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September 16

What is the correct pronunciation of George Pólya's last name?

What is the correct pronunciation of George Pólya's last name? (I consider the way he pronounced his name to be the correct pronunciation.) —Preceding unsigned comment added by 98.114.146.57 (talk) 04:20, 16 September 2009 (UTC)

here is it [1]--84.220.118.109 (talk) 06:08, 16 September 2009 (UTC)

Hungarian pronunciation: [ˈpoːjɒ ˈɟ͡ʝørɟ͡ʝ]. — Emil J. 10:35, 16 September 2009 (UTC)

How many combinations are there?

We have a lock box on our house similar to this: http://www.buyasafe.net/S6-Supra-Pushbutton-Lockbox-Surface-Mount./M/B000M7OXPY.htm?traffic_src=froogle

My question is, how many possible combinations are there?

Here are a few things to consider: 1) There are 10 possible digits (numbers 0 through 9). 2) The order the numbers are pressed doesn't matter. 3) We chose a 4-digit combination; however, we could have selected anywhere from a 0 - 10 digit combination.

To break the 4 digit combination, the odds would be 10 x 9 x 8 x 7 = 5,040, right? However, when you consider we could have selected any number of digits, how does this affect the total number of possible combinations?

Thanks in advance for your help! —Preceding unsigned comment added by 209.206.158.57 (talk) 17:02, 16 September 2009 (UTC)

Your 5,040 isn't right, because that distinguishes different orders. There are 5,040 4-number permutations using 1-10 w/out repetition. You want to divide that by 4! = 24, meaning there are only 210 4-number combinations.

Now, if you add up the number of 0-number combinations (1) + the number of 1-number combinations (10) + the number of 2-number combinations (45) + . . . + the number of 10-number combinations (1), you'll get 2^10 = 1,024.

One way to see why it's 2^10 is that each possible code either uses or doesn't use each of the 10 numbers. Thus, you either use '1' in your code, or you don't - 2 options. You either use '2' or you don't - 2 options. . . You either use '10' or you don't - 2 options. Multiply all those 2's together, boom.

Does that answer your question? -GTBacchus^(talk) 17:19, 16 September 2009 (UTC)

Sorry if I'm being a bit slow, but I don't see how that's right. Surely a code using a '5' say, could use it more than once? As a disproof (I think!), imagine a similar lock using two digits, which can either be 1,0 or not used. That leaves 6 combinations, which isn't 2^2.--Leon (talk) 17:33, 16 September 2009 (UTC)

No, I've used locks like this. When you press a button, it stays pressed, and having all n buttons down at once, with no others down, is how it opens. Thus, with two buttons, labelled '1' and '2', the 4 possible codes are {}, {1}, {2}, and {1,2}. We're really just looking at the power set of {1,...,n}. -GTBacchus^(talk) 18:45, 16 September 2009 (UTC)

I'm sure GTBacchus is right. But for an edit conflict, I would have posted a similar answer. The person who asked the question stipulated that the order the numbers are pressed in doesn't matter. This implies a simple mechanism which doesn't care if a button is pressed once or many times. If you have two digits, you have two 1-digit combinations, not six. RupertMillard (Talk) 17:53, 16 September 2009 (UTC)

Apologies: I neglected to check the link, and I presumed that a number could be selected multiple times!--Leon (talk) 18:53, 16 September 2009 (UTC)

Here's a puzzle. How many distinct, valid combinations are years since 1900? Harder - how many are dates? (eg. 0916) How many combinations does this total for you to try to crack the safe if you knew the combination was a date or a year? I think the answer is the days of our age, but I'm not certain. RupertMillard (Talk) 18:55, 16 September 2009 (UTC)

To your second question... combinations representing dates can involve 2, 3, or 4 buttons (1/10, 1/23 and 12/03 being the first occurrence of each). I count 82 combinations that can represent 149 different dates. If a thief can try a combination in 5 seconds, that means 410 seconds, which is just under 7 minutes to run through all date combinations. I would note that 0916 is not actually an option, because there's no zero button. If there were a 0 instead of a 10, then only 49 combinations represent dates, which is just over 4 minutes of burgling. -GTBacchus^(talk) 20:10, 16 September 2009 (UTC)

Interpretation help

Hi. I have a small problem with some homework but it's only in understanding what the question means.

"Let $g(z)={\frac {z-i}{z+i}}$ where z is a complex number. Writing $g^{2}(z)$ for $g(g(z))$ and $g^{-1}(z)$ for the inverse function, what is the image of the real line under the successive maps $g$, $g^{2}$, $g^{3}$? What is the image of this set of images under these maps?"

Letting z=x+iy, does the first part just mean determine $g(x)$, $g^{2}(x)$, $g^{3}(x)$? And I don't have a clue for the second part. Any ideas? Thanks. 92.2.21.125 (talk) 19:56, 16 September 2009 (UTC)

They want you to determine the images of R under these linear fractional transformations, that is, the sets A_n:=gⁿ(R)={gⁿ(x): x in R}, for n=1,2, and 3. The other question is not clearly stated indeed; probably they just want gⁿ(A_k), which is trivially A_n+k; or they mean the image of the union of the sets A_k under the maps gⁿ. --84.221.68.30 (talk) 20:19, 16 September 2009 (UTC)

(edit conflict) Finding the "image of the real line" means that, if you plug in every point on the real line to g, and then look at where they all end up, what shape on the complex plane does that give you? For example, the point 0 gets sent to -1, the point 1 gets sent to g(1) = (1-i)/(1+i) = -i, etc. Try plugging in some more real numbers, and see if you can make a sketch of where the real line is going, point by point. Once you have an idea, try and prove your conjecture.

