mm-3969 === Subject: Re: Find smallest area of a cylinder >>#What is the ration between radius and height >>of a cylinder when its surface area at certain volume >>is smallest? >>If instead of smallest we would have to find biggest I'd >>have no problem solving it. >On the contrary, it would be impossible to solve: for a >given volume you can make the surface area as large >as you like. I draw a graph and from it saw that we can only find minimum surface area but not maximum. But drawing graphs won't always be an option so how else could I figure out when we can find maximum and when minimum surface area for an object, if V=const? === Subject: Re: Find smallest area of a cylinder >>between radius and height of a cylinder when its >>surface area at certain volume is smallest? >>If instead of smallest we would have to find biggest >>I'd have no problem solving it. >On the contrary, it would be impossible to solve: >for a given volume you can make the surface area as >large as you like. >> V = PI * r^2 * h >> h = V / (PI * r^2) >So r/h = PI * r^3 / V. >> A = 2 * PI * r^2 + 2 * PI * r * h >>A = 2 * PI * r^2 + 2 * PI * r * V/(PI*r^2) >= 2 * PI * r^2 + 2 * V / r >= br^2 + c/r, >where b = 2 * PI, and c = 2 * V. Note that b/c = PI/V, so >that r/h = br^3 / c. >For a fixed V both b and c are constant. Then >dA/dr = 2br - c/r^2, which is 0 when 2br = c/r^2, >r^3 = c/(2b), and r = [c/(2b)]^(1/3). >The second derivative of A with respect >to r is 2b + 2c/r^3, which is clearly positive, so >r = [c/(2b)]^(1/3) gives a minimum surface area for >the fixed volume V. At this point r/h = br^3 / c = b[c/(2b)]/c = 1/2. I haven't yet covered second derivatives, so can this be solved without their help? So we get r = [c/(2b)]^(1/3) and then what? === Subject: Re: Find smallest area of a cylinder On Wed, 12 Apr 2006 19:35:07 EDT, lucky_luke in alt.math.undergrad: [...] >>For a fixed V both b and c are constant. Then >>dA/dr = 2br - c/r^2, which is 0 when 2br = c/r^2, >>r^3 = c/(2b), and r = [c/(2b)]^(1/3). >>The second derivative of A with respect >>to r is 2b + 2c/r^3, which is clearly positive, so >>r = [c/(2b)]^(1/3) gives a minimum surface area for >>the fixed volume V. At this point r/h = br^3 / c = b[c/(2b)]/c = 1/2. > I haven't yet covered second derivatives, so can this > be solved without their help? So we get > r = [c/(2b)]^(1/3) and then what? Look at the derivative when r is close to [c/(2b)]^(1/3) on each side. Recall that dA/dr = 2br - c/r^2, which is 0 when r = [c/(2b)]^(1/3); what happens if you decrease r a bit? 2br decreases, and c/r^2 increases (because r^2 gets smaller), so dA/dr decreases and therefore must become negative. What happens if instead you increase r a bit? 2br increases, and c/r^2 decreases, so dA/dr increases and therefore must become positive. That is, as r increases from less than [c/(2b)]^(1/3) through [c/(2b)]^(1/3) to greater than [c/(2b)]^(1/3), dA/dr goes from negative to 0 to positive. This means that A itself goes from decreasing to flat to increasing, which clearly means that it has a minimum at r = [c/(2b)]^(1/3). Brian === Subject: Total Sum of Squares I am working on showing that SS[total] = SS[Residual] + SS[Regression] Sum[(y - mu[Y])^2] = Sum[(y - y(hat))^2] + Sum[(y(hat) - mu[Y])^2] I am really just looking for advice on the method I tried, and see if anyone knows what went wrong. And would be willing to tell me where I went wrong. I tried expanding it a few times, making substitutions for y(hat) and well, things got messy real fast. Anyway, I decided to take the derivate with expression for y(hat), and then substituted that in the right hand side of the equation. The result is so close, but not quite right. Sum[(y - mu[Y])^2] = {Sum[(y - mu[Y])^2]}/2 I checked my algebra with Mathematica and it seemed to check. So, I guess my question is, do you think I am overlooking something (which is giving me the factor of 2 difference) or is this just a coincidence resulting from fallacious processes? === Subject: Re: Elementary event more likely to happen-lost >2. >P(B) = 3 / 20. Is correct way to think of this >probability that out of 20 experiments of >throwing a die 3 outcomes will be event B? >>you could consider it like a fair 20 faced die with >>three faces marked with B. >>but really all it is is a fraction 3/20 or 0.15 , >>b will occur 0.15 of the time (or 15% of the time) > Out of all tosses? >If you toss it many times, B will occur about 15% >of the time; the more times you toss it, the closer >to 15% the proportion of B's is likely to be. >It's really no different from tossing a fair coin. >The probability that it comes up heads is 1/2, but >you surely must realize that if you toss the coin >10 times, you may very well *not* get exactly 5 >heads. Indeed, if you toss the coin 1,000,000 >times, you are very, very unlikely to get >exactly 500,000 heads. But your *percentage* of >heads is very unlikely to be less than 49.985% or >more than 50.015%. So how is that different from me interpreting that 3/20 means that if we toss a die with with number 2 three times more likely to happen, then the desired face up will happen around 3 times if we toss it 20 times? >The tosses are completely independent. What >happens in the first 15 tosses has absolutely >no effect on the probability of getting B on the >16th toss. So chances of of tossing 10 heads and prob of tossing 10 heads after you have already tossed 9 heads are equal? === Subject: Re: Elementary event more likely to happen-lost On Wed, 12 Apr 2006 21:04:21 EDT, something_about_math in alt.math.undergrad: >>2. >>P(B) = 3 / 20. Is correct way to think of this >>probability that out of 20 experiments of >>throwing a die 3 outcomes will be event B? >you could consider it like a fair 20 faced die with >three faces marked with B. >but really all it is is a fraction 3/20 or 0.15 , >b will occur 0.15 of the time (or 15% of the time) >> Out of all tosses? >>If you toss it many times, B will occur about 15% >>of the time; the more times you toss it, the closer >>to 15% the proportion of B's is likely to be. >>It's really no different from tossing a fair coin. >>The probability that it comes up heads is 1/2, but >>you surely must realize that if you toss the coin >>10 times, you may very well *not* get exactly 5 >>heads. Indeed, if you toss the coin 1,000,000 >>times, you are very, very unlikely to get >>exactly 500,000 heads. But your *percentage* of >>heads is very unlikely to be less than 49.985% or >>more than 50.015%. > So how is that different from me interpreting that > 3/20 means that if we toss a die with with