Banned User
Joined: 3/10/2004
Posts: 7698
Location: Finland
The classical proof that there are infinitely many primes is a proof by contradiction: Let's assume that there's a largest prime. If we multiply all the primes up to that largest prime and add 1, we get a number which is not divisible by any of the primes, and thus the assumption we made is false: There was a prime which is larger than the one we assumed was the largest. However, consider this: 2*3*5*7*11*13 + 1 = 30031, which is a composite number. Isn't this a contradiction to the proof? It clearly doesn't hold that the product of the first n primes plus 1 is a prime. How to explain this apparent contradiction?
Skilled player (1882)
Joined: 4/20/2005
Posts: 2160
Location: Norrköping, Sweden
Warp wrote:
The classical proof that there are infinitely many primes is a proof by contradiction: Let's assume that there's a largest prime. If we multiply all the primes up to that largest prime and add 1, we get a number which is not divisible by any of the primes, and thus the assumption we made is false: There was a prime which is larger than the one we assumed was the largest. However, consider this: 2*3*5*7*11*13 + 1 = 30031, which is a composite number. Isn't this a contradiction to the proof? It clearly doesn't hold that the product of the first n primes plus 1 is a prime. How to explain this apparent contradiction?
I might as well try this too. :) Let's say that you've found all the primes from 1 to n. Construct the product of them and add 1 (call this number k). Assume that there is no prime larger than n (we want to get a contradiction with this assumption). There are two possibilities: 1: k is prime: this completes the proof, because we found a prime strictly larger than n, giving us a contradiction under our assumption. 2. k is composite (as in your question): If this is the case, then k can be factorized into primes. But k can not be divisible by any prime from 2 to n, so the smallest prime factor of k (let's call it v) satisfies v>n, which also gives us a contradiction, because now we've found a prime v strictly larger than n.
Experienced player (616)
Joined: 11/30/2008
Posts: 650
Location: a little city in the middle of nowhere
Although 30031 is composite, it is not divisible by any number in the set of primes that you defined [2,3,5,7,9,11,13] as it's factors are 59 and 509. By using the proof in this manner, you assumed that 13 was the largest prime in existence, but contradicted yourself by finding two larger primes: 59 and 509. This is in fact what we set out to do in the first place. EDIT: too late already answered
Measure once. Cut twice.
nfq
Player (92)
Joined: 5/10/2005
Posts: 1204
One thing I've wondered a long time is how to for example calculate seconds to minutes or hours or frames to minutes or hours. For example if there's a TAS that is 238465 frames long, how would i calculate how long that is in hours? By dividing it thrice with 60, I get 1,1040046296296etc but how am I supposed to know how many minutes and seconds "0.1040046296296etc" hours is?
Skilled player (1882)
Joined: 4/20/2005
Posts: 2160
Location: Norrköping, Sweden
nfq wrote:
One thing I've wondered a long time is how to for example calculate seconds to minutes or hours or frames to minutes or hours. For example if there's a TAS that is 238465 frames long, how would i calculate how long that is in hours? By dividing it thrice with 60, I get 1,1040046296296etc but how am I supposed to know how many minutes and seconds "0.1040046296296etc" hours is?
Let's say that a TAS is x frames long. To calculate its length in hours (h), minutes (m) and seconds (s), do this: h = floor(x/216000) m = floor((x mod 216000)/3600) s =(x mod 3600)/60
nfq
Player (92)
Joined: 5/10/2005
Posts: 1204
Randil wrote:
Let's say that a TAS is x frames long. To calculate its length in hours (h), minutes (m) and seconds (s), do this: h = floor(x/216000) m = floor((x mod 216000)/3600) s =(x mod 3600)/60
dude... i have no idea what you mean by "floor" and "mod". we should have sticked with the 60 base system, then maybe this wouldn't be so hard to calculate.
Skilled player (1882)
Joined: 4/20/2005
Posts: 2160
Location: Norrköping, Sweden
"floor" and "mod" aren't as complicated as they may sound. :) *mod means the remainder after division. For example, 15mod6=3 because 15 gives the remainder 3 after division with 6. *floor means that you round down to the nearest integer. So floor(1.2324)=1, and floor(7.8123)=7. Instead of using these formulas, we can reason like this: Let's take your example in this case, a movie of length 238465 frames. To find hours, we'll do as you did, and divide by 60^3, which gives us (as you've already said) 1,1040046296296 hours. To translate what 0,1040046296296 hours is in minutes, we multiply this number by 60. We now get 60*0,1040046296296 = 6,240277777776 minutes. So 6 minutes. To find out how many seconds 0,240277777776 minutes is, we multiply this number by 60 and get 60*0,240277777776 = 14,4166... seconds. So the movie length is 1 hour, 6 minutes and 14.4 seconds.
nfq
Player (92)
Joined: 5/10/2005
Posts: 1204
ok, cool, i understand now.