The second part of the question seems to be saying, do the same thing, but instead of just the real line as your starting set, use the set $\mathbb {R} \cup g(\mathbb {R} )\cup g^{2}(\mathbb {R} )\cup \cdots$. Does that help at all? -GTBacchus^(talk) 20:29, 16 September 2009 (UTC)

OK I'm with the first part of the question now - just determine the sets (/regions of the complex plane) produced by the results of $g(x)$, $g^{2}(x)$, $g^{3}(x)$. That was mainly what I was thinking before, it's just that the wording '... the successive maps...' seemed to imply you might do $g^{3}(g^{2}(g(x)))$ but clearly not. You both seem to be going for the $g^{n}(\mathbb {R} \cup g(\mathbb {R} )\cup g^{2}(\mathbb {R} )\cup \cdots )$ idea for the second part so I'll try that. Thank you both! 92.2.21.125 (talk) 20:43, 16 September 2009 (UTC)

But notice that what is the image of the real line means: what set is g(R) --you should find that it's a circle. So you do not have to find what is the image of every single real point g(x). Also, recall that the composition of linear fractional transformations is easily computed multiplying the associated 2x2 matrices of coefficients. Here g³ sounds like a sort of well known map ;-) --84.221.68.30 (talk) 21:00, 16 September 2009 (UTC)

Of course nobody's going to try every single real point. We're mortal. I think he's supposed to figure out whether it's a circle; that's why it's homework. Let's not give away so much, eh? -GTBacchus^(talk) 21:18, 16 September 2009 (UTC)

For extra credit (and the pleasure of a Eureka moment) visualise and describe the action of g(z) on the Riemann sphere. Gandalf61 (talk) 09:42, 17 September 2009 (UTC)

Ooh, nice! Thank you. :) -GTBacchus^(talk) 14:44, 17 September 2009 (UTC)

I've just worked through g(R) and, once simplified, have $g(x)={\frac {x^{2}-1}{x^{2}+1}}-{\frac {2ix}{x^{2}+1}}$. Now I can show that the modulus of this is 1 but when I graph this, I most certainly do not get a circle. Where am I going wrong? 92.2.18.79 (talk) 14:24, 17 September 2009 (UTC)

Right, for $g^{2}(x)$ I have $g^{2}(x)=(1-i){\frac {x+1}{x-1}}$ but don't know where to go from here. Taking the modulus still leaves it in terms of x, which doesn't seem helpful. Any suggestions? Thanks 92.2.18.79 (talk) 14:48, 17 September 2009 (UTC)

Yeah, double-check your work. That's not g^2(x). -GTBacchus^(talk) 14:55, 17 September 2009 (UTC)

My mistake. $g^{2}(x)=-i{\frac {x+1}{x-1}}$. So does this just mean that every point on the real line, excluding x=1, is mapped onto a point on the imaginary line? 92.2.18.79 (talk) 15:06, 17 September 2009 (UTC)

Yes, you've just shown that $g^{2}(\mathbb {R} )\subset i\mathbb {R}$. If you can show inclusion in the other direction, then you'll have $g^{2}(\mathbb {R} )$! (That's an exclamation point at the end, and not a factorial, of course.) -GTBacchus^(talk) 15:24, 17 September 2009 (UTC)

OK, $g^{3}(x)=x$ and so every point on the real line is mapped onto itself. Now for the final part of the question, I just need to check the circle. I managed to get $g(x)={\frac {x^{2}-1}{x^{2}+1}}-{\frac {2ix}{x^{2}+1}}$ and that its modulus is one. So is the equation of this circle given by $(x-1)^{2}+y^{2}=1$? 92.2.18.79 (talk) 18:31, 17 September 2009 (UTC)

If the modulus of every point on the circle is 1, then where should the center of the circle be? -GTBacchus^(talk) 19:04, 17 September 2009 (UTC)

Hmm, I sense you mean the origin. This is the first time I've ever considered the modulus of the points on the circle, rather than the distance of the points o the circle from the centre, to determine the centre. Thanks GTBacchus. 92.2.18.79 (talk) 19:11, 17 September 2009 (UTC)

You're quite welcome. Complex analysis is fun; enjoy! -GTBacchus^(talk) 19:15, 17 September 2009 (UTC)

Expectation

Resolved

So, I have emailed my TA a question once and his response indicated he did not understand my question at all. I just got back from asking him other questions in person and he again did not understand what I was even asking, though I explained it several times. The professor does not have any office hours. So, I come here.

I'm doing a simple problem and I need help in understanding. The problem says to show if you have random variables X and Y then

E(max(X, Y)) = EX + EY - E(min(X, y))

They even give a hint to show that X + Y = max(X, Y) + min(X, Y). Well, this is a problem that is intuitively simple but I don't know how to explain it. I know if I am dealing with 2 real numbers x and y, then x + y = max(x, y) + min(x, y) and the proof is easy. So, it seems obvious to me from this that the same equation with random variables must be true. But, just saying it is obvious does not prove it. This is my only difficulty with this problem. Clearly, if I prove the equation with random variables, I can just take the Expectaion of both sides and use linearity and then subtract the min from both sides.

One thought is to let A = X + Y and B = max + min, new random variables. Then, P(A = a) = P(X + Y = a) = P(X = a - Y)... I don't know. We have only talked about functions of one random variable, but not functions of multiple random variables.StatisticsMan (talk) 20:48, 16 September 2009 (UTC)

According to Random variable#Formal definition X and Y are both functions. Taemyr (talk) 21:15, 16 September 2009 (UTC)

I know they are functions. What does that have to do with my question? StatisticsMan (talk) 21:22, 16 September 2009 (UTC)

ecThis: you already have all you need. Recall that X and Y are in particular functions Ω →R, so at any ω in Ω, X(ω) and Y(ω) are certain real numbers, and for them you do know the equality:

X(ω) + Y(ω) = max( X(ω), Y(ω)) + min( X(ω), Y(ω) );

this equality for all ω in Ω just means the equality between functions

X + Y = max(X, Y) + min(X, Y);

so you conclude as you said by linearity of the expectation. --pma (talk) 21:24, 16 September 2009 (UTC)

Well, I guess it has everything to do with my question! And, I believe I understand this. Thanks a lot! StatisticsMan (talk) 02:41, 17 September 2009 (UTC)

September 17

Formula for placing things equidistant along a circle circumference

As the subject implies, i need to find a formula to place items an equal distance apart on the circumference of a circle. Specifically i need 24 equidistant locations. I dont know quite yet how i will define the circle, probably with parametrics? But ive looked on google for an answer to this and there doesnt seem to be any. Please help?

Thanks! 137.81.115.58 (talk) 01:26, 17 September 2009 (UTC)

You can define a circle with radius R in terms of the angle θ around the origin with the parametric equations x = Rcosθ, y = Rsinθ. Evenly spaced points along the circle are at evenly spaced angles, so we divide the full 2π angle of the circle up into 24ths. The points with coordinates (Rcos(2πn/24), Rsin(2πn/24)) with n ranging from 1 to 24 satisfy the conditions you're looking for. Rckrone (talk) 02:44, 17 September 2009 (UTC)

See Root of unity. Bo Jacoby (talk) 08:58, 17 September 2009 (UTC).

