Re: ga8- Integers found that form the sqrt(2) or transcendental numbers.

From: Guillermo Arango (abril10_at_geo.net.co)
Date: 12/31/04 

Date: Fri, 31 Dec 2004 18:31:17 +0000 (UTC)

On 30 Dec 2004, Timothy Little wrote:
>Guillermo Arango wrote:
>> 14142135623730950488016887242097... / 10000000000000000000000000000000... = ?
>>
>> If you do not accept the two numbers that I have written down as
>>integers, you must proof why they are not Integers to you.
>
>Actually, the burden of proof is on you, since you're the one claiming
>that such integers exist.
>
>However, it is simple to prove that they are not integers. One proof:
>every integer is finite. For any finite number, your representation
>denotes something larger. Therefore it denotes something infinite,
>and hence does not represent an integer.
>
>Or another proof: No integer is both even and odd. By the proof you
>quoted, your numbers must be both even and odd. Therefore they are
>not integers.
>
>Either one suffices, and if you don't accept those there are plenty
>more.
>
>
>There is an "out" though: you could define a set consisting of both
>finite and infinite decimal representations, together with arithmetic
>operations based on manipulating the digits much as usual integers are
>added, subtracted, divided and so forth. This seems to be what you're
>trying to do. However, you can't expect the world to accept calling
>your objects "integers", since that name has already been used for
>hundreds of years to name objects with quite different properties.
>Your objects would include the integers, but would not be the same
>thing.
>
>You'll have to watch out for traps inherent in expecting such objects
>to obey the usual laws of arithmetic for integers, since they don't.
>You'll have to decide for objects like a = 1000... , whether a+1 or
>10a are different numbers from a or not. Any choice leads to
>complications, which is fine so long as you're aware of them and
>ensure that your definitions don't lead to self-contradiction.
>
>Once you've worked out the arithmetic of the objects you've defined,
>you can then define a set of objects representing ratios of them, in
>much the same way as the rationals are developed from the integers.
>Then you can try to find an isomorphism between a subset of your
>ratios and a subset of the reals. At that point, you can say "these
>ratios are the same things as these real numbers". If you made good
>choices earlier, you might even be able to show that sqrt(2) is a
>member.
>
>That's a lot of work. Maybe some of it has been done already, you
>could search the Web or written literature to see if anyone else has
>done similar work you could build on. In the end, it's up to you to
>decide whether it's interesting enough to pursue.
>
>
>- Tim

ANSWER:

Dear Timothy:
 
Thank you for your comments they are really math based. You are right I am the one that has to provide the proof mathematically. This is what I am trying to do.
 
Responding to your observations:
 
"However, it is simple to prove that they are not integers. One proof:
every integer is finite. For any finite number, your representation
denotes something larger. Therefore it denotes something infinite,
and hence does not represent an integer."
 
The definition "every integer is finite" means that integers have no decimal expansions and that
they are exact as I understand it. However, the set of integers is infinite, this means there is no last integer,
therefore the number of digits can be infinite but on the real line one integer is smaller or bigger than another,
making them finite.
The integers I find using the formula are finite in the number of digits they have. Since they construct irrational
numbers such the sqrt(2) in the formula, gives me a reason to say this is why I am saying that irrational
numbers are also finite. which is what I am trying to prove.
 
your other observation:
 
"Or another proof: No integer is both even and odd. By the proof you
quoted, your numbers must be both even and odd. Therefore they are
not integers."
 
Not at all, they are either even or odd. The confusion comes when you multiply the initial integer by a common
integer for the numerator and denominator. At this point the numerator changes depending if the common
integer for numerator and denominator is even or odd. but each numerator in each fraction is even or odd,
never both. The reason why I multiply the fraction was to show that there are many fractions that make an irrational number, but only the irreducible fraction complies to prove the rationality of the real numbers.
 
thank you again for your observations, you have hit it in the direction I am trying to get.
 
Guillermo
 
 

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