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December 9

6th grade math

42 minutes/ 15 seconds —Preceding unsigned comment added by Chefsae (talk • contribs) 00:13, 9 December 2008 (UTC)[reply]

Convert the minutes to seconds and then it should be easy. --Tango (talk) 00:51, 9 December 2008 (UTC)[reply]

Or use Google. Algebraist 04:06, 9 December 2008 (UTC)[reply]

Oh yeah, that'll teach 'm a lot - NOT -hydnjo talk 04:13, 9 December 2008 (UTC)[reply]

Sorry about the bite but sometimes (IMO) it's better to provide some insight to method rather than a numerical answer. -hydnjo talk 04:21, 9 December 2008 (UTC) [reply]

Knowledge that Google has an inbuilt calculator system is a useful method. I use it all the time. Algebraist 15:06, 9 December 2008 (UTC)[reply]

42 x 4 = 168 ~~(in seconds)~~

Top Epert

It is tempting to say 2.8 minutes per second ;) Gandalf61 (talk) 10:19, 9 December 2008 (UTC)[reply]

Or perhaps just point out that if you arive at 168, that is not the answer in seconds. Taemyr (talk) 11:47, 9 December 2008 (UTC)[reply]

Stop answering other people's questions? Why not? People come here to get answers, not to be forced to work out the problem themselves (otherwise they could ask their teacher (who probably won't tell them the answer and hence they come here)). Yes, I would agree that it is not a good idea to give them answers to exams or homework, but this is trivial homework and in any case, he will have to learn the principles sooner or later.

Topology Expert (talk) 10:53, 9 December 2008 (UTC)[reply]

But you didn't tell the OP anything useful, you just told them what 42*4 was, any calculator can do that. It's the method that is important. It also doesn't help that you got the answer wrong (it should dimensionless), and that you didn't use the standard method (you spotted that 15 seconds was a quarter of a minute and used that to simplify it, which is exactly what I would have done, but it is far better to learn the standard method which would be converting everything into the smaller unit since then you don't have to deal with so many fractions). --Tango (talk) 13:47, 9 December 2008 (UTC)[reply]

168 should be dimensionless; I agree. I just wasn't thinking (and was really frustrated with my keyboard; excuse me for my mistakes until my keyboard is fixed).

But, the OP asked for the answer (not for the method). Shouldn't we give him the answer rather than what he probably would not want? I have seen people insulting volunteers because they weren't giving them the answer. These OP's should not do this in the first place, but my point is that this is a casual reference desk and should not have to force other people to follow rules. If rules are imposed on others, I don't think many people will ask questions here (don't get me wrong; I agree giving the method is the right way but we should rather answer the question).

Topology Expert (talk) 14:25, 9 December 2008 (UTC)[reply]

See Wikipedia:Reference desk/Guidelines#What the reference desk is not (the 4th point). I think in some cases it's better to teach than to answer. This is such a case. Zain Ebrahim (talk) 14:32, 9 December 2008 (UTC)[reply]

Yes I have read those guidelines a lot and I completely agree that teaching is better than answering. But if people want an answer (like say I wanted to know the antiderivative of the square of the cosine function) give it to them rather than writing information which the OP will probably ignore. Don't give him hints and such when most probably the OP wants the answer (unless of course he says 'explain the method' which I am sure people will ask if they really want the explanation rather than the answer). Guidelines are guidelines but we are not here to impose something on people who want a straightforward answer.

Topology Expert (talk) 17:15, 9 December 2008 (UTC)[reply]

So you don't agree that teaching is better than answering? Algebraist 17:17, 9 December 2008 (UTC)[reply]

No I strongly agree but if someone does not want to be taught, don't force them to learn (I mean there are people who just want the answer for their purposes in life so why force them to learn?).

Topology Expert (talk) 17:21, 9 December 2008 (UTC)[reply]

So that we don't have to answer the same question next week when he wants to know what 25kg/20g is? If you don't like the idealistic reasoning, go for the selfish one instead. --Tango (talk) 18:21, 9 December 2008 (UTC)[reply]

OK, good point but it is unlikely that he will come back after this discussion.

