mm-2329 === Subject: Re: People are stupid Well, who's going to empty the > waste bins and vacuum the carpets? Eventually, maybe we'll learn to treat garbage men and cleaners with > respect, too... Wise words. How fair are our evaluation functions for people? I doubt little, look at the people who rule entire countries. Those are supposed to be top of the crop, eh? -- Eray === Subject: Re: People are stupid !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> Well, who's going to empty the >> waste bins and vacuum the carpets? >> Eventually, maybe we'll learn to treat garbage men and cleaners with >> respect, too... Wise words. How fair are our evaluation functions for people? I doubt > little, look at the people who rule entire countries. Those are > supposed to be top of the crop, eh? At the top of the crop is ergot. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Two little integrals Hello everybody. I'm stucked with two integrals. I was trying to solve an integral by trigonometrical substitution: 1 Integral of: ------------ (36-x^2)^(4/2) This is the integral of one over the square root of ((36 - x^2)^4) Well... so I did all the trigonometric substitution and I get the following integral: 1 --- Integral of: (secant of theta)^3 dtheta 216 My problem is that I cannot find an easy way to do the integral of (sec x)^3. I've tryed every trigonometric property I know, integration by parts, etc... and I get cyclic results that bring me nowhere. Can someone tell me an easy method to solve this integral? My calculator gets me a very long answer, like 6 additions so, is there any way or are there any trigonometric properties to solve in an easier way this integral? If not, how does the calculator gives so big answers? An the other one that I've tryed for days and I don't get even near to the response (by the way I've tryed this one in 4 calculators and none was able to do it) is: Square root of: [sin(square root of x)] --> This one is not urgent, it came outfor curiosity only, but the other one must be solvable, but, how!? === Subject: Re: Two little integrals 1 > --- Integral of: (secant of theta)^3 dtheta > 216 Denoting I := int (1/cos^3(t)).dt you can integrate by parts using J := D(tan(t)) = 1/cos^2(t) : I = 1/cos(t) . int(J).dt - int (D(1/cos(t)) . int(J).dt).dt = sin(t)/cos^2(t) - int (sin(t)/cos^2(t) . sin(t)/cos(t)).dt = sin(t)/cos^2(t) - int (sin^2(t)/cos^3(t)).dt So that: 2I = sin(t)/cos^2(t) + int(1/cos(t)).dt === Subject: Re: Two little integrals > Hello everybody. I'm stucked with two integrals. > I was trying to solve an integral by trigonometrical substitution: > > 1 > Integral of: ------------ > (36-x^2)^(4/2) > > This is the integral of one over the square root of ((36 - x^2)^4) > > Well... so I did all the trigonometric substitution and I get the > following integral: > > 1 > --- Integral of: (secant of theta)^3 dtheta > 216 > 1/(36-x^2)^(4/2) = 1/(36-x^2)^2 =(1/(x-6)^2 + 1/(x+6)^2)/144 + (1/(x+6) - 1/(x-6))/864 Which can easily be integrated term by term with no sustitutions needed. The method is called partial fractions, q.v. === Subject: Re: Two little integrals I get the following integral: 1 > --- Integral of: (secant of theta)^3 dtheta > 216 Denoting I := int(1/cos^3(t)).dt : 2/cos^2(t) = int(sin(t)/cos^3(t)) + C = sin(t).I - int(cos(t).1/cos^3(t)).dt + C = sin(t).I - tan(t) + C === Subject: Re: Two little integrals >Hello everybody. I'm stucked with two integrals. >I was trying to solve an integral by trigonometrical substitution: 1 >Integral of: ------------ > (36-x^2)^(4/2) This is the integral of one over the square root of ((36 - x^2)^4) > Which is just the integral of 1 / (36 - x^2)^2. Why not rewrite it as: 1 / ((6 + x)^2*(6 - x)^2 ) and use partial fractions? >Well... so I did all the trigonometric substitution and I get the >following integral: 1 >--- Integral of: (secant of theta)^3 dtheta >216 My problem is that I cannot find an easy way to do the integral of (sec >x)^3. I've tryed every trigonometric property I know, integration by >parts, etc... and I get cyclic results that bring me nowhere. > I haven't checked your calculations, but Int(sec^3(x)) is usually done by parts twice which gives the integral in terms of itself, from which you can solve for it. The first time let u = sec(x) and dv = sec^2(x)dx and the second time let u = tan(x) and dv = sec(x)tan(x)dx >An the other one that I've tryed for days and I don't get even near to >the response (by the way I've tryed this one in 4 calculators and none >was able to do it) Well, that settles it then eh? Square root of: [sin(square root of x)] --> This one is not urgent, it >came outfor curiosity only, but the other one must be solvable, but, >how!? > That one doesn't have a elementary antiderivative. --Lynn === Subject: