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Topic: what is the value of 1-1+1-1+1......... (Read 1585 times) |
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shake_de_shock
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what is the value of 1-1+1-1+1.........
« on: Oct 3rd, 2005, 7:10pm » |
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Im a newbie user......Hi guys.....
Ok consider this:
let x=1-1+1-1+1.........inf
then x=1-(1-1+1-1.........inf)
thus adding we get 2x=1
or x = 0.5
Is this true
my 2 cents
shake_de_shock
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towr
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Re: what is the value of 1-1+1-1+1.........
« Reply #1 on: Oct 3rd, 2005, 11:05pm » |
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There isn't a limit in the normal sense.
take
x0=1
xn=xn-1-1 if n is odd
xn=xn-1+1 if n is even
then x is limit n-> inf xn
for every n xn is either 0 or 1. The sequence doesn't converge.
The average (sumi=0n xi/n) does converge though, and this does converge to 0.5. I think that's called the Cesaro limit.
see also here
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| « Last Edit: Oct 3rd, 2005, 11:27pm by towr » |
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Icarus
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Re: what is the value of 1-1+1-1+1.........
« Reply #2 on: Oct 4th, 2005, 7:16pm » |
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Another concept is that it converges to the set {0, 1}. These are the limit points of the sequence of partial sums (i.e, the set of all points to which some subsequence of the sequence of partial sums converges).
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shake_de_shock
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Re: what is the value of 1-1+1-1+1.........
« Reply #3 on: Oct 6th, 2005, 6:40am » |
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thanks mates.....but since im a grade 11 schoolboy i didnt get one bit of what u said. But still you had the courtesy to reply 
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towr
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Re: what is the value of 1-1+1-1+1.........
« Reply #4 on: Oct 6th, 2005, 9:20am » |
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I'm surprised you're tackling infinity then.
The usual way to look at an infinite sum, is to aproach is as a sequence
x0 = 1 = 1
x1 = 1-1 = 0
x2 = 1-1+1 = 1
x3 = 1-1+1-1 = 0
...
x = xinf = 1-1+1-1+1....
As you can see, the sequence keeps switching between 1 and 0. And there is no reason to assume that behaviour would be different at infinity.
Contrast this with
y0 = 1 = 1
y1 = 1 - 1/2 = 1/2
y2 = 1 - 1/2 + 1/3 = 5/6
y3 = 1 - 1/2 + 1/3 - 1/4 = 19/24
...
y = yinf = 1 - 1/2 + 1/3 - 1/4 + 1/5....
Here the difference between two successive numbers keeps getting smaller, in other words the sequence converges. Eventually the difference between two successive numbers is 0 and you reach the value of y.
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Icarus
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Re: what is the value of 1-1+1-1+1.........
« Reply #5 on: Oct 6th, 2005, 8:47pm » |
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on Oct 6th, 2005, 9:20am, towr wrote:| Eventually the difference between two successive numbers is 0 and you reach the value of y. |
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A clarification is needed here. The difference between two successive numbers is never zero, and you don't actually every reach the value of y. However, you can make the difference as small as you like by adding more terms, and there is a unique value for y that the sequence approaches. That is the value we define the sequence as having.
Actually, the sequence towr chose is not really a good introduction into the concept of infinite sums, as the sum is difficult to find. A more common choice for introduction is:
1 + 1/2 + 1/4 + 1/8 +...
If let Sn be the sum of the first n+1 terms, then we see that
S0 = 1
S1 = 1 + 1/2 = 3/2
S2 = 1 + 1/2 + 1/4 = 7/4
S3 = 1 + 1/2 + 1/4 + 1/8 = 15/8
More generally, Sn = 2 - 2-n
By choosing higher and higher values of n, we can get closer and closer to the value 2 for the sum, but never actually get there.
Now, pretend that it is actually possible to sum up the entire infinite sum, and let
S = 1 + 1/2 + 1/4 + ...
Can we figure out what the value of S should be? Since each of terms is positive, adding more terms should increase the value. So we can expect that S > Sn for each finite n. But also, since each Sn < 2, no matter how high we take n to be, it does not seem reasonable that somehow S will jump over 2 to be something bigger. So we would also expect that S <= 2.
Thus we expect that 2-2-n < S <= 2 for all n. The only number that satisfies this condition for all values of n is 2 itself.
