zerodivisors?


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towr
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Re: 10adic numbers
« Reply #1 on: Sep 10^{th}, 2003, 12:59am » 
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it's a bit similar to two's compliment binary notation. An unsigned byte also has 00000001 = 11111111 so 123.45 would be ...999876.55 in analogy for every digit take the complement (9d), and add one to the last digit


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Icarus
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Re: 10adic numbers
« Reply #2 on: Sep 10^{th}, 2003, 7:05pm » 
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Yes  It is the 10's complement instead of the 2's compliment. Of course, if you look at the 2adics, negation is exactly the 2's compliment. So all this time you thought computers were doing approximate real calculations, instead they're doing approximate 2adic calculations!


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towr
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Re: 10adic numbers
« Reply #3 on: Sep 10^{th}, 2003, 11:21pm » 
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on Sep 10^{th}, 2003, 7:05pm, Icarus wrote:So all this time you thought computers were doing approximate real calculations, instead they're doing approximate 2adic calculations! 
 Actually I've been thinking of other things most of the time (And of course one thing doesn't exclude the other either. What computers approximate depends on how we interpret what they do, in reality they're just juggling packets of electrons in a way which happens to act like numbers) btw, I'm not sure what you mean with (2) do you mean the normal decimal representation of 1/7 flipped at the decimal point? looking at ...999.3 + 1 = 0.3. it seems that the 10addic ...999.3 doesnt' have the same value as the normal 3.999.. , unless + is something else entirely. Since 3.999... + 1 = 5, and flipping that would give 0.5, not 0.3


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James Fingas
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Re: 10adic numbers
« Reply #4 on: Sep 11^{th}, 2003, 6:16am » 
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10adic representation of 1/7 x * ...00007.0 = ...00001.0 = ...999^{1}01.0 doing reverse long division (heh), we get: x = 3.0 + ...999999980.0/7 x = 43.0 + ...999999700.0/7 x = 143.0 + ...999999000.0/7 x = 7143.0 + ...999950000.0/7 x = 57143.0 + ...999600000.0/7 x = 857143.0 + ...994000000.0/7 x = 2857143.0 + ...980000000.0/7 x = ...2857142857143.0 Odd. Very odd.


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Icarus
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Re: 10adic numbers
« Reply #5 on: Sep 12^{th}, 2003, 3:20pm » 
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Isn't it though? And yes, James is correct about what I meant. Not only the integers, but all rational numbers are part of the 10adics (and all other nadics)  and addition, subtraction, multiplication, and division are the same on them as their are in the Reals. (Thus the zerodivisors of part (3) will be irrational 10adics.) After the rest is solved, I'll give a follow up that may help explain the familiar pattern, and the deviation!