number 2 > three times more likely to happen, then the desired > face up will happen around 3 times if we toss it > 20 times? If you toss the thing 20 times, it would not be terribly surprising to get the B face only once, or four or five times. If *many* of us toss it 20 times each, however, and we average the numbers of times we get the B face, that average is very likely to be very close to 3. >>The tosses are completely independent. What >>happens in the first 15 tosses has absolutely >>no effect on the probability of getting B on the >>16th toss. > So chances of of tossing 10 heads and prob of > tossing 10 heads after you have already tossed > 9 heads are equal? Yes. You're starting from scratch with each toss. Brian === Subject: Re: Elementary event more likely to happen-lost So set has an impossible event only if we include it in a set. If an experiment is to roll a die, and if outcomes are {1,2,3,4,5,6} then this set doesn't have an imposible event. But if outcomes are {1,2,3,4,5,6,7} then set has impossible event - > 7 ? === Subject: Re: Elementary event more likely to happen-lost On Wed, 12 Apr 2006 21:05:18 EDT, something_about_math in alt.math.undergrad: > So set has an impossible event only if we include > it in a set. > If an experiment is to roll a die, and if outcomes > are > {1,2,3,4,5,6} > then this set doesn't have an imposible event. > But if outcomes are > {1,2,3,4,5,6,7} > then set has impossible event - > 7 ? Yes, if it's an ordinary cubical die. And the seven outcomes are of course not equally likely; their probabilities are 1/6, 1/6, 1/6, 1/6, 1/6, 1/6, and 0, in order. Brian === Subject: Dependent events-if I don't understand them till monday I'm F In short, if I don't understand this till monday I'm F*****, so I I truly, madly, deeply hope you will have the patience and inner strength to help me with this. Dependent and independent events really got to me. I understand the basics but can't really connect the dots. I know that definition of a dependent event means that if probability of B occuring after A has already happened is different from P(B) then event is dependent. I know that in P(A | B) we are getting a new sample space B, and a new event, A / B. But when it comes to applying all this to actual examples, I get lost! And I mean really really lost. Perhaps it's best to start with examples. #We toss a coin 10 times. Find probability of tossing 10 heads and prob of tossing 10 heads after you have already tossed 9 heads. First one is P( 10H ) = 1/2 * 10 But why do we multiply with 10? P(9/10) - I don't know #47 % of students play tenis and out of those 47% of students, 78 % of them also play basketball. If we randomly pick a student, what is the probability that randomly picked student plays tenis and also basketball? P(A / B) = P(A)*P(B|A) = 47/100*( 47/100*78 ) New sample space would have 47 outcomes and 78% of those outcomes is P(B|A)? #14% of students play piano and violin,and 67 % students play piano. Find probability that randomly picked student that plays piano, also plays violin? P(A ) = P(A)*P(B|A) = (67+14)/100 * (81/100*) #If the weather is nice Katja goes for a walk 90% of the time, but if it is bad, she stays at home in only 1/3 of the time. Weather report for the end of a week forecasts with 70% probability that weather will be sunny. What is the probability of Katja not going out for a walk? #In Europe, according to statistical data ,43 % of drivers wear seatbelt. We randomly picked two drivers. Find probability that both wear seatbelt. First driver ... event A Second driver ... event B P(A B) = 49/100 * 49/100 But shouldn't new sample space A (if P(B|A) )for second driver be smaller by one since one driver has already been picked out? So in fact B should be dependent event. # # # # # # Product of events A B is represented with Venn diagram: The intersection of sets are those elements that belong to all intersected sets and diagram for A / B is same as that for intersection of sets. But that diagram representation is true only if A and B both contain some common elements. In other words, it's true only when A and B are dependent events. When A and B are independent events, they don't have any common elements. So why we still use that diagram to represent product of two independent events? # # # # # # # # # Next question is about sample space and elementary events. I know what those are and thus this question: We have an experiment of tossing a blue die and a green die - on each die one of the numbers 1 - 6 comes up. One collection of outcomes is {(1,1), (1,2), ..., (1,6), (2,1), ... (6,6)} so 36 outcomes total. One event is that the two dice add to 4: L = {(1,2), (1,3), (2,2), (2,1), (3,1), (2,2)}. Or we have an experiment of tossing just green die. Then one collection of outcomes is L={1,2,3,4,5,6}. One event is that up face is even number A:{2,4,6}. But what if experiment is tossing die twice and face up value of first trow is event E and face up value of second trow is event F and product of the two events is Q = E F? This can't be right since we can either decide that either single throw of die is an experiment or two ( or more ) throws of a die is experiment. If the former is true, then elementary events are face-up values {1,2,3,4,5,6}, but if experiment is two throws of a die then there are 36 outcomes and thus 36 elementary events. But we can't have en experiment where first throw of a die is one event, and second throw is second event. Isn't an event supposed to encapsulate the whole experiment and not just one third of 1/1000 of it? Outcome is defined as possible result of an experiment, so if experiment is 1000 throws of die then one outcome describes some condition after we toss a die 1000 times. And since elementary event is a subset of sample space with single outcome in this subset and since other events are made out of elementary events, then I don't understand why a single toss of die would be an event in an experiment defined as two tosses of a die? Or should I interpret it as if we have two different experiments ( experiment Q is first throw of a die and experiment P is second throw, and this two experiments together make a completely new experiment )? thank you === Subject: Re: Dependent events-if I don't understand them till monday I'm F On Wed, 12 Apr 2006 21:10:50 EDT, something_about_math in alt.math.undergrad: [...] > Dependent and independent events really got to me. > I understand the basics but can't really connect the > dots. I know that definition of a dependent event > means that if probability of B occuring