Re: Two little integrals > Hello everybody. I'm stucked with two integrals. > I was trying to solve an integral by trigonometrical substitution: 1 > Integral of: ------------ > (36-x^2)^(4/2) This is the integral of one over the square root of ((36 - x^2)^4) Well... so I did all the trigonometric substitution and I get the > following integral: 1 > --- Integral of: (secant of theta)^3 dtheta > 216 My problem is that I cannot find an easy way to do the integral of (sec > x)^3. I've tryed every trigonometric property I know, integration by > parts, etc... and I get cyclic results that bring me nowhere. Can someone tell me an easy method to solve this integral? My > calculator gets me a very long answer, like 6 additions so, is there > any way or are there any trigonometric properties to solve in an easier > way this integral? If not, how does the calculator gives so big > answers? An the other one that I've tryed for days and I don't get even near to > the response (by the way I've tryed this one in 4 calculators and none > was able to do it) is: Square root of: [sin(square root of x)] --> This one is not urgent, it > came outfor curiosity only, but the other one must be solvable, but, > how!? The integral of (sec(theta)) ^ 3 is obtained by integration by parts, but it comes up so often that it's worth memorizing. That integral is (1/2) (sec(theta))(tan(theta) - (1/2)ln | sec(theta) + tan(theta) | + C. As far as the second one, I don't think you can do it in terms of elementary terms; at least I don't see a way. Dave === Subject: Re: Relative Cardinality > > Actually, it doesn't. The numbers that mathematics can call by name >> (namely computable numbers) are countable. Why should only computable numbers be called by name? > > Because the others have neither a finite name, nor a finite recipe for > their name, nor a finite recipe for the recipe of their name. > > Try specifying a single non-computable number. That one over there! === Subject: Re: Relative Cardinality <85pstqcwtx.fsf@lola.goethe.zz> <85k6jycu1n.fsf@lola.goethe.zz> > Actually, it doesn't. The numbers that mathematics can call by name >> (namely computable numbers) are countable. Why should only computable numbers be called by name? Because the others have neither a finite name, nor a finite recipe for > their name, nor a finite recipe for the recipe of their name. Try specifying a single non-computable number. That one over there! Pick a real number at random. With probability 1, it will be non-computable. --- Christopher Heckman === Subject: Re: Relative Cardinality !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >>> Actually, it doesn't. The numbers that mathematics can call by name >>> (namely computable numbers) are countable. >> Why should only computable numbers be called by name? >> Because the others have neither a finite name, nor a finite recipe for >> their name, nor a finite recipe for the recipe of their name. >> Try specifying a single non-computable number. >> That one over there! Pick a real number at random. No, you do it. > With probability 1, it will be non-computable. Too bad that there are so few perfect random pickers around. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Relative Cardinality > > All numbers are ideas, it is just that mathematics calls more ideas > numbers than WM does. > > Actually, it doesn't. The numbers that mathematics can call by name > (namely computable numbers) are countable. The unnamed numbers make > up the dark matter in the mathematicians' universe: when they gang up > together, they have weight and measure. Let me rephrase it: Mathematicians call a lot of things numbers that WM doesn't! === Subject: Re: Relative Cardinality Let me digress for a moment: Lemma. If we can't count up to N, then we can't count up to N-1. Proof: This statement is equivalent to: If we can count up to N-1, then > we can count up to N. This is trivially true. No. Use, as a simple model, three chips. How far can you count and > store the result by means of these chips? 1 , 2, end. In order to store > 3, you must forget 1 and 2. Now turn the whole universe into a big > computer. The principle remains the same, the numbers get larger > though. That's right, you're not allowing any numbers with more than 10^100 digits, because you can't write them down in the real world. How pathetic; you haven't even scratched the surface in what numbers are allowed outside this universe. (Comment added later: The hydrogen atom has an infinite (countable) number of stable energy states. Any amount of energy can be created, as long as it's paid back a short time later, so you can store one of a countable number of natural numbers in one atom.) > Er ... Euclid proved that there are an _infinite_ number of primes. Euclid