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September 5

Differentiation

I encountered a differential equation problem recently, and, being unable to solve it, turned to the answers. The first step involved substituting ${\frac {d}{dy}}\left({\frac {dy}{dx}}\right)$ for ${\frac {d^{2}y}{dx^{2}}}\times {\frac {dx}{dy}}$ . I can see that these are equal, but the explanation of the answer says that the first expression is differentiated to give the second using the chain rule. None of my math teachers have been able to figure out how this is done. Could someone give me some help? --superioridad ^(discusión) 10:18, 5 September 2007 (UTC)[reply]

The chain rule would say that we can do this (I think!);

{\frac {df(x)}{dy}}={\frac {df(x)}{dx}}\times {\frac {dx}{dy}}

if you then substitute dy/dx in for f(x), then you'll have;

{\frac {d({\frac {dy}{dx}})}{dy}}={\frac {d({\frac {dy}{dx}})}{dx}}\times {\frac {dx}{dy}}

{\frac {d({\frac {dy}{dx}})}{dy}}={\frac {d^{2}y}{dx^{2}}}\times {\frac {dx}{dy}}

. Richard B 11:43, 5 September 2007 (UTC)[reply]

Quine's method?

Prove that the following proposition is a tautology using Quine's method and standard logical equivalences.

$(P\to Q)\land (Q\to R)\to (P\to R)$

So,

$(P\to Q)\land (Q\to R):(P\to R)$

$(P\to R)\to (P\to R):??$

a) What is Quine's method? Can't find much on it

b) Did I do anything right there —Preceding unsigned comment added by 91.84.143.82 (talk) 12:43, 5 September 2007 (UTC)[reply]

Possibly Quine–McCluskey algorithm is what you need. Try to express implications with or, and and not operators, then use the algorithm to minimize the whole expression. If it minimizes to true then it is a tautology. --CiaPan 13:40, 5 September 2007 (UTC)[reply]

Question about zero

I am trying to find out if the number 0 (zero) is an even number or is it considered to be an odd number. My friends and I are in a major debate about this question. Thank you very much for your time and assistance in this matter. Please forward your answer directly to my email address: (email removed) thanks again. —Preceding unsigned comment added by 4.239.234.26 (talk) 18:13, 5 September 2007 (UTC)[reply]

Apologies, we answer questions right here, as noted at the top of the page. This gives everybody a chance to see the answer (and correct it if necessary). Also if you had looked up Zero here on Wikipedia, you'd have been shown along to 0 (number), in which it is stated that zero is considered an even number. --LarryMac | Talk 18:36, 5 September 2007 (UTC)[reply]

I confess I'm curious about how there could be much debate. Most people would agree that 3, 5, and 7 are odd, and that 2, 4, and 6 are even. Continuing the pattern of separation by two implies that 1 is odd and that 0 is even. A number is even if it is a multiple of 2, and odd if it is not; since 2×0 = 0, zero must be even. Notice it does not matter what integer is doubled, whether even, odd, positive, negative, or zero; thus −6 is also even (2×(−3) = −6), and −7 is odd. (Often negative numbers are not discussed as odd or even, but there are times when they are important; an example is modular arithmetic.) --KSmrq^T 19:25, 5 September 2007 (UTC)[reply]

I believe the logic is that the even numbers are multiples of 2, and the odd numbers are the others. In that case, if you figured a number could only be a multiple of 2 that was at least 2, it would be hard to categorize 0. It's not at all natural to consider 0 even, when working with positive numbers. It's only when working with all integers that it seems obvious. Black Carrot 18:14, 7 September 2007 (UTC)[reply]

Actually, one natural option would be (working in the nonnegative integers) to call positive integer multiples of 2 even, all other positive numbers odd, and 0 neither. It's the same way we deal with primes - primes 2 and up are primes, the rest are composites, 1 is a unit, and 0 is none of those. It's fairly arbitrary that we consider 0 automatically even. Black Carrot 18:16, 7 September 2007 (UTC)[reply]

