PROOF OF BEAL'S CONJECTURE

[Guest]

PROOF OF BEAL'S CONJECTURE

[dodomaze]

Why are all these numbers integers, again? You seem to believe that the product of irrationals cannot be an integer. That would be a false belief. For example, [tex]\sqrt 2 \cdot \sqrt 2 = 2[/tex], or [tex]\sqrt 2 \cdot \sqrt{18} = 6[/tex], or [tex]\sqrt 6 \cdot \sqrt{10} \cdot \sqrt{15} = 30[/tex].

[Awojobi]

I know that the product of irrationals can give rational numbers. However, irrationals in themselves can only be approximations to their exact value, an exact value that cannot be written numerically.

[dodomaze]

So how did you get to this conclusion?

then every term in each of the 2 brackets must be integers and not irrational numbers

[tex]C^z[/tex] and [tex]B^y[/tex] were integers, but after you add fractions as exponents, how do you get any of them to be an integer again?

[Awojobi]

The point I am trying to make is that the first step in the proof is to make sure that the terms in each of the brackets are integers and not irrational numbers. If the first and/or second bracket works out to be irrational, there is no way the product of the 2 brackets can become rational, talk less of an integer.

[dodomaze]

Why? You said you already knew that the product of two irrationals can give an integer.

[Awojobi]

Yes I did. However, the answer cannot produce a rational number just as, even though the product of square root of 2 and square root of 2 = 2, the product of a numerical approximation of square root of 2 and a numerical approximation of square root of 2 is not equal to 2. The square root of 2 is basically not an exact value in reality. It can only be approximated to a level of accuracy that is so desired.

[dodomaze]

You started with the integers [tex]C^z[/tex] and [tex]B^y[/tex], and did algebraic manipulations on them. At no point an approximation was necessary.

Exactly which of your symbols you say it's an approximation? Please be precise and mention exactly which ones.

[Awojobi]

let me illustrate using examples. 27^(4/3) = 81, an integer. However 27^(3/4) is an irrational number. So, all I'm saying is that the second example i.e. the irrational number 27^(3/4) appearing in any of the 2 brackets cannot be considered in the proof. However, the first example of 27^(4/3) can progress to the next stage of the proof because it is an integer. This is what I mean by the terms need to be integers before the next level of the proof can be discussed.

[dodomaze]

So 27^(3/4) = 3^(9/4) is an irrational. Here is another irrational: 3^(3/4). Now I multiply them together and I get: 3^(9/4 + 3/4) = 3^3, an integer.

The algebraic irrationals that you are using are an abstract idea, and have no obligations to reality. They are designed to be the exact root of a polynomial equation with rational coefficients, such as [tex]\sqrt 2[/tex] is the theoretical solution to the equation [tex]x^2 - 2 = 0[/tex]. No one is asking to produce a piece of wood exactly C^(z/n) in length. In Ancient Greece that would have been a problem, but mathematics has advanced since then, by the ingenious procedure of separating itself from reality, to include these fictitious unicorns that are helpful in finding solutions to theoretical problems. Beal's conjecture is a problem in pure mathematics, not in physics.

[Guest]

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

Play the lottery instead.

Good luck!

[Awojobi]

C^z - B^y = (C^z/n)^n - (B^y/n)^n = [C^z/n - B^y/n][(C^z/n)^n-1 + (C^z/n)^n-2 B^y/n + (C^z/n)^n-3 (B^y/n)^2 +... + (C^z/n)^2 (B^y/n)^n-3 + C^z/n(B^y/n)^n-2 + (B^y/n)^n-1]
The above is one of the equations in my proof. Clearly if C^z/n and B^y/n are irrational, the products of the brackets will always give a rational answer of C^z - B^y if algebraic expansion of the brackets are carried out because most of the irrational terms produced in the expansion will simply cancel each other out in pairs. However, if each of the 2 brackets is worked out separately, each will produce an approximate non integer answer to whatever degree of accuracy we require if C^z/n and B^y/n are irrational. If the product is calculated, the answer will also be a non integer answer close to the exact integer answer.
So, my proof rewrites C^z - B^y as the product [C^z/n - B^y/n][(C^z/n)^n-1 + (C^z/n)^n-2 B^y/n + (C^z/n)^n-3 (B^y/n)^2 +... + (C^z/n)^2 (B^y/n)^n-3 + C^z/n(B^y/n)^n-2 + (B^y/n)^n-1].
My proof tries to show that this product cannot equal another integer product, A^x, if C^z/n and B^y/n are irrational. This is why each term in each of the brackets has to be an integer before the proof can proceed further.

[dodomaze]

an approximate non integer answer to whatever degree of accuracy

I already covered this issue. There is no approximation. These are algebraic numbers. They can be added and multiplied just fine, and can potentially produce an integer as a product.

I'm sorry, but all this hand-waving about approximations will not convince anyone serious. If this is the center of your proof, I'm afraid you don't have one.

[Awojobi]

I believe that my reasoning is valid. The product of the non integer numerical approximations of the values of both brackets cannot ever give an integer and so only the product of integer values from both brackets should be considered.

[dodomaze]

Allow me to add one more note. I will write two statements below; please let me know if you agree of disagree with each one.

(a) 0.99999..., with infinite decimals, is exactly the same number as 1.
(b) 0.33333..., with infinite decimals, is exactly the same number as 1/3.

Would you agree of disagree with (a) or with (b)?

