Mathematical Consistency

[Don Blazys]

Everyone seems to have a different, and rather vague explanation,

but this is a simple yes or no question.

 

_______________________________________________________________________________________

 

Given the identity,

 

[math] Ta^x = (Ta)^{\left( \frac{\frac{x*\ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1} \right)} ,[/math]

 

can we let [math]T=1[/math] and thereby multiply [math]a^x[/math] by [math]1[/math]

 

if we cover the right hand side with our fingers,

 

so that all we can see is the left hand side?

 

_______________________________________________________________________________________

 

I say "no".

 

Please, do as I did,

and just answer yes or no without any commentary.

 

Don.

[sanctus]

But Don everyone replied to that question already. The problem is your extrapolation of the answer. You say that if it is "no" then there is a problem with the identity element (i.e. 1), we all say that is a wrong conclusion. The whole thing is that we should write

[math]

Ta^x = (Ta)^{\left( \frac{\frac{x*\ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1} \right)} \forall T\neq 1 ,

[/math]

 

then your question is implicitely replied.

 

So I did like you first comment than reply to your question with no comment (or almost):

 

"NO" because then the equality is not valid.

[Qfwfq]

just answer yes or no without any commentary.

I can't, because I don't know, you need to ask Schrödinger and Wigner's friend.

[Don Blazys]

Quoting Sanctus:

[math] Ta^x = (Ta)^{\left( \frac{\frac{x*\ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1} \right)} \forall T\neq 1 , [/math]

 

Thanks Sanctus!

 

If [math]\forall T, T\neq1,[/math] then according to that perfectly true and perfectly defined identity,

we quite obviously can't let [math]T=1[/math] and thereby multiply [math]a^x[/math] by [math]1.[/math]

 

In fact, according to that perfectly true and perfectly defined identity,

the only way that we can "multiply [math]a^x[/math] by [math]1[/math]" is if we divide both sides by [math]T,[/math]

which gives us:

 

[math] \left(\frac{T}{T}\right)a^x = T\left(\frac{a}{T}\right)^{\left( \frac{\frac{x*\ln(a)}{\ln(T)}-1}{\frac{\ln(a)}{\ln(T)}-1} \right)} .[/math]

 

Don.

[modest]

I can't, because I don't know, you need to ask Schrödinger and Wigner's friend.

 

Until you look under your finger, the RHS both exists and does not exist?

 

:hihi:

 

In fact, according to that perfectly true and perfectly defined identity,

the only way that we can "multiply [math]a^x[/math] by [math]1[/math]" is if we divide both sides by [math]T,[/math]

which gives us:

 

[math] \left(\frac{T}{T}\right)a^x = T\left(\frac{a}{T}\right)^{\left( \frac{\frac{x*\ln(a)}{\ln(T)}-1}{\frac{\ln(a)}{\ln(T)}-1} \right)} .[/math]

 

Don.

 

The only way to multiply [math]a^x[/math] by one in the first identity is to use a second, different, identity? It's like saying "the only way an apple tree will reach 100 meters height is if it is a redwood tree". it doesn't make sense.

 

You can multiply [math]a^x[/math] by [math]T=1[/math] in this identity:

 

[math] Ta^x = (Ta)^{\left( x \ln(a) + \ln(T) \over \ln(a) + \ln(T) \right)} [/math]

 

and many others. None of them are equivalent to your first identity.

 

~modest

[Kharakov]

It's like saying "the only way an apple tree will reach 100 meters height is if it is a redwood tree". it doesn't make sense.

 

It's totally off topic, so I'll just interject quickly and drop it, but I'm not positive that you can't graft an apple branch to a redwood and have your apple tree reach 100 meters (although nutrients would have a different, perhaps unusable, profile for the apple graft). Of course, falling apples would be a problem in this case, so most people probably refrain from grafting apple scions to redwood branches...

[Qfwfq]

Goodness, Newton didn't even know what a close call he had!

 

Until you look under your finger, the RHS both exists and does not exist?

It might or might not be defined for [math]T=1[/math]. If your finger also covers the equal sign, it might or might not be equal to the LHS.

 

Don, how about an alternative to "the Don Blazys experiment" you proposed:

 

First you show the mathy only the one card (with the LHS on it) and ask the question.

 

Then you place the other card beside it and repeat the question.

 

When you ask the question the first time, should the mathy be clairvoyant in order to answer correctly?

[Don Blazys]

Obviously, mathematics should not require that its users and practitioners be clairvoyant.

 

Unfortunately, in its present state, clairvoyance is exactly

what mathematics requires of us, because in its present state,

mathematics remains inconsistent in its notions regarding unity.

