Wikipedia:Reference desk/Archives/Mathematics/2009 July 16

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July 16

Change the subject of equation

How do I solve a V=pie r2h problem for r —Preceding unsigned comment added by 71.244.44.98 (talk) 00:13, 16 July 2009 (UTC)

sorry I meant for h and r=6 and pi=72pi —Preceding unsigned comment added by 71.244.44.98 (talk) 00:15, 16 July 2009 (UTC)

Adding section header I think you should go and talk to your teacher. Rearranging an equation like that is a fundamental bit of algebra that you need to learn properly. --Tango (talk) 12:49, 16 July 2009 (UTC)

Continuity and differentiability

Show that the function

If f [x} = I 2x-3 I . [ x ] , x > 1 [ Greater than and equal to 1]

and f{x} = sin { pi .x/ 2 } , x < 1 [Less than 1]

is continuous but not differentiable at x = 1,

Where pi = 180 Degree, [ x ] is the greatest integer —Preceding unsigned comment added by 122.174.90.103 (talk) 05:27, 16 July 2009 (UTC)

Adding section header --Tango (talk) 12:49, 16 July 2009 (UTC)

If you can let us know what work you have done for this problem, and where you have gotten stuck, then we can help you with solving it. Do you know and understand the definitions of continuous and differentiable? Eric. 216.27.191.178 (talk) 22:22, 16 July 2009 (UTC)

Unique representation of an element of a local ring

What I want to show is that an element u of ${\frac {\mathbb {Z} }{p^{n}}}=\mathbb {Z} _{p^{n}}$ (p is prime, n is a natural number) can be written uniquely in the form $u_{0}+u_{1}p+u_{2}p^{2}+\cdots +u_{n-1}p^{n-1}$ where $u_{i}\in \mathbb {Z} _{p},0\leq i\leq n-1$. I vaguely feel that this is something related to p-adic numbers but I never studied those. What I do know is that ${\frac {\mathbb {Z} _{p^{n}}}{p\mathbb {Z} _{p^{n}}}}$ is the field on p elements. If someone can point me in the right direction I will be grateful.--Shahab (talk) 09:27, 16 July 2009 (UTC)

First, your notation does not make much sense, as $\mathbb {Z} _{p}$ is not a subset of $\mathbb {Z} _{p^{n}}$. Assuming you wanted to write $u_{i}\in \{0,1,\dots ,p-1\}$, existence follows immediately from the fact that any natural number can be written in a base-p representation. As for uniqueness, the sets $\{0,1,\dots ,p-1\}^{n}$ and $\mathbb {Z} _{p^{n}}$ have the same finite number of elements, and we have just established that the mapping from the former to the latter sending each sequence $\langle u_{0},u_{1},\dots ,u_{n-1}\rangle$ to the sum $\sum _{i<n}u_{i}p^{i}\,$ is surjective, hence it is also injective. Or you can simply use the fact that the base-p representation of integers is unique; since $\sum _{i<n}u_{i}p^{i}\,$ as an integer is smaller than pⁿ, and distinct natural numbers below pⁿ are incongruent modulo pⁿ, the representation is also unique qua elements of $\mathbb {Z} _{p^{n}}$. — Emil J. 10:32, 16 July 2009 (UTC)

Thank you. (I am however confused by your statement about my notation.)--Shahab (talk) 13:43, 16 July 2009 (UTC)

As Emil pointed out, your notation was nonsensical. $\mathbb {Z} _{p}$ is the ring of integers modulo p, while $\mathbb {Z} _{p^{n}}$ is the ring of integers modulo pⁿ. There's a natural map from the latter to the former, so it makes sense to treat elements of $\mathbb {Z} _{p^{n}}$ as also being in $\mathbb {Z} _{p}$ if you want to. In elementary terms, if you know an integer is 7 mod 9, then you know it is 1 mod 3. However, there is no natural map from $\mathbb {Z} _{p}$ to $\mathbb {Z} _{p^{n}}$ (for n>1), so it doesn't make sense to regard elements of $\mathbb {Z} _{p}$ as being in $\mathbb {Z} _{p^{n}}$. In elementary terms, if you know an integer is 1 mod 3, then it could be 1, 4, or 7 mod 9, and there's no reason to prefer one above the others. Algebraist 14:14, 16 July 2009 (UTC)

