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Old 01-25-2009   #98 (permalink)
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Re: A Mathematical Emergency.

Quote:
Originally Posted by Don Blazys View Post
To: Modest,

Please allow me to both clarify and elaborate.


Quote:
Originally Posted by Don Blazys View Post
If we are to be consistent in our our use of nomenclature,
then the words "proper factor" must mean the exact same thing as "proper divisor",
because in the case of "proper divisor",
the word "proper" is clearly taken to mean "other than itself".

Also, as mathematicians, if there is one thing that we absolutely must agree on,
it's that we must always remain consistent in our mathematical definitions.

However, in both the "math community" and on the internet,
there is indeed a whole lot of inconsistency, disagreement and confusion
on what the words "proper factor" and "proper divisor" should mean,
which only adds to the "urgency of the mathematical emergency".

Definitions should be both simple and easy to understand,
and since multiplication is "easier" and "more simple" than division,
let us henceforth define "perfect numbers" as I did in post #94,
using "proper factors" rather than "proper divisors".
Well, that's a rather large preamble... and I'm not exactly sure I see the point. Perfect numbers have been defined for over 2,000 years. Factors, for longer. If there's some online confusion and inconsistency... I'm completely out of the loop on that.

Quote:
Originally Posted by Don Blazys View Post
Now, according to the only logically consistent definition of a "proper factor",
the proper factors of 28 are:

1, 2, 4, 7 and 14,

and 28 is called a "perfect number" because:

1+2+4+7+14=28=(1*2*4*7*14)^{\left(\frac{1}{2}\right)}.

Note that the proper factors of 6 "multiply perfectly" as a "first root",
while the proper factors of 28 "multiply perfectly" as a "second root".
If we continue in this manner, then we will find that the proper factors of
the next perfect number, 496, also "multiply perfectly", but as a "fourth root",
and that in fact, all perfect numbers "multiply perfectly" as some "Nth root".
(It's one of the "deeper" reasons why these numbers are called "perfect".)
Thus, the argument that I presented in post #94 is not
"coincidental as far as perfect numbers go", but applies throughout.
First, let me say, that's really cool, Don.

It appears to be true (or, at least, the first few have no counterexample). I've written a little algorithm to test it up to the 7th perfect number which was successful. The roots were 1,2,4,6,12,16,18. My brain is not working well-enough this Sunday morning to figure why that works... any opinion on that would be much-appreciated.

Second, I should say... what you have there is not a definition of a perfect number. It appears I was incorrect using the term "coincidental", but it is, nevertheless, superfluous. A perfect number is an "integer which is the sum of its proper divisors". It's really that simple. You could say the integer (n) is half its divisor function (2n = \sigma(n)), or equal to its restricted divisor function (n = s(n)).
Perfect Number -- from Wolfram MathWorld
But, you cannot say it is "equal to some integer root of the product of its factors". The integer 30 would qualify under that definition as would many others.
(1\cdot2\cdot3\cdot5\cdot6\cdot10\cdot15)^{1/3} = 30 \neq (1+2+3+5+6+10+15)
Moreover, I see nothing about the following equality that prohibits multiplication by unity.
(1+2+4+7+14) = (1\cdot2\cdot2\cdot4\cdot7\cdot14)^{1/2}
I think you want to imply that multiplying the rhs by one would necessitate adding one to the lhs. If that's the case, I don't follow. One is a factor of 28 once.

Quote:
Originally Posted by Don Blazys View Post
Therefore, as serious seekers of both truth and beauty,
we are now compelled to come up with some reasonable model, paradigm,
or explanation, as to why unity should occur as a factor... only once.
Because 1\cdot1\neq28. One is a factor of 28 because (1 • 28) = 28. You could alternatively write (28 • 1) = 28, but that doesn't make 1 a factor twice any more than it doubles up any of the other factors.

Either I'm missing something big or there's nothing about number theory that precludes multiplication by unity. Since multiplication by unity is included in the axioms of natural number arithmetic, I'm pretty confident that it isn't precluded.

~modest


----------------

Last edited by modest; 01-25-2009 at 11:49 AM.. Reason: fixed equation
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