Nonsense. It is anything but arbitrary that zero is even; it is essential! That's one reason I mentioned modular arithmetic. Throughout mathematics we depend on the rules for multiplying, adding, and subtracting evens and odds. If zero were not even, the consequences would be painful indeed. --KSmrq^T 20:03, 7 September 2007 (UTC)[reply]

Absolutely - in the more general context. How many people have heard of modular equivalence, though? For that matter, how many people are really all that familiar with the integers, other than those aspects related to basic arithmetic? Not so many, especially at the level I assume the poster is at. What I said was, considering only the counting numbers (1,2,3... with 0 tacked on as an afterthought) it makes as much sense to consider 0 a separate case, which is where the confusion seems to come from. Black Carrot 23:50, 7 September 2007 (UTC)[reply]

I think that's a stretch. The reason 0 is a special case in the unit/prime/composite thing is that it doesn't have a unique factorization, which is getting reasonably deep into the multiplicative structure of the naturals. For even/odd you don't need to do that; you just notice that the numbers are arranged, as it were, boy-girl-boy-girl. It's about the simplest pattern you can imagine that would ordinarily be called a pattern, and it unambiguously assigns 0 to the evens.

My personal theory as to why so many people are confused about this (even a biology PhD I won't name!) is that they're remembering being taught that 0 is neither positive nor negative, and confusing the two dichotomies. --Trovatore 21:26, 8 September 2007 (UTC)[reply]

Another good example. I would take it as supporting my case, though, not refuting it. My point was that, from the perspective of someone who doesn't yet have a bird's-eye-view of the whole thing, it all seems a bit arbitrary, so why not one more to toss on the pile? Another example that a lot of people have trouble with, until they accept it and move on, is that negative times negative is positive, not negative. Another, from a bit later in schooling, is that .999~ = 1, or that the square root of a negative number exists. That final one is really only justified by the fact that it works, which only experience can show, and that it forms an algebraic closure, which is an advanced concept. I myself didn't know whether zero was even or not until several years ago, because nobody really said whether it was at all clearly, and it wasn't obvious by itself. It wasn't until I saw it rigidly defined as "absolutely any integer multiple of 2, including 0" that I could be sure. Kids have seen enough patterns randomly broken in math by that time that it's hard to be certain unless someone says it clearly. Black Carrot 03:35, 10 September 2007 (UTC)[reply]

I can allow that that may be correct, if you're offering it as an explanation of why many people get this wrong. But it doesn't remotely support your previous claim that "It's fairly arbitrary that we consider 0 automatically even". All it supports is the notion that the confusion is not, after all, too surprising a mistake, given the large number of peculiar-sounding things that students are asked to accept. --Trovatore 17:56, 10 September 2007 (UTC)[reply]

It has sometimes happened that I've been with mathematical nonexperts and have had to address the question of whether 0 is even or odd. They always seem more surprised by my assertion that there is absolutely no controversy about the answer than they are by the answer itself. So I conclude that for nonmathematicians the answer is really not obvious! -- Dominus 16:45, 14 September 2007 (UTC)[reply]

Math Riddle

The following is a simplification of a math riddle I have came across:

"Bob met Alana on the street and forgot how old each of her kids are. She has 3 of them. Alana told Bob that the product of their ages is equal to 36. The sum of their ages is 1 higher than the house number of the white building across the street. Bob went home and couldn't figure it out and phoned Alana. Alana told him that she forgot to give him one more clue: the oldest one is a girl. Upon receiving that clue, Bob immediately figured out their ages."

Using some logic with that last hint, it seems to infer that two of the kids are twins, while the older one is not because the last clue seemed to make a huge difference in Bob solving the riddle. If that is the case, then the possible combinations are 1-1-36, 2-2-9 and 3-3-4.