did not talk of infinity. There are more than any given number, > he said. That's what infinity means. No. By this definition the values of the natural numbers would become > infinite. Not this @#$%@ again. Every natural number is finite, since it only has a finite number of digits. But there are an infinite _number_ of them. That's like saying that {4, 6, 8} is a set of prime numbers, because the size of that set (3) _is_ a prime number. > But they don't by definition of natural number. Actually, they don't by _your_ definition of a natural number (it has to be less than 10^10^100). The definition of natural number allows more digits than yours does. > Actually, he didn't say, There are more > than any given number, because he didn't know English. A Greek word to use was available: apeiron. He did not use it > deliberately. You missed the point of what I said. You were arguing over the exact words he said, but you weren't even close because the words were in English. I figured everyone would know there was a 8-) at the end of that sentence. > What do you find wrong with that proof? Do you really think it necessary to demonstrate such things here? That's not what I asked. I asked what YOU found wrong with that proof. The proof is correct, in principle. But it is impossible to form > numbers which require more than 10^100 different digits which cannot be > derived from a simple rule. What do you mean by a simple rule ? > If that's what you've really meant, then you should have said so, > instead of wondering about whether numbers of the form 111...111 give > you infinitely many primes, since you won't be able to write them down > anyway. Not write them down, but I would know each digit of that number. You need to record the number of digits in the number 111...111, though, because 11 and 111 differ only in the number of digits. That means you can't deal with numbers with that simple rule which are larger than 10^10^10^100. But this means that, if you came upon the number N (> 1) written among to be N or 111...111 (with N digits)? well-defined, because you can get more than one number out of a single representation. > Therefore, for any other number I could find out which of them is > larger. This is the criterion of existence for a measure of > largeness, i.e., a number. > Mathematics was designed to work outside of the physical universe. Nothing does work outside of the physical universe. How can you be sure of this? All the physics that has been developed has been for use in this universe, based on observations and trying to come up with a system that fits those observations. There's no problem with my statement, because I included the phrase was designed to. > Therefore > there cannot exist a set with more than 10^100 elements and we cannot > count up to 10^10^100. Well, this will make life much more difficult for you as a physicist, > since physics uses lots of statistical mathematics (which allows sets > of arbitrarily large size, even sets of size 10^10^10^10^10^10^100) to > make certain calculations easier. Those numbers do certainly exist. So you've stated that there are numbers that exist that you can't write down. But this has been your point throughout the discussion. So you've contradicted yourself here. > But how would you satisfy the > set-theoretic requirement that each element must be distinguished by at > in the whole universe? I don't need to follow the second requirement with my mathematics, which is the point of the whole thread. _You_ need to tell me this, because you've insisted on it. --- Christopher Heckman === Subject: Re: Relative Cardinality <85d5q0dwt0.fsf@lola.goethe.zz> It cannot be distinguished from all other numbers. Of course it can. All numbers which are not equal to > sqrt(2) are either larger than it or smaller than it, > are distinguished from it, and are separated from it > by a nonzero amount. Like 0.999... and 1.000... are separated by a non-zero amount? No, those numbers are not separated by a nonzero amount. The separation between those numbers is 0. > Or is it a larger amount? Is what? The separation between sqrt(2) and x which is different from sqrt(2)? If x is different from sqrt(2), then |x-sqrt(2)| is larger than 0. > How much more is it? Please specify. How can I do that unless I know what x is? All I can tell you is that if x is different from sqrt(2), then the difference is larger than 1/10^n, where n is a finite natural number. No matter what x you choose, it will have the property that there exists some n such that |x - sqrt(2)| > 1/10^n. Furthermore, if x is different from sqrt(2), then there exists another number y which is different from both x and sqrt(2), and which lies between the two of them. (In fact there are infinitely many such