A Mathematical Emergency.

CraigD · 10-12-2008

Quote:

Originally Posted by Don Blazys

When you say that "any unambiguously defined collection of symbols" can be used to create a self consistent axiomatic and/or postulational system of logic, you are, of course, absolutely right.

But, is that what we are doing when we let a particular independent variable such as "x" represent "any real number"? In other words, are we practicing what we preach?

Yes.

Starting with a minimal set of postulates that describe the natural numbers $\mathbb{N}^*$ (or any set ${a: a>b; a, b \in \mathbb{Z}}$ ) such as the Peano postulates, we can construct the real numbers $\mathbb{R}$ . The construction of $\mathbb{R}$ is more complicated (to be precise, requires more formal statements) than constructing the other common number sets (the integers $\mathbb{Z}$ from the naturals $\mathbb{N}$ , the rationals $\mathbb{Q}$ from $\mathbb{Z}$ , or the complex numbers $\mathbb{C}$ from $\mathbb{R}$ ), but is not mysterious or untrustworthy. In my opinion, the most unsettling consequence of formal systems are Gödel's incompleteness theorems, which apply to any formal system including $\mathbb{N}^*$ and the operations of addition and multiplication. Therefore, restricting a formal system to $\mathbb{N}^*$ – that is, forbidding using it to define other number systems - gains no improvement in consistency and completeness over a formal system including $\mathbb{C}$ .

I think I can sympathize with Don’s sentiments, however. I like writing expressions using the “most primitive” symbols with which they can practically be written. Because of the ease with which $\mathbb{Z}$ and $\mathbb{Q}$ can be constructed from $\mathbb{N}^*$ , it’s practical, and to me, aesthetically pleasing to restrict expressions traditionally involving $\mathbb{Q}$ using only $\mathbb{N}^*$ and 2 operations under which $\mathbb{N}^*$ is not closed: addition and division. For example:

[1] $q = (a-b) \div c : q \in \mathbb{Q}; a,b,c \in \mathbb{N}^*$

allows a degree 1 equation such as

[2] $a +bc =d$

To be written

[3] $\frac{a_a-b_a}{c_a} +\frac{a_b-b_b}{c_b}\cdot \frac{a_c-b_c}{c_c} = \frac{a_d-b_d}{c_d}$

where: $a,b,c \in \mathbb{Q}; a_a,b_a,c_a,a_b,b_b,c_b,a_c,b_c,c_c,a_d,b_d,c_d \in \mathbb{N}^*$

with little added complexity. There are both disadvantages (if there were not, why would any use any numbers other $\mathbb{N}^*$ ?) and advantages to such the technique of this rewriting. Especially using electronic computers, the disadvantages are easily surmountable.

Where the technique appears to me to collapse – or at least exceed my current skill, intuition, and imagination - is in attempting to extend it to $\mathbb{R}$ , which is necessary for equations other than degree 1, such as

[4] $ab^c=d$

its equivalent

[5] $\log(a) +c\log(b)= \log(d)$

and

[6] $a = \ln(b)$

The problem is that I know of no finite general expression for $\mathbb{R}$ in terms of naturals, as there is for $\mathbb{Q}$ as given in [1]. Though I’ve not proven it, I’m fairly confident no such expression exists. I think many proofs, such as various proofs of the irrationality of various real numbers, could be fairly easy adapted to such a proof.

Don Blazys · 10-14-2008

To: cgaigD.

Good stuff! Another reason why I prefer to study non-negative integers is because they seem to have the most interesting properties. Concepts such as "prime number", "perfect number", "common factor" etc. simply lose their meaning when applied to numbers other than non-negative integers. Even transendentals such as (Pi) and (e) have properties that connect them to the non-negative integers in ways that are nothing less than profound.

As for Godels "incompleteness theorem", well, it also demonstrates that the "game of mathematics" will go on forever, and that there will always be new frontiers to explore, and new challenges to overcome. Thus, we really shouldn't want it any other way because guys like you and I would simply die of boredom in a "mathematical universe" where everything was well known.

Now, let's move on to my "Proof Of The Beal Conjecture". To me, the proof is simply an unavoidable consequence of "cohesive terms", and "cohesive terms", in turn, are merely an unavoidable consequence of logarithms. Here's why:

If we assume that all variables are positive integers, all three terms are co-prime and all three terms are factorable into q equal parts, then defining unity as 1=(T/T) where T>1, we have:

((T/T)a^(x/q))^q+((T/T)b^(y/q))^q=((T/T)c^(z/q))^q, ____________________________(1)

which is really not very informative, because eliminating the outermost parenthesis and simplifying results in:

a^x+b^y=c^z. ____________________________(2)

However, if we re-write (1) using "cohesive terms", then we have:

(T(a/T)^(((x/q)ln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)))^q + ____________________________(3)

(T(b/T)^(((y/q)ln(b)/(ln(T))-1)/(ln(b)/(ln(T))-1)))^q =

(T(c/T)^(((z/q)ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)))^q

where, amazingly enough, it is now possible to eliminate the outermost parenthesis in two different ways. One way does not involve the cancelled T's, and results in:

T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) + ____________________________(4)

T(b/T)^((yln(b)/(ln(T))-1)/(ln(b)/(ln(T))-1)) =

T(c/T)^((zln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)).

