PROOF OF BEAL'S CONJECTURE

[dodomaze]

the sum of numbers that consist of at least one irrational number is an irrational number

[tex](2-\sqrt 2)[/tex] and [tex]\sqrt 2[/tex] are two irrationals; when you add them together, you get 2. I have given similar examples for multiplication.

no one has challenged this by, for example, coming up with a counterexample.

You may be asking for too much. Producing your counterexample would require giving a counterexample to Beal's conjecture, which would prove the conjecture false! You can't possibly declare "if no one can show the conjecture is false, then my proof must be true". I imagine it was not your intention to ask for that much, I just want you to realize what you're asking for. Smaller counterexamples (of irrationals that add or multiply to an integer) have already been given.

As for that paragraph I wrote, that you believe supports your position: all I intended to illustrate was:
(a) irrationals, and approximations to them, are two different things; the latter are rational.
(b) it's not the same to have "an arbitrary number of decimals" than to have "infinite decimals". You can have as many decimals as you want in an approximation, but they will be a finite number of them. An irrational would have to be represented with infinite decimals and, if that makes you happy, could never actually be written. That doesn't mean that the ideal, perfect irrational, does not exist. It just means that decimal representations are inadequate to describe them.
(c) At the risk of being annoying by repeating the obvious: if irrationals did not exist at all, the set of real numbers would not be bigger than the set of rationals.

If there is a specific sentence on that paragraph that truly bothers you, by all means let me know. The paragraph is large, and it's difficult to see which part of it needs clarification.

P.S.: If we reverse roles for a moment, and I claim that 1/3 + 1/3 + 1/3 cannot be an integer, because each of them can only be approximated as 0.333333 (to some number of decimals)... Would you be able to correct me? Or are rationals a problem too? Or only rationals that have prime factors other than 2 and 5 in their denominator? (As you see, simply a "feature" of the decimal representation.)

Another P.P.S, sorry, I can't help myself:

e.g. a line not really having a thickness, these are the kind of concepts that have hindered the progress of some fields of mathematics, in my view.

Nothing prevents anyone from imagining a line with 0.1 units of thickness: in other words, a rectangle which is infinite in one dimension but bounded in the other. Some would just call it "the area between two parallel lines". Mathematicians have defined concepts because they are useful. It would be more difficult to describe a line without thickness, having only lines with thickness.

And let's not mention circles, or squares with a diagonal in them, because all of these (completely imaginary, nor made of wood) objects would have an irrational measure somewhere. This bugged the ancient Greeks to no end.

[Awojobi]

You are limiting yourself to a trivial example of proving that an irrational number + an irrational number can be a rational number. For the particular cases I present in each bracket , this cannot be. The answer will always be an irrational number, unless the problem is solved by expansion of brackets where the irrational numbers cancel out in pairs, similar to the trivial example you present.

[Awojobi]

All my proof has done is to rewrite one side of the Beal equation as a product of 2 brackets. Since I want to prove that this product cannot be equal to a product of integers A^x, is it not obvious that I have to be dealing with products of the 2 brackets that produce only integer values, hence every term in each of the 2 brackets need to be integers? To me, it is quite obvious. Concerning the other proof, or should I say disproof, I was talking about which it seems you misrepresented, I was talking about no one will be able to come up with a counterexample that shows a Beal's conjecture equation with 3 different exponents being manipulated to a new example with 2 exponents that are the same.

[dodomaze]

It seems to me that the same argument could be used to prove that 5 minus 3 cannot be equal to 2, since, for some n,

[tex]5 - 3 = (5^{1/n} - 3^{1/n}) \cdot (5^{(n-1)/n} + 5^{(n-2)/n} \,3^{1/n} + 5^{(n-3)/n} \,3^{2/n} + \ldots + 5^{2/n} \,3^{(n-3)/n} + 5^{1/n} \,3^{(n-2)/n} + 3^{(n-1)/n})[/tex]

This is unrelated to Beal's conjecture, because I chose x=y=z=1, but the argument is the same.

[Awojobi]

I don't get what you're trying to illustrate, not least because in my proof, I wouldn't get into such a dilemma because I make n>2. So, you need to specify the values n cannot be.

[dodomaze]

[tex]n=3[/tex] will do just fine. The question is, does

[tex]5 - 3 = (5^{1/3} - 3^{1/3}) \cdot (5^{2/3} + 5^{1/3} \,3^{1/3} + 3^{2/3})[/tex]

imply that 5 - 3 can never be 2?

[Awojobi]

The question you are asking is not the question you should be asking. The question you should be asking is if each bracket can work out to be an integer such that the product of the 2 brackets gives 2. So the two possible integers are 2 and 1 which give a product of 2. None of the brackets can work out to be 2 or 1, hence the right hand side of your equation cannot work out to be 2. It can only produce an answer that tends to 2, depending on what accuracy you require.

