Wikipedia:Reference desk/Archives/Mathematics/2008 February 4

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[edit] February 4

[edit] HEY IT`S ME THE PHYSICS MAGAZINE GUY THIS TIME WITH AN ALGEBRA QUESTION

From an Algebra Magazine. 1.Find the consecutive integers whose product is 156. Find two consecutive integers whose product is 156. Find two consecutive even integers whose product is 288. Find two consecutive odd integers whose product is 783.

I also have some word problems Michelle is 5 years younger than Cindy.Janet is years less than twice Cindy`s age.Kenny is 10 years old.The ratio of Janet`s age to Cindy`s age equals the ratio of Michelle`s age to Kenny`s age.How old are Cindy,Michelle and Janet.

A rectangular poster has an area of 190 square inches.The height of the poster is 1 inch less than twice it`s width.Find the dimensions of the poster.

Daryl earns twice as much per hour as Andy,and John earns 6 dollars more per hour than Andy.June earns 16 dollars per hour.The ratio of John`s hourly earnings to Andy`s hourly earnings is the same as the ratio of John`s hourly earnings to Andy`s hourly earnings is the same as the ratio of Daryl`s hourly earnings to June`s hourly earnings.How much does Daryl earn per hour.

A small rocket is launched upward from ground level.The height of the pocket from the ground is given by the quadratic equation h=-16t2 plus 144t where h is the height of the rocket in feet and t is the number of seconds since the rocket was launched.How many seconds will it take for the rocket to return to the ground.

given by the quadratic equation h0 —Preceding unsigned comment added by Yeats30 (talk • contribs) 00:13, 4 February 2008 (UTC)

Try taking the square root of the numbers. --Salix alba (talk) 00:33, 4 February 2008 (UTC)
Well I'd strongly encourage you to try them yourself - for example with no. 1 make the first number x and the second number y, then x*y = 156 (from "the product is 156") & y = x + 1 (from "the numbers are consecutive") and you can solve them like any pair of simultaneous equations...the others all work similarily - decide what the variables are and make equations with them, then try to solve those equations. So for the second question you might represent michelle, cindy, janet & kenny's ages by the letters m, c, j & k respectively. Then "michelle is 5 years younger than cindy" becomes m = c - 5, and so on and so forth. It's not that hard once you get equations. Trimethylxanthine (talk) 00:47, 4 February 2008 (UTC)
While solving simultaneous equations is a good general method, in these cases, I would expect it to be easier to just take the square root and use trial and error with nearby integers. --Tango (talk) 17:58, 4 February 2008 (UTC)
True true, but that can't be applied to the word questions? Or am I missing something?? Trimethylxanthine (talk) —Preceding comment was added at 00:27, 5 February 2008 (UTC)
They're only word questions because that's the way they're formulated. It would be perfectly easy to formulate them all as a system of equations. -mattbuck 00:36, 5 February 2008 (UTC)
I meant taking square roots for them, which is why I added in. Trimethylxanthine (talk) 06:29, 6 February 2008 (UTC)
These are homework questions. See the science desk. Ignore him. --98.217.18.109 (talk) 23:32, 7 February 2008 (UTC)

[edit] Complex Base

It's possible to write the set of complex numbers in decimal-like form using a complex base, like 2i, and the appropriate number of digits, in that case 0, 1, 2, and 3. Given a base, how do you determine whether it will produce a consistent number system in that way? Black Carrot (talk) 00:52, 4 February 2008 (UTC)

Yeah, it is. See quater-imaginary base — Kieff | Talk 01:17, 4 February 2008 (UTC)

I'm not sure you actually read my question. Black Carrot (talk) 07:34, 4 February 2008 (UTC)

Sorry, you're right. I read it as two questions ("is it possible, if then how to determine"). My bad. :( — Kieff | Talk 08:24, 4 February 2008 (UTC)

Not sure if I understood - but if the base is a+ib where a and b are integers then the new number system will be workable - ? (it's just solving linear equations of order 2)87.102.90.249 (talk) 12:57, 4 February 2008 (UTC) Don't ask me how to find out how many digits to use...87.102.90.249 (talk) 13:11, 4 February 2008 (UTC)