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James Fingas
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Re: 10adic numbers
« Reply #6 on: Sep 16^{th}, 2003, 10:26am » 
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Hmm ... I've been trying to generate a pair that multiplies to give zero, but as you point out, they both must be irrational. So I can only generate the first few numbers before the computations get too difficult. However, it seems like you can generate two numbers of length N so that the lowest N places of their product are all zeros. Here is an attempt to show why you can always do this: Consider trying to generate two numbers, A and B so that A*B = 0. We will do this in stages, generating the last digit of A and B, then the secondtolast digit, and so on. At each stage i, we will have two decimal numbers A_{i} and B_{i} so that A_{i}*B_{i} is a multiple of 10^{i}. To generate the (i+1)^{st} digit, perform the following operations: 1) Generate all possible combinations of {p,q}, where 0 [le] p < 10, 0 [le] q < 10. p will be the (i+1)^{st} digit for A, and q will be the (i+1)^{st} digit for B. There will be 100 such combinations. 2) Calculate the product C_{p,q} = (A_{i} + p*10^{i})(B_{i} + q*10^{i}) for each combination. 3) Choose a combination {p,q} so that C_{p,q} is a multiple of 10^{i+1}. For this method to work, you must choose the first digit in both A and B to be nonzero. It is possible for A and B to have any number of trailing zeros, but putting trailing zeros in keeps this method from working properly. This step indicates for which N Nadic zerodivisors can be found: composite N. Now all I have to show is that there is always a {p,q} so that C_{p,q} is a multiple of 10^{i+1}. This should not be hard to do: C_{p,q} = (A_{i} + p*10^{i})(B_{i} + q*10^{i}) = A_{i}B_{i} + qA_{i}*10^{i} + pB_{i}*10^{i} + pq*10^{2i} We will consider the (i+1)^{st} digit of this product. In general, the last term does not matter, since 2i [ge] i+1 for all i [ge] 1, and I have tried this method up to i=9 with no problem. The first term will have some arbitrary digit r in the (i+1)^{st} place (it could be any number from 0 to 9). The second and third terms allow us to add to this digit qA_{1} + pB_{1}. This method assumes that both of these are nonzero, and it can readily be shown that {A_{1},B_{1}} must be {2a,5} for some integer 0 < a < 5. It can readily be verified that for any choice of A, and any digit r, that we can pick a {p,q} that work. Here is a comprehensive listing to prove it, assuming that A_{1} = 2a and B_{1} = 5. a r 1 2 3 4 0 0,0 0,0 0,0 0,0 1 2,1 1,1 4,1 3,1 2 4,0 2,0 3,0 1,0 3 1,1 3,1 2,1 4,1 4 3,0 4,0 1,0 2,0 5 0,1 0,1 0,1 0,1 6 2,0 1,0 4,0 3,0 7 4,1 2,1 3,1 1,1 8 1,0 3,0 2,0 4,0 9 3,1 4,1 1,1 2,1 Note that when q can be chosen, q+2 can also be chosen, and when p can be chosen, p+5 can also be chosen. Therefore, there are an infinite number of zerodivisor pairs to find. Get going on it!


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Icarus
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Re: 10adic numbers
« Reply #7 on: Sep 16^{th}, 2003, 6:07pm » 
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That does it! A somewhat simpler statement of the reason why padics have no zerodivisors for prime p: The product of the least significant nonzero digits of A and B provides a digit (modulo p) in their product AB. For the product to be zero, this digit is zero. Therefore there must exist integers a, b with 0 < a, b < p, such that ab [equiv] 0 mod p. This is possible if and only if p is composite. Note that this also shows that unlike with the reals, different bases in the padics give different number systems (actually it only shows that primes and composites give differing number systems, but it is not difficult to show that all of them are different). Followup Now let [subn][bbr] be the set of all "bidirectional numbers" base n. That is, each element of [subn][bbr] has a decimal expression that may extend infinitely in either direction without terminating. Since all possible decimal expressions are included, both the reals and the nadics are subsets of [subn][bbr]. It is not possible to reasonably define multiplication between a decimal that extends infinitely to the left and one that extends infinitely to the right. But rules for adding and subtracting decimals are unphased by this bidirectionality. However, since the reals are a subset of [subn][bbr], we know that sometimes multiple decimal expressions represent the same number (i.e. ...0001.000... = ...000.999... ). To make this explicit: two bidirectional decimals u & v are equal if you can show u  v = 0. Are there nonreal examples of differing bidirectional decimal expressions which are equal? More generally, what conditions must hold for two bidirectional expressions to be equal? Since the exposition of bidirectional numbers given above is more than a little weak in spots, here is a more mathematical definition: Let [smiley=bba.gif][subn] be the nadics for some fixed integer n > 1. Consider the set [smiley=bba.gif][subn] [times] [bbr] = { (a, x)  a [in] [smiley=bba.gif][subn], x [in] [bbr] }. We can define a relation on this set by (a, x) ~ (b, y) [bigleftrightarrow] a  b = y  x. (Note that this requires that a  b and y  x are both terminating decimals.) It is easy to see that ~ is an equivalence relation: (i) Reflexive: (a, x) ~ (a, x) since a  a = 0 = x  x. (ii) Symmetric: (a, x) ~ (b, y) [bigleftrightarrow] a  b = y  x [bigleftrightarrow] b  a = x  y [bigleftrightarrow] (b, y) ~ (a, x). (iii) Transitive: If (a, x) ~ (b, y) and (b, y) ~ (c, z), then a  c = a  b + b  c = y  x + z  y = z  x. So (a, x) ~ (c, z). The equivalence classes of ~ are the sets [a, x] = {(b, y)  (b, y) ~ (a, x)}. Define the set [subn][bbr] = {[a, x]  (a, x) [in] [smiley=bba.gif][subn] [times] [bbr]}. Addition is defined in [subn][bbr] by [a, x] + [b, y] = [a+b, x+y]. It is easy to see that this definition provides the same sum regardless of which representatives (a, x) and (b, y) from each equivalence class is used. [subn][bbr] can be thought of as a superset of both [smiley=bba.gif][subn] and [bbr] by the mappings [smiley=bba.gif] [to] [subn][bbr] : a [mapsto] [a, 0] and [bbr] [to] [subn][bbr] : x [mapsto] [0, x], which are onetoone and preserve addition. From now on, I will suppress these maps and treat a and x as actual elements of [subn][bbr], with [a, x] = a + x. If r is any terminating decimal, then a + x = (ar) + (x+r). This can be used to choose a value for a with no fractional part, and a value for x with no nonfractional part (i.e., 1 > x [ge] 0). Combining them gives a "bidirectional" decimal expansion for a + x. [edited to correct the error Rezyk points out below.]