For this reason, S is defined to be 2:
1 + 1/2 + 1/4 + 1/8 + ... = 2
Now consider your series, and pretend it can be summed:
S = 1 - 1 + 1 - 1 + 1 - 1 + ...
Can we similarly determine a single value for S? The answer is no. In the other series ("series" = "infinite sum"), the partial sums (sums of the first k terms for various k) got closer and closer to a single value, 2, as more terms were added in. In this series, this is not true. The partial sums keep switching back and forth between 1 and 0. They never settle down to be just one or the other. Thus we cannot choose a single value for S.
Your algebra trick assumes such a value exists in the first place. Since you assumed something which was false, any results you derive from it are spurious.
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JocK
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Re: what is the value of 1-1+1-1+1.........
« Reply #6 on: Nov 22nd, 2005, 2:09pm » |
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on Oct 3rd, 2005, 7:10pm, shake_de_shock wrote:
let x=1-1+1-1+1.........inf
then x=1-(1-1+1-1.........inf)
thus adding we get 2x=1
or x = 0.5
Is this true
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As much as:
let x = 1 - 2 + 4 - 8 + 16 - 32 + ...
x = 1 - 2 (1 - 2 + 4 - 8 + 16 - 32 + ... )
x = 1 - 2 x => 3 x = 1 => x = 1/3
is true.
Actually, in some areas of theoretical physics it is quite common to evaluate diverging series in similar manners.
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| « Last Edit: Nov 22nd, 2005, 2:41pm by JocK » |
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xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Eigenray
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Re: what is the value of 1-1+1-1+1.........
« Reply #7 on: Nov 22nd, 2005, 2:44pm » |
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on Nov 22nd, 2005, 2:09pm, JocK wrote:
As much as x = 1 - 2 + 4 - 8 + 16 - 32 + ... = 1/3
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No, no, much less true than that. There's a perfectly good topology on Q where this one holds.
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Icarus
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Re: what is the value of 1-1+1-1+1.........
« Reply #8 on: Nov 22nd, 2005, 5:58pm » |
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Indeed there is a nice infinite family of such topologies - one for each power of 2.
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Re: what is the value of 1-1+1-1+1.........
« Reply #9 on: Nov 23rd, 2005, 3:48am » |
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on Nov 22nd, 2005, 2:44pm, Eigenray wrote:
No, no, much less true than that. There's a perfectly good topology on Q where this one holds. |
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Can you elaborate?
Why would the result
1 - 2k + 22k - 23k + 24k ... = 1/(1+2k)
hold for k = 1 (or any k > 0), but not for k = 0 ?
(sorry, I'm not a mathematician - looking at this like a physicist I'm affraid...  )
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| « Last Edit: Nov 23rd, 2005, 3:52am by JocK » |
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xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Eigenray
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Re: what is the value of 1-1+1-1+1.........
« Reply #10 on: Nov 23rd, 2005, 2:18pm » |
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on Nov 22nd, 2005, 5:58pm, Icarus wrote:| Indeed there is a nice infinite family of such topologies - one for each power of 2. |
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Could you give two (distinct) such topologies?
on Nov 23rd, 2005, 3:48am, JocK wrote:
Of course! For any rational x, write x = 2np/q, with p,q odd, and n an integer. Set ord(x) = ord2(x) = n. Observe that ord(x+y) >= min(ord(x),ord(y)). It follows that
|x-y|2 = 2-ord(x-y)
defines a metric on Q; the topology it induces is called the 2-adic topology. More topologically, we can define
mk = { x : ord(x) >= k },
and take these to be a local basis for the topology at 0. A basis for the 2-adic topology on Q is therefore { x + mk, x [in] Q, k [in] Z }.
(m1 is the maximal ideal of the local ring m0=Z(p); in fact m0 is what's known as a discrete valuation ring, the valuation ring of Q wrt ord. The completion of Z(p) wrt this topology is the p-adics, Zp, which is the valuation ring of Qp, the completion of Q (and also the field of fractions of Zp).)
So what does this mean? xn -> 0 iff for all neighborhoods U of 0, xn is eventually in U. In terms of the given basis, this means that for xn eventually stays in each mk, i.e., eventually xn >= k. That is, xn -> 0 iff ord(xn)-> infinity. Note also that the intersection of all mk is {0}, so this topology is Hausdorff. (These things are also obvious from the metric.) Also note that the field operations on Q are continuous wrt this topology.
