Re: "Collatz 3n+1 conjecture is unprovable" paper


Craig Feinstein wrote:
"(1) Given a number n divisible only by 2, the Collatz algorithm's
first
step will find a number m such that m<n and m is divisible only by 2.

(2) Eventually, any such process must end at 1.


You don't need to know specifically what Collatz does."

This is not a rigorous mathematical proof that 32 causes Collatz to
halt at one. It doesn't even talk about 32.

I think you have fallen into an eye-of-the-beholder fallacy.

Is there any such thing as showing "directly or indirectly" that
32 is not divisible by any prime other than 2?

If I demonstrate this, will you "read into" my proof that 32 = 2^5
and therefore "directly or indirectly" I showed what the Collatz
sequence for 32 looks like at each step?

Here are just two of the ways I think prove that 32 is "some"
power of 2, without establishing that 32 = 2^5.

First: Square 2 repeatedly until the result exceeds 32; we will
get 4, then 16, then 256. Divide 256 by 32 and you get zero
remainder; because 256 is a power of 2, so is its divisor 32.

Second: Make a list of all primes less than 32, e.g. by say
Eratosthenes Sieve:

2,3,5,7,11,13,17,19,23,29,31

Of these, only 2 exactly divides 32.

Representation of 32, by the Fund. Thm. of Arithmetic,
as a unique product of prime powers consists simply of
some power of 2.

(No other prime can appear with positive exponent, since no
prime factor of 32 can exceed 32.)

regards, chip

.

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