after A has > already happened is different from P(B) then event is dependent. > I know that in P(A | B) we are getting a new sample space B, and a new event, A / B. But when it > comes to applying all this to actual examples, > I get lost! And I mean really really lost. > Perhaps it's best to start with examples. > #We toss a coin 10 times. Find probability of tossing > 10 heads and prob of tossing 10 heads after you > have already tossed 9 heads. > First one is P( 10H ) = 1/2 * 10 No, it isn't: it's (1/2)^10, i.e., (1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) = 1/1024. When events A and B are independent, as is the case for two coin tosses, for instance, P(A & B) = P(A) * P(B). Here there are ten events instead of just two, but the principle is the same. > But why do we multiply with 10? We don't. > P(9/10) - I don't know You mean P(10H | previous 9 were H)? It's the same thing: the tosses are independent of one another. > #47 % of students play tenis and out of those 47% > of students, 78 % of them also play basketball. > If we randomly pick a student, what is the > probability that randomly picked student plays tenis > and also basketball? > P(A / B) = P(A)*P(B|A) = 47/100*( 47/100*78 ) This is doing it the hard way. You're told that 78% of 47% of the students play both tennis and basketball; since 0.78 * 0.47 = 0.3666, 36.66% of the students play both sports, and the probability of picking one who does is 0.3666. Still, you can do it that way. If A is the event that the student plays tennis, and B is the event that the student plays basketball, then certainly P(A) = 0.47. If you already know that you have one of these tennis-playing students, you know from the problem statement that there's a 78% chance that he also plays basketball, so P(B|A) = 0.78. The formula then yields P(A / B) = 0.47 * 0.78 = 0.3666, as before. > New sample space would have 47 outcomes and > 78% of those outcomes is P(B|A)? Not quite. If there are n students altogether, there were n outcomes originally. If you know that A has occurred, you know that you got one of the 0.47n tennis-playing students, and your new sample space has 0.47n possible outcomes. 78% of them are in the intersection of A and B, so P(B|A) = 0.78. > #14% of students play piano and violin,and 67 % > students play piano. Find probability that randomly > picked student that plays piano, also plays violin? > P(A ) = P(A)*P(B|A) = (67+14)/100 * (81/100*) No: the lefthand side should be P(A / B). It helps to say what you're doing. I presume that A is the event that the chosen student plays the piano, and B is the event that he plays the violin. Then we're told that P(A / B) = 0.14 (14% play both instruments). We're also told that P(A) = 0.67. The basic formula is P(A / B) = P(A) * P(B|A), so we have 0.14 = 0.67 * P(B|A), and hence P(B|A) = 0.14/0.67 ~= 0.21. (It's exactly 14/67, of course.) > #If the weather is nice Katja goes for a walk > 90% of the time, but if it is bad, she stays at > home in only 1/3 of the time. Weather report for > the end of a week forecasts with 70% probability > that weather will be sunny. What is the probability of > Katja not going out for a walk? Let H be the event that Katja stays home, and let S be the event that the weather is sunny. We're told that P(~H|S) = 0.90, so P(H|S) = 0.10. (I'm using '~' for 'not'. In other words, if it's sunny there's only a 10% chance that Katja stays home.) We're also told that P(H|~S) = 1/3 and that P(S) = 0.70. We want P(H). The event of her staying home is the union of the two disjoint events 'she stays home and the weather is nice' and 'she stays home and the weather is bad', so P(H) = P(H / S) + P(H / ~S). Try using the basic formula and the information in the previous paragraph to calculate P(H / S), P(H / ~S), and then the desired P(H). > #In Europe, according to statistical data ,43 % > of drivers wear seatbelt. We randomly picked two > drivers. Find probability that both wear seatbelt. > First driver ... event A > Second driver ... event B > P(A B) = 49/100 * 49/100 > But shouldn't new sample space A (if P(B|A) )for > second driver be smaller by one since one driver has > already been picked out? So in fact B should be > dependent event. You are correct. However, the number of drivers in Europe is so large that this makes no practical difference. Just to keep things simple, imagine that there are one million drivers in Europe; then 430,000 of them wear seatbelts. Once you've picked one of those 430,000 people, the probability of picking another is 429,999/999,999 = 0.429999429999..., which is barely different from 0.43. (In fact, there's probably more error in the original figure of 43%.) > # # # # # # > Product of events A B is represented with Venn diagram: > The intersection of sets are those elements that > belong to all intersected sets and diagram for > A / B is same as that for intersection of sets. > But that diagram representation is true only if > A and B both contain some common elements. > In other words, it's true only when A and B are > dependent events. > When A and B are independent events, they don't > have any common elements. So why we still use that > diagram to represent product of two independent > events? Technically it covers all cases; you just have to remember that in any given specific situation, any one (or more) of the regions in the diagram might actually be empty. > # # # # # # # # # > Next question is about sample space and elementary > events. I know what those are and thus this question: > We have an experiment of tossing a blue die and a green > die - on each die one of the numbers 1 - 6 comes up. One > collection of outcomes is {(1,1), (1,2), ..., (1,6), > (2,1), ... (6,6)} so 36 outcomes total. One event is that > the two dice add to 4: L = {(1,2), (1,3), (2,2), (2,1), > (3,1), (2,2)}. You don't want (1, 2) and (2, 1): they sum to 3. > Or we have an experiment of tossing just green die. > Then one collection of outcomes is L={1,2,3,4,5,6}. > One event is that up face is even number A:{2,4,6}. > But what if experiment is tossing die twice and face > up value of first trow is event E and face up value of > second trow is event F and product of the two > events is Q = E F? It isn't clear what you're asking here; I'm going to make a guess, but can you restate it, perhaps with a *specific* example that bothers you? Consider the experiment of rolling a die twice. There are 36 possible outcomes, the ones that you listed above for the blue and green dice. There's nothing wrong now with defining the event L to be that the total