MandrakeRoot

Lets reason in a 2d world, where your planet is just a circle. Without any loss of generality we may suppose that the meteores strike in a specific point (Since you assume meteors appear randomly everywhere and there directions are also uniformly distributed), due to symmetry argumentation. According to your assumptions the "hit angle" will be uniformly distributed also and therefore counting symmetry arguments the 1/2-1/2 response is surely the better one. Mandrake

=> what about scientific curiosity ? (Isnt "usefullness" a rather subjective notion and moreover how do you determine if something will or will not be usefull in the future ?) Just a general remark, => no offense meant Mandrake

If i am not mistaken ice actually moves and shaped the landscape (moved rocks) in the ice ages. So i dont think that a scy scraper and roads or whatever other structure will resist to those forces over time. I think that this new civilization will only find debris or maybe some structures that were somehow preserved, but surely not entire cities or stuff like that. I really doubt that any government is prepared for such an event. Why invest billions of dollars into something that will probably never happen in our livetimes ? Mandrake

Dont be sorry. Nobody was offended. It is hard to teach general "proof methods", since that doesnt really exists. It really depends on the problem, how you should deroll your proof. Mostly that will come with time, seeing many proofs and doing many of them. You could try attacking the above problem directly though, bringing all terms into one big fraction and seeing what you can do about it. But personally i dont think that will be the most elegant way to demonstrate the equality. Mandrake

Induction is a valid method of proof. Why not use it ? There is nothing low skill about that method, it is just an elegant way of proving things. If there are other ways you are free to explore them. A transparent proof can be quite handy though and induction usually leads to quite transparents proofs. In the above question you can probably manipulate all terms and do a direct proof though, but the induction proof is probably more insightfull. Mandrake

Let me do one for you => 2c Assuming that with foreign you mean disjoint; assuming A and B are disjoint and P(A)=P(B) > 1/2. Their union is surely contained in the set of all events Omega and since they are disjoint we can write P(Omega) >= P(A union B) = P(A) + P(B) > 1, which is a contradiction. Hence when P(A)=P(B) > 1/2, the two sets cannot be disjoint and must have common compoments. Even : P(A intersection B) > 0 ! (try understanding why) Mandrake

What do you mean with foreign ? Don't you mean disjoint ? What have you tried for the first exercise for example ? Mandrake

Simple rules for real valued differentiable functions : (f(x) g(x))' = f'(x) g(x) + f(x) g'(x) (f(x) + g(x))' = f'(x) + g'(x) [f(g(x))]' = f'(g(x))g'(x) Try showing their correctness with the definition of the derivative : Remember that a function f is differentiable at x if the following limit exists [math]\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}[/math] If so then the value is denoted f'(x) Mandrake

That is so annoying. You are right indeed that that is unacceptable from a textbook ! Also that is not really a proof, since you might as well say it before stating the theorem. Something like : The following result follows easily from theorem 2.1.2, .... Stating the result as a pseudo proof doesnt really add anything to the book. Books should be largely self contained in my opinion Mandrake

I agree that it is rather tiring to look back all the time what was again theorem 3.3.1 when reading a book. Mandrake

The way i understood it was that the passengers went in only when the previous one is seated (somewhere). Like in a stack where you can only pop off one thing at a time ! Mandrake

In fact it is easy : Take the problem of n seats and n passengers of which the first is crazy like in the above problem for n = 100. n = 2 => This is easily seen that the probability that passenger two can sit on his seat is 1/2. Now Assume that the probability that passenger n can sit on his seat is 1/2, then for n+1 we obtain : 1) passenger 1 sits on his own seat 1/(n+1) 2) passenger 1 sits on any seat except for seat (n+1) : probability (n-1)/(n+1) Now rename seat and remark this is the same problem as problem with only n passengers . We obtain all in all therefore the probability that passenger (n+1) can sit on his seat with (n+1) seats is 1/(n+1) + (n-1)/(n+1)*1/2 = 1/2 So indeed it is 1/2, like the programs show Mandrake

I wrote a little matlab program which runs quite fast and indeed you are right. There is a flaw in the reasoning i gave above. If person 1 takes seat k then all people 2,...,k-1 can sit normally in their place. Person k will choose one seat from the 100 - k + 1 remaining etc... So the probability would be the sum over k of (person k takes the seat of 100 + person k does not take the seat of 100 * person k + 1 takes the seat of 100 etc...) I dont have time to write it out completely but i guess that would turn out to be 1/2 like the programs seems to show.... Mandrake

I clearly agree with you. Surely since lemmas are often only there too make the proof of a theorem more readable. So when two mathematicians would proof the same ("big") theorem they might end up with a different number of lemmas. Mandrake

Yeah but it is pretty annoying since you have also T3.5 etc... Once you have numbered your properties and later you discover another you need to run to 3.5 etc... (I have no idea how that went historically though, but i imagine that people first talked commonly of T3 and T4, before defining T3.5) I would say that numbering is not a good idea, naming could be though Mandrake

Work out parenthesis ! (a + bi)(c + di) = ac - bd + (bc + ad) i Mandrake PS : Here a,b,c and d are real ofcourse

The probability that person 100 can not sit on his seat is 1/100 + 99/100*1/99 + 99/100 * 98/99 *1/98 + ..... (probability that person one takes his seat + the probab that he takes another and person 2 takes his seat etc....) There are 99 terms making this 99/100 and thus the probability that he will sit on seat 100 1/100 . Mandrake

Most important theorems have a name like "Hahn Banach Theorem" or "Banach-Steinhaus" etc... The other theorems are theorems you will have to cite explicitely in talking with your fellow-mathematicians (In a discussion it is often clear what kind of result you need) Mandrake

Just show that the two sets are equal. 1) Let A be any subset of R, then [math]A^c = \{x \in \mathbb{R} \; : \; x \nin A\}[/math]. [math](A^c)^c = \{x \in \mathbb{R} \; : \; x \nin A^c\}[/math]. Now every x in A is in this set and every x in this set is in [math]A^c[/math] 2) Show that if [math]x \in (\bigcup_{i \in I} A_i)^c[/math] then it is in [math]\bigcap_{i \in I} A_i^c[/math] and vice versa. The first part would be an argument of the type, that if x is in the complement of your union, then it is not in the union, hence in none of the A_i, therefore it would be in all of the complement of A_i and thus also in the intersection. The other way around goes along the same lines. Mandrake

Yeah i didnt quite catch what you are doing. Are you trying to plot the evolution of a sequence ? What is your conjecture exactly ? If you formulate it very precisely you might be able to proof it ! Try formulating it like : "Let x be .....". If ..... ,then ..... or something like that. Mandrake

Ok you got me there. I was talking of mathematical scientific circles. Though i dont see why knowing an arcminute is Pi/108000 radians would pose a problem ? It stays the same object right ? Mandrake

I think there has been a debate already on wether or not numbers exists. They are usefull to help us understand real-life stuff but they do not really exist in "reality". In no way negative numbers are equal to positive ones but if you were to define negative numbers you would need only the existence of positive numbers in order to do so. Mandrake

Yeah you are probably right on that dave. It was what i was trying to say though not very clearly Mandrake

Possible. I have never once seen anybody use this system. Mandrake