y). - Randy === Subject: Re: Relative Cardinality <85d5q0dwt0.fsf@lola.goethe.zz> It cannot be distinguished from all other numbers. Of course it can. All numbers which are not equal to > sqrt(2) are either larger than it or smaller than it, > are distinguished from it, and are separated from it > by a nonzero amount. Like 0.999... and 1.000... are separated by a non-zero amount? Or is it > a larger amount? How much more is it? Please specify. Oh @#@#%, not this again! In summary, for any n (which is the number of decimal places in 0.999...999 and 1.000...000): 0.999...999 < 0.999... <= 1.000... = 1.000...000, so | 1.000... - 0.999... | <= |1.000...000 - 0.999...999 | = 10^(-n). In the real number system, 0.999... and 1.000... are the same number, because there is no positive real number less than 10^(-n) for all n. In other extended real number systems, they may be different, because there may be positive numbers less than 10^(-n) for all n (such as 1/omega in the surreal numbers). But you shouldn't be even asking the question, because you can't write down 0.999... in the physical world, because it has an infinite number of digits (which is at least 10^10^100 as well). --- Christopher Heckman === Subject: Re: Relative Cardinality <85pstqcwtx.fsf@lola.goethe.zz> <85k6jycu1n.fsf@lola.goethe.zz> <85pstqbacg.fsf@lola.goethe.zz Take the number where the nth digit in the binary representation is > defined by the result of running the halting function on the nth > turing machine. That specifies a precise single non-computable > number. So I think you mean the set of finitely describable or > definable numbers (in some formal language), which are still > countable of course. Pretty much, yes. But doesn't this amount to the same after G.9adelian > encoding? Not quite sure what you mean there, computable real numbers are different beasts then propositions encoded in natural number arithmetic. But no, computable has a very specific meaning and you can define some non-computable numbers precisely (that is exactly the way you prove that not all real numbers are computable). I'm not an expert on these matters though, and these things tend to be very VERY subtle. Jiri === Subject: Re: Simple problem > Here's a variation of a familiar problem: > > I man walks due North a distance 'x'. He then walks due East > exactly one mile. He then walks South the distance 'x'. He > has now returned to this starting point. > > What are the minimum and maximum possible values for 'x'? The minimum, which exists only as a limit, is zero. The maximum, which exists only as a limit and assumes the ability to walk on water, is the distance between the poles. === Subject: Re: Simple problem !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Here's a variation of a familiar problem: >> >> I man walks due North a distance 'x'. He then walks due East >> exactly one mile. He then walks South the distance 'x'. He >> has now returned to this starting point. >> >> What are the minimum and maximum possible values for 'x'? The minimum, which exists only as a limit, is zero. Why only as a limit? Start on a circle of 1 mile circumference (or half a mile, or a third of a mile) near one of the poles. Walking one mile East is perfectly possible and lands you at the same position. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Simple problem Here's a variation of a familiar problem: I man walks due North a distance 'x'. He then walks due East > exactly one mile. He then walks South the distance 'x'. He > has now returned to this starting point. What are the minimum and maximum possible values for 'x'? [...] > -- > http://www.crbond.com Minimum: Starting point (1/(2*pi)) miles south of north pole -> x=0 > Maximum: Start at south pole -> x=pi*R_earth-(1/(2*pi)) mile The eastward part of his walk is a small circle around the north pole or > around the south pole. You don't even need a small circle, because if you start at the South Pole and walk north x miles and east 1 mile, you are still x miles from the South Pole -- you haven't changed your latitude. So the minimum and maximum answers are 0 and pi*R. Now, if the OP had given THE STARTING POSITION, then the sort of analysis above would be necessary; you would have to go around a pole an integral number of times. --- Christopher Heckman === Subject: A Reverse Integer Triangle, Consecutive Integer Sums, And Potential Primes FWIW (or not, as the case may be): Please note: for space considerations, only the text is included here ; the tables are located at http://www.maplenet.net/~reriker/triangleprimes.html Odd prime numbers can only be formed by the sum of no more than two consecutive integers. Stated another way, the sum of three or more consecutive integers are not prime (they are either divisible by