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 Post subject: Re: Irrational numbersPosted: Thu, 19 Apr 2012 00:33:58 UTC
 S.O.S. Oldtimer

Joined: Mon, 28 Dec 2009 00:16:28 UTC
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You do not have a sum of two algebraic number. By your assumption, sqrt(2) - r = -pi is not algebraic, so you cannot claim it is simply because you gave it a different name.

Consider my incorrect sqrt(2) proof. Just because I have sqrt(2)q does not mean I can treat sqrt(2) as an irrational number, because by assumption it is not. I can also incorporate some definitions, but it still would not work, because of my assumption. Do you see what I mean? I think you have exactly the same problem, you are violating your assumption to get a contradiction.

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 Post subject: Re: Irrational numbersPosted: Thu, 19 Apr 2012 01:23:43 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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glebovg wrote:
You do not have a sum of two algebraic number. By your assumption, sqrt(2) - r = -pi is not algebraic, so you cannot claim it is simply because you gave it a different name.

Consider my incorrect sqrt(2) proof. Just because I have sqrt(2)q does not mean I can treat sqrt(2) as an irrational number, because by assumption it is not. I can also incorporate some definitions, but it still would not work, because of my assumption. Do you see what I mean? I think you have exactly the same problem, you are violating your assumption to get a contradiction.

I don't think you understand the rules of how proofs work. I *am* in-fact allowed to make any assumption I want about my numbers when I do not know what the truth is ahead of time. Without knowing if is rational or not, I can assume it is, and if I derive a contradiction from this fact by using nothing but other facts which have been already established, this is a valid proof by contradiction. (Note all my other steps are justified by the facts/lemma I had before hand) As a result my proof is 100% valid, this is how EVERY proof by contradiction works. You're just noticing the contradiction sooner than the proof does, that does NOT invalidate it as a form of proof. I use sound logic and a set of previously verified assumptions with the sole exception of the one hypothesis I'm testing, and derive a false statement, which is what is more commonly referred to as "proof by contradiction".

Now let's examine your attempt at a proof:

glebovg wrote:
Can anyone explain what is wrong with my reasoning?

Suppose and let . Also, let and assume . If I show that xc must be an integer, and I know there does not exist c such that or is an integer. Then, xc cannot be an integer, a contradiction.

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 Post subject: Re: Irrational numbersPosted: Thu, 19 Apr 2012 02:07:39 UTC
 S.O.S. Oldtimer

Joined: Mon, 28 Dec 2009 00:16:28 UTC
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Actually, I was trying to prove something else. I just used different numbers in the original post. I think your proof is very simple and short. This is probably why I thought it was incorrect. Could you post a different proof with, say, a product of irrational numbers, but where you need to do more work. I would really appreciate it.

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 Post subject: Re: Irrational numbersPosted: Thu, 19 Apr 2012 02:35:09 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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glebovg wrote:
Actually, I was trying to prove something else. I just used different numbers in the original post. I think your proof is very simple and short. This is probably why I thought it was incorrect. Could you post a different proof with, say, a product of irrational numbers, but where you need to do more work. I would really appreciate it.

There is no general theorem for irrational products, a product of irrationals may be rational and it may not be, depending on the two numbers in question. If you cannot be more specific, then there is nothing more than can be said by anyone.

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 Post subject: Re: Irrational numbersPosted: Thu, 19 Apr 2012 03:08:20 UTC
 S.O.S. Oldtimer

Joined: Mon, 28 Dec 2009 00:16:28 UTC
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Suppose we have . How would you prove it is rational/irrational?

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 Post subject: Re: Irrational numbersPosted: Thu, 19 Apr 2012 03:42:56 UTC
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glebovg wrote:
Suppose we have . How would you prove it is rational/irrational?

Algebraic numbers form a field, so if that product is rational, is algebraic, hence that number is not rational.

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 Post subject: Re: Irrational numbersPosted: Thu, 19 Apr 2012 05:07:53 UTC
 S.O.S. Oldtimer

Joined: Mon, 28 Dec 2009 00:16:28 UTC
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It is much harder to prove is rational/irrational, right?

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 Post subject: Re: Irrational numbersPosted: Thu, 19 Apr 2012 05:15:25 UTC
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glebovg wrote:
It is much harder to prove is rational/irrational, right?

I believe that is still an open question.

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 Post subject: Re: Irrational numbersPosted: Thu, 19 Apr 2012 14:09:14 UTC
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