I can not figure out whether the clue about the white building is relevant.

Could someone point me in the right way?

Thanks. Acceptable 21:59, 5 September 2007 (UTC)[reply]

The idea behind this riddle is that it is much easier for Bob than it is for us -- Bob knows what the number is, and we don't -- but we can work out the number, and their ages, based on Bob's reactions (whether or not he could figure it out at some point.) Try to work out what Bob is thinking, and cross off possibilities as you go. That list you've got will not help you. Gscshoyru 22:05, 5 September 2007 (UTC)[reply]

Is the clue about the house number relevant? Acceptable 22:10, 5 September 2007 (UTC)[reply]

Yes. He couldn't solve it after those two hints -- what does that mean? Gscshoyru 22:14, 5 September 2007 (UTC)[reply]

Is there one definite answer or are there multiple answers? Acceptable 22:20, 5 September 2007 (UTC)[reply]

Just one. Gscshoyru 22:24, 5 September 2007 (UTC)[reply]

Oh, then in that case it means that before the "oldest one is a daughter clue", there must have been multiple answers? Acceptable 22:28, 5 September 2007 (UTC)[reply]

So that means if I find the sums of all of the possible combinations of the 3 ages, then the answer is the one with the repeating sum? Acceptable 22:32, 5 September 2007 (UTC)[reply]

Yes. Gscshoyru 22:44, 5 September 2007 (UTC)[reply]

Is the answer 9-2-2? Acceptable 23:13, 5 September 2007 (UTC)[reply]

Sorry, Didn't realize it was a question. Zain Ebrahim 07:44, 6 September 2007 (UTC)[reply]

Okay if anyone wants the answer it's on [1] at the bottom somewhere. Just Ctrf+F "insurance salesman". Zain Ebrahim 10:56, 6 September 2007 (UTC)[reply]

That link gives me a "page not found" message. DuncanHill 10:59, 6 September 2007 (UTC)[reply]

Try it now. I'm not quite sure how to link to other websites. Zain Ebrahim 11:02, 6 September 2007 (UTC)[reply]

Thanks, that works now. DuncanHill 11:03, 6 September 2007 (UTC)[reply]

Thanks guys. Acceptable 23:04, 6 September 2007 (UTC)[reply]

If I understand this right, the solution hinges on the idea that two people can't be born in the same year and still have one that's older. Even if there's twins, there's an older one, and there doesn't have to be twins. — Daniel 02:28, 7 September 2007 (UTC)[reply]

Wow. I've known this riddle for a while and have never considered this point. This can be remedied by rephrasing it as "the one with the greatest age", though this sounds artificial. Of course, clarifying that ages are taken to be integers is necessary for the riddle to make sense, and is all the more important with this phrasing. -- Meni Rosenfeld (talk) 11:26, 7 September 2007 (UTC)[reply]

"...the oldest one is a girl" tells us that 1-6-6 is wrong and 2-2-9 is right (without that clue it could be either). I don't quite see how the problem proposed by Daniel would be remedied by using "...the one with the greatest age is a girl". Unless I'm completely missing something here.
Zain Ebrahim 11:36, 7 September 2007 (UTC)[reply]

Is Daniel's problem unclear, or is it my solution? 1-6-6 isn't wrong because it could be that one was born exactly 6 years ago, and one was born 6 years and 10 months ago. This one would be the eldest, though both their ages are 6. With my phrasing, there will be no single person of greatest age and thus 1-6-6 is indeed wrong. -- Meni Rosenfeld (talk) 11:41, 7 September 2007 (UTC)[reply]

If you really want to continue down this road, you'd want to say "the one with the greatest age, when expressed as a truncated integer," or some such. But come on, guys, it's a riddle, not an exercise in how finely you can slice a problem (with or without Occam's Razor). --LarryMac | Talk 11:56, 7 September 2007 (UTC)[reply]

