ruveyn1

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Posts posted by ruveyn1


Nope.
I had a little precalc 30 years ago.
Do two things.
1 Learn what a proof is
2. Write you proof out carefully and have a professional mathematician vet your proof for errors.
If you can prove FLT by elementary means and you are under 45 years of age, I guarantee you will get a Fields Medal.

Marc, I just confirmed that Ruveyn is in fact Bob Kolker, aka BaalChatzaf :
http://www.objectivistliving.com/forums/index.php?showtopic=12579#entry172290
He has called Leonard Peikoff a "mathematical and scientific ignoramus" http://www.objectivistliving.com/forums/index.php?showtopic=5924&page=3
He has been trolling Oist sites for a long time and I wouldn't expect it to change any time soon...
Define Troll. I observe the rules of the forum.
And during my little tete a tete with L.P on the David Brudnoy program (beamed out at 50 kw clear channel) I demonstrated that at the time he did not know beans about mathematical logic which was the topic of discussion.
Who knows? Maybe he learned something since. That was 30 years ago.
ruveyn1

Alright. Now lets see if we can attach this thing.
Imagine walking into a building at the door. The building is X by X. Walk across the diagonal to the corner opposite the door. Climb the stairwell crossing back across the diagonal to the corner over the door. This is (3X^{2})^{^1/2}
The next solution where I corrected my original formula, is for a 2 story, 3 story, 4 story etc. The stairwell starts in the same corner, only it climbs to the top corner over the door of the destination floor .
(X^{2}+X^{2}+((exponent2)X)^{2})^{1/2}
Do you have any idea of what constitutes a rigorous proof?

As to Collantz Conjecture that should work for any odd number.
T(n) = n/2 if n is even
T(n) = (3*n + 1)/2 if n is odd.
T(n) = (5*n + 1)/2 if n is odd.
T(n) = (7*n + 1)/2 if n is odd.
T(n) = (9*n + 1)/2 if n is odd.
...
Prove that and you will become an Immortal.

I recall AR saying through Rearden, something along the lines of, Jim and his friends don't matter. I think that applies well to this topic.
As to Obama 'living for power'  what? In AS, Jim knew what he was doing and the code he was living by was wrong, he just didn't care to think about it. I don't think that is the case with Obama. He seems like a sincere guy who truly believes that his actions are moral. If true, that doesn't excuse him from all the crap he's done but it doesn't make him an evil, powerhungry monster, either. It makes him Andrei, from WTL one of my favorite characters.
Nice guys do not lie under oath or take an oath in vain. Especially as serious an oath as Presidents must affirm

Do the math.
No. YOU do the math. You claimed the proof, now show it step by step so we can check to see if it is right.

He won't. Obama has read the history books too. Look what happened to Julius Caesar when it even appeared he would crown himself king for life. Brutus who was like a son to him, turned on him. Et tu Brute?
Tyrant Obama would become an instant target if he tried to rule permanently.
ruveyn
In addition to which: If Obama attempts to stay in power, the Army will step in and put an end to that pretending. There will be a coup or putsch from the military who will not obey orders from someone not constitutionally in power. The U.S. has been fortunate to avoid military takeovers, but a bold attempt at tyranny will be put down in a matter of days if not hours. The folks in the Army take their Oath seriously

(x^{2}+x^{2}+x^{x2})^{1/2 }should show that x^{n}+y^{n}=z^{n} has no nonzero solutions x,y,z for ALL n>2. I do not have the mathematical forte to convert it at this time. It is true, for the same reason that the hypotenuse of a square is incommensurate to its sides.
I have a bookmark for consideration of Collatz conjecture. If fame were my goal, . . . who knows.
By the way, 5>(3*n+1/2)=8, not 4, altsohough 8, 4, 2, 1 does finish the hypothesis.
sorry about the error. A trivial corollary: once a collatz sequence hits a power of two, the game is over.
You say you have a proof : "(x^{2}+x^{2}+x^{x2})^{1/2 }should show that x^{n}+y^{n}=z^{n} has no nonzero solutions x,y,z for ALL n>2. I do not have the mathematical forte to convert it at this time. It is true, for the same reason that the hypotenuse of a square is incommensurate to its sides."
Claims are cheap. Valid proofs are dear: Show the proof or be still.

