Mathematical Consistency

[sanctus]

Don, thing is that if I say "no" you will use it to draw your false conclusions, if I say "yes" then most likely you think I did not get your point ;-).

 

So making 2 questions out of your one, but keeping the only yes/no answers.

 

Can we let T=1 in[math] Ta^x[/math]? Yes

 

Can we let T=1 in [math]\left(T a\right)^{\frac{\frac{x \ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1}}[/math]? No

 

 

And a bonus yes/no answer:

 

Is there a contradiction/problem? No

[Qfwfq]

so, which is it?

Don, I believe that was just an absent minded typo on Craig's part. I believe he mean "can't" in the second case.

[Don Blazys]

Quoting Sanctus.

Can we let T=1 in [math] Ta^x [/math] ? Yes

 

Not if there is a "post it" with [math] =\left(T a\right)^{\frac{\frac{x \ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1}} [/math] written on it,

stuck to your computer screen, or to the back of your computer, or on a wall in the next room,

or even on some refrigerator in some house in some far distant land !

 

Don.

[sanctus]

That is exactly what I meant, now you use my answer as a support to your false conclusion, while it is no support...

 

But what you should criticize is the equal sign, that is what is not always/permantently valid...

 

So since the right-hand side is indefinite for T=1 then the identity is just not valid (means the equal sign is not right but should be a [math]\neq[/math]), but you STILL can set T=1 on the left-handside provided you write a defined term on the right hand side...

[Don Blazys]

But, I can't change the [math]=[/math] sign to [math] \neq[/math] if that "post it" is on someone else's refrigerator in some far distant land!

 

Anyway, what do you say are the permanent (and therefore consistent)

domains of the variables [math]T, a, x[/math] in the term [math]Ta^x[/math] ?

 

Don.

[sanctus]

That is why you shou should write your famous identity as (see one of my odler posts in this thread) [math]

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

[/math]

 

then problem solved, even if the post-it is on another planet ;-)

[Don Blazys]

We agree!

 

The symbols [math]\forall T\neq 1[/math] are a wonderful reminder that [math]T\neq 1[/math].

 

Now, if I write:

 

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

 

on a "post it", cut it in two, and mail you only the part that says: [math]Ta^x[/math],

 

then will we still agree that [math]T\neq 1[/math]?

 

Or will you then look at your part, (that says only [math]Ta^x[/math])

 

and assume that [math]T=1[/math] is now allowable?

 

Don.

[sanctus]

Don, an identity is only an identity as long as you have an equal sign squeezed between 2 things. You want to use the 2 things separately...

 

So [math]T a^x[/math] perfectly allows for T=1. But if you allow T=1 you just never can derive the right handside.

[CraigD]

Now, if I write:

 

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

 

on a "post it", cut it in two, …

We all agree that this expression (identity) is true. What its written on and any cutting up an mailing of it is just a matter of paper, computer memory, etc, and unimportant to the identity.

 

I’m seriously unable to grasp your point, Don. What’s so special about

 

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

 

? How is it more important, than, say

 

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

 

, or any other of the infinite number of expression we can derived from [math]T a^x[/math], for which any expression involving [math]T[/math], [math]a[/math], or [math]x[/math] can have any value for which the derived identity is undefined?

[Qfwfq]

And now....

 

Here comes the Devil's Advocate!

/forums/images/smilies/devilsign.gif

 

What’s so special about

 

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

 

? How is it more important, than, say

 

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

Actually, the significance of the [imath]a=1[/imath] case in the second identity has a more obvious cause, just like the case of [imath]Ta=1[/imath] in the first: there is no finite exponent to which [imath]1[/imath] could be raised, giving a result other than [imath]1[/imath]. This means that the trouble in the first for [imath]T=1[/imath] is due to the exponent alone; indeed it does not occur in the similar case:

 

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

 

In fact the one above can be obtained from this one through simple steps on the exponent, that involves dividing numerator and denominator by [imath]\ln T[/imath].

