Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: minor revisions of Proof of the No Odd Perfect (exce
pt 1)  #1366 Correcting Math

Replies: 1   Last Post: Jan 4, 2014 9:16 PM

 Messages: [ Previous | Next ]
 plutonium.archimedes@gmail.com Posts: 18,572 Registered: 3/31/08
minor revisions of Proof of the No Odd Perfect (exce
pt 1)  #1366 Correcting Math

Posted: Jan 4, 2014 4:00 PM

minor revisions of Proof of the No Odd Perfect (except 1)  #1366 Correcting Math

The proof is in, and it is unsinkable but I need to keep on making minor adjustments for that ultra-clarity.

____________________________________________
Proof that No Odd Perfect Number (except 1) Exists
____________________________________________

For the odd numbers 3 or larger we have:

Suppose k is an arbitrary odd perfect number which means the sum of its
reciprocal cofactors is equal to 2. Algebraically it is written as 2k/k = 2, such as in 2(6)/6 = 2, but better written to focus on the numerator of (2k) with its component cofactors as this:

2(numerator)/k = 2.

In the proof we need specifics of the numerator and not just 2k/k = 2.

The equation 2(numerator)/k = 2 is more general than 2k/k =2, and more general in that we add piecewise the cofactors that makes up the numerator.

Now k is a sum of reciprocals of cofactors obeying this equation

(f1/k + f2/k) + (f3/k + f4/k) + . . + (fn/k + fm/k) = 2t/k.

Since the numerator sum must be even because we add up pairs of odd numbers as cofactors, for in an odd k, all its divisors are odd and adding odd with odd is even so the numerator is finally 2t.

Now in this proof we add the cofactors of the equation incrementally.

So we can list the cofactors as this list:

(f1/k + f2/k) = (2s)/k where s is minimally reduced.

(f3/k + f4/k) = (2r)/k where r is minimally reduced.

(f5/k + f6/k) = (2u)/k where u is minimally reduced.

(f7/k + f8/k) = (2v)/k where v is minimally reduced.

Out to (fn/k + fm/k)

Now we made the assumption or supposition k was Odd Perfect meaning it obeys 2(numerator)/k = 2, and we finally apply that to the fact that the numerator is this summation of 2s + 2r + 2u + 2v + . . (fn/k +fm/k) = 2t

We cannot say 2(2s)/k is equal to 2 nor can we say 2(2s+2r)/k is equal to 2
nor can we say 2(2s + 2r + 2u)/k is equal to 2 but as we add more and more of the terms of the numerator we are getting closer and closer to 2(2t)/k =2.

Remember, we assumed k as Odd Perfect and in that assumption k satisfies 2(2t)/k = 2 since the 2t is independent of the 2numerator/k=2. Now we have the contradiction at hand for we have 2*2t/k = 2 and that is 2t/k = 1 and that is 2t = k where k is a odd number and t if not odd is reduced to odd and is a smaller odd number than k and the contradiction is obvious in that no odd number multiplied by an even is another odd number.

I save the exception of the odd number 1 for last. The number 1 has cofactors of 1/1 + 1/1 in that 1 is a divisor and is a dividend as in division or multiplication:

In division
a/b = c where a is numerator (dividend), b is divisor and c is quotient

In multiplication
a x b = c where a is multiplicand, b is multiplier, c is product

The odd number 1 is a cofactor of 1 for we cannot in mathematics leave 1 as a singularity of 1 being only a multiplier and not a multiplicand or only as a divisor and not a dividend. All divisions that end up as a quotient being another integer has both a dividend and divisor and 1 cannot be a singularity.

So 1 obeys the rules of Perfect Number as (1/1 + 1/1) = 2 and thus 1 is odd perfect, and the only odd perfect number.

QED

--

Recently I re-opened the old newsgroup of 1990s and there one can read my recent posts without the hassle of mockers and hatemongers.

Archimedes Plutonium

Date Subject Author
1/4/14 plutonium.archimedes@gmail.com
1/4/14 Wizard-Of-Oz