Dozens Demigod
Double sharp
Dozens Demigod
Joined: Sep 19 2015, 11:02 AM
For my 900th (thirty squared) post, here is the explanation of the p-adics I promised to give previously.

In this particular case, though, Lagrange in his advocacy of a prime base had a point (though he could not have known this), because the p-adics only work with all the desired properties for prime p. So instead of using decimal, I'm going to write the rest of this post in quinary.

{5} (default quinary)

We know how to count in quinary fairly well, of course: 1, 2, 3, 4, 10, 11, 12, 13, 14, 20, and so on. Whenever we need to increment a 4, it turns into a zero and carries one to the place right before it. Every place corresponds to a different power of five: thus "143" means one times five squared, plus four times five to the first power, plus three times five to the zeroth power. Of course, five to the first power is just five, and five to the zeroth power is one.

We also know fairly well what finite quinary expansions mean. For example, 0.1 must be one fifth, as that position one place to the right of the radix point corresponds to five to the negative first power: that's the reciprocal of five, which is one fifth. Then we can keep going further and further to the right of the radix point.

Infinite expansions, however, are a little more tricky. What does 0.444... mean, with the fours going off to infinity? It must be the sum of 0.4, 0.04, 0.004, and so on to infinity, but what is that?

Well, we can look at the interval between 0 and 1. If I mark the point 0.4 on that interval, I've divided it into two parts of four-fifths and one-fifth each. Going to 0.44, we then take the smaller interval and divide it into two parts of four-fifths and one-fifth each. And then we go to 0.444, and divide that tinier interval again into two parts of fourth-fifths and one-fifth each. The differences between these numbers and unity, {0.1, 0.01, 0.001, ...}, dwindle exponentially to zero.

The sequence {0.4, 0.44, 0.444, ...} can be said to converge to 1, in the precise sense that you can get a number as close to 1 as you like by picking a number sufficiently far along in the sequence. Since the partial sums of the infinite series 0.4 + 0.04 + 0.004 + ... thus approach 1, the series itself has the sum of 1. Since by definition the sum of the series can also be written 0.444..., we must have 0.444... = 1.

Now, of course, this makes many people uncomfortable, I suppose because it seems strange that we should be able to allow the digit-1's in the sequence {0.1, 0.01, 0.001, ...} to go away by taking the limit. Some might point out that if we take the sequence {1, 10, 100, 1000, ...}, the digit-one also moves farther and farther away, so that eventually it should vanish. Of course, this doesn't work in the real numbers, since the numbers {1, 10, 100, 1000, ...} increase in absolute value without limit and so get further and further from zero.

But is there a metric on the rationals such that the numbers {1, 10, 100, 1000, ...} really approach zero as one would naïvely think they should?

{a}

Yes indeed! This is called the 5-adic metric, and it works if you replace the 5 with any other prime. The general idea is that you can express any rational number r as some power of five 5n times another rational number p/q where neither p nor q is divisible by five and they do not share a common factor. Then we define the 5-adic absolute value &#124;r&#124;5 as follows:

&#124;r&#124;5 = 5&#8722;n
&#124;0&#124;5 = 0

In the simplest case, this means that the powers of five really do approach zero in the 5-adic metric because &#124;5&#124;5 = 5&#8722;1, &#124;25&#124;5 = 5&#8722;2, &#124;125&#124;5 = 5&#8722;3, and so on. The negative powers of five don't approach anything however, since &#124;1/5&#124;5 = 51, &#124;1/25&#124;5 = 52, and &#124;1/125&#124;5 = 53, and so on. (The values of the absolute value are interpreted as real numbers, not 5-adic numbers, of course, because the other choice results in a circular definition.) This means that numbers can now expand infinitely to the left, because they converge that way; but no longer to the right, because they don't converge that way.

Previously, the choice of 5 as our radix had nothing to do with the structure of numbers, and its being crowned "the king of numbers" was an empty title with no real power. Now, the lordship of 5 actually means something, and it is in the very fabric of this new metric on the rationals. (But it should also be noted that any other prime number could equally well take on that lordship.)

