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January 31

Alternative solution to Birthday Paradox

The standard way of illustrating the birthday paradox is to first calculate the probability that no two people out of N people in a classroom have the same birthday, then subtract this from 1. Is there a way of tackling this without first considering the converse?Jonpol 00:00, 31 January 2007 (UTC)[reply]

You can count the expected number of pairs of people who share a birthday as (n choose 2)/365, and apply Markov's inequality. It won't get you the exact probability of two people sharing a birthday, or the correct threshhold for when to expect two people to do so, but it will get within a small constant factor of the right bound. —David Eppstein 04:02, 31 January 2007 (UTC) Edited to add: to be more precise, this only upper bounds the probability of a shared birthday. To lower bound it you need some other argument involving independence. —David Eppstein 04:09, 31 January 2007 (UTC)[reply]

Thanks David, Just so I've understood: e.g., (20 choose 2) would be 190 (?), then divide this by 365. If I go up to 28, then (28 choose 2) would be 378, which would lead to an actual logical paradox, since it's greater than 365, and would imply greater than certainty. Presumably, that's where "Markov's inequality" comes in, which I will now have to research.Jonpol 05:35, 31 January 2007 (UTC)[reply]

That's about right. But it's not a paradox because 378/356 isn't a probability, it's the average number of twin pairs you would expect to see per 28-person group. —David Eppstein 06:39, 31 January 2007 (UTC)[reply]

math question

do u know any two consecutive even intergers that go into 528?

This appears to be a homework question. So I won't give the actual numbers. But there certainly are two consecutive even numbers that multiply to equal 528. The way I did it was to simply use a calculator and play around with a couple of numbers until it worked. All the best. - Akamad 06:10, 31 January 2007 (UTC)[reply]

Break it down into prime factors (528 = 2⁴ × 3¹ × 11¹) and then analyze those to see what you can get. − Twas Now ( talk • contribs • e-mail ) 06:24, 31 January 2007 (UTC)[reply]

Yes, eye due no too consecutive even "intergers" that go into 528. In fact, I know four in a row. Do you know how to read directions, like those at the top of this page that say we will not do your homework for you? This is so trivial a problem that a brute force attack can quickly succeed. Sheesh. --KSmrq^T 06:49, 31 January 2007 (UTC)[reply]

In addition to brute force using a calculator, you can solve the problem algebraically. Start by asking if there are two numbers n and n+2 that multiplied together are 528. Dugwiki 17:24, 31 January 2007 (UTC)[reply]

...although the question as stated just asks for n and n+2 that are factors of ("go into") 528 - it does not say that 528 must be the product of these two factors. As KSmrq points out, this gives you a wider choice of solutions. Gandalf61 17:43, 31 January 2007 (UTC)[reply]

Another thing you can do to home in on the correct answer by taking the square root of 528 (=22.97825059), so you know the answers are right above and below 23... Sjakkalle (Check!) 15:06, 1 February 2007 (UTC)[reply]

Why solving problems like, "P versus NP" and others

What is the use of solving problems like, P versus NP and others like,

Does these solutions make anything possible, what is impossible now? :( --V4vijayakumar 14:02, 31 January 2007 (UTC)[reply]

At least the "Navier-Stokes existence and smoothness" is directly applicable to fluid flow (apparently) - but all knowledge brings rewards - even if it is only understanding - solving one of these makes it possible to 'move on' - that for me would be enough. By the way can I add another to the list (bring it to your attention) #Counting problem - if you could solve this - then maybe I could tell you if it has any use - currently I can't...87.102.7.133 12:50, 31 January 2007 (UTC)[reply]

By analogy I could say what use would someone living in a mud hut 1000BC thing there would be for pythagorus's theorem, the golden mean and trigonometry - to them they would seem useless and irrelevant pastimes - but today they have taken on great importance.87.102.7.133 12:56, 31 January 2007 (UTC)[reply]

(Plus of course there is a monetary reward and limited fame. money makes many things possible)87.102.7.133 14:36, 31 January 2007 (UTC)[reply]

Coming up with a way to prove these things often involves coming up with a new way of doing something in mathematics. The applications might be much farther reaching than just the proof itself. (Also, it often or usually takes years for new ideas in pure math to trickle down into practical usage.) - Rainwarrior 16:45, 31 January 2007 (UTC)[reply]

Yes, those problems were selected as Millennium Prize problems not because they are hard, but because we could do so much with them. "YM existence and mass gap" is integral the understanding of quantum chromodynamics and gauge theories, the Navier-Stokes describe fluid flow, Hodge conjecture for linking algebraic topology of a non-singular complex algebraic variety. Mathematics is abstract, and is about mathematics. However physics is mathematics, and we use that quite well. The three I listed here have to do with physics. [Mαc Δαvιs] X (How's my driving?) ❖ 18:14, 31 January 2007 (UTC)[reply]

You never know what something might be useful for. As an example, for centuries number theory was considered "pure mathematics" with no practical applications. Today, number theory research is the basis for all modern encryption techniques. --Carnildo 21:04, 31 January 2007 (UTC)[reply]

Distributive property of gcd and lcm?

Hello,

I read this weird problem : prove that in $\mathbb {N} :\gcd(\operatorname {lcm} (x,y),z)=\operatorname {lcm} (\gcd(x,z),\gcd(y,z))$ and you can only use the basic definition of greatest common divisor and least common multiple. Now that is a weird statement, because normally I would be able to use prime factorization, but now I can't. But... if I have to use something right because lattices aren't always distributive, right?

Sorry for my somewhat weird question and thanks,Evilbu 16:58, 31 January 2007 (UTC)[reply]

Hi, I'd try to prove it like this: show that LHS divides gcd(x,z ) and gcd(y,z ) and is minimal (with respect to division ordering) with this property Jakob.scholbach 00:34, 1 February 2007 (UTC)[reply]

How are the "basic definitions" you have worded? Consider the least common multiple, m = LCM(x,y); we could say x|m, y|m, and if n is any number divisible by both x and y then m|n. For the greatest common divisor, g = GCD(x,y), we could say g|x, g|y, and if h is any number that divides both x and y then h|g. If we must use m≤n and h≤g our task will be more awkward. --KSmrq^T 01:51, 1 February 2007 (UTC)[reply]

First, get rid of lcm (what is gcd(x, y)·lcm(x, y)?)

Once you have done that, write gcd(x, y) as just (x, y); this will save space and make things more obvious. Complete the following:

k(a, b) = ?
((a, b), c) = ?
(a, b)(c, d) = ?

After that, it's just algebra. –EdC 06:11, 1 February 2007 (UTC)[reply]

What is the increase percent

I start out with 1 of something and it increases to 30. What would the incease percent?

The value increased by (30 - 1) × 100 / 1 = 2900%, reaching 3000% of the initial value. — Kieff | Talk 20:42, 31 January 2007 (UTC)[reply]

I'll add $ signs to make it a little clearer - I always like thinking of "real" objects. It increases from $1 to $30. The original value is defined to be 100%, so $30 is 3'000% of the original value. This is calculated by ($30 / $1) x100% = 3'000%. However, the increase is the difference in the percentage values, which is 3'000% - 100% = 2'900%. --h2g2bob 01:41, 1 February 2007 (UTC)[reply]

Why? When I read "increased 2900%" I mentally translate it to "increased thirtyfold" and wonder (i) whether the measurement was meaningfully fine enough to say for sure even that the increase was more than 2800% and less than 3000%, let alone between 2895% and 2905% which is my criterion for using so many digits, and (ii) what is keeping you from writing "30 times" if that's what you mean. —Tamfang 08:25, 2 February 2007 (UTC)[reply]