But if he were to declare himself the dictator one morning, I have no doubt that he'll do so very "charismatically" and most Americans will simply adore him for it.
He won't. Obama has read the history books too. Look what happened to Julius Caesar when it even appeared he would crown himself king for life. Brutus who was like a son to him, turned on him. Et tu Brute?
Tyrant Obama would become an instant target if he tried to rule permanently.
ruveyn

It doesn’t matter whether there is 1 cube, or a gazillion cubes, the answer remains that a diagonal of a cube contains a factor of the square root of 3, or an integer was not used to derive the answer.
The real task was showing that x^n + y^n = z^n did not have nonzero solutions x, y, z for ALL n > 2.
That stymied the world's best mathematicians for over 300 years.
Here is a problem which if you can solve it will make you as famous as Dr. Wiles.
It is the proof of the Collatz conjecture.
I will define a function on the integers as follows:
T(n) = n/2 if n is even
T(n) = (3*n + 1)/2 if n is odd.
Here is the problem: show that starting with any integer N if you keep applying the function T you will eventually get to 1.
Example.. N = 7
7, 11, 17, 26, 13, 20, 10, 5, 4, 2, 1
That problem has been open since around 1950. Paul Erdos one of the great great mathematicians was of the opinion that mankind was not ready to solve that problem.
It has already been show for N < 10^20 this is true, but a proof for ALL N remains to be found.
All attempts to prove that the general proposition is undecidable have also failed.
ruveyn1

There are no values without life; "value" does not exist in the absence of life.
Well yes. Isn't that rather obvious. Only live sentient beings can do valuing. And where do I go from this?
What useful conclusion does this logically imply?

How does Rand Paul mange to take an excretion break while philibustering?

Just about any sane and functional human being alive knows what a contradiction is and rejects the possibility of contradictions actually being fact.
There must be something "hard wired" genetically into members of our species which account to that.
O.K. It is not instinct (Rand forbid!), so what is it?

I have not read the proof cited above, but I consider it possible that Fermat might have had, or someone else might discover, a much more elegant and insightful proof that is much shorter than 109 pages. I'm sure it wouldn't be the first time that a long proof could be replaced by a much shorter one, but I don't have any examples at hand to cite.
I really doubt it. Some of the greatest mathematicians of all time have had a go at it, and Wiles was the one who finally did it after 300+ years. I will believe there is a short elementary proof when I see it with my own eyes.l

What Fermat described as a "truly marvelous proof" that would not fit in the margin where he made his notation, has not been discovered.
It probably does not exist.

Zeus did not represent an idea like Chavez does.
Sure he did. He represented the power of lightning and he ruled other other gods (more or less). At one time, Greeks prayed to Zeus, made offerings and such like.

Only if the movement is not based on an idea but only on a man. Personality cults die with the leader.
Belief in Zeus seems to have died out. Why?

Good news. If only there were a hell!
Seen on FB: "In his new state as fertilizer, ... Hugo Chavez finally accomplished his promise of stimulating Venezuelan agriculture."
It does not matter. Chavez is gone. Wishing pain on the dead is an exercise in futility. The dead can no long feel. The dead, as persons, no longer exist.
ruveyn1

Sorry to be dark but do you think I'm paranoid to worry that Obama might not step down at the end of his term? If the economy comes unraveled or the government goes bellyup and the nation finds itself in the midst of a serious crisis when the next election cycle rolls around, I can just see him saying something like "we can't afford partisan bickering at a time like this" or "we can't afford to let the Tea Party terrorists bankrupt the economy by cutting 1% of the Federal budget" and pushing to suspend the next election, either by executive order or by an act of a Democratcontrolled congress. I say this because I think the man lives for power and I have a very hard time seeing him relinquishing it.
Obama has no legal choice but to step down at the end of his current term. Even the batsh*t crazy democrats in the house will not tolerate a coup d'etat and the army surely will not support Obama if he tried something blatantly unconstitutional.
Congress has no legal basis for preventing the election of the next president.
There is an historical precedent. With the country split by a Civil War Lincoln did not declare an emergency. He stood for an election which he could have lost. In fact, he expected to lose.
ruveyn1