The distance between two numbers x and y in the 5-adic metric is then &#124;x &#8722; y&#124;5, so that shift-invariance of distance is preserved.

Under this metric, the 5-adics (and indeed any of the p-adics) form a complete ultrametric space, meaning that if d(x,y) is the distance between x and y, then d(x,z) &#8804; max{d(x,y), d(y,z)}; this is known as the "strong triangle inequality", since it implies the triangle inequality as well. This grants it a few strange and characteristic properties: all triangles are acute isosceles or equilateral, every point inside a ball is its centre, and interescting balls are contained wholly in each other.

Nevertheless, some of the familiar properties we know and love are retained. It is still true that the absolute value is multiplicative, so that &#124;x&#124;p&#124;y&#124;p = &#124;xy&#124;p, and it satisfies the triangular inequality as previously stated, so that &#124;x + y&#124;p &#8804; &#124;x&#124;p + &#124;y&#124;p.

{5}

In the 5-adic numbers, there is no longer a need for a negative sign. If we consider the calculations {10 &#8722; 1 = 4; 100 &#8722; 1 = 44; 1000 &#8722; 1 = 444; ...}, we notice that the minuends approach 0, so that the differences must approach 0 &#8722; 1 = &#8722;1. Then if we take this to the limit, every place borrows from the place before it, so that ...000 &#8722; 1 = ...444. An infinite sequence of fours extending to the left of the quinary point is then the 5-adic expansion for &#8722;1.

Similarly, further negative integers are {...444, ...443, ...442, ...441}, and so on, all with an infinite number of fours trailing into the distance after a brief prelude. (This is the same idea as two's complement in binary, although here it should probably instead be called "five's complement", since this is quinary.)

The 5-adic numbers that do not need a radix point to represent are called the 5-adic integers, which are labelled Z5. (For other primes p, we replace 5 with that p.) Z5 is also the set of all rationals r such that &#124;r&#124;5 is non-negative. Zp is always a commutative ring with the two binary operations of addition and multiplication when p is prime. This means that addition and multiplication are both commutative and associative, that they both have identities (0 and 1 respectively), that every member has an additive inverse, and that multiplication distributes over addition. Its topology is that of a Cantor set.

Of course, all members of Z are members of Z5 as well. But that is not all! Consider the rational number one-half. Its simplest representation in the format given above is 50 * 1/2, so &#124;1/2&#124; = 50 = 1, which is non-negative. So one-half is a 5-adic integer, and so we should not need any digits to the right of the radix point to represent it. And indeed, we have 1/2 = ...22223. When we add ...22223 to itself, the units column produces a 6, which cascades a carry down all the columns to the left that produce fours to infinity; thus the result is ...00001.

We can similarly check that 1/3 = ...13132, with the sequence "13" repeating to infinity. In fact, all rational numbers are 5-adic integers as long as their denominators in simplest form are not divisible by 5. So {1/2, 1/3, 1/4} are all 5-adic integers, but 1/5 is not.

Allowing digits to the right of the radix point (but only finitely many of them) admits such rationals, and results in Q5, the 5-adic numbers. When p is prime, this is a field: otherwise, we end up with zero divisors if it has multiple distinct prime divisors, or it reduces to the field generated from its constituent prime if it happens to be a prime power. Being a field means that in addition to the properties of a commutative ring, we also always have multiplicative inverses for every element except the additive identity. (We also demand for a field but not a ring that the additive identity is different from the multiplicative identity, but this is true for Zp as well as Qp, and is indeed true for the commutative ring Z as well.) The topology of Qp is that of a Cantor set with one point missing.

