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Title: 10-adic numbers Post by Icarus on Sep 9th, 2003, 5:09pm There are a couple of ways of defining the 10-adics [smiley=bba.gif][sub1][sub0]. Conceptually, the easiest is to just "flip" traditional decimal notation: A 10-adic 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 10-adics 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 10-adics). Thus, while irrational 10-adics may appear to be "infinitely big", the concept is without meaning for them. The 10-adics do not require a "unary minus": ...999 + 1 = 0, so -1 = ...999. Some challenges, in increasing difficulty: (1) For any 10-adic number x, describe how to calculate -x. (2) Find 1/7 as a 10-adic number. More generally, for integer N, describe how to calculate 1/N. (3) Show that the 10-adics possess zero-divisors (i.e. non-zero numbers whose product is zero). (4) Show that the 2-adics (same construction, but with base 2) do not possess zero-divisors. (The 2-adics form a field.) For what values of N do the N-adics possess zero-divisors? |
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Title: Re: 10-adic numbers Post by towr on Sep 10th, 2003, 12:59am it's a bit similar to two's compliment binary notation. An unsigned byte also has -00000001 = 11111111 so [hide] -123.45 would be ...999876.55 in analogy for every digit take the complement (9-d), and add one to the last digit[/hide] |
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Title: Re: 10-adic numbers Post by Icarus on Sep 10th, 2003, 7:05pm Yes - It is the 10's complement instead of the 2's compliment. Of course, if you look at the 2-adics, negation is exactly the 2's compliment. So all this time you thought computers were doing approximate real calculations, instead they're doing approximate 2-adic calculations! |
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Title: Re: 10-adic numbers Post by towr on Sep 10th, 2003, 11:21pm on 09/10/03 at 19:05:14, Icarus wrote:
(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 10-addic ...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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Title: Re: 10-adic numbers Post by James Fingas on Sep 11th, 2003, 6:16am 10-adic representation of 1/7 x * ...00007.0 = ...00001.0 = ...999101.0 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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Title: Re: 10-adic numbers Post by Icarus on Sep 12th, 2003, 3:20pm 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 10-adics (and all other n-adics) - and addition, subtraction, multiplication, and division are the same on them as their are in the Reals. (Thus the zero-divisors of part (3) will be irrational 10-adics.) 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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Title: Re: 10-adic numbers Post by James Fingas on Sep 16th, 2003, 10:26am 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 second-to-last digit, and so on. At each stage i, we will have two decimal numbers Ai and Bi so that Ai*Bi is a multiple of 10i. 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 Cp,q = (Ai + p*10i)(Bi + q*10i) for each combination. 3) Choose a combination {p,q} so that Cp,q is a multiple of 10i+1. For this method to work, you must choose the first digit in both A and B to be non-zero. 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 N-adic zero-divisors can be found: composite N. Now all I have to show is that there is always a {p,q} so that Cp,q is a multiple of 10i+1. This should not be hard to do: Cp,q = (Ai + p*10i)(Bi + q*10i) = AiBi + qAi*10i + pBi*10i + pq*102i 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 qA1 + pB1. This method assumes that both of these are non-zero, and it can readily be shown that {A1,B1} 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 A1 = 2a and B1 = 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 zero-divisor pairs to find. Get going on it! |
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Title: Re: 10-adic numbers Post by Icarus on Sep 16th, 2003, 6:07pm That does it! A somewhat simpler statement of the reason why p-adics have no zero-divisors for prime p: The product of the least significant non-zero 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 p-adics 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). Follow-up 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 n-adics 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 non-real 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 n-adics 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 (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 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 one-to-one 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 = (a-r) + (x+r). This can be used to choose a value for a with no fractional part, and a value for x with no non-fractional 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.] |
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Title: Re: 10-adic numbers Post by Barukh on Sep 23rd, 2003, 2:23am 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 p-adic numbers. For instance: on 09/12/03 at 15:20:01, Icarus wrote:
It took me time to understand that although p-adic integers have the same representation as normal integers, p-adic rational do not have the same representation as normal rationals! Here are some observations I've made about p-adic numbers. First, some definitions. p-adic integer is a p-adic number without fractional part. p-adic number is regular (terminating) if it has only finite number of non-zero digits. p-adic number is periodic if its integer part has a finite period. 1. An integer is a regular p-adic integer. 2. If q is relatively prime to p, then 1/q is a periodic p-adic integer. For instance, 1/3 = ...66667 as a 10-adic 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 p-adic number (not an integer). For instance, 1/6 = ...333.5 as a 10-adic number. 4. If p and q have a common divisor, and p/q is regular, then 1/q is a regular p-adic number. For instance, 1/4 is 0.25 as a 10-adic number. Although rationals are part of p-adic 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 p-adic field. Then, what is the subset of irrationals that can be represented as a p-adic number for some p? For instance, is there a p-adic number q such that q2 = 2? |
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Title: Re: 10-adic numbers Post by Icarus on Sep 23rd, 2003, 6:33pm 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 "p-adic". (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 p-adic 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 p-adics, it turns out to be an infinite-dimensional 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 p-adics 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 p-adic 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 p-adics for prime p. I realized on my own that the construction also works for non-prime p, except that you get zero-divisors instead of a field. However I have since seen references to "n-adics" 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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Title: Re: 10-adic numbers Post by Barukh on Sep 25th, 2003, 12:13pm The following statement can be proved: For p a prime number, and q < p a natural number, [sqrt]q is a p-adic number, iff q(p-1)/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 7-adic 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 p-adic 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. |
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Title: Re: 10-adic numbers Post by Icarus on Nov 10th, 2003, 6:43pm 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,
(I'm not going to post this proof in the 0.999... = 1 thread!) Note that this result makes it easy to calculate the 10-adic 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...
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 10-adics. |
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Title: Re: 10-adic numbers Post by Rezyk on Nov 10th, 2003, 7:28pm 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 09/16/03 at 18:07:54, Icarus wrote:
Should that be "a-b = y-x" instead? Otherwise it seems like (4, .999...) ~ (3, 0), which is unintuitive. 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) |
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Title: Re: 10-adic numbers Post by Icarus on Nov 10th, 2003, 8:22pm on 11/10/03 at 19:28:50, Rezyk wrote:
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:
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 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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Title: Re: 10-adic numbers Post by ivancho on Oct 11th, 2004, 3:36pm [smiley=parallel.gif] Quote:
Actually, those 2 differences don't have to be terminating decimals, as your own example of Quote:
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 non-rational real equal to a non-rational 10-adic.. 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 = 91/3 = 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 10-adic integers [smiley=bigto.gif] they must have the form (m*10k)/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 10j - 1 [smiley=bigto.gif] our fraction is of the form N/(10j - 1), which is definitely repeating in real decimal expansion with period j. It is quite easy to see that the 10-adic 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:
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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Title: Re: 10-adic numbers Post by Icarus on Oct 11th, 2004, 8:08pm 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 n-adics, 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. |
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Title: Re: 10-adic numbers Post by ivancho on Oct 11th, 2004, 10:11pm 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.. |
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Title: Re: 10-adic numbers Post by PiotrGrochowski on Oct 7th, 2014, 11:23am on 09/09/03 at 17:09:09, Icarus wrote:
Icarus, To find negative of 957236: 1000000-957236=42764 -957236=...99999042764 1/7 in 10-adic is ...42857143 |
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Title: Re: 10-adic numbers Post by PiotrGrochowski on Oct 8th, 2014, 10:26am Icarus says: Quote:
No, it's not allowed in 10-adic 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 10-adic numbers are read right to left. For example: 20736 is sixone three thousand sevenone zero two |
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Title: Re: 10-adic numbers Post by PiotrGrochowski on Oct 8th, 2014, 10:31am on 09/09/03 at 17:09:09, Icarus wrote:
1001/7=143, 1000000001/7=142857143, etc... 1/7=...142857142857142857142857142857142857142857142857143 |
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