So, when we say
1 - 2 + 4 - . . . . = 1/3,
we mean exactly that (in this topology), the limit of the partial sums is 1/3:
lim (1-(-2)n)/3 = 1/3.
But
1/3 - (1-(-2)n)/3 = (-2)n/3 [in] mn,
and therefore this difference converges to 0.
On the other hand, it's pretty clear that the partial sums of
1 - 1 + 1 - 1 ....
can't possible converge to anything in any (Hausdorff) topology.
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Why would the result
1 - 2k + 22k - 23k + 24k ... = 1/(1+2k)
hold for k = 1 (or any k > 0), but not for k = 0 ? |
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The exact same reason
1 - x + x2 - x3 + . . . = 1/(1+x)
holds for |x|<1, but not for |x|=1.
Note |2k|2 = 2-k < 1 iff k>0.
The p-adic norms |.|p are normalized so that for each x,
[prod] |x|p = 1,
where the product is over all (equivalence classes) of field norms on Q (where |.|oo is the archimedian (euclidean) norm, corresponding to the so-called "prime at infinity").
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| « Last Edit: Nov 23rd, 2005, 2:28pm by Eigenray » |
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Re: what is the value of 1-1+1-1+1.........
« Reply #11 on: Nov 23rd, 2005, 8:32pm » |
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on Nov 23rd, 2005, 2:18pm, Eigenray wrote:
Could you give two (distinct) such topologies? |
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Coitainly!
Just replace "2" in your definition by a fixed k. (To make Jock's series converge, k must be a power of 2, but in general it can be an arbitrary positive integer.) The p-adics do not require that p be prime. But you only get a field for prime p. That alone assures that the topology for non-prime p does not match that for prime p. But it can be shown that the topologies for each positive integer are unique.
------------------------------------------------------------
A simpler means of defining the k-adics is as formal decimal sequences in base k, which are allowed to extend infinitely to the left, but must be finite to the right (as opposed to real number sequences, in which the opposite is true). Addition and multiplication are carried out in the normal fashion. So, for example, in the 10-adics,
...999.1 + ...111.2 = ...000.3 = 0.3.
Of course, in this definition, it is not clear that they contain the rational numbers, but rules for division exist in the n-adics just as they do in the reals, and they allow you to divide integers. For example, 1/7 = ...2857142857142857143 in the 10-adics. (You may note that the repeating sequence, 285714 is identical to the repeating sequence for 1/7 in the reals base 10. This is true for all rational numbers. Unfortunately, the non-repeating decimal portion is not the same.)
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| « Last Edit: Nov 23rd, 2005, 8:57pm by Icarus » |
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Re: what is the value of 1-1+1-1+1.........
« Reply #12 on: Nov 24th, 2005, 10:15am » |
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on Nov 23rd, 2005, 8:32pm, Icarus wrote:| Just replace "2" in your definition by a fixed k. (To make Jock's series converge, k must be a power of 2, but in general it can be an arbitrary positive integer.) |
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Unfortunately, the definition I gave in terms of ord will not work in general (indeed, it cannot work, as the completion of any field wrt a valuation is again a field). The fact that p-ord(.) gives a norm depends heavily on the fact that ord is a valuation on Q; the triangle inequality comes from
ord(x+y) >= min( ord(x), ord(y) ).
They'll be trouble trying to define ord6(2) and ord6(3), for example.
The second construction does generalize to give the Krull topology on Z, where a local basis of 0 is given by, for each r, the set of multiples of kr, written (kr) = krZ.
Note that if k, k' have exactly the same prime factors, then k | k'a for some a, and similarly k' | kb for some b. Since then (kr) contains (k'ra), and (k'r) contains (krb), it follows that k,k' induce the exact same toplogy.
Now, it's easy to see that the ring operations on Z are continuous wrt this topology (more generally, one could use any filtration of ideals in any ring). One then takes the completion: formally, the quotient of the ring of Cauchy sequences of Z by the ideal of Cauchy sequences "converging" to 0.
But this is easily seen to be equivalent to the construction using left-infinite base-k expansions.