Note that 24 = 6x2x2 - you can easily divide the circle into 6 arcs, and then half those arcs twice. When you can define how you want to describe the circle it will be easy to give a method for the process.87.102.94.154 (talk) 11:12, 17 September 2009 (UTC)

help me ...it will safe my future..??

sir/mam....which extra book i adopt ?? i am the student of commerce stream..i choose maths because i want to become good C.A.

you are good to go rambharose..best of luck

You are going to need to be more specific. --Tango (talk) 17:56, 17 September 2009 (UTC)

Quite so :o) ~~ Dr Dec (Talk) ~~ 18:30, 17 September 2009 (UTC)

You seem like a nice person, though I'm not sure who 'rambharose' is. I'm going to guess that you need an introductory maths book for a commerse coarse, possibly including accountancy and economic maths? (and maybe statistics?) Maybe someone else would be good enough to give a suggestion?83.100.251.196 (talk) 19:12, 17 September 2009 (UTC)

Rambharose means in trust of Rama(God) in Hindi:)--Shahab (talk) 18:42, 19 September 2009 (UTC)

Writing Roman Numerals

Hello

Hope someone can answer this as I can't find it in the system. I understand how roman numerals work but get stuck at large numbers, ie one million is M with a bar, but how do you write 2 million or 10 million etc ?

--Bruceglasgow (talk) 18:42, 17 September 2009 (UTC)

Two million derives from one million just as two derives from one. If one million is M, two million is easy to generate: MM. Alternately, parentheses may be used to represent "times 1000", such that two million is (MM). Ten million requires some extrapolation from our Roman numerals article, but might be fairly represented as (X), or a longer form. — Lomn 19:21, 17 September 2009 (UTC)

Equation of oscillator - flat section

A question arises from another persons question about oscillations of things travelling through a spherical body exerting and inverse square force law Wikipedia:Reference_desk/Science#Gravity_at_the_center_of_a_body.

My question begins: If a spherical centre part is removed then within this part the field force is zero. (as per Shell theorem), assuming a body is dropped into the whole hollow sphere (assumming a negligable cylindrical hole for it to travel in or whatever) - then it seems to me that the body will accelerate towards the centre, then when it reaches the empty part no field is experienced and the body will move at constant velocity, until it exits the spherical hole at which point the deacceleration occurs, and so on. Thus it oscillates.. but with a part at which the acceleration is zero...

So the equation of motion (eg acceleration vs time) I would expect to be expressable as a fourier series (ie a series of powers of sines or similar).. Since the function is periodic. But at the times at which the particle is inside the spherical void the acceleration is 0, as are the derivatives of the accelerations.. This seems to contradict the taylor-maclaurin type theorems (ie that a function at any point can be constructed from the function at a given time and all it's derivatives.

My questions are:

• Can the function be expressed as a series of sines, ie a fourier series
  • If not what sort of function
• What goes wrong with the taylor-maclaurin series here? (or what is my error?)83.100.251.196 (talk) 19:22, 17 September 2009 (UTC)

Taylor-Maclaurin-type results apply to analytic functions. This function is not analytic. The Fourier series of position as a function of time would converge to that function. This is one of the difference between Fourier series and power series. Michael Hardy (talk) 19:57, 17 September 2009 (UTC)

The page analytic function seems to define itself as "A function is analytic if and only if it is equal to its Taylor series in some neighborhood of every point." (which is somewhat circular in respect to your statement, but I'm not arguing this).

The actual function of this oscillation can be expressed as a fourier series, and so also as a power series (is there an exception here?). I imagine that the square wave is a simple(r) example to discuss.

Is there another way of showing that a given power series (or fourier series) will not be analytic (other than trying the taylor maclaurin series).

Does the fact that eventually a differential of the function includes a discontinuity mean that it will be non-analytic ? - it seems to me that this will be the case, but can't imagine a proof.83.100.251.196 (talk)

You can't have a power series that's not analytic, since a power series is analytic by definition. You're right that a function has to be smooth in order to be analytic. To prove it, any power series is smooth (assuming it converges), so if a function isn't smooth, it's not equal to any power series. The Taylor series article mentions that the converse isn't true: there are smooth functions that aren't analytic.

As for the Fourier series, sine functions are analytic, and so any finite sum of sine functions is analytic, but that doesn't necessarily extend to an infinite sum. So the partial sums of the Fourier series are analytic, but they may converge to something that's not analytic, which is the case with the function being discussed. Rckrone (talk) 21:41, 17 September 2009 (UTC)

A power series is analytic by definition: not quite so; to be precise, you need to prove that at any point within the convergence disk of your power series, the function represented by the series is also representable by a power series centered at that point: not difficult, but still a computation. --84.221.68.30 (talk) 22:34, 17 September 2009 (UTC)

ok thanks so far, For the square wave the 'x¹' term would be of the form 1+1/3+1/5+1/7... etc that relates to pi , (I haven't checked the other terms x^3 etc) , but it looks like it can be expressed as a infinite power series (is this right?).

I'm not very familiar with the radius of convergence - though at first sight it looks like the square wave will converge for all values of x .. ah not for the differentials.

I'm fairly familiar with smooth functions though.

It would be great if a few points could be clarified - specifically about the square wave (I'm confused because I think it is non-analytic, yet expressible as a power series):

• Can the square wave be expressed in the form a₀ + a₁x¹ + a₂x² + a₃x³ + etc all the way to infinity. (as I suspect) - but this form does not converge - but oscillates and diverges does this relate to it being non-analytic?
  • Is this form a power series, or is this the definition of a power series more complex.
• The radius of convergence of the square wave - infinite ? but only for the fourier series not for the differentials, starting to partially understand.
• Apart from lacking smoothness (ignoring smooth non-analytic functions for now) is there another test for lack of analyticalisity in a function? (apart from the obvious taylor maclaurin method)? The test doesn't have to be absolute - just able to reject a set of equations.. Thanks/83.100.251.196 (talk) 23:39, 17 September 2009 (UTC)

I think you might be confusing approximating the square wave with a sequence of analytic functions (such as with partial Fourier series), and a single power series that converges to the square wave. There's no power series that converges to the square wave, which is to say that it's not analytic. We know this because it's not smooth, and all analytic functions are smooth. We can however approximate the square wave with analytic functions, for example with the Fourier series. The partial sums of the Fourier series give us a sequence of different analytic functions that approach the square wave (at most points).

