S.O.S. Mathematics CyberBoard

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:



You start off fine, making your contradiction-hopeful hypothesis, namely that your number is rational, just as I do. However at this point you call upon the integers a, b, c such that a<xc<b. OK, fine, not relevant at all, but certainly such integers exist--a=[x]-1, b= [x]+1, c=1 is one set of choices that works. Then you claim that showing xc necessarily being an integer and knowing that neither of the summands can be an integer are incompatible statements. But that's not true, not for any choice of x as the sum of two other numbers. It is an invalid inference because your logic is unsound, this has nothing to do with your assumption set, it has to do with faulty reasoning, which is independent of the assumptions you make. You do not use a set of previously verified assumptions, one of the assumptions present in your "proof" was that because there is no integer, such that is an integer or then automatically there is no such integer for their sum, which you call "x". However, why should we believe this? Can you quote a theorem which states that the sum of two numbers, neither of which is an integer cannot be an integer? I can tell you that there is no such theorem, because that is ALSO FALSE. And there is the problem, instead of having a bunch of true statements you've already verified combined with one statement you're trying to test the veracity of, you have MULTIPLE unverified statements, so even if you derive a contradiction, there's no way of figuring out how many of your untested assumptions are false, you only know that there is AT LEAST ONE, and in your case we can see that there are at least two. As a result your whole argument is useless in pursuit of the goal.

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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.

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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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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It is much harder to prove is rational/irrational, right?

I believe that is still an open question.

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Thanks for your help.