of the two rolls is 4 and the event A to be that the first roll is even: L = {(1, 3), (2, 2), (3, 1)}, and A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}. L contains 3 of the 36 equally likely outcomes, so P(L) = 3/36 = 1/12, and A contains 18 of them, so P(A) = 18/36 = 1/2. (Note that this is exactly what you'd expect: the probability of getting an even number on a single die roll is certainly 3/6 = 1/2, and the second die doesn't affect whether the outcome is in the event A or not.) > This can't be right since we can either decide that > either single throw of die is an experiment or two > ( or more ) throws of a die is experiment. In this example the experiment is two rolls, but one of the events is defined in terms of what happens on the first roll only. [...] Brian === Subject: Re: Dependent events-if I don't understand them till monday I'm F > In short, if I don't understand this till monday > I'm F*****, so I I truly, madly, deeply hope you > will have the patience and inner strength to help > me with this. > Dependent and independent events really got to me. > I understand the basics but can't really connect the > dots. I know that definition of a dependent event > means that if probability of B occuring after A has > already happened is different from P(B) then event is dependent. > I know that in P(A | B) we are getting a new sample space B, and a new event, A / B. But when it > comes to applying all this to actual examples, > I get lost! And I mean really really lost. > Perhaps it's best to start with examples. The probablity of tossing snake eyes in 1/36, (unless you're a White House advisor, in which cast the probablity of tossing snakes eyes is likely more but never less than one.) Now you toss one di. It's not an ace. Thus the probablity of tossing snake eyes is zero. On the other hand, if it is an ace, then the probablity of tossing snake eyes is 1/6, the probablity that the second toss will yeild up an ace. > #We toss a coin 10 times. Find probability of tossing > 10 heads and prob of tossing 10 heads after you > have already tossed 9 heads. > First one is P( 10H ) = 1/2 * 10 > But why do we multiply with 10? > P(9/10) - I don't know Your notation is cryptic. The probablity of tossing 10 heads in a row is 2^-10. The probablity of tossing 10 heads in a row after tossing 9 heads in a row is 1/2, the probablity of a head upon the last toss. === Subject: Rings and valid expressions There have been these discussions about my non-polynomial factorization research where the replies go on and on, when it can be about dumb things like square roots having two values (others arguing against, me arguing for), and considering the big picture, I think quite a few of them may be clipped by talking some about rings and expressions valid within them. There seems to be confusion about the distributive property when you have an expression not valid in a particular ring, like consider the simple example: In the ring of algebraic integers let 7*(f(x) + 1) = x + 7 ---and you have a contradiction as the expression is NOT valid in that ring, as it requires f(x) = x/7 so notice that SAYING you're in the ring of algebraic integers, doesn't mean you actually are. Here you're pushed out of the ring. Notice though that the distributive property still works so you have 7*f(x) + 7 = x + 7 and it's just that f(x) = x/7, so the value you get from a function does not impact the distributive property. So 7 still multiplied through. With more complex factorizations you can get more complex behavior so that when you move to 7*C(x) = (f(x) + 7)(g(x) + 1) where f(0) = g(0) and C(x) is a polynomial you can start in the ring of algebraic integers, have the equations valid in the ring of algebraic integers, but still get a result not valid in that ring. That's where things get subtle and where it's possible for malicious people to confuse the issue. Now the argument that shows that you're forced out of a particular ring, like the ring of algebraic integers, for certain expressions isn't complicated, and relies on the distributive property still. Just like with 7*(f(x) + 1) = x + 7 where you can SAY you're in the ring of algebraic integers, and find something that may look to a naive person like you defied the distributive property, it's still true that 7 multiplied through, as the distributive property simply says that if you multiply a group, you multiply each of the elements within that group, so f(x) and 1 get multiplied by 7. But you don't see 7 times x on the right side. So? It doesn't change the distributive property. With more complicated expressions you get more complicated behavior, so that with 7*C(x) = (f(x) + 7)(g(x) + 1) where f(0)=g(0) = 0 you can find C(x), f(x) and g(x) such that you are pushed out of the ring if you follow logically where the 7 must have multiplied through. What makes this situation frustrating is that the mathematics is simple to the point of triviality, and I think most of you can comprehend that a function can be outside of a ring, so that someone arguing against the distributive property who jumped up and down about 7*(f(x) + 1) = x + 7 because you can't SEE the 7 multiplied times x, would not be very convincing. But with slightly more complicated expressions where the proof is simple, people seem willing to try and SEE the 7, and when they don't see it, they are convinced by posters like Rick Decker or W. Dale Hall, who make it their business to be confusing on this issue. The value of the functions does not change the distributive property. That is just immutable. If you are mathematicians, given a proof, you will follow the proof. When posters claim examples that contradict a proof, you will look to find what's wrong with their claims versus simply tossing out the proof as if mathematical proof meant nothing to you. Doubt is a good thing. It should lead you to depend on what is absolute, and in mathematics that is mathematical proof. The proof is simple with 7*C(x) = (f(x) + 7)(g(x) + 1) on the complex plane, where f(0) = g(0) and C(x) is a polynomial if you accept that the distributive property does not care about the value of what is being multiplied, then the value of the function does not change the distributive property. That's the main point. Accept it or challenge. If you accept it, then it must be true that if the value of the function does not matter then ANY VALUE can be used, and therefore, it follows logically that x=0 is valid as a useful value. Accept that logical step, or challenge. Now then, if it does not matter what value is used, so x=0 is