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November 24

set of equivalence relations

I want to know ,how to define mapping to prove : let ~ be a equivalence relation on the Ring R,we Say that it is compatible (with ring operations) iff a~b implies a+c = b+c , ac = bc ,ca = ba for all a,b,c belongs to R. prove that there is one to one correspondence between the ideals of R and the set of compatible equivalence relations on R . — Preceding unsigned comment added by 182.187.115.152 (talk) 07:05, 24 November 2012 (UTC)[reply]

Hint: Look at the kernel of the map that takes elements of R to their equivalence class:-)Phoenixia1177 (talk) 08:54, 24 November 2012 (UTC)[reply]

i donot get about this mapping ,please explain more — Preceding unsigned comment added by 182.187.116.9 (talk) 13:26, 24 November 2012 (UTC)[reply]

Statistics

According to the WIkipedia page about how many people are homosexual, it says it's 6% in the UK. So if I'm in a room of 40 people (including me), and the chance of any one person being homosexual is 6%, what is there chance that homosexuals are not a minority in the room, so at least 20 people in the room are homosexual.

I haven't done stats in quite a while, and I've forgotten how, so could you please explain how you get to the number. Many thanks for any help you can provide. 143.210.122.148 (talk) 18:12, 24 November 2012 (UTC)[reply]

You could use the binomial theorem and add together the sum of (20 NcR 40) 0.06^20*0.94^20 + (21 NcR 40) 0.06^21*0.94^19 + ...... + (40 NcR 40) .06^40*.94^0 .--Gilderien Chat|List of good deeds 19:54, 24 November 2012 (UTC)[reply]

Hi Gilderien, thanks for your response! I vaguely remember something like that in A-level maths. It doesn't really matter if I don't understand. So I have to type all of that into a calculator? There must be an easier way, no? Typing (20 NcR 40).06^20*0.94^20 + (21 NcR 40).06^21*0.94^19 + ... + (40 NcR 40).06^40*.94^0 would take ages! 143.210.122.148 (talk) 20:10, 24 November 2012 (UTC)[reply]

I doubt in most cases where this would make any difference that the events are independent. The chance of all 40 people in a room being homosexual are far higher than one would get from the binomial distribution. Otherwise there would be very few gay bars for instance. If the OP is included the chances are 0 if they are not homosexual (considering it to be a yes/no type of thing). There is no point the OP including or excluding themselves in the 20 as they know about themselves. Dmcq (talk) 20:42, 24 November 2012 (UTC)[reply]

By the way if you are really insistent on treating them as independent the best way then you can treat it as being approximately a Normal distribution. The mean is np and variance np(1-p), the standard deviation is the square root of this. If you work this out you have on average about 2.4 (plus the OP if that way inclined) and 20 is (20-2.4)/1.5 =11.7 standard deviations away, so the chances are really really tiny that there are any gay bars, I doubt there's any in the world Dmcq (talk) 20:57, 24 November 2012 (UTC)[reply]

Gilderien's formula is a simple line in many math programs, for example PARI/GP:

? sum (n=20, 40, binomial(40,n)*0.06^n*0.94^(40-n))

%1 = 1.556140404291757270521948976 E-14

That's 1 in 64261553600308. Explanation: The chance that a given group of n people are all homosexual is 0.06^n. The chance that none of the other 40-n are homosexual is 0.94^(40-n). The number of ways to choose n out of 40 people is called many things, for example (n NcR 40) by Gilderien, and binomial(40,n) in PARI/GP. See Combination and Binomial coefficient. The chance of exactly n homosexuals in total is binomial(40,n)*0.06^n*0.94^(40-n). The chance of at least 20 homosexuals is the sum of that for n = 20 to 40. This assumes "the chance of any one person being homosexual is 6%" also applies to you. PrimeHunter (talk) 02:40, 25 November 2012 (UTC)[reply]

Continuing Dmcq's approach. The probability that the OP is homosexual is not 6% because he (or she) is not a random person. He knows! If he is a homosexual he needs to know if 19 or more out of 39 random persons are homosexual. If he is not a homosexual he needs to know if 20 or more out of random 39 persons are homosexual. The expected number of homosexuals among 39 random persons is 39*0.06=2.34 homosexuals, and the standard deviation is √(39*0.06*(1−0.06))=1.4831 homosexuals. Now 19=2.34+11.2332*1.4831 and 20=2.34+11.9075*1.4831. The probabilities are the Q-function values Q(11.2332) and Q(11.9075). They are both zero according to [1]. Bo Jacoby (talk) 15:35, 25 November 2012 (UTC).[reply]

Thanks greatly to everyone that responded. I know there are gay bars, but I meant just if it is totally independent. Ok, so it was wrong to say including me, as I know about me, and I'm not homosexual. Most of you have gone far beyond my understanding of statistical maths; I don't at all understand the 11.7 or the Q-functions. But I do get the 1 in 64261553600308 and PrimeHunter's explaination. Thanks to you all! 143.210.122.124 (talk) 17:52, 25 November 2012 (UTC)[reply]