Actually, I have already mentioned that ages must be integers. But I think I disagree with your point. As a mathematical exercise, this problem is trivial. Its only virtue is the challenge of noticing the mathematical details hidden in the real-world explanations. One part of this is extracting "there is a unique greatest number" from "there is an oldest child" (which itself is extracted from "the oldest is a girl"). But in fact this deduction is not correct, so we are cheating our listeners if we use the original phrasing and expect them to make it. -- Meni Rosenfeld (talk) 12:03, 7 September 2007 (UTC)[reply]

Oh okay, I get it now. Sorry, Meni, I'm really having a slow day here. Btw: listeners?
Zain Ebrahim 12:22, 7 September 2007 (UTC)[reply]

The best word I could think of for "the people to whom we are posing the riddle". I'd be glad to hear better suggestions. -- Meni Rosenfeld (talk) 12:36, 7 September 2007 (UTC)[reply]

I don't know if it's better but . . . Audience? :) Zain Ebrahim 12:39, 7 September 2007 (UTC)[reply]

Call me crazy, but "readers" has always been standard in newspaper and magazine articles. Black Carrot 18:11, 7 September 2007 (UTC)[reply]

Oh, I get it. I didn't mean the readers of the reference desk, but rather people (friends, etc.) to whom we (as in, anyone) might pose the riddle in person. -- Meni Rosenfeld (talk) 18:33, 7 September 2007 (UTC)[reply]

Aahhh... Indeed. <rubs chin thoughtfully> Black Carrot 23:46, 7 September 2007 (UTC)[reply]

Mad mathematician

Hi all, hope you can help. There was a BBC programme recently about a mathematician who spent his declining years (I think in the '30s of the last century) trying to solve some famous mathematical problem. He would write to his publisher announcing that he was close to a solution, only to write again a few weeks later to say he had failed, this behaviour repeated several times, and he eventually ended in an insane asylum (possibly in Switzerland). I must admit that I wasn't really paying attention at the time, but if anyone could identify the chap I would be grateful - it's been nagging away at the back of my mind ever since! DuncanHill 22:00, 5 September 2007 (UTC)[reply]

Sounds like Georg Cantor to me. Gscshoyru 22:06, 5 September 2007 (UTC)[reply]

I'm HUGELY impressed! Thank you :) DuncanHill 22:18, 5 September 2007 (UTC)[reply]

Could it be this? PrimeHunter 22:23, 5 September 2007 (UTC)[reply]

Yes that's it, here it is on the BBC website [2]. Thanks again. DuncanHill 22:25, 5 September 2007 (UTC)[reply]

If it was Cantor, then someone has the story wrong, or at least is giving it the wrong nuances. Cantor was never insane, only depressed. I don't mean to minimize depression; I know it's a cruel and debilitating illness, but in my amateur estimation it's unlikely to lead to crankish behavior. Cantor was greatly disappointed by his inability to prove the continuum hypothesis, so it's conceivable the story derives from that.

Kurt Gödel, on the other hand, did go a bit further round the bend, but again I don't know of any problem that he kept claiming to be on the verge of solving.

Then there's Alexander Abian. Who knows, maybe we just didn't understand him. (Certainly we at least didn't understand him.) --Trovatore 22:36, 5 September 2007 (UTC)[reply]

As I said, I didn't really pay much attention to the programme, so it is entirely possible that they said he died in a san, and I misrembered this as being an asylum. DuncanHill 22:45, 5 September 2007 (UTC)[reply]

According to the article, Cantor did in fact end up in a sanatorium. And he wasn't being a crank -- he truly thought he had the answer a few times... though we know that's impossible. Gscshoyru 22:44, 5 September 2007 (UTC)[reply]

Well, no we don't really "know that's impossible". We know that it's impossible to prove or refute from ZFC. That's not what Cantor was trying to do (in fact ZFC had not even been formulated). --Trovatore 22:48, 5 September 2007 (UTC)[reply]