valid, then it is just a matter of noting that at x=0 you have 7*C(0) = (0 + 7)(0 + 1) showing that 7 multiplied through only one, and the proof is done. Examples equivalent to 7*(f(x) + 1) = x + 1 do NOT REFUTE A MATHEMATICAL PROOF!!! I think part of the problem here is that many of you do not comprehend what a mathematical proof actually is, so you think that proofs can be broken, or created, or they are fragile things that can shake with arguing. But what proofs are, are points of truth that do not require faith. You don't have to believe in some person or entity. You don't have to trust or worry about what education level one person has versus another. All those are points irrelevant to a proof, so when people question my not having a degree in mathematics as my degree is in physics, you do not have to worry about that if you know what mathematics is about, as what you really need is the proof. With the proof social crap is irrelevant. It doesn't matter how many people argue with me. It doesn't matter what your gut feeling tells you. Trace the logical steps from a truth, and the conclusion that follows MUST BE TRUE. See my definition of mathematical proof: http://mymath.blogspot.com/2005/07/definition-of-mathematical-proof.html The arguing would have been over years ago if mathematicians followed proofs as I have the proofs, people just lie about them. Given a proof, argument is just a waste of time, because a proof cannot be refuted. You may have heard of proofs being refuted. You may have heard of proofs failing, but then, they were not proofs!!! If someone has proof that you are dead, but it turns out you are not dead (as none of you are) then did they really have proof? Was it failed proof? Did their proof collapse? Can someone create a proof that you are dead and then someone else refute it? NO!!! They just didn't have proof. James Harris === Subject: Re: Rings and valid expressions > I think part of the problem here is that many of you do not comprehend > what a mathematical proof actually is Interesting! I myself think the *whole* problem here is that you're really, really dumb. That would also probably explain why you can't hold a job, wouldn't it? === Subject: Re: Rings and valid expressions > I think part of the problem here is that many of you do not comprehend > what a mathematical proof actually is > Interesting! I myself think the *whole* problem here is that you're > really, really dumb. > That would also probably explain why you can't hold a job, wouldn't it? One aspect of Harrisiana is observing the assembled multitudes of JSH addicts, those of use who can't keep our eyes off the ongoing train wreck. A subspecies of observer is the person who makes personal remarks -- attacking not JSH's *behavior*, but his *person*. It's the lowest form of argument -- my opponent is a scumbag, therefore his ideas are wrong. But in some cases, it's more of a psychological clue to the poster, rather than JSH. After all, JSH is who he is -- and whatever else he's done, he's turned a lot of us on to the algebraic integers, which is a pretty cool object that doesn't get talked about much in algebra class. Why come on here and criticize his job, or his life ... and to hint that you have personal information about him? As far as I'm concerned, if you'd out JSH you'd out me. I take it personally. Because there's a certain minimal level of civility I like in anonymous forums. Say someone's ideas are stupid, fine. Call someone an asshole, well do so if you must, but too much of that makes for a bad discussion group. But imply that you have personal knowledge of someone's life, and seek to embarass them or attack their idea based on that ... that, sir, is way over the line. I'd like you to consider that. === Subject: Re: Rings and valid expressions > There have been these discussions about my non-polynomial factorization > research where the replies go on and on, when it can be about dumb > things like square roots having two values (others arguing against, me > arguing for), and considering the big picture, I think quite a few of > them may be clipped by talking some about rings and expressions valid > within them. Again, either through malice or ignorance, you misrepresent the position of your detractors! Nobody, as far as I've seen, has argued that the equation x^2 = 2 doesn't have two solutions. Instead, people have told you that the NOTATION sqrt(2) represents the positive solution, while -sqrt(2) represents the negative. Why do we need such notation you ask? Well, it's a short-hand that simplifies exposition. Imagine having to write, the positive square root of 2, everytime you wanted to talk about the square root of 2? Why is this so hard for you to understand? > There seems to be confusion about the distributive property when you > have an expression not valid in a particular ring, like consider the > simple example: > In the ring of algebraic integers let > 7*(f(x) + 1) = x + 7 > ---and you have a contradiction as the expression is NOT valid in that > ring, as it requires > f(x) = x/7 > so notice that SAYING you're in the ring of algebraic integers, doesn't > mean you actually are. Well, if you understood the definition of a ring, you wouldn't be making this mistake. The place where you are confused is CLOSURE UNDER MULTIPLICATION. Specifically, if x,y are in a ring R, then the product x*y must be in R. There's nothing that says that the converse must be true. In other words, it is not necessarily true that if x*y is in R, then x and y must be in R. For example, 3 is an integer, 3/7 is not an integer, yet the product 7 * 3/7 is in the ring of integers. How exactly does this apply to your equation above? Well, if f(x) = x/7, then there are many values of x where f(x) + 1 would be in the algebraic integers, and many values where f(x) + 1 wouldn't be in the algebraic integers. > Here you're pushed out of the ring. But you're not necessarily in the ring to begin with! > Notice though that the distributive property still works so you have > 7*f(x) + 7 = x + 7 > and it's just that f(x) = x/7, so the value you get from a function > does not impact the distributive property. > So 7 still multiplied through. > With more complex factorizations you can get more complex behavior so > that when you move to > 7*C(x) = (f(x) + 7)(g(x) + 1) > where f(0) = g(0) and C(x) is a polynomial > you can start in the ring of algebraic integers, have the equations > valid in the ring of algebraic integers, but still get a result not > valid in that ring. I'll save my comment on this for below, where you