 

As a teacher in the high school where I work once pointd out,

a somewhat similar situation did exist in geometry for many hundreds of years,

so there is a sort of historical precedent for what we are now facing.

 

You see, just as several centuries ago, Euclids fifth axiom (parallel postulate)

severely limited the scope, effectiveness and usefullness of geometry,

notions involving multiplication by unity such as "the unit coefficient"

and "the identity element of multiplication" now severely limit

the scope, effectiveness and usefullness of number theory, set theory,

algebra, calculus and all mathematical branches that stem from them!

 

And just as the works of Bolyai, Lobachevsky, Gauss, Riemann and Einstein

demonstrated that notions such as "the straight line" and "parallel lines"

have no actual basis in reality, our freedom to express the newly discovered identity:

 

[math] Ta^x = (Ta)^{\left( \frac{\frac{x*\ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1} \right)} [/math]

 

clearly shows that all notions predicated on "multiplication by unity"

also have no actual basis or foundation in either reality, truth or logic.

 

I can show that a perfectly consistent (and superior) calculus can be developed

that maintains variables which are perfectly defined by their own unique domains

and therefore does not allow any form of direct multiplication by unity whatsoever.

 

Thus, we are on the verge of significant advancements in mathematics

and thus it is the duty and moral obligation of every intelligent person

to publicize and disseminate the "Don Blazys experiment" throughout the world,

for it is only when the math community is sufficiently embarrased

that they will stop teaching our young people to automatically (and often erroneously) write:

 

[math]a^x[/math]

 

instead of:

 

[math] T\left(\frac{a}{T}\right)^{\left( \frac{\frac{x*\ln(a)}{\ln(T)}-1}{\frac{\ln(a)}{\ln(T)}-1} \right)} [/math]

 

when they encounter:

 

[math] \left(\frac{T}{T}\right)a^x .[/math]

 

At the very, very least, our young people should know and understand

that eliminating cancelled variables is not (as they are now being taught)

something to be done automatically, but rather constitutes an elimination of information

that may or may not be necessary in order to solve some particular problem,

such as "Beal's Conjecture" or "Fermat's Last Theorem".

 

Don.

[sanctus]

You just don't want to get it it seems to me. You use an equality outside its domain of validity to say that the identity element is inconsistent!

Please comment on this :-)

[Qfwfq]

Don, you got the story of Euclid's fifth a bit confused there, it was a more general breakthrough than only geometry. Despite this, I don't get your conclusions.

 

...when the math community is sufficiently embarrased

that they will stop teaching our young people to automatically (and often erroneously) write:

 

[math]a^x[/math]

 

instead of:

 

[math] T\left(\frac{a}{T}\right)^{\left( \frac{\frac{x*\ln(a)}{\ln(T)}-1}{\frac{\ln(a)}{\ln(T)}-1} \right)} [/math]

 

when they encounter:

 

[math] \left(\frac{T}{T}\right)a^x .[/math]

The third expression is meaningful [imath]\forall T\ne0[/imath] which is for more values of [imath]T[/imath] than the second one, so there are values in which one expression is defined and not the other. It is all a matter of what is more useful for solving a particular problem, it strikes me odd you saying folks should do that and not the other.

 

At the very, very least, our young people should know and understand that eliminating cancelled variables is not (as they are now being taught) something to be done automatically,

Concerning the third expression up there, I was always taught that we may replace it with the first but it might be necessary to keep in mind the difference in domain of validity. I would expect young people are still being taught like this.

[Turtle]

...

You see, just as several centuries ago, Euclids fifth axiom (parallel postulate)

severely limited the scope, effectiveness and usefullness of geometry,

notions involving multiplication by unity such as "the unit coefficient"

and "the identity element of multiplication" now severely limit

the scope, effectiveness and usefullness of number theory, set theory,

algebra, calculus and all mathematical branches that stem from them!

...

Don.

 

i think you have have got yourself between gödel's hammer(s) and a hard place don. now there's severe limitation!! :hammer:

Gödel's incompleteness theorems

(boldenation mine )

...In mathematical logic, a theory is a set of sentences expressed in a formal language. Some statements in a theory are included without proof (these are the axioms of the theory), and others (the theorems) are included because they are implied by the axioms.

 

Because statements of a formal theory are written in symbolic form, it is possible to mechanically verify that a formal proof from a finite set of axioms is valid. This task, known as automatic proof verification, is closely related to automated theorem proving. The difference is that instead of constructing a new proof, the proof verifier simply checks that a provided formal proof (or, in some cases, instructions that can be followed to create a formal proof) is correct. This process is not merely hypothetical; systems such as Isabelle are used today to formalize proofs and then check their validity.