« Last Edit: Nov 10^{th}, 2003, 8:01pm by Icarus » 
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Barukh
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Re: 10adic numbers
« Reply #8 on: Sep 23^{rd}, 2003, 2:23am » 
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Icarus, thank you very much for this incredibly interesting thread! It is a field that I was totally ignorant of... It took me some time to understand even the simpliest things about padic numbers. For instance: on Sep 12^{th}, 2003, 3:20pm, Icarus wrote:Not only the integers, but all rational numbers are part of the 10adics (and all other nadics)  and addition, subtraction, multiplication, and division are the same on them as their are in the Reals. (Thus the zerodivisors of part (3) will be irrational 10adics.) 
 It took me time to understand that although padic integers have the same representation as normal integers, padic rational do not have the same representation as normal rationals! Here are some observations I've made about padic numbers. First, some definitions. padic integer is a padic number without fractional part. padic number is regular (terminating) if it has only finite number of nonzero digits. padic number is periodic if its integer part has a finite period. 1. An integer is a regular padic integer. 2. If q is relatively prime to p, then 1/q is a periodic padic integer. For instance, 1/3 = ...66667 as a 10adic number. 3. If p and q have a common divisor, and p/q is periodic (as a normal rational), then 1/q is a periodic padic number (not an integer). For instance, 1/6 = ...333.5 as a 10adic number. 4. If p and q have a common divisor, and p/q is regular, then 1/q is a regular padic number. For instance, 1/4 is 0.25 as a 10adic number. Although rationals are part of padic numbers, the reals are not. For instance in one of the sources, I found the following: The number e, defined as the sum of reciprocals of factorials, is not a member of any padic field. Then, what is the subset of irrationals that can be represented as a padic number for some p? For instance, is there a padic number q such that q^{2} = 2?