The other way does involve the cancelled T's, and results in:

T^q(a/T)^((xln(a)/(ln(T))-q)/(ln(a)/(ln(T))-1)) + ____________________________(5)

T^q(b/T)^((yln(b)/(ln(T))-q)/(ln(b)/(ln(T))-1))=

T^q(c/T)^((zln(c)/(ln(T))-q)/(ln(c)/(ln(T))-1)).

Now, common sense tells us that T=a, T=b and T=c must be allowable. After all, they are clearly allowable in (1), and since (1) and (3) are identities, whatever is true for (3) must also be true for (1). Moreover, since the cancelled T's can not be eliminated by "crossing them out", the only possible way to eliminate them is to let either T=a, T=b or T=c.

Now, let's say that we wanted to eliminate the cancelled T's by letting T=a, and let's assume that q=7.

Immediately, we find that in order to do that, we must first "cross out" the logarithms themselves by letting x=1 in (4), and x=q=7 in (5), thus resulting in:

T(a/T) + ____________________________(6)

T(b/T)^((yln(b)/(ln(T))-1)/(ln(b)/(ln(T))-1)) =

T(c/T)^((zln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)),

and

T^7(a/T)^7 + ____________________________(7)

T^7(b/T)^((yln(b)/(ln(T))-7)/(ln(b)/(ln(T))-1)) =

T^7(c/T)^((zln(c)/(ln(T))-7)/(ln(c)/(ln(T))-1)).

T=a is now clearly allowable, and results in:

a(a/a) + ____________________________(8)

a(b/a)^((yln(b)/(ln(a))-1)/(ln(b)/(ln(a))-1)) =

a(c/a)^((zln(c)/(ln(a))-1)/(ln(c)/(ln(a))-1)),

and

a^7(a/a)^7 + ____________________________(9)

a^7(b/a)^((yln(b)/(ln(a))-7)/(ln(b)/(ln(a))-1)) =

a^7(c/a)^((zln(c)/(ln(a))-7)/(ln(c)/(ln(a))-1)).

Here we find that neither (8), nor (9) allows us to express ordinary
a^2+b^2=c^2 "Pythagorean triples" because we had to let x=1 and x=q=7 in order to eliminate the logarithms preventing T=a. Thus it must be the case that x=q=2, because clearly, "Pythagorean triples" exist, and our equations must be able to reflect them.

This proves both the BC and FLT and renders our equations perfectly in line with the truth, not to mention all empirical data, and observed results.

Logarithms don't lie, and as you can see, the very existence of "cohesive terms" automatically shows (and proves) that the BC and FLT are true, because all we really did was substitute them for the poorly defined "non-cohesive terms".

Thus, if some journal publishes even one teensy little "cohesive term", then the truth (and proof) of the BC and FLT will become apparent to anyone, the moment they use it to state the those problems.

Don.

CraigD · 10-14-2008

Quote:

Originally Posted by Don Blazys

Now, let's move on to my "Proof Of The Beal Conjecture"...

Between

[1] $a^x+b^y=c^z$

and

[2] $\left(T\left(\frac{a}{T}\right)^{E_a}\right)^q +\left(T\left(\frac{b}{T}\right)^{E_b}\right)^q = \left(T\left(\frac{c}{T}\right)^{E_c}\right)^q$

there’s an intermediate step not shown

[1.5] $\left(a^x\right)^q+\left(b^y\right)^q= \left(c^z\right)^q$

which is incorrect, because, except for special cases such as $q=1$ ,
$A+B=C \not\rightarrow A^q+B^q=C^q$
eg: $1+2=3$ but $1^2 +2^2 \not= 3^2$ .

Even if this error is avoided, as it is in your A Simple Proof of the Beal Conjecture, worse trouble arises when you assert

Quote:

Originally Posted by Don Blazys

Now, common sense tells us that T=a, T=b and T=c must be allowable.

Because all of these proofs depend on the identity

$a^x = T^n \left(\frac{a}{T}\right)^{\left( \frac{x\log_Ta -n}{\log_Ta -1}\right)}$

, you’re essentially asserting that “common sense tells us” division by zero is allowed in ordinary arithmetic of real numbers. Division by zero in ordinary arithmetic, however, is not allowed. Also (it goes almost without saying) "common sense tells us" is not an allowed theorem in a rigerous proof.