None of the Qp are actually algebraically closed, and they are much further from being so than R is; R only requires a quadratic closure (adjoining a solution of x2 + 1 = 0) to become the algebraically closed C, but the closure of Qp to form Q&#773;p has infinite degree, and even that is not metrically complete. Its metric completion is called Cp, the p-adic complex numbers, and there we finally can stop. In fact, Cp for any p is just C, the standard complex numbers, with an exotic metric replacing the usual one; this is different from the relationship between Qp and R, which are much more different.

Well, this is just a quick explanation, and I skipped over a lot of things, but I hope it gives a reasonable idea of what these are. Please feel free to comment if something's not clear or mistaken!

wendy.krieger
wendy.krieger
P-adic numbers contain a set of numbers that are constructable, because the isomorphism requires this.

One can construct the chords of every polygon, whose sides divide (p+1)/2 or (p-1)/2. One can construct the cyclotomic numbers for CZ(p^n(p-1)/2). The powers of any number raised to the pth power, converge on the sevenite-tails. For 5, these are 000000, 000001, 4431212, 013233, and 4444444. These also represent the fifth roots of 1, to 6 places. Either one of '2' or '3' can stand for i, so 0013234 ^ 8 = 000031, as you should expect from (1+i)^8 = 16. D.S. gives ...00301131300030330421304240422331102414131141421404340423140223032431212 for this, Here is the other one!

Code: Select all

&#91;D&#58;\&#93;rexxtry say translate&#40;"00301131300030330421304240422331102414131141421404340423140223032431211", "01234", "43210"&#41;
44143313144414114023140204022113342030313303023040104021304221412013233
........................... D&#58;\save\cdata\batch\rexxtry.cmd on WIN32

Likewise, there are iso-sevenite tails corresponding to the forms like 3. 5 does not have a sevenite-tail for the golden ratio, but 11 does: 0.3.4, the square gives 0.3.5, and one finds the square root of 5 by 0.6.7 = 2phi-1. (base 11 = 6.7³ = 4.0.0.5). The standard methods for finding sevenites works for finding the extended values of the chords.

We find in base 13, that i=01550155.

Ordinary long division works, if you progress from the right.

The rules of polygonal isomorphism are observed.

Dozens Demigod
Double sharp
Dozens Demigod
Joined: Sep 19 2015, 11:02 AM
wendy.krieger @ Feb 16 2017, 01:31 PM wrote:P-adic numbers contain a set of numbers that are constructable, because the isomorphism requires this.
{a}

I'm not entirely sure what Wendy means by this. Of course, some of the p-adic numbers are constructible, but the specific ones that appear depend on p, and there is no finite value of p that gets them all in. For example, all square roots of positive integers are constructible, but the 5-adic numbers do not include &#8730;2, &#8730;3, &#8730;5, or &#8730;7 (this is obvious from Hensel's lemma). They do include &#8730;11, though.
wendy.krieger @ Feb 16 2017, 01:31 PM wrote:One can construct the chords of every polygon, whose sides divide (p+1)/2 or (p-1)/2.  One can construct the cyclotomic numbers for CZ(p^n(p-1)/2).  The powers of any number raised to the pth power, converge on the sevenite-tails.  For 5, these are 000000, 000001, 4431212, 013233, and 4444444.  These also represent the fifth roots of 1, to 6 places.  Either one of '2' or '3' can stand for i, so 0013234 ^ 8 = 000031, as you should expect from (1+i)^8 = 16.  D.S. gives ...00301131300030330421304240422331102414131141421404340423140223032431212 for this,
These are not the fifth roots of 1, but the fourth roots. In general, the roots of unity in Qp are the (p &#8722; 1)th roots of unity for odd p, and are ±1 for p = 2.
wendy.krieger @ Feb 16 2017, 01:31 PM wrote:Likewise, there are iso-sevenite tails corresponding to the forms like 3.  5 does not have a sevenite-tail for the golden ratio, but 11 does: 0.3.4, the square gives 0.3.5, and one finds the square root of 5 by 0.6.7 = 2phi-1. (base 11 = 6.7³ = 4.0.0.5).  The standard methods for finding sevenites works for finding the extended values of the chords.