n if n is an odd number of consecutive integers or by n/2 if n is an even number of consecutive integers). The sum of consecutive integers can be shown by setting up a reverse integer triangle. When set up this way, the sum of n consecutive integers is the value in column 1 for each value of n in the column headed by n (e.g., the sum of the first 12 consecutive integers is 78). Other sums of consecutive integers similarly show up in columns (e.g., the columns headed by -1,-3,-6,-10 ; please note that the absolute value of the column headings correspond to the value of the sum of the first n consecutive integers). A lot of interesting patterns appear to emerge. Which leads to a preliminary conjecture concerning the column headed by -2: If an odd number, m (m > 1), has no factors within m rows of row 1 (by row 1, I mean the row in the column headed n), it has no factors in the column (column -2) If an odd number, m (m > 1), has factors within m rows of row 1, it will have multiple factors in column -2 If m (m > 1) is contained within the column -2 and has no previous factors (other than 1), m is prime Other than an intellectual curiosity, perhaps this is of no value. However, my suspicion is that it might be possible to search for primes in column -2 faster because the brute force calculation may be scaled down (e.g., if it holds that if m does not have any factors within m rows of row 1, it is not necessary to divide any of the numbers beyond row [1 + m] by m). Although I have not investigated it much at this point, I suspect this may also be a possibility for other columns (e.g., column -4, column -5, etc.). guished by at > in the whole universe? I don't need to follow the second requirement with my mathematics, which is the point of the whole thread. _You_ need to tell me this, because you've insisted on it. --- Christopher Heckman === Subject: Re: Relative Cardinality <85d5q0dwt0.fsf@lola.goethe.zz> It cannot be distinguished from all other numbers. Of course it can. All numbers which are not equal to > sqrt(2) are either larger than it or smaller than it, > are distinguished from it, and are separated from it > by a nonzero amount. Like 0.999... and 1.000... are separated by a non-zero amount? No, those numbers are not separated by a nonzero amount. The separation between those numbers is 0. > Or is it a larger amount? Is what? The separation between sqrt(2) and x which is different from sqrt(2)? If x is different from sqrt(2), then |x-sqrt(2)| is larger than 0. > How much more is it? Please specify. How can I do that unless I know what x is? All I can tell you is that if x is different from sqrt(2), then the difference is larger than 1/10^n, where n is a finite natural number. No matter what x you choose, it will have the property that there exists some n such that |x - sqrt(2)| > 1/10^n. Furthermore, if x is different from sqrt(2), then there exists another number y which is different from both x and sqrt(2), and which lies between the two of them. (In fact there are infinitely many such y). - Randy === Subject: Re: Relative Cardinality <85d5q0dwt0.fsf@lola.goethe.zz> It cannot be distinguished from all other numbers. Of course it can. All numbers which are not equal to > sqrt(2) are either larger than it or smaller than it, > are distinguished from it, and are separated from it > by a nonzero amount. Like 0.999... and 1.000... are separated by a non-zero amount? Or is it > a larger amount? How much more is it? Please specify. Oh @#@#%, not this again! In summary, for any n (which is the number of decimal places in 0.999...999 and 1.000...000): 0.999...999 < 0.999... <= 1.000... = 1.000...000, so | 1.000... - 0.999... | <= |1.000...000 - 0.999...999 | = 10^(-n). In the real number system, 0.999... and 1.000... are the same number, because there is no positive real number less than 10^(-n) for all n. In other extended real number systems, they may be different, because there may be positive numbers less than 10^(-n) for all n (such as 1/omega in the surreal numbers). But you shouldn't be even asking the question, because you can't write down 0.999... in the physical world, because it has an infinite number of digits (which is at least 10^10^100 as well). --- Christopher Heckman === Subject: Re: Relative Cardinality <85pstqcwtx.fsf@lola.goethe.zz> <85k6jycu1n.fsf@lola.goethe.zz> <85pstqbacg.fsf@lola.goethe.zz Take the number where the nth digit in the binary representation is > defined by the result of running the halting function on the nth > turing machine. That specifies a precise single non-computable > number. So I think you mean the set of finitely describable or > definable numbers (in some formal language), which are still > countable of course. Pretty much, yes. But doesn't this amount