Besides smoothness I don't know what other criteria there are to decide if a function is analytic besides just comparing it to its Taylor series. Maybe someone else who knows more about this has something. Rckrone (talk) 01:23, 18 September 2009 (UTC)

Yes exactly that, I know see why attempting to make the corresponding power series for the fourier series fails - because the terms for x are all infinite (or non converging) beyond x¹.

I get the principle now thanks.

Resolved

83.100.251.196 (talk) 11:25, 18 September 2009 (UTC)

The taylor series for g(z) defines the fourier series for f(x) = g(e^ix) . Bo Jacoby (talk) 10:04, 18 September 2009 (UTC).

I'll have a look at that too, thanks.83.100.251.196 (talk) 11:25, 18 September 2009 (UTC)

This corresponds to the rather unusual case when the nth (exponential) Fourier coefficient of f is zero for every negative n. In general, the Laurent series for g defines the Fourier series for f. However, in practice the Laurent series corresponding to the Fourier series of a typical periodic function will have both radii of convergence equal to 1, meaning that it's degenerate (its annulus of convergence is empty). — Emil J. 16:34, 18 September 2009 (UTC)

September 18

Substitution in the integral

Suppose we have the integral of the form $\int _{0}^{h(x)}f(g(t))dt$ and then we do a substitution of the form $g(t)=u$. what will happen for the integral limits, 0 and $h(x)$ ? —Preceding unsigned comment added by Re444 (talk • contribs) 15:32, 18 September 2009 (UTC)

I think of it by imagining a little "t=" next to each limit. You want those to turn into "u=", so figure out what u is when t=0, and similarly when t=h(x). Does that do it? -GTBacchus^(talk) 15:36, 18 September 2009 (UTC)

Oh thanks, I really need that! It solved all my problems. Re444 (talk) 18:53, 18 September 2009 (UTC)

A really brainy sum !!!!!

My friend sent me this question via mail and i tried various options for getting this sum.Even after 3 hrs of futile trying I failed pls help.The question as follows

A farmer was dividing his property among his children. This was what the farmer told to his first son “Take as many number of cattle as you can care for and your wife may take one-ninth of the remaining number.” To his second son he said “Take one more than what the first son took and your wife will have one ninth of the cattle remaining after you have taken.” This applied to the remaining sons too i.e. each son would take one more than the next oldest brother and their wives would have one ninth of the remaining after their husbands have got their share.

After the cattle were divided, the farmer proceeded to divide the gold bars he had among the children. Each gold bar was valued at 3.5 times each cow's value. The gold bars were divided such that each couple had equally valued inheritance. What are the number of cows, gold bars and the number of sons the farmer had?

My deductions :

1. Obviously the wife of last son will not get any cows as there should be no cow left.
2. The number of cows of each couple consists of three terms.All terms except the constant term follow a particular pattern except the integer.
3. The difference between the number of cows(assuming the cost of a cow is 1 unit ) of each couple should be equal to a multiple of 7.

so pls help...3 to 4 hrs of working on this didnt help much :-(164.100.170.4 (talk) 16:14, 18 September 2009 (UTC)

Does the wife round down or up?83.100.251.196 (talk) 16:56, 18 September 2009 (UTC)

The wives don't round. That's part of the problem, is that there's always a multiple of nine left when the wife chooses.

Nice problem. This'll take a minute... -GTBacchus^(talk) 19:37, 18 September 2009 (UTC)

what does that mean???

Simplest non-trivial solution: 2 sons, 135 cows, 2 bars of gold. The first son takes 63 cows, leaving 72. His wife takes 8. The second son gets the remaining 64 cows, and the second wife, none. That's 71 cows for the first couple, and only 64 for the second. The 2 bars of gold make it even.

Of course, if there's only 1 son, it's trivial, and (2 sons, 16 cows, no gold) works, but isn't very interesting. There are other solutions, with 2 sons. Not sure about 3 or more. -GTBacchus^(talk) 20:13, 18 September 2009 (UTC)

(after ec) The problem has several possible solutions. For example:

• Number of sons = 1; Number of cows = arbitrary; number of gold bars = arbitrary. The first son takes all the cows, and his wife gets none.
• Number of sons = 2; Number of cows = 16; Number of goldbars = any even number, say 2p. First son takes 7 cows, his wife gets (16-7)/9 = 1; second son gets 7+1=8, and his wife gets none. Both couples have 8 cows each, so the gold bars are divided equally.

I am confident that there are other solutions with more sons, but I haven't worked them out yet. Note though, that the number of gold bars will always be non-unique, because for any solution with n sons, one can always add np to the number of gold bars while meeting all the conditions of the problem. Abecedare (talk) 20:15, 18 September 2009 (UTC)

Ah, good point about the gold. You can always throw more gold around, as long as the extra is divisible by the number of couples. For more unique solutions, one would add the condition that the couple with the most cows gets no gold. -GTBacchus^(talk) 20:18, 18 September 2009 (UTC)

There are no solutions with 3 sons. The number of cows the first son takes would have to be congruent to both 6 and 0, modulo 64, which is impossible. -GTBacchus^(talk) 20:24, 18 September 2009 (UTC)

I can post the general solution for 2 sons, but I don't want to ruin any more of the fun for others... feel free to post to my talk page for it. -GTBacchus^(talk) 20:28, 18 September 2009 (UTC)

Hmmm, I found a few solutions with three sons!

• Number of cows = 24; Goldbars = 3p; First son takes 6 cows, and his wife takes (24-6)/9=2; Second son takes 7 cows and his wife takes (24-6-7-2)/9=1; Third son takes the remaining 8 cows. Each couple has 8 cows, and can divide the gold evenly.
• Number of cows = 1543; Gold bars = 3p+52. First son takes 454 cows, and his wife gets (1543-454)/9 =121 cows. Second son takes 455 cows and his wife gets (1543-454-455-121)/9 =57 cows. Third son takes the remaining 456 cows. So first couple has 575 cows, second couple 512 cows and third couple has 456 cows. So first couple gets p goldbars, second couple p+18 goldbars, and third couple p+34 goldbars, to even the inheritence.