Banned User
Joined: 3/10/2004
Posts: 7698
Location: Finland
What do they teach at grade school nowadays?
Experienced player (616)
Joined: 11/30/2008
Posts: 650
Location: a little city in the middle of nowhere
Fucking nothing that's what. we get given one of these and are told how to do anything with them. It's basically a computer. word processing, spreadsheets, automatic solve function for any relationship ect. I preferred it when you just used a standard scientific calculator, and actually learned how to do it. I had to look on Wikipedia for all my course material, so I actually understand what's going on. You literally don't get taught anything at school.
Measure once. Cut twice.
Banned User
Joined: 3/10/2004
Posts: 7698
Location: Finland
It really worries me how around the world schooling systems are dumbing down the teaching of math, especially in the US. It seems that the current trend is to dismiss the old way of calculating eg. additions, multiplications and divisions mechanically by hand as old-fashioned and obsolete. Instead, especially in the US, the modern trend is to try to make pupils learn to do math "intuitively". Some books and even some schools completely skip teaching the accurate mechanical way of calculating those things, and instead resort to what amounts to "if it's too difficult to quickly calculate in your head, just use a calculator". There's nothing wrong in learning to do math in your head intuitively. However, when this completely replaces the old good accurate way of doing it on paper, that's when things go wrong. People fail to learn the most basic principles of math, even for such simple things as addition and multiplication. I think the schooling systems are missing the point. All they think is "nowadays you don't need to mechanically calculate things on paper by hand, that's just a waste of time". What they fail to realize is that understanding how simple math works helps understanding how more complicated math works. If you are not fluent using the most basic tools, you can't use them for more complicated tasks.
nfq
Player (92)
Joined: 5/10/2005
Posts: 1204
IQs started dropping after they started teaching the theory of evolution in schools (students started acting like monkeys and going apeshit), and after the sexual hippie revolution IQ started dropping even more, that's why they had to make schools and tests dumber so that kids could pass the tests.
Experienced player (616)
Joined: 11/30/2008
Posts: 650
Location: a little city in the middle of nowhere
IQs started dropping after they started teaching the theory of evolution in schools
Where the hell did you get that information from?
Measure once. Cut twice.
Banned User
Joined: 3/10/2004
Posts: 7698
Location: Finland
nfq wrote:
IQs started dropping after they started teaching the theory of evolution in schools (students started acting like monkeys and going apeshit), and after the sexual hippie revolution IQ started dropping even more
http://en.wikipedia.org/wiki/Post_hoc_ergo_propter_hoc
that's why they had to make schools and tests dumber so that kids could pass the tests.
http://en.wikipedia.org/wiki/Wrong_direction Deterioration of general knowledge is caused by dumbing down education, not the other way around.
arflech
He/Him
Joined: 5/3/2008
Posts: 1120
Warp wrote:
What do they teach at grade school nowadays?
I never learned "floor" or "mod" in school until I took a class on programming in college (I had looked them up in high school), although the related concepts of "front-end estimation" and "remainder" were taught in grade school (early 1990s, small-town USA). I don't know how old you are andymac, but the use of the ClassPad makes me weep for Amurrikkka.
i imgur com/QiCaaH8 png
Experienced player (616)
Joined: 11/30/2008
Posts: 650
Location: a little city in the middle of nowhere
While we're on the topic, I'm having exams at the moment, and exams nowadays are formatted differently. There is a non-calculator section and a calculator section. This means you learn about two fundamental aspects of mathematics. how to do a little bit of theory, and how to bang numbers into your calculator. The sad truth is that if you don't know how to use your calculator, you will fail the exam. Even if you know all the theory, and can understand the most complex of concepts. The calculator assumed section I believe is worth 60% of your entire mark. What the fuck are they trying to tell us. That you don't have to know anything? That the singularity is coming? (laugh it's funny dammit). That damn calculator is detrimental to your education and your pockets. And now I'm going to fail my exam because I accidentally left it at school and can't study with it.
Measure once. Cut twice.
Joined: 7/2/2007
Posts: 3960
Maybe they're trying to tell you that practically, knowing how to plug in numbers is an important part of real-life mathematics and physics. Don't get me wrong -- theory is vital. Without theory you're just doing everything by rote. But if you can't apply that theory, then there's very little you can do with it.
Pyrel - an open-source rewrite of the Angband roguelike game in Python.
Experienced player (616)
Joined: 11/30/2008
Posts: 650
Location: a little city in the middle of nowhere
Yes it's true that you need to be able to use tools such as these, however, the course is orientated around calculator specific functions. We don't learn about the general usage of tools to calculate things for us, we learn about how the classpad can do it. We don't get taught how to think, we get taught how to use the classpad.] Better get back to studying, I'm procrastinating by typing this.