Sitting down with a peice of paper after finishing the book, I soon found it filled with numbers. Let me share a few with you.
Let's start with the squares.
1^{2}=1, 2^{2}=4, 3^{2}=9, 4^{2}=16, 5^{2}=25, 6^{2}=36, 7^{2}=49, 8^{2}=64, 9^{2}=81, 10^{2}=100, 11^{2}=121, 12^{2}=144, 13^{2}=169, 14^{2}=196, 15^{2}=225, 16^{2}=256, 17^{2}=289, 18^{2}=324, 19^{2}=361, 20^{2}=400
Moving onto the cubes we find,
1^{3}=1, 2^{3}=8, 3^{3}=27, 4^{3}=64, 5^{3}=125, 6^{3}=216, 7^{3}=343, 8^{3}=512, 9^{3}=729, 10^{3}=1000, 11^{3}=1331, 12^{3}=1728, 13^{3}=2197, 14^{3}=2744, 15^{3}=3375, 16^{3}=4096, 17^{3}=4913, 18^{3}=5832, 19^{3}=6859, 20^{3}=8000
Raising to the power of 4 yields,
1^{4}=1, 2^{4}=16, 3^{4}=81, 4^{4}=256, 5^{4}=625, 6^{4}=1296, 7^{4}=2401, 8^{4}=4096, 9^{4}=6561, 10^{4}=10000, 11^{4}=14641, 12^{4}=20736, 13^{4}=28561, 14^{4}=38416, 15^{4}=50625, 16^{4}=65536, 17^{4}=83521, 18^{4}=104976, 19^{4}=130321, 20^{4}=160000
Looking to the power of 5^{ths},
1^{5}=1, 2^{5}=32, 3^{5}=243, 4^{5}=1024, 5^{5}=3125, 6^{5}=7776, 7^{5}=16807, 8^{5}=32768, 9^{5}=59049, 10^{5}=100000, 11^{5}=161051, 12^{5}=248832, 13^{5}=371293, 14^{5}=537824, 15^{5}=759375, 16^{5}=1048576, 17^{5}=1419857, 18^{5}=1889568, 19^{5}=2476099, 20^{5}=3200000
And lastly at the power of 6^{ths}.
1^{6}=1, 2^{6}=64, 3^{6}=729, 4^{6}=4096, 5^{6}=15625, 6^{6}=46656, 7^{6}=117649, 8^{6}=262144, 9^{6}=531441, 10^{6}=1000000, 11^{6}=1771561, 12^{6}=2985984, 13^{6}=4826809, 14^{6}=7529536, 15^{6}=11390625, 16^{6}=16777216, 17^{6}=24137569, 18^{6}=34012224, 19^{6}=47045881, 20^{6}=64000000
This does give quite a array of numbers.
==================================================================================================================================
Let's take this a step further.
Starting with the squared differences,
41=3, 94=5, 169=7, 2516=9, 3625=11, 4936=13, 6449=15, 8164=17, 10081=19, 121100=21, 144121=23, 169144=25, 196169=27, 225196=29, 256225=31, 289256=33, 324289=35, 361324=37, 400361=39
The cubed differences are as follows,
81=7, 278=19, 6427=37, 12564=61, 216125=91, 343216=127, 512343=169, 729512=217, 1000729=271, 13311000=331, 17281331=397, 21971728=469, 27442197=547, 33752744=631, 40963375=721, 49134096=817, 58324913=919, 68595832=1027, 80006859=1141
The differences raised to the 4^{th} power are,
161=15, 8116=65, 25681=175, 625256=369, 1296625=671, 24011296=1105, 40962401=1695, 65614096=2465, 100006561=3439, 1464110000=4641, 2073614641=6095, 2856120736=7825, 3841628561=9855, 5062538416=12209, 6553650625=14911, 8352165536=17985, 10497683521=21455, 130321104976=25345, 160000130321=29679
The differences raised to the 5^{th} power are,
321=31, 24332=211, 1024243=781, 31251024=2101, 77763125=4651, 168077776=9031, 3276816807=15961, 5904932768=26281, 10000059049=40951, 161051100000=61051, 248832161051=87781, 371293248832=122461, 537824371293=166531, 759375537824=221551, 1048576759375=289201, 14198571048576=371281, 18895681419857=469711, 24760991889568=586531, 32000002476099=723901
And lastly, the differences raised to the 6^{th} power are,
641=63, 72964=665, 4096729=3367, 156254096=11529, 4665615625=31031, 11764946656=70993, 262144117649=144495, 531441262144=269297, 1000000531441=468559, 17715611000000=771561, 29859841771561=1214423, 48268092985984=1840825, 75295364826809=2702727, 113906257529536=3861089, 1677721611390625=5386591, 2413756916777216=7360353, 3401222424137569=9874655, 4704588134012224=13033657, 6400000047045881=16954119
=========================================================================================================================================
Summarizing:
Squared:
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39
Cubed:
1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, 469, 547, 631, 721, 817, 919, 1027, 1141
Quadratic:
1, 15, 65, 175, 369, 671, 1105, 1695, 2465, 3439, 4641, 6095, 7825, 9855, 12209, 14911, 17985, 21455, 25345, 29679
To the 5^{th},
1, 31, 211, 781, 2101, 4651, 9031, 15961, 26281, 40951, 61051, 87781, 122461, 166531, 221551, 289201, 371281, 469711, 586531, 723901
To the 6^{th},
1, 63, 665, 3367, 11529, 31031, 70993, 144495, 269297, 468559, 771561, 1214423, 1840825, 2702727, 3861089, 5386591, 7360353, 9874655, 13033657, 16954119
=========================================================================================================================================
Reviewing the squared sequence, the assigned difference can be represented as:2*X+1.
2*0+1=1, 2*1+1=3 2*2+1=5, 2*3+1=7, 2*4+`=9, 2=5+1=11, 2*6+1=13, 2*7+1=15, 2*8+1=17, 2*9+1=19, 2*10+1=21, etc.
What is not as obvious is for the cubed sequence: X^{2}+(X+1)(2*X+1)
0^{2}+(0+1)(2*0+1)=1, 1^{2}+(1+1)(2*1+1)=7, 2^{2}+(1+1)(2*1+1)=7, 2^{2}+(2+1)(2*2+1)=19, 3^{2}+(3+1)(2*3+1)=37, 4^{2}+(4+1)(2*4+1)=61, etc.
Stepping to the next sequence, what becomes clearer is X^{3}+((X+1)(X^{2}+(X+1)(2*X+1)). This gives us:
0^{3}+(0+1)(0^{2}+(0+1)(2*0+1))=1, 1^{3}+(1+1)(1^{2}+(1+1)(2*1+1))=15, 2^{3}+(2+1)(2^{2}+(2+1)(2*2+1))=65, 3^{3}+(3+1)(3^{2}+(3+1)(2*3+1))=175, etc.
Extrapolating to the 5^{th} power: X^{4}+(X+1)((X^{3}+(X+1)((X^{2}+(X+1)(2*X+1))) yields:
0^{4}+(0+1)(0^{3}+(0+1)((0^{2}+(0+1)(2*0+1)))=1, 1^{4}+(1+1)(1^{3}+(1+1)((1^{2}+(x1+1)(2*1+1))=31, 2^{4}+(2+1)(2^{3}+(2+1)((2^{2}+(2+1)(2*2+1)))=211, 3^{4}+(3+1)(3^{3}+(3+1)((3^{2}+(3+1)(2*3+1)))=671, etc.
And for this example, taking it finally to the 6^{th} power: X^{5}+(X+1)((X^{4}+(X+1)((X^{3}+(X+1)((X^{2}+(X+1)(2*X+1))))
0^{5}+(0+1)(0^{4}+(0+1)(0^{3}+(0+1)(0^{2}+(0+1)(2*0+1))))=1, 1^{5}+(1+1)(1^{4}+(1+1)(1^{3}+(1+1)(1^{2}+(1+1)(2*1+1))))=63, 2^{5}+(2+1)(2^{4}+(2+1)(2^{3}+(2+1)(X^{2}+(2+1)(2*2+1))))=665, 3^{5}+(3+1)(3^{4}+(3+1)(3^{3}+(3+1)(3^{2}+(3+1)(2*3+1))))=3367, etc.
Could Fermat have induced from a pattern such as this, that for a power greater than 2, A^{n}+B^{n}=C^{n} could not have been resolved into an integral relationship?
fixed 761 from 671 on the 5th power 3^{4,}, & parenthesis fixed on various others.
It does not follow. Just because something fails for the first gazillion does not mean it will not succeed for gazillion plus 1.
You cannot, in general disprove non existence of something just by exhibiting a finite number of failures.
Mathematicians very rarely work by empirical induction (not to be confused with arithmetic induction which is something completely different).