Namely, given a Cauchy sequence xn of integers, for each r, xn-xm is eventually divisible by kr, i.e., xn is eventually constant mod kr. This constant defines the leftmost r digits of the expansion. It's easy to see that for a given sequence, all such approximations are consistent, and that the expansion is independent of the choice of sequence in a given equivalence class. Conversely, given an infinite base-k expansion, simply truncate it to n digits to get xn; then xn is clearly Cauchy.
There's yet another construction, which is also useful. Zk is defined as the "inverse (projective) limit" of the groups Z/(kr). It is the set of "consistent" sequences (s1,s2,s3,...), where si is in Z/(ki), and for i<j, sj = si mod ki. This is again obviously equivalent: sn is just the rightmost n digits of the expansion.
That defines the k-adic integers, Zk. Allowing finitely many digits to the right of the "decimal" point is what algebraicists call localizing: we take a certain set, and declare them to be units, i.e., give them inverses (if you "localize at P," i.e., make everything outside a prime ideal P a unit, you kill all ideals not contained in P, while preserving the poset of ideals [0,P]; in algebraic geometry this does actually correspond to local behaviour at a point in affine space).
If k is prime, Zk is a domain, which lives in its field of fractions. If Zk isn't a domain, we have to be more careful, because allowing zero-divisors to have inverses is bad. But we can localize at the set of powers of k: we consider all x/kr, where x is a k-adic integer; this just shifts the decimal point over r spots.
In fact, Zk will be a domain iff k is a prime power. For example, if k=6, note that
Z/6r ~= Z/2r x Z/3r.
From the inverse limit construction, it's then clear that as a ring,
Z6 ~= Z2 x Z3.
But then it can't be a field: (0,1)*(1,0)=(0,0) corresponds to zero-divisors in Z6. (The way these elements are found is by the Chinese remainder theorem: pick xn mod 6r to be congruent to 0 mod 2n, and 1 mod 3n; check this gives a well-defined 6-adic x. Similarly find y corresponding to (1,0). Then xy=0.)
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To make Jock's series converge, k must be a power of 2. |
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But then you get the same topology.
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JocK
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Re: what is the value of 1-1+1-1+1.........
« Reply #13 on: Nov 24th, 2005, 10:41am » |
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Thanks for the additional explanations Eigenray and Icarus.
However, I have to admit I don't master all the terminology you use (ring, ord(), Q, k-adic integers, equivalence classes of fields, etc.) Also, it appears you use the word 'metric' in a different (generalised?) meaning than the one I am used to. (I am familiar with metric tensors, metrics of 'curved space' etc.) I simply fail to see how metrics could ever come into the picture when it comes to convergence of certain series.
Is it possible to cast the above 'machomath' in straightforward analytical form? (If only for a specific example?)
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| « Last Edit: Nov 24th, 2005, 10:41am by JocK » |
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Eigenray
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Re: what is the value of 1-1+1-1+1.........
« Reply #14 on: Nov 24th, 2005, 3:40pm » |
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The regular absolute value is an example of a norm on Q: it is a map |.| : Q -> R satisfying
(1) |x| >= 0; moreover |x|=0 iff x=0
(2) |xy| = |x||y|
(3) |x+y| <= |x| + |y|.
Given a norm, we define convergence:
xn converges to x iff lim |xn-x| = 0.
(A metric is slightly more general than a norm: it gives the distance between two points as a symmetric positive definite function d:Q2 -> R satisfying the triangle inequality; given a metric, xn converges to x iff d(xn,x) converges to 0.)
There are several things we can show just using the definition of norm.
A limit, if it exists, is unique: Suppose xn converged to both x and y, with r=|x-y|>0. By definition, we must have both |xn-x| < r/2 and |xn-y| < r/2 for n sufficiently large. But then
r = |x-y| <= |x-xn| + |xn-y| < r,
a contradiction.
Addition and multiplication are continuous: If xn -> x, and yn -> y, then because
|(x+y)-(xn+yn)| <= |x-xn| + |y-yn|,
it follows xn+yn -> x+y. Similarly,
|xy - xnyn| <= |xy-xny| + |xny - xnyn| = |y||x-xn| + |xn||y-yn|,
which goes to 0 because |xn| is bounded. Thus xnyn -> xy.