mention it again. > That's where things get subtle and where it's possible for malicious > people to confuse the issue. > Now the argument that shows that you're forced out of a particular > ring, like the ring of algebraic integers, for certain expressions > isn't complicated, and relies on the distributive property still. > Just like with > 7*(f(x) + 1) = x + 7 > where you can SAY you're in the ring of algebraic integers, and find > something that may look to a naive person like you defied the > distributive property, it's still true that 7 multiplied through, as > the distributive property simply says that if you multiply a group, you > multiply each of the elements within that group, so f(x) and 1 get > multiplied by 7. > But you don't see 7 times x on the right side. > So? It doesn't change the distributive property. > With more complicated expressions you get more complicated behavior, so > that with > 7*C(x) = (f(x) + 7)(g(x) + 1) > where f(0)=g(0) = 0 > you can find C(x), f(x) and g(x) such that you are pushed out of the > ring if you follow logically where the 7 must have multiplied through. Here's my simple refutation of your claim that 7 must have multiplied through: I have two functions, f(x) + 7 and (g(x) + 1)/7. I multiply them together and call their product C(x). That is, C(x) = (f(x) + 7)(g(x) + 1)/7. you've said above, you conclude that, 7 multiplied through f(x). But you are obviously wrong. If my simple example doesn't refute your claim, then why doesn't it? what a mathematical proof actually is, so you think that proofs can be > broken, or created, or they are fragile things that can shake with > arguing. Well, if you actually took any real math classes, or read a book, then you would know what a proof actually is. Right now, it's painfully obvious that you haven't a clue. > But what proofs are, are points of truth that do not require faith. > You don't have to believe in some person or entity. You don't have to > trust or worry about what education level one person has versus > another. > All those are points irrelevant to a proof, so when people question my > not having a degree in mathematics as my degree is in physics, you do > not have to worry about that if you know what mathematics is about, as > what you really need is the proof. Ah, but an education in mathematics implies exposure to various proofs of various levels of difficulty. A person with such an education will have constructed many proofs themselves. They understand what constitutes proof and they won't easily get tripped-up over simple things like definitons, and their application (unlike some people around here). > With the proof social crap is irrelevant. It doesn't matter how many > people argue with me. It doesn't matter what your gut feeling tells > you. > Trace the logical steps from a truth, and the conclusion that follows > MUST BE TRUE. Correct... the logical conclusion is that you are wrong! > There have been these discussions about my non-polynomial factorization >> research where the replies go on and on, when it can be about dumb >> things like square roots having two values (others arguing against, me >> arguing for), and considering the big picture, I think quite a few of >> them may be clipped by talking some about rings and expressions valid >> within them. > Again, either through malice or ignorance, you misrepresent the position > of your detractors! Nobody, as far as I've seen, has argued that the > equation x^2 = 2 doesn't have two solutions. Instead, people have told > you that the NOTATION sqrt(2) represents the positive solution, while > -sqrt(2) represents the negative. > Why do we need such notation you ask? Well, it's a short-hand that > simplifies exposition. Imagine having to write, the positive square > root of 2, everytime you wanted to talk about the square root of 2? Why > is this so hard for you to understand? I think you really know the answer to that question. === Subject: Re: Rings and valid expressions : There seems to be confusion about the distributive property when you... I have only noticed confusion on the part of one poster. : In the ring of algebraic integers let : 7*(f(x) + 1) = x + 7 : ---and you have a contradiction as the expression is NOT valid in that No, this is false. Check the prevous thread for a couple of good old fashioned counterexamples. Justin === Subject: Re: Rings and valid expressions > There have been these discussions about my non-polynomial factorization > research where the replies go on and on, when it can be about dumb > things like square roots having two values (others arguing against, me > arguing for), For God's sake James stop making claims that are patently false. Nobody is disputing that there are two real solutions to x^2 = 2. What if I made up a function called foo, where foo(x) is the non-negative member of the set of reals that satisfy foo(x). Then foo(2) stands for the one and only number that is satisfies x^2 = 2, and is also positive. What on earth is your misunderstanding or objection to this point? and considering the big picture, I think quite a few of > them may be clipped by talking some about rings and expressions valid > within them. > There seems to be confusion about the distributive property when you > have an expression not valid in a particular ring, like consider the > simple example: > In the ring of algebraic integers let > 7*(f(x) + 1) = x + 7 > ---and you have a contradiction as the expression is NOT valid in that > ring, as it requires > f(x) = x/7 I don't know what you mean by the symbol '/'. In a ring, the only operations that are assumed are + and *. There is also a - (minus) operation defined, because the ring under + is an additive group, which means that every element has an additive inverse. So we can add, and we can subtract. But you have to be very careful about trying to divide, because in a ring, not everything has a multiplicative inverse. For example in the ring of integers mod 6, which consists of {0,1,2,3,4,5} with addition and multiplication mod 6, you can not casually divide by 3. For example in Z_6, we have 3 * 2 = 3* 4, yet you cannot divide by 3 to conclude that 2 = 4. Division by 3 is not defined. I believe you are a little confused on this point, since you continually try to divide in rings without justification. > Here you're pushed out of the ring. You attempted to define that phrase pushed out of the ring the other day, without success. You seem to think it means that you have a set that is not closed under addition and multiplication. So a statement equivalent to what I