 

Many theories of interest include an infinite set of axioms, however. To verify a formal proof when the set of axioms is infinite, it must be possible to determine whether a statement that is claimed to be an axiom is actually an axiom. This issue arises in first order theories of arithmetic, such as Peano arithmetic, because the principle of mathematical induction is expressed as an infinite set of axioms (an axiom schema).

 

A formal theory is said to be effectively generated if its set of axioms is a recursively enumerable set. This means that there is a computer program that, in principle, could enumerate all the axioms of the theory without listing any statements that are not axioms. This is equivalent to the existence of a program to enumerate all the theorems of the theory without enumerating any statements that are not theorems. Examples of effectively generated theories with infinite sets of axioms include Peano arithmetic and Zermelo–Fraenkel set theory.

 

In choosing a set of axioms, one goal is to be able to prove as many correct results as possible, without proving any incorrect results. A set of axioms is complete if, for any statement in the axioms' language, either that statement or its negation is provable from the axioms. A set of axioms is (simply) consistent if there is no statement such that both the statement and its negation are provable from the axioms. In the standard system of first-order logic, an inconsistent set of axioms will prove every statement in its language (this is sometimes called the principle of explosion), and is thus automatically complete. A set of axioms that is both complete and consistent, however, proves a maximal set of non-contradictory theorems. Gödel's incompleteness theorems show that in certain cases it is not possible to obtain an effectively generated, complete, consistent theory....

First incompleteness theorem

Gödel's first incompleteness theorem states that:

 

Any effectively generated theory capable of expressing elementary arithmetic cannot be both consistent and complete. In particular, for any consistent, effectively generated formal theory that proves certain basic arithmetic truths, there is an arithmetical statement that is true,[1] but not provable in the theory (Kleene 1967, p. 250).

The true but unprovable statement referred to by the theorem is often referred to as “the Gödel sentence” for the theory. It is not unique; there are infinitely many statements in the language of the theory that share the property of being true but unprovable. ...

[Don Blazys]

Quoting Sanctus:

You just don't want to get it it seems to me.

You use an equality outside its domain of validity to say that the identity element is inconsistent!

Please comment on this :-)

 

The title of this thread is "Mathematical Consistency" where the word

 

consistent is synonymous with the words "constant" and "permanent".

 

Now, if in the term: [math] Ta^x [/math] we permanently define the coefficient [math] T [/math] as: [math] T\in\{0,1,2,3...\} [/math],

 

then we can't ever divide that term by [math] T [/math] because the result [math] \left(\frac{T}{T}\right)a^x [/math] would require [math] T\in\{1,2,3,4...\} .[/math]

 

Likewise, if in the term: [math] Ta^x [/math] we permanently define [math] T [/math] as: [math] T\in\{1,2,3,4...\} [/math],

 

then we can't ever derive and express the identity: [math] Ta^x = (Ta)^{\left( \frac{\frac{x*\ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1} \right)} [/math] because that would require [math] T\in\{2,3,4,5...\}. [/math]

 

Thus, we can not consistently (permanently) define the coefficient [math] T [/math] as allowing a "unit coefficient" or "identity element"

 

without giving up our precious freedom to derive the perfectly defined term: [math] (Ta)^{\left( \frac{\frac{x*\ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1} \right)} [/math] from [math] Ta^x .[/math]

 

Simply put, if our freedom to derive [math] (Ta)^{\left( \frac{\frac{x*\ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1} \right)} [/math] from [math] Ta^x [/math] is a permanent, God given freedom,

 

then there is no way that we can ever even think of allowing "multiplication by unity", which is only a fairy tale anyway!

 

Don.

[Don Blazys]

Quoting Qfwfq:

I was always taught that we may replace it with the first but

it might be necessary to keep in mind the difference in domain of validity.

I would expect young people are still being taught like this.

 

 

 

Our young people should be taught that replacing [math] \left(\frac{T}{T}\right)a^x [/math] with [math] a^x [/math]

 

might be okay if you are using math to bake cookies, but that replacing [math] \left(\frac{T}{T}\right)a^x [/math] with [math] T\left(\frac{a}{T}\right)^{\left( \frac{\frac{x*\ln(a)}{\ln(T)}-1}{\frac{\ln(a)}{\ln(T)}-1} \right)} [/math]

 

will allow you to solve hitherto intractable problems such as "Beal's Conjecture" and "Fermat's Last Theorem"!