I find another puzzling and unsupported assertion in your proof, but given the severity of the preceding, will avoid discussing it in this post.

PS: If I failed to make it clear in post #15 and later posts, note that $\log_Ta$ is a more compact way to write $\frac{\ln(a)}{\ln(T)}$ , and avoids the unnecessary introduction of a base for the logarithm function, such as e=2.718... as in the case of the natural logarithm, $\ln$ .

Don Blazys · 10-14-2008

To: CraigD.

(a^x)^q+(b^y)^q=(c^x)^q.

does not equal my original equation:

((T/T)a^(x/q))^q+((T/T)b^(y/q))^q=((T/T)c^(z/q))^q,

so it can't be an "intermediate step not shown".

Now, here is the most simple version of my proof.

__________________________________________________ __________________________________________________

"The Beal Conjecture In A Nutshell". (Hypography Version) By: Don Blazys.

For positive integer variables, if:

(T/T)a^x+(T/T)b^y=(T/T)c^z=

T(c/T)^((zln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1))

and

((T/T)a^(x/2))^2+((T/T)b^(y/2))^2=((T/T)c^(z/2))^2=

(T(c/T)^(((z/2)ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)))^2,

and a, b and c do not have a common factor, then x,y and z are not all greater than 2.

Proof: The poorly defined terms that do not involve logarithms tell us that T=c is always allowable while the well defined terms that do involve logarithms tell us that T=c is allowable if and only if we first eliminate ("cross out") those logarithms by letting x=1 and x=2 respectively. Thus, in order to make all of the terms consistent at T=c, we must let x=1 and x=2 respectively, which proves the conjecture.

__________________________________________________ __________________________________________________

As you can see, the observation that T=c must be allowable in no way suggests that "division by zero" must also be allowable. Indeed, the proof works precisely because "division by zero" is not allowable!

CraigD · 10-15-2008

Quote:

Originally Posted by Don Blazys

(a^x)^q+(b^y)^q=(c^x)^q.

does not equal my original equation …

Indeed, I can’t seem to reliably read your text without first transcribing it into a usual rendering, and keep making mistakes.

As I’ve limited reading time, I’ll either need to stop reading it in detail, or take more time between replies, or, Don, you could post either images of handwriting, or transcribe it yourself or with an assistant’s help into LaTeX.

Quote:

Originally Posted by Don Blazys

Now, here is the most simple version of my proof. …

I don’t follow at all.

Take a simple example, such as a=6,b=3,c=3,x=3,y=3,z=5,T=2. Both of the “ifs” in the assertion are true, but the “then” is not. The assertion is disproven by example.

Don, you’ve omitted the condition that a, b and c must not have any common factors other than 1.

Quote:

Originally Posted by Don Blazys

As you can see, the observation that T=c must be allowable in no way suggests that "division by zero" must also be allowable. Indeed, the proof works precisely because "division by zero" is not allowable!

If $T=c$ , then

$T^n \left(\frac{c}{T}\right)^{\left( \frac{ z\frac{\ln(c)}{\ln(T)} -n}{\frac{\ln(c)}{\ln(T)} -1}\right)}$

may be rewritten

$c^n \left(\frac{c}{c}\right)^{\left( \frac{ z \frac{\ln(c)}{\ln(c)} -n}{\frac{\ln(c)}{\ln(c)} -1}\right) }$

and simplifies to

$c^n 1^{\left( \frac{z -n}{1 -1}\right)}$

$c^n 1^{\left( \frac{z -n}{0}\right)}$

How, then, is allowing $T=c$ not also allowing division by zero?

Don Blazys · 10-16-2008

To: CraigD.

Since the first two versions of the proof on my website do state that a, b and c are to be assumed co-prime, I figured that I could safely omit making that statement yet again when putting the proof "in a nutshell". Anyway, if you take a look at it now, then you will see that I put it back in.

Now your "example":

a=6, b=3, c=3, x=3, y=3, z=5, T=2.

works out to:

6^3+3^3=3^5,

where the terms are not co-prime because they contain the common factor 27.

However, if we divide out that common factor, then the equation becomes:

2^3+1^3=3^2,

and applying these particular values to the variables in the proof, we can see that it is the particular value T=3, (not T=2) that would, at T=c, necessitate the exponent z=2 in the factored case.

You see, since the proof makes use of variables, it demonstrates that for any similar numerical example that is rendered co-prime, there will always exist some value T, such that letting either:

T=a, T=b or T=c,

will automatically necessitate either:

x={1,2}, y={1,2} or z={1,2}.

Without the "cohesive terms" retaining the cancelled T's, our symbolic representations of co-prime terms would appear as:

a^x+b^y=c^z,

and would therefore not reflect reality.