[Awojobi]

I disagree with both statements. A more correct statement is to say that 0.99999.... tends to 1 but is never equal to 1. Also, 0.33333... tends to 1/3 but never equals 1/3.

[dodomaze]

Any mathematician would tell you that both statements are true. It's difficult to convince people on these issues, so I'll try to bring the part of them that most affect your work.

We have been using the word "approximation", and I suspect with a somewhat vague meaning. Let me know if the following statements make sense to you.

(a) An "approximation" (at least in the sense needed for your document) is a decimal number that, no matter how many digits you take from it, will never be equal to an irrational. (Spoiler: as long as the number of digits of the approximation is finite, this is true).
(b) The reals are a larger set than the rationals; there are many number that are reals but not rationals (namely, the irrationals). (I'm being sneaky: the statement is meant to insinuate that such irrational numbers actually exist; if they didn't, the reals and the rationals would be the same set.)

From (a) it follows: an "approximation" is necessarily a rational number. (Which is why they are never equal to an irrational.) Any number with a finite decimal part, no matter how long, is a rational number. For example, 1.4142135623 is, in this sense, an approximation of [tex]\sqrt 2[/tex], equal to the ratio of two integers: 14142135623/10000000000. As you see, if the number of decimals is finite, we can always do that. To be fair, there are rationals with an infinite number of digits - but periodically repeating, like 0.3333333... , where the 3 repeats forever, or like 1/700 = 0.00142857142857... where "142857" repeats forever. But 0.333333 = 333333/1000000 is not equal to 1/3; only an approximation to 6 decimals. I can make this approximation as good as I can (by taking as many digits as I want), but, since I always take a certain number of digits and then stop, that is not equal to 1/3.

I think at the root of the problem of your proof is this idea: you believe that irrationals can only exist as (rational) approximations... and I tried to explain in (b) that they do exist, just not in the set of rationals.

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

One brief addendum about those pesky periodic decimals. Consider the following:
Let [tex]x = 0.33333 \ldots[/tex]; it's not entirely clear what is this animal with a tail of dots, but I hope to convince you that, if I multiply it by 10, I get:
[tex]10x = 3.33333 \ldots = 3 + x[/tex]
From which it's not difficult to solve for [tex]x[/tex] and obtain [tex]x = \frac 1 3[/tex]. The same can be done for 0.99999... where the "..." notation is intended to represent that multiplication by 10 does not leave you with one less decimal digit, because there is an infinity of them.

[Awojobi]

The second paragraph in your last write up backs up everything I am saying. Traditional mathematics try to use infinity as if it is some mysterious concept. As far as I see it, there is nothing mysterious about it. It just means something that carries on forever without terminating. The fact that it doesn't terminate means that in real life, and to use an example, square root of 2 written as a decimal number multiplied by square root of 2 written as a decimal number cannot ever be equal to 2, simply because square root of 2 can only be written as an approximation in real life. It cannot ever be written as an exact decimal value. My proof deals with integers, hence only integers MUST be considered and not decimal approximations.

[dodomaze]

Well, that is the problem in your proof precisely. Numbers exist regardless of their decimal representation; the latter is just an historical artifact.

I believe you are confusing two different concepts: a number, and its representation as a collection of decimals digits. The latter, if expressed to a fixed number of decimals, is a rational number. The former, the number itself, comes in many flavors and some of them are outside the rationals, hence not the same thing as a finite collection of decimal digits. The algebraic irrationals themselves are ideally perfect entities, designed, by definition, to solve polynomial equations. [tex]\sqrt 2[/tex] is, by definition, the positive solution of the equation [tex]x^2 = 2[/tex]. Approximations to it are something else; the number exists as this perfect entity regardless of any approximation. If only approximations existed, then only the rational numbers would exist.

Many concepts in mathematics are designed to be perfect, and exist only in the mind. A line in the plane, for example, has no width and does not care about the grain of the paper you draw it on. The latter notions belong only to the representations we poor humans need to communicate with each other. The line itself is intended to be perfect; its a mental concept unconcerned by reality. So is [tex]\sqrt 2[/tex], designed by definition (along with its evil twin [tex]-\sqrt 2[/tex]) to be the only two numbers that, when squared, give exactly the integer 2. By definition. Your proof is attempting to redefine existing mathematical concepts.

---

If you are so kind, could you quote the text you refer to, when you say

The second paragraph in your last write up backs up everything I am saying.

The second paragraph, as I count them from top to bottom,

We have been using the word "approximation", and I suspect with a somewhat vague meaning. Let me know if the following statements make sense to you.

does not seem to be the one. I would like to clear up the misunderstanding, so if you can quote the exact part of the text that you believe supports your position, that would help.

[Awojobi]

The paragraph I was referring to is the paragraph that starts with 'From (a)' and ends with 1/3. Concerning all these concepts you're talking about, 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. Beal's conjecture deals specifically with integers. I have written one side of the Beal equation as a product of 2 brackets, each containing algebraic terms. If I want this product to equal A^x, an integer, then none of these algebraic terms in the second set of brackets should be irrational because the sum of numbers that consist of at least one irrational number is an irrational number. Also, the difference of 2 irrational numbers in the firs set of brackets, will always be irrational. Multiplying the approximate values in the 2 brackets together can not ever produce an integer. This is why all the terms in each of the 2 brackets need to be integers, right from the onset. My proof explains why the examples of Beal's conjecture that have 3 different indices can always be manipulated such that 2 of the indices will be the same. I state this in my proof and no one has challenged this by, for example, coming up with a counterexample. There are none.