Science Forums 2

modest · 01-29-2009

Quote:

Originally Posted by Don Blazys

To: Modest,

When you wrote that I "accidentally missed all major points of contention and critisism",
you forgot that I am addressing the many points that were made, one at a time,
in the order that they were made, because that's the best way to avoid further confusion.
Please be patient. I will get to each and every point, I promise.

I do appreciate that.

It’s all too often on forums such as this that valid rebuttals are made for some given argument only for the person making the argument to move the goalpost ignoring the scope of the rebuttal entirely. It’s an informal logical fallacy that gets a lot of play around here which has no doubt made me somewhat skeptical of posts that take the form “In time we’ll get to the main issue raised, but first...”. I appreciate your assurances that this is not the case and apologize for assuming otherwise.

I think you're bringing up a lot of interesting and worthwhile things, but I don't see anything addressing my original question from a week ago. I'm referring in particular to this:

Quote:

Originally Posted by Don Blazys

Thus, if we allow both "multiplication by unity" and "multiplication of unity",
then $6$ can be factored as:

$6=3*2*1*1$ , where $3+2+1+1=7$

and the entire concept of a "perfect number" simply collapses!

It reads to me like you are saying that if we are allowed to factor 28 as (1 • 1 • 2 • 2 • 7) or $(1^2 \cdot 2^2 \cdot 7) = 28$ rather than the normal factorization $(2^2 \cdot 7) = 28$ then its positive proper divisors would be {1,1,2,4,7,14} rather than the normal {1,2,4,7,14} which then wouldn't work for describing a perfect number. I don’t understand how you get that set of divisors. How do you get from the factorization:

(1 • 1 • 2 • 2 • 7) = 28

to the summing of positive proper divisors:

s(28) = 1 + 1 + 2 + 4 + 7 + 14 = 29

I fully accept that I may be misunderstanding something about factoring that makes the above obvious, but as it stands, it appears simply to be mistaken.

If I were to look for the positive proper divisors of a number in order to check if it is perfect, I would find all positive integers which divide the number without leaving a remainder. Those numbers would make a set from which I would exclude the number itself and then sum the set.

As this relates to factorization, if I were to decompose the number 28 into a factorization:

$(2^2 \cdot 7) = 28$

and I wanted to relate that to the positive divisors of 28, then I would say a divisor of 28 (D) is:

$D = (a^x \cdot b^y)$

where a=2, b=7, x={0,1,2}, and y={0,1}. This would mean any of the following are divisors of 28 given the factorization above:

$2^0 \cdot 7^0 = 1$
$2^0 \cdot 7^1 = 7$
$2^1 \cdot 7^0 = 2$
$2^1 \cdot 7^1 = 14$
$2^2 \cdot 7^0 = 4$
$2^2 \cdot 7^1 = 28$

I don’t know how normal it is to consider (1 • 1 • 2 • 2 • 7) a possible factorization of 28. It seems rather redundant, but that does not mean it is wrong or impossible or inconsistent with number theory. Relating this factorization,

$(1^2 \cdot 2^2 \cdot 7) = 28$

to the divisors of 28, I would say,

$D = (a^x \cdot b^y \cdot c^z)$

where a=1, b=2, c=7, x={0,1,2}, y={0,1,2}, and z={0,1}. This would seem to mean that the result of any of the following would be possible divisors of 28 given the factorization (1 • 1 • 2 • 2 • 7)=28:

$1^0 \cdot 2^0 \cdot 7^0 = 1$
$1^0 \cdot 2^0 \cdot 7^1 = 7$
$1^0 \cdot 2^1 \cdot 7^0 = 2$
$1^0 \cdot 2^1 \cdot 7^1 = 14$
$1^0 \cdot 2^2 \cdot 7^0 = 4$
$1^0 \cdot 2^2 \cdot 7^1 = 28$
$1^1 \cdot 2^0 \cdot 7^0 = 1$
$1^1 \cdot 2^0 \cdot 7^1 = 7$
$1^1 \cdot 2^1 \cdot 7^0 = 2$
$1^1 \cdot 2^1 \cdot 7^1 = 14$
$1^1 \cdot 2^2 \cdot 7^0 = 4$
$1^1 \cdot 2^2 \cdot 7^1 = 28$
$1^2 \cdot 2^0 \cdot 7^0 = 1$
$1^2 \cdot 2^0 \cdot 7^1 = 7$
$1^2 \cdot 2^1 \cdot 7^0 = 2$
$1^2 \cdot 2^1 \cdot 7^1 = 14$
$1^2 \cdot 2^2 \cdot 7^0 = 4$
$1^2 \cdot 2^2 \cdot 7^1 = 28$