In your other argument, you are using T^n, which is not contained in any version of my proof, because, (as explained in post #42) it would imply a general factorability of all terms into n>2 equal parts, which is clearly impossible.

Therefore, let us re-state your argument using the actual terms in my proof.

If T=c, then:

T(c/T)^((zln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)) _______________________(1)

and:

(T(c/T)^(((z/2)ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)))^2 _________________(2)

may be re-written as:

c(c/c)^((zln(c)/(ln(c))-1)/(ln(c)/(ln(c))-1)) ________________________(3)

and:

(c(c/c)^(((z/2)ln(c)/(ln(c))-1)/(ln(c)/(ln(c))-1)))^2 ________________(4)

which simplifies to:

c(1)^((z-1)/(o)), _____________________________________________(5)

and:

(c(1)^(((z/2)-1)/(0)))^2. ______________________________________(6)

However, these terms don't actually "allow division by zero" as you believe they do because division by zero simply can't be allowed and must therefore be avoided.

Now the only way to avoid division by zero in the above equations is to let:

z=1 in (1), and z=2 in (2), which results in:

T(c/T), __________________________________________________ _____(7)

and:

(T(c/T))^2 __________________________________________________ _(8)

We can now let T=c without encountering any division by zero, which results in:

c(c/c)=c __________________________________________________ ___(9)

and

(c(c/c))^2=c^2. _____________________________________________(10)

So you see, T=c must be allowable, and is allowable, (and in no way suggests that division by zero is allowable) provided that we let z=1 and z=2 before we let T=c.

The fact that avoiding division by zero thus restricts the values of z to either z=1 or z=2 proves both the BC and FLT.

Don.

Qfwfq · 10-16-2008

Would it not be more helpful to Craig to provide him with a proper link to your website?

CraigD · 10-16-2008

Quote:

Originally Posted by Qfwfq

Would it not be more helpful to Craig to provide him with a proper link to your website?

Alas, no. I’ve seen Don’s website (early in the thread, I provided a link to it, and later, to a specific page), but all of its pages appear to use the same unformatted text he’s used in posts at hypography.

CraigD · 10-16-2008

Quote:

Originally Posted by Don Blazys

…
c(1)^((z-1)/(o)), _____________________________________________(5)

and:

(c(1)^(((z/2)-1)/(0)))^2. ______________________________________(6)

However, these terms don't actually "allow division by zero" as you believe they do because division by zero simply can't be allowed and must therefore be avoided.

Now the only way to avoid division by zero in the above equations is to let:

z=1 in (1), and z=2 in (2), which results in...

Requiring z=1 or z=2 doesn’t avoid division by zero in (3) through (6). (5) becomes

$c 1^{\frac{0}{0}}$

which continues to contain a division by zero.

You appear to be persisting in the claim you made in post #27 that division by zero is permitted (not indeterminate), provided what is being divided (the numerator) is also zero. This simply isn’t true for ordinary arithmetic.

In a nutshell, the whole gist of your proofs consists of introducing a term, $T$ to the equation

$a^x +b^y =c^z$

to transforming it to

$T\left(\frac{a}{T}\right)^{\left( \frac{x\log_Ta -1}{\log_Ta -1}\right)} +T\left(\frac{b}{T}\right)^{\left( \frac{y\log_Tb -1}{\log_Tb -1}\right)} = T\left(\frac{c}{T}\right)^{\left( \frac{z\log_Tc -1}{\log_Tc -1}\right)}$

where, in ordinary algebra, we are required to introduce the condition $T>0$ and $T \not= a, b, c$ . Here, you insist that “common sense” prohibits stating the $T \not= a, b, c$ . You then go on to argue that, because $T = a, b, c$ must be allowed, division by zero can only be avoided by assuring that what is divided by zero is also zero.

Your claim is simply and well known to be wrong. Your proofs are wrong.

As you’ve sent your papers to a referred journal, I expect you’ll be informed of the same thing from them, unless a screener refuses to read them because of poor presentation quality. The best advice I can give toward assuring that your submissions are actually read, is to improve their presentation quality. A straightforward way to do this is to obtain an article in the journal to which you’ve submitted your paper, then transcribe your paper to look like it. If you are unable or unwilling to do this electronically, write carefully with pen on paper, and submit it via regular mail.

You are also well advised to use hypography’s [math] code in your posts, which will allow you to render your math expressions via LaTeX.

It’s also a good idea in math to be formal. If you believe “cohesive terms” have value as a mathematically technique, describe the technique formally, without attempting to prove any famous conjecture. The term “cohesive term”, with your intended meaning, appears to exist nowhere but in your own writing. Without a formal description, readers can only understand you by attempting to guess what you mean by the term.