Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.



Final Detailed Constructive proof of No Odd Perfect Number beyond 1 #1449 Correcting Math
Posted:
Jan 20, 2014 1:36 AM


Final Detailed Constructive proof of No Odd Perfect Number beyond 1 #1449 Correcting Math
Alright, I am sorry for the hundreds or more posts of "proof of No Odd Perfect Number" that I posted on the sci.math newsgroup. Sorry because all of them were tentative and not actually a proof. Because I now have the proof, as the below shows. So that all those hundreds of tentative proofs finally lead to the true one. So I guess, then, No Odd Perfect was the hardest of these 6 proofs: Goldbach, Bertrand's postulate, APpostulate, Fermat's Last Theorem FLT, Beal's conjecture, No Odd Perfect, for it took the longest to resolve. Maybe there is a lesson in it, which I am not going to try to seek. Maybe it is the oldest unsolved math problem for a good reason, in that it hides in the shadows and you need to solve it with a minimum facts such as k1. No Odd Perfect turns out to be more difficult of finding that unwanted "2", whereas square root of 2 the unwanted "2" pops up soon. But one of the reasons No Odd Perfect took so long is the contorted and arbitrary definition of what is a factor and what is not. There are still some illogical folk who think the factors of 9 are just 1,3, or think that 1,3,9, when in truth, with a proper logic the factors of 9 are cofactors of 1x9 and then 3x3 for four numbers in full. So that No Odd Perfect could ever be proven if one batch of mathematicians counts 9 as only 2 factors, another batch counts 3 factors and then I count 4 factors.
____________________________________________ Detailed Constructive proof No Odd Perfect Number ____________________________________________
The basic term used is _cofactors_, where a number has its cofactors paired.
Example is 6 and 15:
The number 6 has cofactors of 1 with 6, and 2 with 3 and represented as this:
(1 + 6) + (2 + 3) = 12
The number 15 has cofactors of 1 with 15, and, 3 with 5 and represented as this:
(1 + 15) + (3 + 5) = 24
Now we omit or delete the number itself and 1 factor and for 6 we omit the 1 and 6 and have remaining the 2,3 cofactors so we have k1 when k is the number. For 6, k1 is 5 where 2+3 = 5. For 15, k1 is 14 and where we have 3+5 = 8.
For 18 we have
(1 + 18) + (2 + 9) + (3 + 6) = 39 and for k1 we have 17, but we have 2+9+3+6 = 20
For 20 we have
(1 + 20) + (2 + 10) + (4 + 5) = 42 and for k1 we have 19, but we have 2+10+4+5 = 21
For 9 we have
(1 + 9) + (3 + 3) = 16 and for k1 we have 8, but we have 3+3=6
For 28 we have (1 + 28) + (2 + 14) + (4 + 7) = 56, and for k1 we have 27, and we have 2 + 14 + 4 + 7 = 27
____________________________________ Constructive Definition of a Perfect Number ____________________________________ Now let me define the Perfect Number in general with a formula. Summation of cofactors with the 1 and k omitted must equal k1.
_______________ Construction proof _______________
Take the arbitrary Odd Perfect Odd number larger than 1 and call it k.
Means that the cofactor summation is k1 once the 1 and k cofactors are omitted.
Since k is Odd Perfect, all its cofactors below k1 are odd numbers. That means their sum is an even number.
Here we have two possibilities: either the smallest divisor of k is 2 or is 3.
Can 3 be the smallest cofactor divisor of this arbitrary odd perfect number k, would mean that we can separate the divisors into two groups, one of which is 1/3(k1) and the the remaining terms are 2/3(k1).
Or, is 2 the smallest cofactor divisor, would mean that the two groups are both 1/2(k1).
Here let me focus on the number 945 since it is odd abundant so as to give the reader some bearings of odd abundant and odd deficient numbers.
(1 + 945) omitted (3 + 315) + (5 + 189) + (7 + 135) + (9 + 105) + (15 + 63) + (21 + 45) + (27 + 35) = 974
For 945 the k1 is 944, but in actuality, once we omit the 1 and 945. We have remaining the sum (3 + 315) + (5 + 189) + (7 + 135) + (9 + 105) + (15 + 63) + (21 + 45) + (27 + 35) = 974
I displayed the abundant odd number to compare with the deficient odd number of 15. Few people know that some odd numbers can be abundant.
Now for the impossibilty of the construction of the Odd Perfect. The Odd Perfect has a k1 that is even, for odd +odd is even. The smallest cofactor divisor of the Odd Perfect number must be either 3 or 2, or if you want the smallest to be say 5, the argument works the same. If 3 is the smallest divisor of k, then 1/3(k1) is the sum of factors and 2/3(k1) the remaining sum of factors for k1. That makes for an impossible construction situation, since the sum of k1 is even, and a portion of the factors is to add together to be 1/3* even number k1 and the other portion to be 2/3*even number k1. You cannot divide an even number k1 into two portions of 1/3 and 2/3.
So here we grasp the impossible construction of a even number k1 with 1/3 portion and 2/3 portion of an even number. Now if 5 were the smallest divisor we run into the problem of 1/5 and 4/5 and the same goes for larger odd numbers as the smallest divisor.
The only other alternative is that 2 is the smallest divisor of k1 and that implies that 2 is a divisor of k. But that is impossible for then an odd number k has 2 as a divisor.
So basically, the proof hinges on the fact that we take the arbitrary odd perfect number and force it to have a sum that is even. That means we separate k into k1 where we are guaranteed of a even number sum since pairs of odd numbers is always even. Now, the final operation of construction is ask what the lowest or smallest divisor could possibly be for the k1. Is it 3 or 2 or some larger odd number? We then show it cannot be 3, nor 2 nor any larger odd number.
QED
 Recently I reopened the old newsgroup of 1990s and there one can read my recent posts without the hassle of mockers and hatemongers.
https://groups.google.com/forum/?hl=en#!forum/plutoniumatomuniverse
Archimedes Plutonium