I am guessing there are solutions with even more number of progeny, but I'm donw with this family :-) Abecedare (talk) 20:40, 18 September 2009 (UTC)

Huh, what did I do wrong? Runs off to double-check congruences... -GTBacchus^(talk) 20:42, 18 September 2009 (UTC)

Aaargh. How silly of me. Thanks for the down-a-peg; those are good for the soul. -GTBacchus^(talk) 20:44, 18 September 2009 (UTC)

(ec) It's easier to run it in reverse. Start with no cows on the pile. First wife adds no cows, first son (last really) adds some number of cows that has to be a multiple of 8, call it 8k, so that the second wife can add 1/8 of the existing pile. The second son adds 8k-1, and the new pile (with 17k-1) now has to be a multiple of eight for the third wife, etc. The only exception is that the last son (first really) doesn't need to make the pile a multiple of 8. For each additional son that there is in the family, there are stricter constraints, but always an infinite number of solutions. For example with 2 son there are no constraints on k. With 3 sons, k = 1(mod 8). With 4 sons k = 1(mod 64). With n sons, k = 1(mod 8^n-2) I think (it's definitely a sufficient condition, but it might be looser than that). Then there's the gold bar thing which adds some more constraints. 1 son puts no constraints on k. With 2 sons, k = 1(mod 7). I'm less sure about how this one grows with the number of sons, although I think requiring k = 1(mod 7^n-1) is always safe. So combining these, for n sons, requiring k = 1(mod 7*56^n-2) should always get you solutions I think, but I'm not totally sure.

Although there are infinite solutions for any given number of sons, an easy solution that pops out is k=1. Each couple gets 8 cows, and no gold is required. That would work for any number of sons, except that the cows each son takes has to be non-negative, so it fails for more than 9 sons. Rckrone (talk) 21:19, 18 September 2009 (UTC)

That is easier. I made my variable the number of cows that the first son takes, which took a bit more calculating. -GTBacchus^(talk) 21:26, 18 September 2009 (UTC)

With no rounding I came to the conclusion that there could be any number of sons - though clearly as 'sons' increase the number of cattle increases vaguely 'factorially' (to satisfy the modular relationships). The gold bars don't seem to affect it at all (at any number of sons I can multiply all cattle by 7 to get a integral relationship with number of gold bars..) Did anyone find a reason not to suggest any number of sons?

the issue I have is that there doesn't seem to be a proper answer 83.100.251.196 (talk) 22:00, 18 September 2009 (UTC)

"at any number of sons I can multiply all cattle by 7" : That's not true since the condition that each son recieves one more cow than his elder brother would then not be satisfied.

The analysis by Rcrone above is right on the point: we need to simultaneously satisfy congruency relations with respect to some power of 8 and some of power of 7, and since this is achievable for any power n, (infinitely many) solutions exist for any given number of sons. Abecedare (talk) 22:10, 18 September 2009 (UTC)

Yes, I just came back to correct that.83.100.251.196 (talk) 22:34, 18 September 2009 (UTC)

I think with last wife taking no cows the method fails after 2 sons, (though probably made a mistake) I did. Doing sons wrong way round.83.100.251.196 (talk) 22:50, 18 September 2009 (UTC)

cows=16 sons=2

first son = 7 , wife = 1

second son 8 , wife =0

cow difference = 0, thus no gold

No solution for 3 sons I can see.(reason:algebraic (5k-1)/8 must equal integer for integer k - impossible)

Solutions also exist for cows = 16+34n , fs=16n+7,fw=2n+1,ss=16n+8,sw=0, cow difference=2n = 7n gold.

Ignore 1 son solution since it seems irrelevent to the question. Is that ok?83.100.251.196 (talk) 23:30, 18 September 2009 (UTC)

Doh, me dumbo see correct solution below.83.100.251.196 (talk) 00:09, 19 September 2009 (UTC)

(5k-1)/8 is an integer if k=5+8n for any integer n. In that last solution, remember, ss=fs+1.

General solution for 2 sons: let k be any non-negative integer.

Cows: 16+119k.

Gold: 2k.

Son 1 takes: 7+56k cows.

General solution for 3 sons: let k be any non-negative integer.

Cows: 24+1519k.

Gold: 52k.

Son 1 takes: 6+448k cows.

General solution for 4 sons: let k be any non-negative integer.

Cows: 32+17225k.

Gold: 902k.

Son 1 takes: 5+3584k cows.

-GTBacchus^(talk) 23:51, 18 September 2009 (UTC)

Can anyone write it down for n sons? I'm working on 5... -GTBacchus^(talk) 23:56, 18 September 2009 (UTC)

Empirically observed pattern: first son's take when there are n sons = (9-n) + (8^(n-1))*7k. One can trace back the total number of cows from this, but I'm too lazy to do that at the moment. My conjecture is that this pattern will hold for n<9, and possibly for n=9. Abecedare (talk) 03:36, 19 September 2009 (UTC)

September 19

Field extensions

Hi. I'm working through a p-adic analysis book, and there's a quick review of fields at the beginning of chapter 3. I read the following there:

A field extension K of F is an F-vector space; if it is finite-dimensional, it must be an algebraic extension, and its dimension is called the degree [K:F]. If $\alpha \in K$ has the property that every element of K can be written as a rational expression in $\alpha$, we write $K=F(\alpha )$ and say that K is the extension obtained by adjoining $\alpha$ to F.

Two paragraphs later:

If F is a perfect field, then any finite extension K of F is of the form $K=F(\alpha )$ for some $\alpha \in K$. $\alpha$ is called a primitive element. Knowing a primitive element $\alpha$ of a field extension K makes it easier to study K, since it means that everything in K is a polynomial in $\alpha$ of degree <n.

Super. So... the first paragraph there says that everything in K is a rational expression in $\alpha$. The second says that everything in K is a polynomial in $\alpha$. Those aren't equivalent, are they? What polynomial in $\alpha$ is the same as ${\frac {1}{1-\alpha }}$, for example? What am I missing? Is the polynomial thing something extra that we get when F is perfect? -GTBacchus^(talk) 00:10, 19 September 2009 (UTC)

They're equivalent as long as α is algebraic over K. Then any rational expression in α over K can be rewritten as a polynomial, using the extended Euclidean algorithm. For example, if α is a root of the polynomial x²-x+1, then 1/(1-α)=α. Algebraist 00:59, 19 September 2009 (UTC)

The algorithm at that link seems to apply in a finite field. What if we're talking about an extension of $\mathbb {Q}$, for example? That's a perfect field, because is has characteristic 0, right? -GTBacchus^(talk) 01:02, 19 September 2009 (UTC)

The algorithm works whenever α is algebraic over the ground field (perfection is not relevant, and neither is finiteness). Algebraist 01:09, 19 September 2009 (UTC)

Ok, I'll study that. Thank you. -GTBacchus^(talk) 01:29, 19 September 2009 (UTC)

calculations over three years

I learned one porcelain bowl from the estate of Barbara Hutton was auctioned and sold by Christie's Hong Kong for $22,240,000 in 2006. When I tried to use the inflation calculator provided by the United States Department of Labor to figure out what that sum of money would be today, I ran into a problem. The original sum has to be at least $10,000,000 or less. So if anyone could improvise and help me figure out what $22,240,000 in 2006 would be in today's money, that would be great. Thank you.69.203.157.50 (talk) 05:43, 19 September 2009 (UTC)

Try the inflation calculator on $22,240 and multiply the result by 1,000. Bo Jacoby (talk) 06:14, 19 September 2009 (UTC).