think you mean is that The set of algebraic integers is not a ring. That statement happens to be false; but at least it's a statement using standard math terminology that means the same as what you mean by you're pushed out of the ring. You see, a ring by definition is closed under + and *, so it's impossible BY DEFINITION to be pushed out of a ring. If you could be pushed out of a set, that set would not be a ring. > Notice though that the distributive property still works so you have > 7*f(x) + 7 = x + 7 > and it's just that f(x) = x/7, so the value you get from a function > does not impact the distributive property. Again, I don't know what you mean by the / sign, since you are working in rings, in which division is not necessarily defined for all numbers. And in polynomial rings, it definitely isn't defined unless the coefficients lie in a field. There are some subtle considerations when you study polynomial rings. If you would just pick up an undergrad abstract algebra text and spend a few months with it, it would really interest you. You have so much focus and intensity and obsession with algebra. It would be so cool if you would learn some. You would be fascinated to see how mathematicians actually deal with polynomial rings and various number fields related to them. > So 7 still multiplied through. The phrase multiplied through is also lacking in proper definition. For example in the real numbers, 7 * 5 = 35 so 7 clearly multiplies through both sides. But the real numbers are a field, so you can divide by any nonzero number. You can say pi * 7/pi * 5 = 35 = 35pi/pi so pi divides through. The visible part of a particular expression in a particular notational system for a number is not important > With more complex factorizations you can get more complex behavior so > that when you move to > 7*C(x) = (f(x) + 7)(g(x) + 1) > where f(0) = g(0) and C(x) is a polynomial > you can start in the ring of algebraic integers, have the equations > valid in the ring of algebraic integers, but still get a result not > valid in that ring. What is a result not valid in that ring? You typically write down a sequence of equations, followed by a non-sequitur prose conclusion that make no sense and uses undefined and vague terms. > That's where things get subtle and where it's possible for malicious > people to confuse the issue. It's possible for confused people to confuse the issue. You are confused. You do not understand what is a ring and what is a field; you don't understand what is a polynomial ring; you don't understand the distinction between a polynomial, on the one hand; and the evaluation of a polynomial at a particular value in some ring, on the other. If you would take a few months and learn about these things, you could come here and talk about them. You clearly must spend an hour or two a day, at least, working on this stuff. Why not read through Herstein. It would blow your mind. It's a terrific book. > Now the argument that shows that you're forced out of a particular > ring, like the ring of algebraic integers, That's a contradition in terms. You cannot be forced out of a ring, BY DEFINITION. Because by definition, a ring is closed under + and *. If you can be forced out by algebraic operations, it's not a ring, it's just a set. > where you can SAY you're in the ring of algebraic integers, and find > something that may look to a naive person like you defied the > distributive property, it's still true that 7 multiplied through, as > the distributive property simply says that if you multiply a group, you > multiply each of the elements within that group, so f(x) and 1 get > multiplied by 7. You are prone to run-on sentences. Every time you see a run-on sentence in your own work, you should take the time to unpack it and make every subclause crystal clear. > But you don't see 7 times x on the right side. seeing has nothing to do with it. You don't see a factor of 7 in 35, but it's there. You don't see a factor of pi in 35, but 35 = pi * 35/pi, so pi is a factor of 35 in the field of real numbers. > What makes this situation frustrating is that the mathematics is simple > to the point of triviality, and I think most of you can comprehend that > a function can be outside of a ring, A function can be outside of a ring. Yes indeed. I believe you are glimmering the concept of a ring of polynomials applied to a particular ring. There's a thing called the evaluation homomorphism. That's what that is. It's a ring of polynomials applied to some other ring. > I think part of the problem here is that many of you do not comprehend > what a mathematical proof actually is, so you think that proofs can be > broken, or created, or they are fragile things that can shake with > arguing. James, I've been reading your stuff for several years. It's painfully clear that you do not understand what constitutes a mathematical proof. === Subject: Re: Rings and valid expressions You are 100% correct James. You can not argue with a CORRECT proof. Since you do not have a correct proof, you just continue to make that claim. I can say that I have a proof that there are an infinite number of mersenne primes, but anyone who knows what they are doing can check my proof and tell if it is valid. If not then it is in fact not a proof. Just like your proof is not valid, as many people have pointed out and shown to you many times. === Subject: Re: Get a clue, my eccentricities are normal <4a43dkFrd7a6U1@individual.net> But I found my own short proof of Fermat's Last Theorem Hang on, I thought you'd dropped this claim? > Not anymore: > http://mymath.blogspot.com/2006/03/proof-of-fermats-last-theorem.html > Oh, I'd seen that post, but I interpreted the actual proof that marked > the end of my failed arguments, and the beginning of remarkable > successes as meaning that this proof was to be included in the former > set, not the latter. Both interpretations work, I suppose. I've had it for years now. It was on my old blog for a while. I doubted it for a bit, as I considered various pieces and argued them out. Once I'd figured out some areas that had troubled me before, there was no reason to deny the proof, so I put it on my new math blog. People are, of course, welcome to attack it! Trouble is that's so hard that no one has bothered in years, except me, of course. Oh, people who wonder what all the arguing about square roots and the distributive property and all of that is about can look over the short proof of Fermat's Last Theorem and see where it all comes together. That will also help you understand why posters are so