So, when does
1 - x + x2 - x3 + . . .
converge? The n-th partial sum of this series is
sn=(1 - (-x)n)/(1+x).
Suppose sn -> s; then
|sn+1-sn| <= |sn+1-s|+|s-sn|
must converge to 0. But |sn+1-sn|=|x|n+1, so it is necessary that |x|<1. Conversely, if |x|<1, we claim this converges to 1/(1+x). For, the difference
(-x)n/(1+x) has norm |x|n/|1+x|,
which goes to 0, |1+x| being fixed.
That's the standard norm, with the standard topology. But for any prime p, we can define another norm. Set
|pr m/n|p = p-r
when m,n are not divisible by p. For example,
|12|2 = 1/4, |17/42|2 = 2.
Thus we measure a number by its "p-ness": numbers highly divisible by p are small.
Then one should check that |.|2 satisfies the same three properties (1),(2),(3), and therefore is a norm on Q. But then the reasoning above goes through exactly the same:
1 - 2k + 22k - 23k + . . . = 1/(1+2k)
exactly when |2k|2 = 2-k < 1.
Here's an example where the Archimedean and p-adic norms differ greatly. Consider the series
1 + x + x2/2! + x3/3! + x4/4! + ....
In the usual norm, this converges for all x, because n! grows so fast. But in the p-adics, the exact opposite is the case: n! goes to zero! What is the radius of convergence of this series, in terms of p?
Compare with C, where ez gives a nice entire nowhere-zero function. But in the case of the p-adics (the completion of the algebraic closure of Qp), it can be shown that any power series with an infinite radius of convergence factors as f(x) = Cxn[prod](1-x/r), where the product is over all the (non-zero) roots of f (compare with the Hadamard factorization theorem!!). In particular, if f has no zeroes, it must be constant.
Another difference comes from |.|p being non-Archimedean:
|x+y|p <= max (|x|p, |y|p)
(i.e., all triangles are isosceles). Suppose xn -> 0, and let
sn = x1 + x2 + ... + xn
be the n-th partial sum. For any e>0, we can pick N such that |xn| < e for n>N. Now if m>n>N, then
|sm-sn| = |xn+1+...+xm| <= max (|xn+1|,...,|xm|) < e,
so sn is Cauchy, therefore converges. Thus a series converges iff its terms converge to 0!
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| « Last Edit: Nov 24th, 2005, 4:03pm by Eigenray » |
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Re: what is the value of 1-1+1-1+1.........
« Reply #15 on: Nov 24th, 2005, 8:07pm » |
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You can get around your problems with ord by being a little more creative. The development of the Krull topology you mention is equivalent.
You are correct that powers of 2 give the same topology, and the same 2-adic numbers. I had been under the impression that all composite numbers gave rings with zero-divisors, but had never thought to look at powers of primes. In fact, you can convert 2-adic numbers to 4-adics simply by grouping adjacent pairs of digits - something I should have realized before.
Still, if someone wants a small, but interesting, challenge, try to find a pair of zero-divisors in the 10-adics.
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Re: what is the value of 1-1+1-1+1.........
« Reply #16 on: Nov 25th, 2005, 12:35am » |
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on Nov 24th, 2005, 8:07pm, Icarus wrote:| You can get around your problems with ord by being a little more creative. |
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I think I see what you mean. You can put a metric on Q, where d(x,y)=10-r when x-y = 10rm/n, with m not divisible by 10, and n relatively prime to 10. It won't be multiplicative, and of course you just get the Krull topology again.
But any non-trivial norm on Q is equivalent to either the standard norm or a p-adic norm. I remember the proof being fairly elementary, but I'm not seeing it right now. (Take the smallest n with |n| != 1; then in fact n is prime. If |n|>1, show it's a power of the standard norm; if |n|<1, show it's a power of the n-adic norm.)
But it's easy to show for example there's no truly 10-adic norm, where 10r converges to 0, while |n|=1 for n relatively prime to 10. Since |2||5|=|10|<1, wlog say |2|<1. Then
|5|r = |5r+2r-2r| <= |5r+2r| + |2|r = 1+|2|r
is bounded, so |5|<=1. Since we also have
1 = |2r+5r| <= |2|r + |5|r
for all r, we get |5|>=1 as well. Thus the norm reduces to a power of the 2-adic norm.
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