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Icarus
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Re: 10adic numbers
« Reply #9 on: Sep 23^{rd}, 2003, 6:33pm » 
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If so, then p >= 7. I don't remember off the top of my head what results were out there about which algebraic numbers could be considered "padic". (I only looked into them very lightly, as they were not involved in my area of study at the time when I first came across them.) However, I do recall this: the padic fields (p prime)* are not algebraically complete for any p. ("Algebraically complete" = "every polynomial has at least one root". The Complex numbers are the algebraic completion of the Real numbers.) If one takes the algebraic completion of the padics, it turns out to be an infinitedimensional extension (i.e.  instead of adding a single new number [smiley=i.gif] and discovering that the resulting field [bbc] is complete, as happens with the real numbers, you have to add an infinite number of "[smiley=i.gif]"s to the padics to get algebraic completeness.) Unfortunately matters are still not in hand, because in adding the infinite number of new dimensions, you also lose topological completeness. On the other hand, good fortune strikes when you take the topological completion: it is still algebraically complete. So the padic version of the complex numbers is obtained only after taking both the algebraic and then the topological completion. * The books I looked in dealt only with padics for prime p. I realized on my own that the construction also works for nonprime p, except that you get zerodivisors instead of a field. However I have since seen references to "nadics" elsewhere. What I have never seen are any reference to what I have called "bidirectional numbers" above. I doubt that I am the first to come up with them, though  it's too obvious an extension of the same idea. It is a curious property of these numbers that caused me to pose the "follow up" question.


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Barukh
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Re: 10adic numbers
« Reply #10 on: Sep 25^{th}, 2003, 12:13pm » 
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The following statement can be proved: For p a prime number, and q < p a natural number, [sqrt]q is a padic number, iff q^{(p1)/2} = 1 (mod p). (Putting it differently: q is a quadratic residue modulo p). As this condition holds for p = 7, q = 2, [sqrt]2 is indeed a 7adic number. I've written a program that calculates an arbitrary number of digits for it. Here are a few: [sqrt]2 = ...610450423211441550454 The following table shows for what small p there exists a padic number [sqrt]q: q p  2 7, 17 3 11, 13 5 11, 19 6 19 7 19 8 17 10 13 Still, I don't know what happens if p is composite. Modified to fix a typo.

« Last Edit: Sep 25^{th}, 2003, 11:46pm by Barukh » 
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Icarus
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Re: 10adic numbers
« Reply #11 on: Nov 10^{th}, 2003, 6:43pm » 
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Since this one seems to be dead in the water, I will toss out the particular result that I refered to, and see if anyone wants to prove it: Claim: in the bidirectional numbers [subn][bbr], any decimal expression consisting of a repetition of a sequence of digits repeated infinitely in both directions is equal to 0. For example, in [sub1][sub0][bbr], ...999.999... = 0 ...324132413.24132413241... = 0 etc. I believe (if I ever proved it, I've forgotten), that two decimal expressions are equal in [subn][bbr] if and only if subtracting one from the other according to the standard rules for subtracting decimal expressions leaves either 0, or one of the repeating patterns above that are equivalent to 0. For instance, ...000.999...  ...001.000... ...999.999... = 0 (I'm not going to post this proof in the 0.999... = 1 thread!) Note that this result makes it easy to calculate the 10adic version of any rational. Just find the base 10 decimal expression for the real version of the fraction, subtract it from the expression having the same pattern repeating indefinitely both directions, and then negate by taking the 10's compliment of the result. For example, in the reals, 22/7 = 3.142857142857... ...142857142857.142857...  3.142857... ...142857142854 The 10's complement (replace each digit with 9  the digit, then add 1 to the rightmost digit) is ...857142857146. Therefore 22/7 = ...857142857146 in the 10adics.

« Last Edit: Nov 10^{th}, 2003, 6:54pm by Icarus » 
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Rezyk
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Re: 10adic numbers
« Reply #12 on: Nov 10^{th}, 2003, 7:28pm » 
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I was formulating a response to this a while ago, but I'm not sure what happened to it.. Anyways, there were at least 2 things tripping me up: on Sep 16^{th}, 2003, 6:07pm, Icarus wrote:We can define a relation on this set by (a, x) ~ (b, y) [bigleftrightarrow] a  b = x  y. 
 Should that be "ab = yx" instead? Otherwise it seems like (4, .999...) ~ (3, 0), which is unintuitive. Quote:If r is any terminating decimal, then a + x = (ar) + (x+r). This can be used to choose a value for a with no fractional part, and a value for x with no nonfractional part (i.e., 1 > x [ge] 0). Combining them gives a "bidirectional" decimal expansion for a + x. 
 If this is how one generates all bidirectional decimal expansions, isn't ...999.999... not a valid expansion? (because you are choosing x=.999...=1)