We find in base 13, that i=01550155.
{d}

It certainly ends that way, but the pattern is not repeating. To a few more digits, you get ...6a69555a67b41505474036c688101550155. Also, the choice between that and its additive inverse ...626377726518b7c7858c960644bcb77cb78 is arbitrary; I wouldn't have bothered to give this one here, because it's trivial to find it (just take the "zen's complement" of every digit but the last), but since you enthusiastically showed the additive inverse of the quinary primitive fourth root of unity that I chose to exhibit, I'm giving both here. Not that it is very impressive, of course, given the ease of actually finding it.
wendy.krieger @ Feb 16 2017, 01:31 PM wrote:Ordinary long division works, if you progress from the right.
I suppose I shall have to give an example of long division to produce the 5-adic expansion of 1/3.

{5}

We start with 1, and we must find a multiple of 3 that ends in 1 in quinary, which must be 3 * 2 = 11. Now we must subtract this partial quotient 11 from ...00000 00001, giving ...44444 44440.

Now we have a 4 in the next place, so we need to find a multiple of 3 that ends in 4 in quinary, which must be 3 * 3 = 14. Now we subtract this partial quotient 140 from ...44444 44440, giving ...44444 44300.

Now we have a 3 in the next place, so we need to find a multiple of 3 that ends in 3 in quinary, which must be 3 * 1 = 3. Now we subtract this partial quotient 300 from ...44444 44000.

We now have the same quotient as after the first step, with two more zeroes padded onto the end, so the quotient must be periodic from here on, and the answer is 1/3 = ...1&#773;3&#773;2. In ordinary quinary, you would be subtracting partial quotients from 1.000... to the right instead, and you would get 0.1&#773;3&#773;.
wendy.krieger @ Feb 16 2017, 01:31 PM wrote:The rules of polygonal isomorphism are observed.
Yet another undefined term. Do you have an explanation of this somewhere?

One other cool thing that I forgot to mention in the previous post is that quadratic equations have solutions in Q if and only if they have solutions in R and in Qp for every p (the Hasse principle).

We can always say that a polynomial can have solutions in Q only if it has solutions in Qp for every p, but having such p-adic solutions does not actually guarantee that it has rational solutions; it only means that the possibility is not yet ruled out. We can embed Q in all the Qp, but also in R, which corresponds to p being the "real prime" or the "prime at infinity": I've posted about this before. In this context, R is sometimes called Q&#8734; for this reason.

While the reals might be originally thought of as the completion of the rational numbers, there are really an infinite number of other ones in the form of the different p-adic numbers. The reals are different from all the rest: they are the completion "at infinity" and are the only one of these completions with an Archimedean metric. They are all completions in the sense that the new numbers you add to Q are limits of Cauchy sequences of rational numbers.

This actually explains immediately why some p-adic numbers are not reals, and some reals are not p-adic numbers, because most sequences which converge in some p-adic metric will not converge in another one, or in the real metric; and most sequences which converge in the real metric will not converge in any p-adic metric. In fact, apart from the rationals, which we defined to be there in every Qp as well as R, there are no other real numbers that are in every one of the different Qp's.

In particular, the number e (defined as the sum of the reciprocals of the factorials) is not a p-adic number for any finite value of p, because that sum does not converge in the p-adic metric for any value of p: this is the opposite situation as for the rationals. In fact, just as the various finite primes p are the lords of their respective fields of p-adic numbers and are the key to their geometries, so because the "infinite prime" isn't actually a number, the number e which is unique to the real numbers takes on its duties.

wendy.krieger
wendy.krieger
The p-aidic numbers, come as tBn (tail of base-n numbers) since Q means variously quarterion, Q and rational numbers F [fraction]

The nature of 'what is a prime' has some interesting effects against p-adic numbers, as they do in the regular bases n.

The set Bn, is represented by the closure of the set Z to division by n, that is, if some x is in Bn, then so is x/n. Notionally it represents the number actually expressed in base n. The span of the set of prime divisors, is eg B2B5.