Measure once. Cut twice.
Joined: 3/7/2006
Posts: 720
Location: UK
This isn't really a new thing (at least where I live). All GCSEs and A-levels are taught to make you pass the exam, not to impart some knowledge to you. Hell, in most A-level courses, whenever you're doing any coursework, they usually give you the marking scheme, so if you don't get full marks it's because you're lazy. How's that for education.
Voted NO for NO reason
Joined: 5/2/2006
Posts: 1020
Location: Boulder, CO
It does seem that math requirements are being reduced, even from college level course work. Many engineering majors offer "methods" classes in place of taking calc 3, which teaches use of major specific software. I think its a bad idea to reward pople for not understanding math.
Has never colored a dinosaur.
Skilled player (1882)
Joined: 4/20/2005
Posts: 2160
Location: Norrköping, Sweden
How well do you remember your linear algebra? Given two lines L1: p1+c1*s and L2: p2+c2*t, where p1,p2,c1 and c2 are 3-dimensional vectors and s and t are real numbers, find the shortest distance between L1 and L2.
Joined: 4/16/2005
Posts: 251
I don't remember anything from LA. And for the better. Well... Assuming these are not parallel, I'd make the two planes p1+c1*s+c2*t and p2+c1*s+c2*t, get those in normal vector form (no idea how that one is called in english, v*n+s=0 form), and calculate their distance. Still, I never want to see linear algebra again in my life.
Experienced player (616)
Joined: 11/30/2008
Posts: 650
Location: a little city in the middle of nowhere
Never done linear algebra in any other dimensions except 2. so anyways, here's my solution, even though it is probably far more complex than needs be. I have never used quaternions either, and have not been taught most of this crap: http://cid-de1e7fef4aaf7e0d.skydrive.live.com/self.aspx/.Public/quarternial%20problem.PNG Okay, so basically, assign each vector a quaternion value in the pure imaginary realms, then manipulate one vector so it is the i axis, rid yourself of the i value in the other equation's constant then calculate the distance between the two lines. Find the derivative of the distance between the two lines, find the turning point, and then sub into the distance equation and voila, you have the shortest distance between two lines. There's probably a lot of mistakes, but I think the theory is right. EDIT: I think my theory is wrong. when multiplying by vi, I introduce a real part into the system, and it all goes per shaped, and I completely forget about it.
Measure once. Cut twice.
Skilled player (1882)
Joined: 4/20/2005
Posts: 2160
Location: Norrköping, Sweden
Gorash is indeed right - the easiest way is to move the vector c2 so that it intersects the point p1. Now you have a plane that is spanned by the vectors c1 and c2, and that includes the point p1. By moving the vector c1 so that it intersects p2 we get another plane, and the shortest distance between the lines is the distance between the two planes. After some calculation, we find the distance, d, to be d = |w*(c1-c2)| where w=(c1xc2)/|(c1xc2)| (the vector product of c1 and c2, with length 1) I don't know what andymac is up to exactly, but it looks pretty cool. :P I don't know if that method works, but it would be exciting if it did.
Experienced player (616)
Joined: 11/30/2008
Posts: 650
Location: a little city in the middle of nowhere
I didn't explain very well what I did. lines 1-2: define Vn as a quaternion vector, and x being a real coefficient. 2-4: define functions f(x) and g(x) that they equal your two lines. 4-6: find Vi such that V2xVi=i, which can be found by premulitiplying by V2^-1 6-10: dilate both f(x) and g(x) by a factor of Vi, then translate both functions. This means that one function is the I axis (xi) and the other is now just a jumble. This is where my stuff falls apart, by multiplying g(x) by Vi, you introduce a real part into the system and g(x)xVi, which makes everything turn to shit. 10-13: What i try to do here is shift the g(x) variant along the real axis (which shouldn't change the line at all, because it's imaginary, but it does because I fucked up.) and remove the i term from the constant. Vs is basically the i intercept of the line. The distance now, should be the j,k distance away from the i axis. so basically, the distance from the origin to the point (x,y). This next step I forgot to add: you can now differentiate the difference, find the turning point, and then substitute the x into the original line You have dilated the lines, so you need to use the original lines to find the distance. use the i coefficient of h(x) to substitute into f(x). Then find the distance between f(x) and g(x) at the two points found. question easily done EDIT: I get your reasoning, both planes will be parallel, even if both vectors are parallel (however, then there will only be two lines not planes). therefore if you find a line perpendicular to the planes, and find the intersections between the lines and the planes, the distance between the two will be the shortest distance between the two lines. EDIT2: here is a general (working) solution. by substituting 1/c(x+ck) you get an equation of the form below, with no k therm in the constant, and 1 as the coefficient of k in the other term. This does not change the line shape at all.
Measure once. Cut twice.