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Icarus
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Re: 10adic numbers
« Reply #13 on: Nov 10^{th}, 2003, 8:22pm » 
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on Nov 10^{th}, 2003, 7:28pm, Rezyk wrote:Should that be "ab = yx" instead? 
 Yeah, I accidently reversed it. The idea is define an addition between elements of [smiley=bba.gif][subn] and [bbr], so that a + x [equiv] (a, x). The idea of the equivalence is to identify when a + x = b + y. By subtracting b and y from each side, we get the equivalent relation a  b = y  x. Only in this equation, both subtractions are already defined. So we define the equivalence relation by it. Thanks for pointing it out! I've corrected the post now. Quote:If this is how one generates all bidirectional decimal expansions, isn't ...999.999... not a valid expansion? (because you are choosing x=.999...=1) 
 No. both expressions for "1" are valid in the reals and in [sub1][sub0][bbr], and one will give ...999.999... . Besides, that statement is just one way of generating such repeating expressions. Any function d : [bbz] [to] {0,1,2,3,4,5,6,7,8,9} defines a bidirectional decimal expression, representing the number [sum]_{n=0}^{[supinfty]} d_{n}10^{n} + [sum]_{n=1}^{[supinfty]}d_{n}10^{n}. The first series converges in [smiley=bba.gif][sub1][sub0], the second converges in [bbr]. The purpose of my argument in the previous post was to show that such expressions exist for every element of [subn][bbr].


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ivancho
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Re: 10adic numbers
« Reply #14 on: Oct 11^{th}, 2004, 3:36pm » 
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[smiley=parallel.gif] Quote:Let be the nadics for some fixed integer n > 1. Consider the set = { (a, x)  a , x }. We can define a relation on this set by (a, x) ~ (b, y) a  b = y  x. (Note that this requires that a  b and y  x are both terminating decimals.) 
 Actually, those 2 differences don't have to be terminating decimals, as your own example of Quote:...324132413.24132413241... = 0 
 shows  you are saying that ...324132413  0 = 0  0.24132413241... and it's true, they are both equal to 2413 / ( 1  10000) and neither of them is a terminating decimal. effectively, what you are looking for, is the intersection of [smiley=bfcb.gif] [smiley=sub1.gif] [smiley=subset.gif] [smiley=bbq.gif]_{10} and [1,0] [smiley=subset.gif] [smiley=bbr.gif], where [smiley=bfcb.gif] [smiley=sub1.gif] = { a [smiley=in.gif] [smiley=bbq.gif]_{10}, [smiley=parallel.gif]a[smiley=parallel.gif]_{10} [smiley=le.gif] 1 } This intersection can only be considered as a subset of [smiley=bbq.gif]  anything else is nonsensical: all other numbers in [smiley=bbq.gif]_{10} and [smiley=bbr.gif] are defined as limits of Cauchy (wrt to some metric) sequences of rationals that have nothing in common  so it is absurd to have a nonrational real equal to a nonrational 10adic.. for example, we do have that a = ...60569 is a cube root of 9 in [smiley=bbq.gif]_{10}, but we cannot possibly say that a = 9^{1/3} = 2.0800838.... [smiley=in.gif] [smiley=bbr.gif].. when restricted to [smiley=bbq.gif], we are looking for rational numbers in [1,0], which are 10adic integers [smiley=bigto.gif] they must have the form (m*10^{k})/n, where gcd(m,n) = 1, gcd(10,n) = 1, n>0, k[smiley=ge.gif] 0 and m,n,k [smiley=in.gif] [smiley=bbz.gif]. Since gcd(10,n) = 1, [smiley=exists.gif] j s.t. n divides 10^{j}  1 [smiley=bigto.gif] our fraction is of the form N/(10^{j}  1), which is definitely repeating in real decimal expansion with period j. It is quite easy to see that the 10adic version of that is repeating the same number the other way  so your claim is true. What you're basically creating is the quotient vector space [smiley=bbq.gif]_{10}[smiley=largetimes.gif][smiley=bbr.gif] / {(q, q), q [smiley=in.gif] [smiley=bbq.gif]}. The innocent fact you claim: Quote:If r is any terminating decimal, then a + x = (ar) + (x+r). This can be used to choose a value for a with no fractional part, and a value for x with no nonfractional part (i.e., 1 > x 0). 
 , or translated, [smiley=bigvee.gif] a [smiley=in.gif] [smiley=bbq.gif]_{10}, x [smiley=in.gif] [smiley=bbr.gif], [smiley=exists.gif] r [smiley=in.gif] [smiley=bbq.gif], s.t. xr [smiley=in.gif] [0,1] and [smiley=parallel.gif]a + r[smiley=parallel.gif]_{10} [smiley=le.gif] 1, is a special case of the Weak Approximation Theorem. Asking what's in [0,0] is fine, I suppose  but the rest of that space is truly bizarre. To be honest, both [smiley=bbq.gif]_{10}/[smiley=bbq.gif] and [smiley=bbr.gif]/[smiley=bbq.gif] are weird enough as it is, there's hardly need to combine them