There seems to be a misunderstanding here. I understand the difference between Z_n and Z_{p^n}. What I don't understand is what notation of mine claims that one is contained in the other. Emil says "Assuming you wanted to write $u_{i}\in \{0,1,\dots ,p-1\}$" and I say "$u_{i}\in \mathbb {Z} _{p},0\leq i\leq n-1$". Its the same thing, isn't it? Thanks for bearing with me.--Shahab (talk) 16:45, 16 July 2009 (UTC)

The elements of $\{0,\dots ,p-1\}$ are integers, a typical element is 2. The elements of $\mathbb {Z} _{p}=\mathbb {Z} /p\mathbb {Z}$ are equivalence classes of integers with respect to congruence modulo p, a typical element of $\mathbb {Z} _{3}$ is $\{\dots ,-4,-1,2,5,\dots \}$. (Furthermore, it's not a terribly good idea to rely on this internal representation, as its not unimaginable that someone defines it in another way. One can only be sure how the object looks like up to isomorphism.) — Emil J. 16:56, 16 July 2009 (UTC)

Yes I know that. I was making the assumption that the elements of {0,1,...p-1} are really the equivalence classes {[0],[1],...[p-1]}. That is just a loose way of writing I have seen in books. However this does not clear my doubt as to why you said that I think Z_n is in Z_p^n.--Shahab (talk) 17:18, 16 July 2009 (UTC)

No one mentioned $\mathbb {Z} _{n}$, it was $\mathbb {Z} _{p}$. You wrote (omitting some irrelevant bits): an element u of $\mathbb {Z} _{p^{n}}$ can be written uniquely in the form $u_{0}+u_{1}p+u_{2}p^{2}+\cdots +u_{n-1}p^{n-1}$ where $u_{i}\in \mathbb {Z} _{p}$. Here, you take elements u_i of $\mathbb {Z} _{p}$, and try to interpret them as elements of $\mathbb {Z} _{p^{n}}$ in the sum. — Emil J. 17:30, 16 July 2009 (UTC)

Actually, what I wrote two posts above is misleading. $\{0,\dots ,p-1\}$ was not supposed to stand for a set of integers, but of the ring elements 0, 1, 1 + 1, …, $\underbrace {1+\dots +1} _{p-1}$, which are also commonly denoted by 0, 1, 2, …; the ring in question being $\mathbb {Z} _{p^{n}}$ in the context. — Emil J. 17:40, 16 July 2009 (UTC)

OK I understand what you are saying. In my defense I was reading from this book here(Pg 21) and it does say $u_{i}\in \mathbb {Z} _{p}$. (This book assumes a lot of knowledge which I sadly don't have.) I get your point though. Thanks--Shahab (talk) 17:56, 16 July 2009 (UTC)

On thinking more, I realize that the u_is could be in $\mathbb {Z} _{p}$, if you interpret juxtaposition as the natural action of $\mathbb {Z} _{p}$ on $\mathbb {Z} _{p^{n}}$, rather than as multiplication. Algebraist 16:52, 17 July 2009 (UTC)

How do you intend to define a "natural action" of $\mathbb {Z} _{p}$ on elements u of $\mathbb {Z} _{p^{n}}$ that do not satisfy pu = 0? How is it different from mapping $\mathbb {Z} _{p}$ to $\mathbb {Z} _{p^{n}}$ (using whatever is the "action" on 1), and then doing multiplication in $\mathbb {Z} _{p^{n}}$? — Emil J. 17:48, 17 July 2009 (UTC)

It does satisfy pu = 0. It's just the natural action of Z on $\mathbb {Z} _{p^{n}}$ (i.e. nu is the sum of n copies of u), which descends to the quotient Z/pZ because pZ acts trivially. Algebraist 18:13, 17 July 2009 (UTC)

The annoying thing is that there is no standard notation for the set of the first n natural numbers {0,1..,n-1}. Von Neumann's " n " is extremely elegant but ambiguous outside ordinals (what is n+m or f(n)?). I saw using [n] sometimes, that maybe deserves more popularity...So one is tempted to use $\mathbb {Z} _{n}=\{0,1..,n-1\}$ forgetting about its algebraic structure and its quotient nature. After all we do not use different symbols for $\mathbb {Z}$ as a set or as a ring. I don't know. --pma (talk) 20:52, 16 July 2009 (UTC)