Method to generate symmetries of a point lattice

I am interested in algorithmically generating the symmetries of a 2D or 3D point lattice, given reduced lattice basis vectors. I know that there are a finite number of crystallographic point groups in these dimensions, but I am wondering how those tables of symmetries were generated. Is there an automatic way of discovering lattice symmetries? Victor Liu (talk) 19:39, 19 September 2009 (UTC)

One way you could go about it is to begin at the origin and generate all the lattice points near the origin, expanding your search outward until you have in your collection 3 points that are linearly independent (for a 3D lattice). Then divide all the points in your collection into sets of points that are equidistant from the origin. Any symmetry has to map the points in each set into one another, and any isometry that successfully maps all the points in your collection to other points in the collection is a symmetry of the lattice. That should narrow down the search for what isometries work to something finite. There might be something more clever though. Are you looking to implement this on a computer? Rckrone (talk) 06:56, 20 September 2009 (UTC)

Yes, I plan to implement this on a computer. I thought it would be interesting to be able to generate the standard tables of crystallographic point groups in 2D and 3D, and then see if I can go to higher dimensions. That's the mathematically curious reason. The practical reason is to implement [2] very generally, among other possible applications. Your proposal sounds good. I'm think to use the set of equidistant nearest neighbors from the origin, and trying rotations and flips on them. I still need to convince myself this works. 76.102.161.22 (talk) 08:36, 20 September 2009 (UTC)

made link visible - no reference section on this page83.100.251.196 (talk) 09:47, 20 September 2009 (UTC)

Just the nearest neighbors won't be enough in general, since that set might not span the lattice. Consider for example the 3D lattice Z×2Z×3Z. You may need to look at the sets of points at several different distances so that the union of the sets spans. For the nD case, you could need up to n different sets. Rckrone (talk) 23:18, 20 September 2009 (UTC)

September 20

Collection of six related primes

The numbers 931981, 941981, 961981, 971981, 991981 and 1001981 are all prime. I have verified that no other number of one to four digits can be put in place of 1981 for this to remain true, and crude estimates of how many five- and six-digit numbers there are for which the six implied numbers will all be primes are 'around two' and 'around eight-and-a-half'. (The formula used for the first approximation is 12000*[3.75/log(9700000)]^6, where 12000=90000/7.5 comes from congruence modulo 30 considerations and the square brackets encloses a value derived from the prime number theorem and consideration of the first three primes.) If anybody can spare the time to generate by program all of the five- and six-digit values one can substitute for 1981, I would appreciate it. I imagine there is a database listing of primes that one can simply search.Julzes (talk) 03:52, 20 September 2009 (UTC)

N.B. You don't need 1001981. For a four digit number, say wxyz we can show that 93wxyz, 94wxyz, 96wxyz, 97wxyz and 99wxyz are all prime if and only if wxyz = 1981. In the case of a five digit number then we do need all six cases. For a five digit number, say vwxyz, we find that 93vwxyz, 94vwxyz, 96vwxyz, 97vwxyz, 99vwxyz and 100vwxyz are all prime if and only if vwxyz = 46501. ~~ Dr Dec (Talk) ~~ 11:07, 20 September 2009 (UTC)

Did the sequence "93, 94, 96, 97, 99, 100" come from anywhere in particular? It's not surprising that no other 4-digit numbers can be prefixed by elements of that sequence to make primes, more interesting would be some kind of how many such numbers work for various sequences. Is there any sequence of 6 consecutive prefixes that has a 4-digit number that gives all primes, for example? --Tango (talk) 04:02, 20 September 2009 (UTC)

It doesn't look like it. I couldn't even find a single triple of prime numbers of the form 10,000m + n, 10,000(m + 1) + n and 10,000(m + 2) + n where 1 ≤ m ≤ 97 and 0 ≤ n ≤ 9,999. As for pair of the form 10,000m + n and 10,000(m + 1) + n where 1 ≤ m ≤ 98 and 0 ≤ n ≤ 9,999, well, there were 10,477 of such prime pairs. ~~ Dr Dec (Talk) ~~ 12:22, 20 September 2009 (UTC)

I would say I just took a kind of numerological curiosity in all of the values beginning with 11981 and worked upward. You could say I just stumbled upon it. Obviously the answer to your other question is no by consideration modulo 3. I'm in the midst of formulating another nice question beginning at 100 and working backward, but one could come up with all sorts of related questions. You can sort of figure out what the question I am trying to formulate is if I say that 9 is the first value in sequence, and 1981 is the sixth. Julzes (talk) 04:21, 20 September 2009 (UTC)

Sorry, but I don't understand. I'm probably being quite dense now, but how does reduction modulo 3 tell us anything? For example 10,000m + n = m + n (mod 3). And m + n = 0, 1, or 2 (mod 3). I can't see why any result comes from this. ~~ Dr Dec (Talk) ~~ 12:51, 20 September 2009 (UTC)

A number is divisible by three when the sum of the digits is divisible by 3. (A key part of the standard mathematical toolbox for any base-10 numerologist.) So if you make a sequence of three numbers by increasing the same digit by 1 each time, then one of them must be divisible by three. Staecker (talk) 14:02, 20 September 2009 (UTC)

Ah! I see... That explains why I couldn't find a triple of prime numbers of the form 10,000m + n, 10,000(m + 1) + n and 10,000(m + 2) + n where 1 ≤ m ≤ 97 and 0 ≤ n ≤ 9,999; one of them would always be divisable by 3. ~~ Dr Dec (Talk) ~~ 14:27, 20 September 2009 (UTC)

The first six members in the sequence would be 9, 67, 67, 787, 1981 and 1981 (1009 is prime; 10067, 9967 and 9767 are prime; 100787, 99787, 97787 and 96787 are prime; and then the above is the first that allows 94 as a prefix and it also allows 93. The sequence of prefixes continues 91, 90, 87, 85, 84, 82, 79, 78, 76, 72, 69, etc., with modulo 7, 11 and 13 considerations first necessarily coming in at 95 (taking 88 and 74 off), 92 (81 and 70) and 88 (75), respectively. How would we expect the sequence to grow, rather than asking what its precise values are (the next few terms would be nice, but I expect the growth to be pretty fast)? What does the sequence starting with prefixes of 0 (no prefix), 1, 3, 4, 6, 7, 9, 10, 13, 15, 16, etc., look like? This is probably a more natural question. Is there a good base-independent related question is also something I wonder about.Julzes (talk) 06:17, 20 September 2009 (UTC)

For 5 digits, the only answer is starting 93,94, etc. is

9346501,9446501,9646501,9746501,9946501,10046501.