dedicated in arguing with me. I pulled out one little piece of that proof to come up with the paper that got published and yanked under social pressure. The entire proof is probably more complicated for some reason than just about any human being can fully comprehend without years of study, despite its simplicity in terms of the mechanics. Like, to explain some basic ideas from my research I use 7*C(x) = (f(x) + 7)(g(x) + 1) but in the proof, you have p total factors on the right side and not just 2, which is an infinite level of complexity in one way, as yes, the proof handles primes out to infinity, as the ideas are that powerful. So, for most of you, it's so far beyond what you can begin to comprehend that it's, wait, hey, you can still try Try to even understand how that little piece I just mentioned fits in, if you can. James Harris === Subject: Re: Get a clue, my eccentricities are normal [jstevh@msn.com, on his purported short FLT proof] > ... > Oh, people who wonder what all the arguing about square roots and > the distributive property and all of that is about can look over the > short proof of Fermat's Last Theorem and see where it all comes > together. > That will also help you understand why posters are so dedicated in > arguing with me. Sorry, you got that backwards. As you noted earlier: Trouble is that's so hard that no one has bothered in years, except me, of course. That is, most responders have no idea why you go on so about square roots and the distributive property and aliens from the planet Contrary. They argue against that stuff because it appears wrong, or incoherent, all on its own. The fact that your purported FLT proof relies on that stuff is why _you_ argue about it endlessly, not why they do: if you have to admit that the foundation is just wishful thinking, the purported FLT proof collapses even in your mind. Sucks, but so it goes. === Subject: Re: Get a clue, my eccentricities are normal <4a43dkFrd7a6U1@individual.net >> But I found my own short proof of Fermat's Last Theorem Hang on, I thought you'd dropped this claim? Not anymore: http://mymath.blogspot.com/2006/03/proof-of-fermats-last-theorem.html > Oh, I'd seen that post, but I interpreted the actual proof that marked > the end of my failed arguments, and the beginning of remarkable > successes as meaning that this proof was to be included in the former > set, not the latter. Both interpretations work, I suppose. > I've had it for years now. It was on my old blog for a while. > I doubted it for a bit, as I considered various pieces and argued them > out. > Once I'd figured out some areas that had troubled me before, there was > no reason to deny the proof, so I put it on my new math blog. > People are, of course, welcome to attack it! > Trouble is that's so hard that no one has bothered in years, except me, > of course. > Oh, people who wonder what all the arguing about square roots and the > distributive property and all of that is about can look over the short > proof of Fermat's Last Theorem and see where it all comes together. > That will also help you understand why posters are so dedicated in > arguing with me. > I pulled out one little piece of that proof to come up with the paper > that got published and yanked under social pressure. > The entire proof is probably more complicated for some reason than just > about any human being can fully comprehend without years of study, > despite its simplicity in terms of the mechanics. > Like, to explain some basic ideas from my research I use > 7*C(x) = (f(x) + 7)(g(x) + 1) > but in the proof, you have p total factors on the right side and not > just 2, which is an infinite level of complexity in one way, as yes, > the proof handles primes out to infinity, as the ideas are that > powerful. > So, for most of you, it's so far beyond what you can begin to > comprehend that it's, wait, hey, you can still try > Try to even understand how that little piece I just mentioned fits in, > if you can. What do you want, a fee on top of expenses? I'll have to tape the interview and you'll have to sign a release if you want that. > James Harris === Subject: JSH: So yeah, it's about FLT Yup, I admit it. All the fighting and arguing for years has to a large extent been about the short proof of FLT that I have. Most of you probably knew that, as what else could have the power to grip so many people to argue, literally for years. Why else would the sci.math newsgroup erupt with such fury upon hearing about some paper of mine being published? The paper covered some key ideas used in a couple of sentences in the FLT proof. What else could have the energy to crush an entire mathematical journal, even if it was only an electronic one? What else could have the power to drive so many people to put up all those webpages and work so hard to convince you to ignore me? It's kind of neat how it works. You come up with crucial ideas big enough and important enough, then people don't try to steal them. They try to hide them. James Harris === Subject: Re: JSH: So yeah, it's about FLT > Yup, I admit it. All the fighting and arguing for > years has to a large > extent been about the short proof of FLT that I have. > Most of you probably knew that, as what else could > have the power to > grip so many people to argue, literally for years. > Why else would the > sci.math newsgroup erupt with such fury upon hearing > about some paper > of mine being published? > The paper covered some key ideas used in a couple of > sentences in the > FLT proof. > What else could have the energy to crush an entire > mathematical > journal, even if it was only an electronic one? > What else could have the power to drive so many > people to put up all > those webpages and work so hard to convince you to > ignore me? Maybe I'm missing something, but I see many pages of postings that directly challenge your claims. I don't see others trying to convice anyone to IGNORE you. For those who do ignore you, they don't need convincing, all they need to do is read a few of your postings. You destroy your own credibility. > It's kind of neat how it works. You come up with > crucial ideas big > enough and important enough, then people don't try to > steal them. > They try to hide them. > James Harris === Subject: Re: JSH: So yeah, it's about FLT > Yup, I admit it. All the fighting and arguing for years has to a large > extent been about the short proof of FLT that I have. You are half write. Your proof is short. If were a proof of FLT that would really be something but alas, that half's not true. It's not even half true.