Agreed. I have also found distinguishing between $\mathbb {\mathbb {Z} } /n\mathbb {\mathbb {Z} }$ as a group and as a ring very troublesome. I would like a notation to mean explicitly "the cyclic group of n elements"; I know that chemists use $C_{n}$ but I find that ugly. (For whatever reason, the additive group $\mathbb {\mathbb {Z} }$ comes up rarely enough that the ambiguity there doesn't bother me.) Eric. 216.27.191.178 (talk) 22:17, 16 July 2009 (UTC)

$\langle x\mid x^{n}\rangle$ ? My fourth year geometric group theory lecturer used C_n, he was a pure mathematician through and through. Maelin (Talk | Contribs) 23:18, 16 July 2009 (UTC)

Ah, is $C_{n}$ standard notation when studying symmetry groups? That would explain where the inorganic chemists got it from. I would prefer something with blackboard-bold, though, myself. Unfortunately $\mathbb {C}$ is already taken, and I think I've seen $\mathbb {C} _{p}$ used for something else, too. Perhaps we can introduce $\mathbb {G} _{n}$?

I'll admit to not being fond of group presentations... feels unnecessarily complicated here. Just gut feeling. Eric. 76.21.115.76 (talk) 05:28, 17 July 2009 (UTC)

I don't know about it being a "standard" notation, it's just what he used. I don't remember whether I saw it in the literature. I wouldn't expect a blackboard bold thing though, given that the symmetric groups S_n don't get bolded either. Maelin (Talk | Contribs) 07:31, 17 July 2009 (UTC)

My point is that usually a basic mathematical object admits so many different structures of interest, that having a special notation for it as endowed with each of these structures, is just hopeless. I think that a reasonable choice is to reserve a special a symbol only for the basic, precise set without structure (e.g. deciding that $\mathbb {Z} _{n}:=\{0,1,..,n-1\}$ ) ; then, at each use, one specifies which structures that object is endowed with, and in which category it is to be seen. But we do not have enough standard symbols to distinguish from $\mathbb {Z} _{n}$ as a set, as an ordered set, cyclic group, topological space, dynamical system, topological group, ring,&c. And the same for the elements: I prefer to think that 3 is always 3; not a different thing as element of $\mathbb {N}$, $\mathbb {Z} _{p}$, or $\mathbb {C}$ &c... --pma (talk) 08:10, 17 July 2009 (UTC)

It's safe to think of 3 as referring to the same thing in $\mathbb {Z}$, $\mathbb {C}$, etc, since there are natural embeddings of these various rings into the larger ones. But the element 3 in $\mathbb {Z} _{p}$ really has a totally different meaning since there's no embedding of $\mathbb {Z} _{p}$ into any of these others. Rckrone (talk) 16:47, 17 July 2009 (UTC)

Ah, of course, we can't have separate notation for everything. But I use $\mathbb {Z} /n\mathbb {Z}$-the-ring and $\mathbb {Z} /n\mathbb {Z}$-the-group often enough that I would like notation to distinguish those two cases. Although now that I think of it, the fact that I use it frequently as a ring and as a group is just a consequence of the particular choice of what courses I am taking; someone else might want notation to distinguish the set, someone else might want notation to distinguish the space, etc. Eric. 216.27.191.178 (talk) 19:57, 17 July 2009 (UTC)

Rckrone, yes, but in fact there is a natural embedding of $\mathbb {Z} _{p}$ in $\mathbb {Z} _{q}$ for p<q; it is not a group homomorphism, but it's there. As I see it, what is totally different is the algebraic structure of $\mathbb {Z} _{p}$ compared to $\mathbb {Z}$, but why should this make 3 a different thing as an element of either set? Take for example me: I belong to totally different groups, but I'm always the same person. But of course everybody is free to see it according to his or her way. --pma (talk) 21:49, 17 July 2009 (UTC)

Rope cutting formula?

My math knowledge is hopeless and this particular problem can by solved by most teenagers eventhough iam an adult. Consider that there is a rope that is about 30 thousand foot long. It is cut exactly into two equal length ropes. Those two ropes are cut again into 4 equal ropes. This goes on and on until the ropes are about 3 hundred foot long. The question is what would be the formula that gives the number of cuts so that the rope is around 3 hundred foot long?. May be the formula will not work when the original rope is too long or too short, I have no idea. —Preceding unsigned comment added by 131.220.46.25 (talk) 12:00, 16 July 2009 (UTC)