[dodomaze]

You don't need any approximation to say 5 - 3 = 2. Two things equals to a third are equal to each other (x = y and x = z imply y = z). Specifically,

[tex]5 - 3 = 2[/tex] and [tex]5 - 3 = (5^{1/3} - 3^{1/3}) \cdot (5^{2/3} + 5^{1/3} \,3^{1/3} + 3^{2/3})[/tex] imply [tex](5^{1/3} - 3^{1/3}) \cdot (5^{2/3} + 5^{1/3} \,3^{1/3} + 3^{2/3}) = 2[/tex].

How did all these irrationals ended up being exactly equal to 2? There are no approximations in the first two equalities, and logic implies the third one.

[Awojobi]

You are basing your arguments on expanding the brackets so that pairs of irrationals cancel each other out and then you are left with 5-3=2. You are ignoring the other way of solving where each bracket is worked out and the product calculated. The answer therefore cannot be 2 but tends to 2 depending on the accuracy you require. It's just like the Pythagorean triples which have only integer values. However, we still use Pythagoras theorem to get approximate answers to real life Pythagoras theorem problems. The same reasoning should be used for Beal's conjecture i.e. ONLY INTEGER VALUES SHOULD BE CONSIDERED when finding the product of the 2 brackets, just as is done for the Pythagorean triples.

[dodomaze]

In your proof, you wrote

[tex]C^z - B^y = \left( C^{z/n} - B^{y/n} \right) \cdot \left( (C^{z/n})^{n-1} + (C^{z/n})^{n-2} B^{y/n} + (C^{z/n})^{n-3} (B^{y/n})^2 + \ldots + (C^{z/n})^2 (B^{y/n})^{n-3} + C^{z/n} (B^{y/n})^{n-2} + (B^{y/n})^{n-1} \right)[/tex]

Are you telling me that this is an exact equality, but

[tex]5 - 3 = (5^{1/3} - 3^{1/3}) \cdot (5^{2/3} + 5^{1/3} \,3^{1/3} + 3^{2/3})[/tex]

is not an exact equality?

[Awojobi]

Both sides of the equation can only be equal if every term in each of the brackets are integers.

[dodomaze]

But you claim that the first equation was obtained by algebraic means. You say,

algebraic factorisation of the left hand side of the equation gives

You mean, that algebra sometimes provides an exact equality and sometimes it does not?

[Awojobi]

But of course.

[Awojobi]

An exact equality if one is dealing with integers and not irrationals.

[dodomaze]

Well... no.

Can we at least agree on the existence of irrationals, so that the set of reals is larger than the set of rationals?

[Awojobi]

You are asking me to agree on an entirely new topic which I don't think has any bearing on my proof. A topic that I have never thought about before. Hence, I can't agree or disagree. All I know is that Beal's conjecture deals with products of integers, hence any irrationals must be avoided at the earliest opportunity, just as I have done in my proof.

[dodomaze]

Fair enough. The point I'm trying to bring down is that decimal representations are very inadequate to deal with irrationals, to the point of creating the false idea that they are unreachable except by approximations. The latter is true only if one is limited to work exclusively with rationals.

The deceptive idea above then debilitates the notion of equality, making any form of proof difficult to believe. I gave you the example
[tex](5^{1/3} - 3^{1/3}) \cdot (5^{2/3} + 5^{1/3} \,3^{1/3} + 3^{2/3})[/tex]

which you claim that it can be equal to 2 because of telescopic cancellation, but not as the product of two brackets. Either it is equal to 2, or it isn't; this should not depend on alternative explanations for the same expression.

Something similar occurs in your proof, when you have
[tex]C^z - B^y = \left( C^{z/n} - B^{y/n} \right)^n[/tex]

where the left-hand side is clearly an integer, and the right-hand side is potentially a product of irrationals (finitely many times),
[tex](C^{z/n} - B^{y/n}) \cdot (C^{z/n} - B^{y/n}) \cdot (C^{z/n} - B^{y/n}) \cdot \ldots[/tex] (n times)

You'd have a problem if your interpretation of this equality work sometimes, doesn't work sometimes, depending on convenience.

At the very least, specify clearly which of the "=" signs in your proof represents an exact equality, and change the symbol to something else where they don't. Just by doing this exercise your proof will likely fall apart.

[Awojobi]

The 'equation' you wrote is incorrect and I never wrote it in my proof, hence I will not address anything you've said concerning it. However, I will address the gist of what I think you're talking about.
Let me see if I can get my idea across. Solve the equation (a^0.5 - b^0.5) (c^0.5 + d^0.5) = 3^2 where a,b,c,d are integers. In order to solve this equation, is it not clear that only integer values of a^0.5, b^0.5, c^0.5 ,d^0.5 should be considered and not irrational numbers? If yes, this is the type of reasoning I have used in my proof.

[Guest]

Why bother with a so-called proof...?

Play the lottery instead.

Good luck!

Cmon, dynamic duo ("Batman & Robin"), the truth of ABC conjecture proves the Beal conjecture, and the Collatz conjecture is also true. Study/review the works on this site.

What's the point of your work?

And why reinvent the "wheels" over again?

Play your hunch (lottery pick) and win a bunch ($$$) if you're so lucky.

Good luck!

[Guest]

I think some of the lotteries in the US are rigged.

And the game of life is not fair (rigged too). Go figure!