In order to go to the Moon which is a moral choice you need to establish few scientific factsthat is, identity of objects and causeeffect connections between them which determine their interaction. The fire of the rocket engine is determined event. The will to fire itnot. But will has nothing to do with science.
Making the steps to go to X is a choice. What the steps are is a matter of science and technology. So there is the problem of determining what physical arrangements are necessary for a trip to X. Then there is the question of whether the trip is worth our while or not. The latter question is not a scientific question at all.

The facts:
 man must act in order to live
 man has free will
 therefore man must choose to act in order to live
 man can choose to act for or against his life
 in order to live man must choose to act for his life
The last is an ethical principle. This is empirical evidence that and ethical principle can be derived from the laws or facts of nature.
uestion of how could you know Ayn Rand was wrong when you've never read her? What is your purpose on this site?
Actually it isn't. It is a tautology. If a person takes the steps necessary to maintain his biological functions he may last a bit longer. If he fails to do so he won't last very long. That is simply a consequence of the fact that our biological engine exists far from thermodynamic equilibrium. In simple terms we have to eat, drink and maintain our body temperature in order to live. If we don't we will die. And in the long run no matter what we do, we die. At best we can delay our death a bit.
I simply fail to see how that has any ethical import. What does it have to do with Right and Wrong?

Death can stop any movement.

Given what it took to finally prove FLT is is fairly clear that Fermat did NOT have a proof that would stand up to modern standards of rigor.
If you want to see how Wiles did it look at
http://www.cs.berkeley.edu/~anindya/fermat.pdf
It is only 109 pages long. Fermat claimed his proof fit on the margin of a text book.
ruveyn1
Worries about the future  Obama's actions
in Domestic United States Politics
Posted
He and Jimmy Carter will have their portraits hung top front in the Hall of Disdain.