For 6 digits, there are 4 solutions:

93132667,94132667,96132667,97132667,99132667,100132667

93338149,94338149,96338149,97338149,99338149,100338149

93484393,94484393,96484393,97484393,99484393,100484393

93740131,94740131,96740131,97740131,99740131,100740131

70.90.174.101 (talk) 09:01, 20 September 2009 (UTC)

Hmmm. Then 132667 is the only number of six or fewer digits that also gives a prime when 91 is prefixed, and to get 90 also prefixed to get a prime seven digits are needed at the very least. My 0, 1, 3, etc., alternative question yields analogously a sequence starting 2, 3, 7, 7, 7, 229, and then what?Julzes (talk) 09:31, 20 September 2009 (UTC)

You get to 91 with 2824117 and no other 7 digit sequences. For 8 digits you get 91 with 02374819, 07184533, 08614087, 67173973, and 84167761. In neither case do you ever get to 90. Dragons flight (talk) 14:25, 20 September 2009 (UTC)

The 14th term in your original sequence is probably 9955031915073901. It gives a prime when one of 78, 79, 82, 84, 85, 87, 90, 91, 93, 94, 96, 97, 99, 100 is prefixed. PrimeHunter (talk) 01:33, 21 September 2009 (UTC)

Thanks, all. PrimeHunter, how on Earth did you find that and what are the eighth through thirteenth terms? I'm surprised to find that the number of digits grows as slowly as indicated, too. Is there a good way to estimate the size of terms? Also, just looking at the 1981 case, I'm wondering if there is any sequence of four or fewer digits other than it that produces primes for five of the six prefixes (we've ruled out consecutive ones for numbers without leading zeros). I'm going to look into that the way I have been (almost by hand, with a primality testing module).Julzes (talk) 03:10, 21 September 2009 (UTC)

Correction: Dr. Dec only considered four-digit numbers (937, 947, 967, 977 and 997 are prime). I'll have to limit my search to four-digit sequences if I'm hoping to find 1981 unique in the way I've suggested.Julzes (talk) 03:49, 21 September 2009 (UTC)

0889 turns out to be the first value that's prime for five of the six prefixes (composite for 97).Julzes (talk) 04:23, 21 September 2009 (UTC)

Ok. 3367 gives primes for prefixes 94, 96, 97, 99 and 100, so what I said earlier was not correct. Here is a problem then (which unfortunately won't have my set of six primes as a solution, but might be interesting/challenging): What is the first string of eight consecutive numbers such that there is a unique four-digit value (or string, for a slightly different problem) that can be appended to five of them to make primes, and which can actually be appended to the maximum of six to make primes?Julzes (talk) 05:59, 21 September 2009 (UTC)

There is a reason for my username. The search took 10 cpu minutes after adapting my old prime pattern finder. But it was not designed for this and setting it up to make exhaustive searches for a lot of sequence terms would require more manual time than I'm willing to use. It would require separate source modifications and recompilation for each combination of sequence term and number of digits it searches for this problem. Finding 14 primes with a 16-digit end was lucky. 18 digits would have seemed more likely. An expected chance of solutions below a given size can for example be computed with help from a formula at [3] but it's a little complicated and I didn't do it. The 14th term of your new sequence is probably 1989530586646177 which gives a prime with prefix 0, 1, 3, 4, 6, 7, 9, 10, 13, 15, 16, 18, 21, 22. That was also lucky. PrimeHunter (talk) 15:21, 21 September 2009 (UTC)

Normal approximation to the binomial with continuity correction

So, I am asked to use the normal approximation to the binomial and use the continuity correction with it to make it more accurate. The way my textbook describes how to use the continuity correction is:

The best way to determine whether to add or subtract is to draw a picture like Figure 4.26.

That is, draw the graph of the normal approximation along with a histogram of the binomial distribution. In other words, calculate the binomial distribution EXACTLY so you will know which correction, adding or subtracting 0.5, is closer to the actual value. Is there some way that is not stupid to know how to do the continuity correction? StatisticsMan (talk) 19:46, 20 September 2009 (UTC)

The graph doesn't need to be exact. You just need to work out roughly where the point you are interested in is. You can work out the mean and standard deviation of the binomial distribution and that is enough for a rough sketch. In fact... it's been about 5 years since I've done this so I'm not sure, but don't you just need move further away from the mean? If it is less than the mean, subtract 0.5, if it is more, add 0.5. I'm going to have to go and look this up now... --Tango (talk) 20:00, 20 September 2009 (UTC)

According to our article, continuity correction, you always add... Is it that you always expand the area of interest? Add for a less than, subtract for a more than? Now I'm getting myself thoroughly confused, so I'll let someone else take over! --Tango (talk) 20:06, 20 September 2009 (UTC)

Has some information been deleted from this post? I can't understand the original post as it is, and I can't see how Tango understands enough to make a reply. Have you two discussed this previously, on a talk page for example? ~~ Dr Dec (Talk) ~~ 20:33, 20 September 2009 (UTC)

First, imagine I am talking about a very good approximation here. When the normal probability density function is increasing, then the rectangle (representing binomial dist probability of some certain value) will be above the graph on the left side and the graph will be above the rectangle on the right side. So, to approximate the probability of some number happening, you want to go up an extra .5 to grab the full area of the rectangle you are trying to approximate. On the other hand, if it is decreasing, I think it would go the opposite and you'd subtract .5. Now, this seems to answer my question, but it's not always going to be so simple and such a good approximation as this. What if the normal graph completely contains some rectangle? It could increasing or decreasing at this point.