Assuming you mean 1 cut gives two pieces of 15,000 ft and 1 more cut then gives four pieces of 7,500 ft... You are dividing the rope into 2^n parts after N cuts. 300 ft is 1/100 of 30,000, so you have to choose between 2^6=64 and 2^7=128. -- SGBailey (talk) 13:39, 16 July 2009 (UTC)

The general solution: with each step you are multiplying the number of ropes by 2 (replacing every single rope with 2 ropes half the size of the original). So, after n steps, you have $R=2^{n}$ equal-sized ropes, of length $l={\frac {L}{R}}={\frac {L}{2^{n}}}$, where L is the original length (30 thousand feet, in this case), and you want to solve for n given a particular l. We can rearrange $l={\frac {L}{2^{n}}}$ to get:

$2^{n}={\frac {L}{l}}$

This can be solved for n by the use of a logarithm, to get:

$n=\log _{2}({\frac {L}{l}})$

Since most calculators have a logarithm button, you can just punch this in to get your answer. You may have to use the base conversion formula, if your calculator only has a base 10 logarithm, to get:

$n={\frac {\log _{10}({\frac {L}{l}})}{\log _{10}(2)}}$

For your particular numbers (that is, L=30000 and l=300), you should get about 6.6. This means, as SGBailey alluded to, that after 6 cuts the ropes will be longer than 3 hundred feet, but after 7 cuts the ropes will be shorter 3 hundred feet. That is to say, they will never be 3 hundred feet long. --COVIZAPIBETEFOKY (talk) 13:52, 16 July 2009 (UTC)

Wikipedia seems to be having some difficulties with the mathematical notation. My guess is that the server is just being slow, and the problems will work themselves out eventually. I'm pretty sure I wrote everything correctly. --COVIZAPIBETEFOKY (talk) 13:55, 16 July 2009 (UTC)

If you start with 30000 feet and cut exactly in half, the sequence of lengths you get is

30000.0,15000.0,7500.0,3750.0,1875.0,937.5,468.75,234.375...

so you never quite get 300. 67.117.147.249 (talk) 16:58, 16 July 2009 (UTC)

Subspace complemented -> Hilbert space

Let X be a Banach space. Suppose every closed subspace M of X has a complementary subspace (that is assumed to be closed); i.e., X is a direct sum of M and some other closed subspace. Does it follow that X is a Hilbert space? In other words, construct a subspace that is not complemented when X is not a Hilbert space (i.e., it doesn't satisfy the polarization identity). Either a textbook on Function Analysis or Wikipedia probably has an answer, but I'm lazy :) -- Taku (talk) 22:58, 16 July 2009 (UTC)

Finite-dimensional Banach spaces have complementary subspaces. Algebraist 23:10, 16 July 2009 (UTC)

Sure :) Assume X is infinite-dimensional in addition. (In face, in view of the Hahn-Banach theorem, such a counterexample must be infinite-dimensional). -- Taku (talk) 23:12, 16 July 2009 (UTC)

The answer is yes, but you are not going to find it in every textbook on Functional Analysis. It's Lindenstrauss and Tzafriri's characterization of Hilbert spaces (more precisely: of the Banach spaces that admit an equivalent Hilbert norm; in particular, of course, any finite dimensional space). Notice that "all closed subspace have a complement" is a topological vector space theoretical property, that is not affected by re-norming. In other words, a Banach space has a non-complemented closed subspace if and only if it admits no equivalent Hilbert norm (so your statement has to be slightly corrected, as A. already pointed out). --pma (talk) 07:13, 17 July 2009 (UTC)

Thank you for the information. Right, this is a topological question; not geometric one (in the sense that one can replace the original norm by the equivalent one.) It sounded like this isn't a trivial result. (Somehow I was thinking: since every Banach space contains a copy of, umm, c_0?, you first assume X = c_0, then somehow generalize it. I guess it's probably not that simple.) -- Taku (talk) 10:39, 18 July 2009 (UTC)

Yes it's quite a deep theorem. If you are interested, and you want to work on related problems (there are still open variants of the complementary subspace problem), or if you are just curious to learn the techniques, maybe a good starting point is Lindenstrauss and Tzafriri's 2-volume book (Classical Banach spaces I&II). After the work of Banach, the geometry of Banach spaces had a great, increasing developement, culminating with the impressive boom of the '70s. In that golden period many of the main problems were solved; still the topic remained active in the subsequent decades, and new spectacular results have been proven, especially in connection with measure theory. --pma (talk) 12:41, 18 July 2009 (UTC)