Calculating vector cross products

For this question the two vectors will be ${\vec {A}}={\begin{bmatrix}A_{x}\\A_{y}\\A_{z}\\\end{bmatrix}}$ and ${\vec {B}}={\begin{bmatrix}B_{x}\\B_{y}\\B_{z}\\\end{bmatrix}}$ . Is ${\vec {A}}\times {\vec {B}}={\begin{bmatrix}A_{x}B_{z}-B_{x}A_{z}\\A_{x}B_{z}-B_{x}A_{z}\\A_{x}B_{y}-B_{x}A_{y}\\\end{bmatrix}}$ right? --Melab±1 ☎ 21:38, 24 November 2012 (UTC)[reply]

(ec) No. See the last equation in Cross product#Coordinate notation. Most of your subscripts need to be rethought. Jheald (talk) 21:56, 24 November 2012 (UTC)[reply]

The easiest way to remember it is as a formal determinant:

{\vec {A}}\times {\vec {B}}=\left|{\begin{array}{ccc}{\hat {i}}&{\hat {j}}&{\hat {k}}\\A_{x}&A_{y}&A_{z}\\B_{x}&B_{y}&B_{z}\end{array}}\right|

where

{\hat {i}}

,

{\hat {j}}

,

{\hat {k}}

denote unit vectors in the x, y, and z directions. --Trovatore (talk)

That 3 × 3 determinant may be a useful mnemonic, but there are disadvantages;

clumsy because you have to expand the first row and end up with −j rather than +j every time, when each can be avoided,
abuse of notation: a matrix should have elements all from the same field, not a mix of different fields and vectors of some field,
determinants should be real numbers (by definition), this is a psuedovector.

A better way is

$\mathbf {a} \times \mathbf {b} ={\begin{vmatrix}a_{y}&a_{z}\\b_{y}&b_{z}\end{vmatrix}}\mathbf {i} +{\begin{vmatrix}a_{z}&a_{x}\\b_{z}&b_{x}\end{vmatrix}}\mathbf {j} +{\begin{vmatrix}a_{x}&a_{y}\\b_{x}&b_{y}\end{vmatrix}}\mathbf {k}$

because this doesn't have the same problems, and the columns and basis vectors are just cyclic permutations of x, y, z components. Maschen (talk) 09:50, 25 November 2012 (UTC)[reply]

I think that's harder to remember, though. --Trovatore (talk) 09:59, 25 November 2012 (UTC)[reply]

Actually, there are other advantages to the formal determinant. It makes it obvious that the cross-product is antisymmetric, bilinear, and normal to both the original vectors. Once you have that you're not that far from nailing down the cross-product uniquely.

I disagree with some of your other statements:

First, a matrix is just a way of encoding a linear transformation. There is not in fact any requirement that all entries be from the same field. In fact, lots of matrices have elements that are other matrices.
Determinants don't have to be real numbers. They can certainly be from any field, maybe any (commutative?) ring — I'd have to think about it, or our determinant article may say something in this regard. In any case, the central feature of determinants is that they are multilinear and antisymmetric, which, lo and behold, are also two of the three most important facts about cross-products.

It's certainly true that the cross-product is not a determinant in the usual sense, but I think it's more finickiness than any real mathematical virtue to stick on that point. --Trovatore (talk) 10:13, 25 November 2012 (UTC)[reply]

The cyclic permutations are simply (y, z, i), (z, x, j), (x, y, k), or using a more sensible notation for the basis vectors (y, z, e_x), (z, x, e_y), (x, y, e_z) where (i, j, k) = (e_x, e_y, e_z).

An alternative is the cross product matrix:

\mathbf {a} \times \mathbf {b} =[\mathbf {a} ]_{\times }\mathbf {b} ={\begin{bmatrix}\,0&\!-a_{3}&\,\,a_{2}\\\,\,a_{3}&0&\!-a_{1}\\-a_{2}&\,\,a_{1}&\,0\end{bmatrix}}{\begin{bmatrix}b_{1}\\b_{2}\\b_{3}\end{bmatrix}}

Which may be easier? Maschen (talk) 10:06, 25 November 2012 (UTC)[reply]

See my remarks above. I've re-thought my original weak response; I think the determinant answer is the best one, and both the ones you mention are inferior to it. --Trovatore (talk) 10:14, 25 November 2012 (UTC)[reply]

Very well, just commenting. Best, Maschen (talk) 10:17, 25 November 2012 (UTC)[reply]

$(A\times B)_{i}=\varepsilon _{ijk}A_{j}B_{k}$

where $\varepsilon _{ijk}$ is the Levi Civita pseudotensor. Count Iblis (talk) 04:14, 26 November 2012 (UTC)[reply]

For myself, the mnemonic I find instinctive for the cross product of two column vectors A and B, like the two up in the question, is to start with the first component, for that component forget the first row of the the two vectors, and make a "cross" out of the second and third row just like the determinant of a 2x2 matrix, so A_yB_z - A_zB_y. Then, for the next component, slide that "cross" down with wrap-around, so it still starts just the below the row you now want the component for, so: A_zB_x - A_xB_z. Then for the last component, slide the "cross" down one more (it's actually wrapped around completely now), so A_xB_y - A_yB_x.

Yes there are of course more sophisticated ways to think about the cross-product. But if I actually have to work one out from two column vectors, the above is still what I think to myself in my head while I'm doing it. Jheald (talk) 12:29, 26 November 2012 (UTC)[reply]