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Icarus
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Re: 10adic numbers
« Reply #15 on: Oct 11^{th}, 2004, 8:08pm » 
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You are right  at the time I was pushing for the concept of these numbers as a decimal expansion infinite in both directions, in order to make it more understandable to a broader audience. Because I was focusing so much on the decimal representation, I failed to note the obvious: for any set of nadics, the intersection with the reals is the rationals, so all that is required for the equivalence is that both differences be rational. Since few people here have much exposure to the concept of equivalence relations, I felt the need to try to explain things on a lower level, without bringing in mod notation. Regardless of other oddities in [bbq]_{[subn]} [oplus] [bbr], I find the idea that all decimal expressions consisting of a pattern repeated indefinitely in both directions should represent 0 to be intriguing, and in its own way, beautiful.

« Last Edit: Oct 11^{th}, 2004, 8:11pm by Icarus » 
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ivancho
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Re: 10adic numbers
« Reply #16 on: Oct 11^{th}, 2004, 10:11pm » 
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true, true... hmm... possible connection to entropy? I mean, the only sequences with zero surprise in them are the ones that are repeating a pattern in either direction... in fact, it's quite possible, a shift map on the product ring as multiplication by 10.. I wish I went to more of those lectures.. suddenly, university seems like centuries ago..

« Last Edit: Oct 11^{th}, 2004, 10:18pm by ivancho » 
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PiotrGrochowski
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Re: 10adic numbers
« Reply #17 on: Oct 7^{th}, 2014, 11:23am » 
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on Sep 9^{th}, 2003, 5:09pm, Icarus wrote: (1) For any 10adic number x, describe how to calculate x. (2) Find 1/7 as a 10adic number. More generally, for integer N, describe how to calculate 1/N. 
 Icarus, To find negative of 957236: 1000000957236=42764 957236=...99999042764 1/7 in 10adic is ...42857143


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PiotrGrochowski
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Re: 10adic numbers
« Reply #18 on: Oct 8^{th}, 2014, 10:26am » 
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Icarus says: Quote: ...999.999... = 0 ...324132413.24132413241... = 0 
 No, it's not allowed in 10adic base. Fractions can't extend infinitely to the right, although they can be extended infinitely to the left. 1/6 is NEVER 0.1666... BUT ...33333333.5. Also 10adic numbers are read right to left. For example: 20736 is sixone three thousand sevenone zero two


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PiotrGrochowski
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Re: 10adic numbers
« Reply #19 on: Oct 8^{th}, 2014, 10:31am » 
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on Sep 9^{th}, 2003, 5:09pm, Icarus wrote:More generally, for integer N, describe how to calculate 1/N. 
 1001/7=143, 1000000001/7=142857143, etc... 1/7=...142857142857142857142857142857142857142857142857143


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