Topology Expert (talk) 18:29, 9 December 2008 (UTC)[reply]

Sometimes it is not easy to understand if somebody just needs an answer for his job (like e.g. a politician that wants to know what's 12% of 100 people) or also would like to understand the method. It could be added in the directions of the Reference Desk, at the point "How do I word my question for the best results?": to specify whether one needs a dry answer, or an explanation, or both (or neither, just for completeness sake).--PMajer (talk) 19:12, 9 December 2008 (UTC)[reply]

I agree. Could we word it there? My main reasoning is that several people (in the past) do not seem to care about the working out, just the answer and it is not good to impose something on someone (run around in circles by giving hints) if he/she does not want it (just wants the answer).

Here's one way:

{\frac {42{\text{ minute}}}{15{\text{ second}}}}={\frac {42{\text{ minute}}}{(1/4){\text{ minute}}}},

so the minutes cancel and you get

{\frac {42}{1/4}}=42\times 4=168.

Michael Hardy (talk) 01:01, 10 December 2008 (UTC)[reply]

Proof

Hi. I'm trying to prove that, for n being an integer greater than 2, $n^{n+1}>(n+1)^{n}$ . I was thinking I could try using proof by contradiction and so far I have the following.

$(n+1)^{n}>n^{n+1}\,\!$

$(n+1)^{n}>n^{n}n\,\!$

$[{\frac {n+1}{n}}]^{n}>n$

This is clearly true for n=3. As < $math>x$ gets</math> bigger, ${\frac {n+1}{n}}$ gets closer to one so the RHS gets smaller, whereas the LHS just gets bigger, so the final line is incorrect so the original assumption must be incorrect. Is this enough for a proof? Thanks 92.2.197.63 (talk) 19:24, 9 December 2008 (UTC)[reply]

yes but you do not need to put it as a proof by contradiction; just use the other inequality. You can also observe that

(1+1/n)^{n}

is an increasing sequence converging to the number

e

; thus between 2 and 3 for all n=1,2.. --PMajer (talk) 19:35, 9 December 2008 (UTC)[reply]

That it converges to e means it is true for all sufficiently large n, but it doesn't tell you that it's true for all n>2. --Tango (talk) 19:40, 9 December 2008 (UTC)[reply]

Well, the opposite of "less than" is "greater than or equal to". I'm not convinced you have proven that the last line is false for all n>2 - while the fraction gets smaller, the exponent gets bigger and the fraction is greater than one, so the whole thing together could feasibly get bigger. Proof by induction might be a better approach. --Tango (talk) 19:40, 9 December 2008 (UTC)[reply]

Note that I wrote increasing, that's why it is between 2 and 3 for all n=1,2.. (both LHS and RHS are increasing of course)--PMajer (talk) 19:47, 9 December 2008 (UTC) If the point was, why it is increasing, you can do it by the arithmetic mean vs geometric mean inequality. --PMajer (talk) 19:52, 9 December 2008 (UTC)[reply]

But in fact if you didn't already know that $(1+1/n)^{n}$ is increasing, then you can do a more direct (even though less natural) proof using the geometric mean vs arithmetic mean inequality this way: consider $n+1$ data $x_{1},..,x_{n+1}$ , of which $n-1$ are equal to $n+1$ , and the other two being equal to $\scriptstyle {\sqrt {n+1}}$ . Then their sum is $\scriptstyle n^{2}-1+2{\sqrt {n+1}}$ , that is less than $\scriptstyle n(n+1)$ , say for any $n>3$ . So (for n>3) their arithmetic mean is less than n. Their product is $(n+1)^{n}$ , their geometric mean is $\scriptstyle (n+1)^{\frac {n}{n+1}}$ , which is less than the arithmetic mean by the quoted inequality, and you conclude. As I wrote, for the case $n=3$ we just have proven "less or equal", but say it is not equal for some reason. --PMajer (talk) 20:51, 9 December 2008 (UTC)[reply]

As noted above, it suffices to show that (1+1/n)ⁿ<3. Since the first two terms of the Taylor series for exp(1/n) are 1+1/n, we have that 1+1/n < exp(1/n). Therefore (1+1/n)ⁿ<exp(1) < 3. McKay (talk) 09:24, 11 December 2008 (UTC)[reply]

Good. Notice that this is less elementary (since requires some calculus while the other is just a four operations thing) but much simpler. More advanced theory make problems simpler. I though the query was for an elementary proof however. Note:

\scriptstyle \exp(x)>1+x\ \forall x\neq 0

comes in fact from the convexity of

\exp(x)

; to be precise, the first two terms of the Taylor expansion at x=0 would not imply the inequality, without convexity (being a convex function, it is everywhere above its first order expansion at any point). --PMajer (talk) 16:30, 11 December 2008 (UTC)[reply]

Small but not finite fields

I'm trying to understand some basic ideas about fields for field's sake.