How do you define "consistency" of a number system? Is it that you can get all integral numbers as a sum of powers of the base weighted with digits? Do you also require uniqueness?  --Lambiam 13:14, 4 February 2008 (UTC)
I can't help wondering that since real integer number bases produce unique results then complex number bases (using integers) will as well (eg base 4+2i using the numerals 0,1,2,3,1+i,2+i,3+i probably these numbers should go up to 19..) since there is no distinction algebraically between real and complex numbers.. No proof or refutation I can provide right now.87.102.90.249 (talk) 14:11, 4 February 2008 (UTC)I was wrong.87.102.90.249 (talk) 15:02, 4 February 2008 (UTC)
Well, decimal expansions of real numbers aren't unique (see 0.999...), so chances are a similar approach to complex numbers won't be either. --Tango (talk) 17:55, 4 February 2008 (UTC)
Yes, but representations in non-integer bases can fail to be unique in more interesting ways.
For example, in base 3/2, 2 does not have a finite representation, but it does have at least two quite different non-terminating representations. A greedy algorithm gives 2=10.01000001001001... whereas we also have 2=0.111111....
And if we use the golden ratio as a base in the ordinary way, then 1=0.11, 2=1.11=10.01 etc. because the base satisfies x2=x+1. (Note that the golden ratio base avoids this non-uniqueness by introducing the additional restriction that the digit string "11" is not allowed in a representation). Gandalf61 (talk) 11:20, 5 February 2008 (UTC)

Here's an attempt at an answer. Let b be the base and say |b| > 1. Because we can always move the "decimal" point around (i.e. scale by powers of the base) it suffices to consider points inside a shape of our choice which contains a neighborhood of the origin. The fundamental reason why base 2i with digits 0,1,2,3 works is that there exists such a shape which can be covered by four copies of itself scaled down by 2i and translated by 0,1,2,3: namely the rectangle defined by -4/5 \le \Re(z) \le 16/5 and -8/5 \le \Im(z) \le 2/5. In general, since the unit disc (for example) can be covered by finitely many scaled copies of itself regardless of the scaling factor, any base b with |b| > 1 can be used, but the minimum number of digits for arbitrary b looks like it might be a difficult geometric covering problem. We can get a lower bound of \left\lceil|b|^2\right\rceil from area alone, though. -- BenRG (talk) 18:30, 4 February 2008 (UTC)

That makes sense. It looks like the same argument would produce a system base bi for b the square root of any natural number. Black Carrot (talk) 02:51, 5 February 2008 (UTC)
Yes, I think that lower bound is tight for pure imaginary bases. It's not tight in general because I'm pretty sure that base 1 + ε (ε real) won't work with less than 4 digits. The disk covering problem gives an upper bound of n+1 digits for |b| ≤ 1/r(n). (The +1 is because we need a zero digit, which gives us the additional restriction that one of the small disks must be contained in the large disk; otherwise the large disk can't contain a neighborhood of the origin.) Among other things this implies an approximate asymptotic upper bound of \frac{2\pi}{3\sqrt{3}}|b|^2 \approx 1.21|b|^2 digits, though I guess that was pretty obvious to begin with (as was the lower bound).
I just noticed the article Complex base systems, which contains a link to a recent arXiv article on exactly this subject. It seems to only consider "proper" systems, meaning those in which almost every complex number has only one representation (which of course implies that |b|2 is an integer). Actually, on the basis of the IFS fractals in that paper, I'm starting to think that that lower bound is also an upper bound except on parts of the real line. That would be disappointingly boring. I was hoping that the boundaries between minimum-digit regions would be fractal. -- BenRG (talk) 21:53, 5 February 2008 (UTC)

[edit] solution of Euclide

i visted clay mathematics institute web site and i read this, Euclid gave the complete solution for that equation,(x^2+y^2=z^2).my question is how could we give such solution?by picking up an arbitrary values? or by finding operators, like we put,y=(a+s),x=(b+s),then z=(b+a+2s),where,a,b are constants and s,is variable.then we fixed a,and ,b,and start chossing differents value of ,s? is that how we find the solutions?or the general way is different?209.8.244.39 (talk) 12:01, 4 February 2008 (UTC)

You may want to take a look at Pythagorean triple. If you then have a more specific question we'll be happy to help. -- Meni Rosenfeld (talk) 12:06, 4 February 2008 (UTC).thank you very much.i read it.but why cannot we apply the same way on Hilbert's tenth problem?88.116.163.226 (talk) 13:22, 4 February 2008 (UTC)
Because Hilbert's tenth problem is about solving arbitrary Diophantine equations, while the Pythagorean equation a^2+b^2=c^2\;\! is just one simple example. The analysis that works for it needn't work in the general case. -- Meni Rosenfeld (talk) 15:33, 4 February 2008 (UTC)

[edit] how to pose and solve problems involvo=ing squares and square roots

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Your question is somewhat vague, but perhaps our article on quadratic equations might help? —Ilmari Karonen (talk) 14:15, 4 February 2008 (UTC)