to the same after G.9adelian > encoding? Not quite sure what you mean there, computable real numbers are different beasts then propositions encoded in natural number arithmetic. But no, computable has a very specific meaning and you can define some non-computable numbers precisely (that is exactly the way you prove that not all real numbers are computable). I'm not an expert on these matters though, and these things tend to be very VERY subtle. Jiri === Subject: Re: Simple problem > Here's a variation of a familiar problem: > > I man walks due North a distance 'x'. He then walks due East > exactly one mile. He then walks South the distance 'x'. He > has now returned to this starting point. > > What are the minimum and maximum possible values for 'x'? The minimum, which exists only as a limit, is zero. The maximum, which exists only as a limit and assumes the ability to walk on water, is the distance between the poles. === Subject: Re: Simple problem !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(5eZ41to5f%E@'ELIi $t^ VcLWP@J5p^rst0+('>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > Here's a variation of a familiar problem: >> >> I man walks due North a distance 'x'. He then walks due East >> exactly one mile. He then walks South the distance 'x'. He >> has now returned to this starting point. >> >> What are the minimum and maximum possible values for 'x'? The minimum, which exists only as a limit, is zero. Why only as a limit? Start on a circle of 1 mile circumference (or half a mile, or a third of a mile) near one of the poles. Walking one mile East is perfectly possible and lands you at the same position. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: Simple problem Here's a variation of a familiar problem: I man walks due North a distance 'x'. He then walks due East > exactly one mile. He then walks South the distance 'x'. He > has now returned to this starting point. What are the minimum and maximum possible values for 'x'? [...] > -- > http://www.crbond.com Minimum: Starting point (1/(2*pi)) miles south of north pole -> x=0 > Maximum: Start at south pole -> x=pi*R_earth-(1/(2*pi)) mile The eastward part of his walk is a small circle around the north pole or > around the south pole. You don't even need a small circle, because if you start at the South Pole and walk north x miles and east 1 mile, you are still x miles from the South Pole -- you haven't changed your latitude. So the minimum and maximum answers are 0 and pi*R. Now, if the OP had given THE STARTING POSITION, then the sort of analysis above would be necessary; you would have to go around a pole an integral number of times. --- Christopher Heckman === Subject: A Reverse Integer Triangle, Consecutive Integer Sums, And Potential Primes FWIW (or not, as the case may be): Please note: for space considerations, only the text is included here ; the tables are located at http://www.maplenet.net/~reriker/triangleprimes.html Odd prime numbers can only be formed by the sum of no more than two consecutive integers. Stated another way, the sum of three or more consecutive integers are not prime (they are either divisible by n if n is an odd number of consecutive integers or by n/2 if n is an even number of consecutive integers). The sum of consecutive integers can be shown by setting up a reverse integer triangle. When set up this way, the sum of n consecutive integers is the value in column 1 for each value of n in the column headed by n (e.g., the sum of the first 12 consecutive integers is 78). Other sums of consecutive integers similarly show up in columns (e.g., the columns headed by -1,-3,-6,-10 ; please note that the absolute value of the column headings correspond to the value of the sum of the first n consecutive integers). A lot of interesting patterns appear to emerge. Which leads to a preliminary conjecture concerning the column headed by -2: If an odd number, m (m > 1), has no factors within m rows of row 1 (by row 1, I mean the row in the column headed n), it has no factors in the column (column -2) If an odd number, m (m > 1), has factors within m rows of row 1, it will have multiple factors in column -2 If m (m > 1) is contained within the column -2 and has no previous factors (other than 1), m is prime Other than an intellectual curiosity, perhaps this is of no value. However, my suspicion is that it might be possible to search for primes in column -2 faster because the brute force calculation may be scaled down (e.g., if it holds that if m does not have any factors within m rows of row 1, it is not necessary to divide any of the numbers beyond row [1 + m] by m). Although I have not investigated it much at this point, I suspect this may also be a possibility for other columns (e.g., column -4, column -5, etc.).