The sets are class-2 infinites, these intersect at the rate of the sum of the least logrithms of the primes. eg here are the logrithms of the powers of 2, 3, 5 that make 120 and 60. We total the smaller value of each, and derive a cascade-value.

Code: Select all


&nbsp; &nbsp; base 60 &nbsp; &nbsp; base 120
&nbsp;
2 &nbsp; 0.338587 &nbsp; &nbsp; 0.434348
3 &nbsp; 0.268324 &nbsp; &nbsp; 0.229475
5 &nbsp; 0.393088 &nbsp; &nbsp; 0.336175

&nbsp; intersection &nbsp;0.904237

For a given number, say 120^86, the common intersection is something that divides 120^(86*0.904237).

The common intersection of 10 and a binary base, is 0.30103. A ruler of millimeters in kms, would contain 10^(6*0.30103) = 64 common marks. Although the decimal will eventually swallow any binary number, the binary numbers are producing them at the rate of the cube or more.

In the set Bn, the numbers that regular to the base (ie have prime divisors that are a subset of those of n), are units, and the numbers that are co-prime to n are 'prime'. That is, the set of primes in B120 are 7, 11, 13, 17, 19, 23, 29, 31, ....

In the set tB120, these same numbers are units, since the inverse of these numbers are also in tB120, eg 7 = V2V2 V2V2 V2V3, and 11 is 10 V910 V910 V911. But the primes become 2, 3, 5.

The nature of a prime, then is to reduce the number towards 0. That is, if p is some prime of S, and S a set, the pS is a proper subset of S: there are numbers in S not in pS. Since the operation of p,q,r... is to reduce S to a set pqrS, the one definite member of the set is 0.

In t2B120 (tail of 2 places of base 120), we see that the effect of 2^6 and 3^2 and 5^2 together will reduce any t2B120 element to 0000.

The set t2B120 is a composition of t6B2 t2B3 t2B5.

The sevenites, and polygon-tails etc, in base 120 resolves into the co-even number of divisors (the remainder when an even number is subtracted). 120 has 45 sevenite tails, these are 3 binary * 3 trinary * 5 pentagonal. An example is 9123.

This is a generalised form that 9376 in t4B10, squares to 9376 etc But 3568, has a fifth power (which returns to ending in 8) as 0013568. Any number ending in 8, on sucessive fifth powers, ends in progressively more digits of this. eg 68**5 = 933568.

Dozens Demigod
Double sharp
Dozens Demigod
Joined: Sep 19 2015, 11:02 AM
As usual, I doubt if Wendy knows what she is talking about, because one of the first things you learn about the quaternions (besides what they are) is that their symbol is H (for their discoverer Hamilton), not Q (which means the rationals). F denotes a field; if a subscript follows, it represents the order. All this is very standard mathematical terminology, so I am not surprised that she rejects it.

The main difference between the reals and all the other possible completions of Q (the p-adic numbers) is that the reals do not care what base you express them in, whereas you get a different p-adic number system depending on what p is, and those systems are only fields if p is a prime power (and if it is a prime power, then it is isomorphic to that generated by the original prime).

The rest of Wendy's post contains so much invented terminology of dubious value that I backspaced my entire attempt to make sense of it. Whatever it means, the fact that it does not seem to have any relationship with any concepts in mainstream mathematics does not speak well for it being actually useful.

Obsessive poster
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Joined: Sep 10 2011, 11:27 PM
Double sharp @ Feb 20 2017, 11:01 AM wrote: The rest of Wendy's post contains so much invented terminology of dubious value that I backspaced my entire attempt to make sense of it. Whatever it means, the fact that it does not seem to have any relationship with any concepts in mainstream mathematics does not speak well for it being actually useful.
The responsibility for defining and explaining terminology, ideas, and mathematical techniques that are unfamiliar, peculiar, even downright idiosyncratic, should rest upon the perpetrator, not on the innocent victims upon whom such oddities are inflicted. It is the height of arrogance to invent one's own private language for something, and simply assume, because one has accustomed oneself to it by talking to oneself in the murky isolation of one's underground lair, that it then magically becomes intuitively self-evident to everyone else. But this is Wendy's habit, and her conceit. If she is incapable of relating what she is talking about to standard concepts and language, then we need not consider it indicative of anything more than a terminal case of glossolalia.