[edit] How to find generator of a group

The article Digital Signature Algorithm contains the line: Choose g, a number whose multiplicative order modulo p is q. This may be done by setting g = h(p–1)/q for some arbitrary h (1 < h < p-1), and trying again if the result comes out as 1. Here p-1 is a multiple of q. I am translating this as follows: If q|(p-1) then h(p–1)/q generates the multiplicative subgroup of F^*_p of order q provided it is not 1. Can someone give me the proof for this fact. Thanks.--Shahab (talk) 15:34, 4 February 2008 (UTC)

Is q assumed to be prime? If so, it's fairly straightforward: You have gq = hp − 1 = 1, so the order of g must divide q, and is thus either 1 (if g=1) or q. -- Meni Rosenfeld (talk) 15:40, 4 February 2008 (UTC)
q is prime. Thanks. I hadn't thought of Fermat's result. —Preceding unsigned comment added by Shahab (talk • contribs) 15:54, 4 February 2008 (UTC)

[edit] Top Ten List

Recently, I've done a poll on an online forum asking other users what their Top Ten favorite The Land Before Time characters are (the list is in a hierarchy, meaning #10 on their list is their 10th most favorite, and #1 is their favorite character. The positions are not equal). I've recieved about 30 replies. Now, I want to calculate the Top Ten favorite characters overall. What would be the best way to calculate the popularity of each individual character? I'll add that I have ample time to go through each poll and check each character one-by-one, if need be. --Ye Olde Luke (talk) 23:57, 4 February 2008 (UTC)

This sounds like Preferential voting, but I've no idea how they add it up sorry - the article might help just a little :) Trimethylxanthine (talk) 00:40, 5 February 2008 (UTC)
Well, there are a many many ways you could do it.
1 - Whoever has most #1 appearances is #1, person with 2nd most is #2, etc. Once all #1s are exhausted, rank by #2s.
2 - Assign values to the positions, (eg 1 for 1st, up to 10 for 10th), then rank by lowest score, divided by the number of appearances.
3 - Assign values as above, but rank by number of appearances first, then score.
4 - Rank by # of appearances, then by # of 1sts, etc.
There are many more ways you could do it, but those are the ones which seem obvious to me. -mattbuck 00:41, 5 February 2008 (UTC)

I like the 2nd and 3rd options. I might do one of those.

One other thing. Since the main characters appear in all 13 movies, while most guest stars only appear in one, the recurring characters have a bit of an advantage, seeing as more people are likely to have seen them. Should this be factored in? --Ye Olde Luke (talk) 00:52, 5 February 2008 (UTC)

You could divide the scores by the # of movies they appeared in I guess, or just create a separate list for guest stars. -mattbuck 00:58, 5 February 2008 (UTC)

Wait, there is one problem with your #2 solution. You're assuming there are only ten characters to choose from. Since there are many, many characters, I can't rate the lowest score as the most popular. Some character who's #1 on a single person's list would beat a character who is #2 on five different lists, even though the latter character is obviously more popular. --Ye Olde Luke (talk) 00:58, 5 February 2008 (UTC)

You have a point. I went to a lecture once on voting systems. Nothing is perfect, no matter how you score it. You could rank #1 as 10pts, #2 as 9, ranking by highest score, and this would fix it, but likely throw up further problems. -mattbuck 01:03, 5 February 2008 (UTC)
At least you got me thinking in the right direction. I can probably work out any bugs from here. Thanks!
Say, are you the user that I introduced to this place? His name was similar to yours. --Ye Olde Luke (talk) 01:07, 5 February 2008 (UTC)
Seems so! -mattbuck 09:26, 5 February 2008 (UTC)
Incidentally, the 'no perfect voting system' result is Arrow's impossibility theorem. AndrewWTaylor (talk) 11:27, 5 February 2008 (UTC)
Most individual rankers have a clear notion of which items are their #1, #2 and perhaps #3 favourites, but by the time they get to #8 or so, there is generally not much perceived difference between one item and the next. A method to obtain an aggregate ranking that takes this into account is to assign a value of 1/1 to #1, 1/2 to #2, 1/3 to #3, and so on. The item with the highest total value is #1 in the aggregate ranking, and so on. Example: three items A, B and C and three rankers, giving input 1:A,2:B,3:C, 1:B,2:A,3:C, 1:B,2:C,3:A. Then A gets 1/1 + 1/2 + 1/3 = 11/6, B gets 1/2 + 1/1 + 1/1 = 15/6, and C gets 1/3 + 1/3 + 1/2 = 7/6. B is highest, then A, then C, giving aggregate ranking 1:B,2:A,3:C.  --Lambiam 02:17, 5 February 2008 (UTC)
Oh, all right. I'll do that. Thanks! --Ye Olde Luke (talk) 02:28, 5 February 2008 (UTC)