So basically, I guess I want to understand the fields that lie between k and k(t) where k is the algebraic closure of a finite field, and t is a transcendental. I think that if k < F < k(t):

if [F:k] is finite, then F=k, since k is algebraically closed
if [k(t):F] is finite, then F is k-isomorphic to k(t) by Luroth.

Are there any other fields F between k and k(t)?

A second family I want to understand are the ones that contain k(t), but not k(s,t). I think that means I want to understand fields L with k(t) < L, but the transcendence degree of L over k is 1. I think I know three such isomorphism classes of fields:

Some field L k-isomorphic to k(t), basically k(t):k(t^2) can also be considered L:k(t). One could even call L=k(t^(1/2))
The limit of that idea, L = union k(t^(1/2^n))
Apparently the field of fractions L of k[x,y]/(y^2-x^3-x-1) is a third such field, possibly excluding p=31

Are either of the last two also isomorphic to fields F above?

Is there a maximal L? Maybe just up to isomorphism: is there a field L such that if L' is a field of transcendence degree 1 over k, then L' is k-isomorphic to a subfield of L?

Thanks, JackSchmidt (talk) 20:17, 9 December 2008 (UTC)[reply]

There are no other fields F between k and k(t). We don't even need the assumption that k is algebraically closed. Suppose F is such a field, and contains a rational expression f=p(t)/q(t). Then t satisfies the polynomial fq(t)-p(t)=0 over k(f), so k(t) has finite degree over k(f) and hence over F. Algebraist 20:23, 9 December 2008 (UTC)[reply]

On your last question, surely you just want the algebraic closure of k(t)? Algebraist 20:26, 9 December 2008 (UTC)[reply]

Thanks.

So between k and k(t) there is only k and copies of k(t). The biggest field I need to consider is the algebraic closure of k(t), maybe call it K. Assuming that my sources are right, and that neither k(t^(1/2^oo)) nor that curve thing are isomorphic to k(t), then that gives two fields properly between k(t) and K.

Is there some way to understand the other fields L between k(t) and K? What are the minimal over-fields of k(t) that are not themselves just copies of k(t)? JackSchmidt (talk) 20:50, 9 December 2008 (UTC)[reply]

Your source is certainly correct that k(t^(1/2^oo)) is not k-isomorphic to k(t), since it is not even finitely generated over k. I can't immediately see why the curve thing shouldn't be isomorphic to k(t); we need the fact that for polys f and g, we never have f²=g³+g+1, which seems to require actual work. Algebraist 20:58, 9 December 2008 (UTC)[reply]

The curve thing is to me just some field, but apparently it has some geometric objects attached to it, and those geometric objects have something called a genus, and their genus differs from the genus of the geometric objects associated to k(t). And apparently the genus of a geometric object associated to a field only depends on the field and not on the object. I've just taken this on faith from reading articles like birational geometry, articles it links to, and the Springer versions.

Do you think k(curve) is a finite extension of k(t)? You call one of the variables (x or y) "t", and then the other one satisfies some equation in k(t)[s]? I guess I'm asking if k(t)[s]/(s^2-t^3-t-1) is the field of fractions of k[x,y]/(y^2-x^3-x-1). JackSchmidt (talk) 21:16, 9 December 2008 (UTC)[reply]

It is. Algebraist 21:19, 9 December 2008 (UTC)[reply]

Cool, so looking for the minimal over-fields might actually be interesting but somewhat impossible. k(curve) is isomorphic to both a degree 2 and a degree 3 extension of k(t), so its minimal in two different ways, and of course k(t):k(t^p) makes k(t) be minimal over itself in infinitely many ways. A little more insanity: k(curve) as a degree 2 extension is Galois over k(t), but as a degree 3 extension is not (I don't think).

Maybe I just want to understand finitely generated subfields of L? Maybe even in just some retarded way like "A finitely generated subfield of the algebraic closure of Q is isomorphic to Q[x]/(f(x)) for some irreducible polynomial f, and there is some relatively nasty but practical algorithm for determining when two irreducible polynomials define the same field.". Perhaps there is something similar for f.g. subfields of L? (Thanks again, I might even be able to understand more of the sources now). JackSchmidt (talk) 21:29, 9 December 2008 (UTC)[reply]