 

Think about it... as I write this, millions of poor unsuspecting young people are being "taught" that

 

it is somehow "okay" to write [math] Ta^x [/math] where [math] T\in\{0,1,2,3...\} [/math], [math] a\in\{0,1,2,3...\} [/math] and [math] x\in\{0,1,2,3...\} [/math]

 

while in reality, defining those three different variables using the same domain is kind of like defining eagles, lions and sharks as as mammals!

 

Worse yet, those so called domains aren't even stable, and are in fact, downright schizophrenic for if we now divide [math] Ta^x [/math] by [math] T [/math],

 

then the result [math] \left(\frac{T}{T}\right)a^x [/math] requires that we now change the domains to: [math] T\in\{1,2,3,4...\} [/math], [math] a\in\{0,1,2,3...\} [/math] and [math] x\in\{0,1,2,3...\} [/math].

 

Now, compare the "cookie term" [math] Ta^x [/math] to the "Blazys term" [math] (Ta)^{\left( \frac{\frac{x*\ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1} \right)} [/math] whose variables are all perfectly defined by their own unique domains:

 

[math] T\in\{2,3,4, 5...\} [/math], [math] a\in\{1,2,3,4 ...\} [/math] and [math] x\in\{0,1,2,3...\} [/math] and note that dividing [math] (Ta)^{\left( \frac{\frac{x*\ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1} \right)} [/math] by [math] T [/math]

 

results in: [math] T\left(\frac{a}{T}\right)^{\left( \frac{\frac{x*\ln(a)}{\ln(T)}-1}{\frac{\ln(a)}{\ln(T)}-1} \right)} [/math] where the domains remain stable (not schizophrenic) and you tell me which is the better, more informative,

 

and more "cohesive" algebraic term... [math] Ta^x [/math] or [math] (Ta)^{\left( \frac{\frac{x*\ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1} \right)} [/math] ?

 

Don

[Don Blazys]

Quoting Turtle:

i think you have have got yourself between gödel's hammer(s) and a hard place don.

now there's severe limitation!!

 

I agree.

 

The basic (one exponentiation, one multiplication) algebraic terms [math] Ta^x [/math] and [math] (Ta)^{\left( \frac{\frac{x*\ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1} \right)} [/math]

 

demonstrate that "the larger the domains of the variables, the less defined those variables will be" and vice versa.

 

This is somewhat analagous to Gödel's "the more complete the postulational system of logic, the less consistent that system will be",

 

and to Heisenberg's "the more certain the position of the particle, the less certain that particle's momentum will be".

 

 

Those were some heavy cats, but even before them, Gauss once remarked that "mathematics is largely a point of view".

 

Don.

[Qfwfq]

...if you are using math to bake cookies

That's what you seem to be doing mate.

[CraigD]

Let me try to summarize Don’s argument in this and other threads in which he’s discussed the equation

 

[math] T a^x= \left(T a\right)^{\frac{\frac{x \ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1}} [/math]

 

as best I’ve been able to understand it.

 

Because [math]\frac1{\log T}[/math] is undefined for [math]\log T = 0[/math], [math]T = 1[/math], Don argues that expressions of the form [math]T a^x[/math], (a general term for single argument polynomial functions, which are widely used in math) should not be permitted when [math]T=1[/math]. Her further argues, with great passion, that by permitting students to consider [math]T a^x[/math] when [math]T=1[/math], teachers have done a terrible disservice to them, retarding the progress of math and science in general. He claims that “the math community” is aware of this, find it “embarrassing and humiliating”, and conceal awareness of it from “math majors” through some sort of indoctrination / brain washing, resulting in them being “defrauded, duped and swindled”.

 

I’m certain Don is wrong in all these arguments and claims.

 

First, mathematically, let’s consider what the “excluded domain” – the values that [math]T[/math] and [math]a[/math] cannot have for the equation to be defined – means.

 

It’s convenient to describe an equation like this – an identity, that is, an equation that is true for many values in its domain - as a equivalence relation between functions with the same arguments, eg:

 

[math]f_1(T,a,x) = T a^x[/math]

 

[math]f_2(T,a,x) = \left(T a\right)^{\frac{\frac{x \ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1}}[/math]

 

[math]f_1(T,a,x) = f_2(T,a,x)[/math]

 

In short, if the equation is true, [math]f_1[/math] and [math]f_2[/math] are two different algorithms that return the same number given arguments with the same value.

 

If one of the function is not defined for some argument values for which the other is, as is the case with these two, those arguments are outside of (excluded from) the equation’s domain. This equation’s excluded domain includes [math]a \le 0[/math], [math]T \le 0[/math], and [math]T = 1[/math], all due to undefined values of [math]f_2[/math].