If someone comes up with a good idea, and articulates it clearly, and moreover demonstrates it in action, Wendy can not get away with claiming ownership of the idea, simply because some indecipherable bit of argot she gabbled at some point in the past vaguely sounds like it's related.
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wendy.krieger
wendy.krieger
DoubleSharp's description of p-adic numbers shows that they are a class-one cascade set, where the only primes are those that divide b, and the balance are units. Since class-one is itself isomorphic with the integers, and the construction is uni-directional cascade, this is not improbable.

The set corresponds to the base-tails of integers in base b, the cascade is caused by increasing the length of the cascade. Some number that has a realisation in this, follows different paths, so for example, 1/7 = '3', '33', '013', '2033', etc, but the longer one exposes the path of 1/7, the more it diverges from other numbers in this path.

Since the paths are self-counting, that is '2033' is itself a number d268, it follows that you can't tell the difference between 268 and 1/7 until you reach a five-digit tail. The former is '02033', the latter is '12033'. This is the nature of teelic infinities. The paths of 268 and 1/7 converge on the single point '2033'.

Since ultimately you can calculate the rationals p/q, except where 5¦q, it means that there is a proper intersection between the base-tails and the rationals, but this is a proper intersection. There are rationals that can not be represented in this form.

In fact, the set of numbers represented in a base-tail set have a construction, and therefore are a subset of the "algebraic fractions" (ie algebraic integers ÷ Z). These are further subsetted to where 5 does not divide the most-significant term in the lowest form.

Kanonier
Kanonier
Doube Sharp said

"Infinite expansions, however, are a little more tricky. What does 0.444... mean, with the fours going off to infinity? It must be the sum of 0.4, 0.04, 0.004, and so on to infinity, but what is that?"

It is 0.99999 ... in decimal = 9 (0.111111....) and the recurring 0.1111.. is the reciprocal of 9 in base ten - so therefore we have 9 x (1/9) = 1.
For quinary 0.4444... = 4(0.1111...) and quinary 0.1111... is the reciprocal of 4 in quinary notation, so 4(1/4) = 1.

What is then hard here? All we have to do is to relate these reciprocals to the fact that for number base b the reciprocal of (b-1) takes form 0.1111.... [and reciprocal of (b+1) takes form (b-1)(0.01 01 01... usw)].

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Dozens Demigod
Double sharp
Dozens Demigod
Joined: Sep 19 2015, 11:02 AM
Of course there is nothing hard about it if you've seen it before, but it does require you to rigorously define what an infinite sum means (the limit of its partial sums), which we hitherto had not defined. That is what I was referring to: this is where the metric comes in, to let you define the convergence of 0.444...5, and with a different metric it might conceivably be that 0.444...5 doesn't converge, as in the 5-adics.

This is, in fact, where all these infinite expansions come from in the first place, so using them to justify 0.444...5 = 1 and so on is, while pedagogically sound, a little circular. If we wanted to really justify it, we need to define infinite expansions and identify them with the limit of its associated partial sums, and then we can say things like 0.111... = 1/9 and 0.090909... = 1/11 in decimal.

(Very late P.S., 7 Nov 2017: Kanonier may be surprised to know that the recognition that rational numbers are precisely those whose decimals repeat indefinitely came only from Leibniz in 1677, even though infinite decimals were first used by Al-Samaw'al in 1172. In any case, the main point is that the idea that this series converges needs justification, as it depends on the Archimedean norm that we are explicitly uprooting for the p-adics.)