No, we have not talked about this elsewhere and no nothing was deleted. StatisticsMan (talk) 21:49, 20 September 2009 (UTC)

I see... I think I understand what you were getting at now. Thanks! ~~ Dr Dec (Talk) ~~ 22:00, 20 September 2009 (UTC)

If I recall correctly from many years ago, whether you add or subtract depends on the exact wording of the question being asked. This is why a little sketch (just of the two rectangles in the binomial histogram each side of the chosen point on the normal curve) will aid understanding of whether to add or subtract the correction. Neither the normal nor the binomial distribution has any turning points other than the mean, so both curve and histogram are increasing to the left of the mean and decreasing to the right. There is no need to do any calculations for the sketch. It is just an aid to careful thought about the wording of the question. Dbfirs 07:16, 21 September 2009 (UTC)

September 21

Set-builder notation and primes

Would

$P=\{1<p\in \mathbf {N} :\forall q\in \mathbf {N} \neq (p,1),[p/q]\notin \mathbf {N} \}$

be legitimate notation for the set of primes? I'm wondering what the boundaries for set-builder notation are, especially with regard to the pre-colon component. Thanks. —Anonymous Dissident^Talk 13:27, 21 September 2009 (UTC)

The pre-colon part is fine, but the post-colon part has trouble. You could write,

$P=\{1<p\in \mathbf {N} :\forall q\in \mathbf {N} \setminus \{1,p\}(p/q\notin \mathbf {N} )\}$

but it would be more clear if you just rewrote the whole thing, for example,

$\{p\in \mathbf {N} :\forall q\in \mathbf {N} (q{\mathbin {|}}p\to (q=1{\text{ or }}q=p))\}$.

— Carl (CBM · talk) 15:18, 21 September 2009 (UTC)

You only need to check for divisibility by numbers up to $[{\sqrt {p}}]$, so why not try something along the lines of

$P=\{p\in \mathbb {N} :\ \forall \ n\in \mathbb {N} ,\ 2\leq n\leq [{\sqrt {p}}],\ [p/n]\notin \mathbb {N} \}?$ ~~ Dr Dec (Talk) ~~ 15:37, 21 September 2009 (UTC)

The two commas make that hard to read; it would be easier if the second one was an arrow (→) of if you use the words if/then. — Carl (CBM · talk) 15:40, 21 September 2009 (UTC)

How should I read an arrow in set notation? ~~ Dr Dec (Talk) ~~ 15:49, 21 September 2009 (UTC)

As "implies", as in

$P=\{p\in \mathbb {N} :\ \forall \ n\in \mathbb {N} (2\leq n\leq {\sqrt {p}}\to p/n\notin \mathbb {N} )\}$

— Carl (CBM · talk) 15:52, 21 September 2009 (UTC)

Your first definition includes 1, and your second definition includes both 0 and 1. This does not agree with the generally accepted definition of primes. — Emil J. 16:43, 21 September 2009 (UTC)

So try $P=\{2\leq p\in \mathbb {N} :\ \forall \ n\in \mathbb {N} (2\leq n\leq {\sqrt {p}}\to p/n\notin \mathbb {N} )\}$ ~~ Dr Dec (Talk) ~~ 16:52, 21 September 2009 (UTC)

You're right; my first one did include 1, and both of our second ones included 0 and 1. That's a pretty common error when trying to write a formula that defines the primes. But at least the idea is clear. — Carl (CBM · talk) 17:47, 21 September 2009 (UTC)

Personally, I would go with: $\{p\in \mathbb {N} |p\,\mathrm {prime} \}$. Why go with a complicated solution when there is a simple one? --Tango (talk) 17:01, 21 September 2009 (UTC)

but yours contains an English word ;) --84.221.69.1 (talk) 17:08, 21 September 2009 (UTC)

English words are fine in set-builder notation, as we are discussing it here. In the context of formal set theory, we might restrict things to formulas in a particular formal language, but that's a different setting. — Carl (CBM · talk) 17:47, 21 September 2009 (UTC)

It is a common mistake among novice mathematicians to try and avoid using English. If the easiest way to express something is to write a paragraph of prose, then write a paragraph of prose. Using a single English word isn't a problem at all. --Tango (talk) 17:52, 21 September 2009 (UTC)

I think the original spirit of Anonymous Dissident's post was more to do with whether the set is actually the same as the set of (positive) prime numbers. In other words, are the assumptions and conditions a reformulation of the definition of a prime number? ~~ Dr Dec (Talk) ~~ 18:06, 21 September 2009 (UTC)

The OP talks about the boundaries of set builder notation, so I think primes are just an example. I think the real answer is that any notation is acceptable as long as it is clear what it means (and it means what you want it to mean, of course). --Tango (talk) 18:23, 21 September 2009 (UTC)

Actually, thinking about it, isn't the usual set notation of the form

$\{{\mbox{a general member of some collection of objects}}:{\mbox{some conditions imposed on that set}}\}?\,$

In that case, the condition 1 < p should go after the colon. ~~ Dr Dec (Talk) ~~ 19:03, 21 September 2009 (UTC)

Given all of the above comments, how does

$P=\{p\in \mathbf {N} \setminus \{1\}:\forall q\in \mathbf {N} \setminus \{1,p\},[p/q]\notin \mathbf {N} \}$

work? To Tango: those below you are correct – I realise that it would be much simpler just to use your form, but I wanted to know if my denotation was feasible. Thanks to all of you for your feedback. —Anonymous Dissident^Talk 05:59, 22 September 2009 (UTC)

Almost there. I have two concerns. One, the notation $[p/q]$ is sometimes used to mean the floor function -- especially right after dividing two integers (if you're a computer scientist, anyhow) -- so its use is liable for confusion. The simple solution is to just drop the brackets, like Carl did. The second concern is a bit more nebulous: some mathematicians use "natural numbers" to mean "nonnegative integers" rather than "positive integers", and taking the former interpretation causes confusion when you consider the case q = 0; if q = 0 then is $p/q$ an integer or not? Ideally, whatever definition you write will work equally well for either definition of "natural numbers". The direct solution is to explicitly specify that q is positive before doing the division. Eric. 71.139.177.245 (talk) 06:45, 22 September 2009 (UTC)

The problem with Tango's solution is not that it contains an English word, it is that it is circular. Taemyr (talk) 06:48, 22 September 2009 (UTC)

If you want an actual formalized expression, you have to precisely specify the syntax you're looking for. Otherwise you may as well be a bit less formal. It's also a bit weird that you're using integer division. Normally I'd describe a prime as a natural number p so that there are no naturals a,b so that a*b=p and a>1 and b>1. 70.90.174.101 (talk) 07:17, 22 September 2009 (UTC)