 

What this means is that [math]f_1[/math] can be used to return a value for arguments in these ranges, but [math]f_2[/math] cannot. It doesn’t mean that [math]f_1[/math] can't be used to return a value for arguments in these ranges, as Don appears to claim.

 

An intuitive meaning of this is more obvious with this substitution:

 

[math]f_2(T,a,x) = (T a)^c[/math]

 

[math]T a^x = (T a)^c[/math]

 

We can see intuitively that if [math]T a = 1[/math] and [math]T \not= a[/math], [math]f_1[/math] “works” – returns many values – while, even if c is defined for those values (which is not) [math]f_2[/math] can return only 1.

 

Notice that the excluded domain in this case is not [math]T=1[/math], but [math]T a = 1[/math]. Reading modest’s derivation of [math]f_2[/math] from [math]f_1[/math] in , we can see this explained in

 

(4)... [math]Ta^x = (Ta)^{\left( \ln(a^x) + \ln(T) \over \ln(a) + \ln(T) \right)} [/math]

 

An important point here is that finding functions that are equal to one for some but not all of their domains does not change the domain of either function alone. So, despite

 

[math](Ta)^{\left( \ln(a^x) + \ln(T) \over \ln(a) + \ln(T) \right)} [/math]

 

being undefined for all [math]T[/math], [math]a[/math] such that [math]T a = 1[/math],

 

[math]Ta^x[/math] is not undefined for these values.

 

Last, let’s consider the claim that the math community – teachers and professional mathematicians and mathematical scientist, presumably – are engaged in a conspiracy to conceal a key truth from math majors or other students. As a former math major, tutor, and college professor, I know this simply isn’t true. Don’s equation isn’t commonly known or discussed by mathematicians and students. It’s not embarrassing or humiliating to anyone I’ve ever known. It does not call into question any conventional math, including the concept of multiplicative identity.

Edited November 30, 2010 by CraigD
fixed "can't" vs. "can" typo

[Don Blazys]

First you say...

 

Quoting Craig D:

What this means is that [math] f_1 [/math] can be used to return a value for arguments in these ranges.

 

and right after that you say...

 

Quoting Craig D:

It doesn’t mean that [math] f_1 [/math] can be used to return a value for arguments in these ranges.

 

so, which is it?

 

Quoting Craig D:

We can see intuitively that if [math] T a = 1 [/math] and [math] T \not= a [/math] , (then) [math] f_1[/math]“works”.

 

We are dealing with non-negative integer variables, so how can [math] T a = 1 [/math] if [math] T \not= a [/math]?

 

Quoting Craig D:

Reading modest’s derivation of [math] f_2 [/math] from [math] f_1 [/math] in ,

we can see this explained in (4)...[math] Ta^x = (Ta)^{\left( \ln(a^x) + \ln(T) \over \ln(a) + \ln(T) \right)} [/math]

 

That identity allows [math]T=1[/math] and is not the identity being discussed in this thread.

 

Quoting Craig D:

Let’s consider the claim that the math community –

teachers and professional mathematicians and mathematical scientist, presumably –

are engaged in a conspiracy to conceal a key truth from math majors or other students.

Don’s equation isn’t commonly known or discussed by mathematicians and students.

It’s not embarrassing or humiliating to anyone I’ve ever known.

 

Well, I really don't think that there is some kind of "conspiracy" going on!

However, each and every math professor that I personally performed the "Don Blazys experiment" on

did, in fact, turn red with embarrassment and does prefer not to talk about it!

Apparently, it is extraordinarily difficult for most professors of mathematics to admit that

according to their so called "logic" and "reasoning", the possibility of allowing a "unit coefficient"

depends entirely on the location of a 3 by 5 card!

 

Quoting Craig D:

Don argues that expressions of the form [math]Ta^x[/math] should not be permitted when [math]T=1[/math].

 

That's not true!

 

I never ever said that!

 

All I am asking is this...

 

Given the identity: [math] T a^x= \left(T a\right)^{\frac{\frac{x \ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1}} ,[/math]

 

can we let [math]T=1[/math] and thereby multiply [math]a^x[/math] by [math]1[/math]

if we cover the right hand side with our fingers so that all we can see is [math]Ta^x[/math]?

 

In other words, can we let [math]T=1[/math] and thereby have a "unit coefficient"

or an "identity element" if we can't see the right hand side of the above identity?

 

This is a very simple yes or no question that doesn't require any commentary whatsoever,

yet, no one dares give me that simple yes or no answer because they are too embarrassed !.

 

Well, I'm not!

 

My simple and straight answer is a resounding "no!".

 

How about you?

 

What is your simple and straight answer?

 

(Please do as I did and answer yes or no only.)

 

Don.