Dozens Demigod
Double sharp
Dozens Demigod
Joined: Sep 19 2015, 11:02 AM
There is also an interesting analogy here, first noted by Dedekind and Heinrich Weber, of positive integers to polynomials and fractions to rational numbers. (Similarly, transcendental functions seem to be analogous to transcendental numbers, and some algebraic numbers seem to be analogous to various elliptic functions.)

Kurt Hensel, the inventor of the p-adics, had the idea that the normal base-n notation for numbers already implicitly tells us something that looks very much like the Taylor series of a function around some value. For after all, when we write "78125", we mean

78125 = 5 + 2 * 10 + 1 * 10^2 + 8 * 10^3 + 7 * 10^4

and just as we can look at a polynomial around some other point than zero, by expanding it in terms of (x-a) instead of x, so we can change the base and write

78125 = 53773{b} = 3 + 7 * 11 + 7 * 11^2 + 3 * 11^3 + 5 * 11^4,

and we get this expansion by taking 78125 mod 11, mod 11^2, and so on. 78125 mod 11 is 3, so we take that remainder and subtract it from 78125, and then the result is divisible by 11. Then we can divide it by 11 and then repeat the process, and so forth.

The only problem with using 10 instead of 11 is that 10 is factorable, while the linear factors (x-a) are not, which is another way in which we can see that the p in p-adic really ought to be a prime if this analogy is going to do what we want it to. So primes correspond to points in the complex plane in this analogy, while the integers correspond to polynomials and the rationals to rational functions.

We can take this idea to fractions as well. If we want to write the expansion of 1/2, we need to take it mod 11. Well, 1/2 = 6 mod 11, so the first term is 6. So we can write 1/2 - 6 = -11/2 = -1/2 * 11, taking away the remainder. Now -1/2 = 5 mod 11, giving us the next digit, and -1/2 - 5 = -11/2 = -1/2 * 11. We will keep repeating this forever without end, and so we will be taking the integers mod 11^k to the limit as k approaches infinity, and we will get

1/2 = 6 + 5 * 11^2 + 5 * 11^3 + 5 * 11^4 + ... = ...5556,

and indeed if you sum ...5556 + ...5556 in undecimal arithmetic you will get 1 because the carries go on to infinity (and this whole thing is made rigorous by the non-Archimedean norm associated with the p-adics that I explained back in February). You will need to include finitely many negative powers of 11 if you want to allow for fractions with powers of 11 in the denominator, like 7/55. And of course, nothing forced us to choose 11: we could have chosen 2, 3, 5, 7, or any other prime at all.

So: just as a polynomial mod (x-a)^n is its Taylor series cut off to the (n-1)st order - in other words, Z/p^n corresponds to C[z]/(z-a)^n, so the p-adic integers Zp correspond to the Taylor series C[[z-a]] and the p-adic rationals Qp correspond to the Laurent series C((z-a)). (I think you might not even have to click the link now to figure out what a Laurent series is! )

Of course, there are many numbers in Qp which are not in Q, as shown above. Every Qp, along with R, contains the rationals, but the rest of what it contains is completely different (for one thing, many Qp contain square roots of negatives, and no Qp contains the number e). But the analogies hold and you can indeed do p-adic analysis just like you can do real analysis.

So the reals and the p-adic numbers act like local examinations of the global rationals, and there is this wonderful thing called the Hasse principle (the local-global principle). As Wikipedia says, it means that "certain types of equations have a rational solution if and only if they have a solution in the real numbers and in the p-adic numbers for each prime p". For instance, if x always has a square root in R and in Qp for every p, then we find that the square root of x must be rational, and that is the global root whose local images we see in the reals and the p-adics. (Indeed, the reals are sometimes thought of just another p-adic corresponding to the "real prime" or the "infinite prime", or, as John Conway would have it, -1. Yes, really; you can read about it here.) Just as we study rational functions locally with Laurent series, so we study the rational numbers locally with p-adic numbers. (John Baez gives a nice exposition in "week218" of This Week's Finds, though it quickly becomes quite a bit heavier!)