 

Now I'll quit being silly and propose another identity:

 

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

 

Which requires [imath]Ta\neq n[/imath]. Note that [imath]n[/imath] needn't be [imath]1[/imath], it could also be [imath]2[/imath], [imath]3[/imath] or whatever.

Edited December 9, 2010 by Qfwfq
slight correction

[modest]

I’m seriously unable to grasp your point, Don. What’s so special about

 

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

 

? How is it more important, than, say

 

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

 

, or any other of the infinite number of expression we can derived from [math]T a^x[/math], for which any expression involving [math]T[/math], [math]a[/math], or [math]x[/math] can have any value for which the derived identity is undefined?

 

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

 

Which requires [imath]Ta\neq n[/imath]. Note that [imath]n[/imath] needn't be [imath]1[/imath], it could also be [imath]2[/imath], [imath]3[/imath] or whatever.

 

Good point.

 

Don, because [math]T=1[/math] is undefined in your identity you advocate disallowing any case of a unit coefficient. By the same reasoning, [math]T[/math] times [math]a[/math] should never be allowed to equal [math]x[/math] in the term:

 

[math]Ta^x[/math]

 

Is that correct? Does the identity,

 

[math]Ta^x = \left( \frac{Ta}{x} \right)^{\left( \ln(Ta^x) \over \ln(\frac{Ta}{x}) \right)} \forall \ Ta \neq x[/math]

 

prove that [math]Ta=x[/math] should always be disallowed any time someone runs into the term [math]Ta^x[/math]?

 

Likewise, Craig's identity shows that [math]a = 1[/math] should always be disallowed in [math]Ta^x[/math] and Q's identity, or this one,

 

[math]Ta^x = \left( \frac{Ta}{n} \right)^{\left( \ln(Ta^x) \over \ln(\frac{Ta}{n}) \right)} \forall \ Ta \neq n[/math]

 

proves that [math]Ta^x[/math] is meaningless for any [math]T[/math] or [math]a[/math]?

 

Do you follow the reasoning? If the domain of any identity applies to all other instances of the terms in the identity then certainly all monomials like [math]T a^x[/math] are meaningless. Yes?

 

~modest

[LaurieAG]

When [math] x=0 [/math], [math] T [/math] becomes undefined regardless of what [math] a [/math] or [math] T [/math] are.

 

Thats the problem with all of these types of identities, you cannot start from zero, pass through zero or end at zero ever.

[Qfwfq]

Yesterday I didn't get quite what I was seeking, lousy effort in a great muddle, due to some bad news. Well, here is a better shot:

 

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

 

This requires [imath]T\neq n[/imath], exactly as I sought. It more directly means that we can make whatever chosen value of [imath]T[/imath] forbidden, by very simply choosing it as the value of [imath]n[/imath].

 

Goodness this LaTeX rendering isn't the best I've seen for understanding the finer details. Here is just the exponent and also the simple one in my previous post:

 

[math]\frac{x\ln a+\ln T}{\ln\frac{T}{n}}\frac{\frac{\ln T}{\ln n}-1}{\frac{\ln T}{\ln a}+1}=\frac{x\ln a+\ln T}{\ln Ta}[/math]

 

It should be easy to check the algebra above.

[modest]

I think this might accomplish T [math]\neq[/math] n as well,

 

[math]Ta^x = \left( \frac{T}{n} \right)^{\left( \ln(Ta^x) \over \ln(T/n) \right)} \forall \ T \neq n [/math]

 

~modest

[Qfwfq]

I think this might accomplish T [math]\neq[/math] n as well,

Sure, except that as usual, in the limit of T approaching n, you get 1 raised to an infinite exponent instead of a different base and exponent with a finite limit (as in Don's version).

[Don Blazys]

Thanks everyone for all the recent posts, and please forgive me for not answering sooner.

My schedule at the school and elesewhere get's really... really busy just before Christmas vacation.

 

As in many of Turtles most exellent threads, in this thread,

we are exploring and investigating issues and possibilities

that have never before been brought up or even considered.

 

Thus we find ourselves in mathematical territory that is virtually unexplored

and as responsible explorers, we must make every effort to interpret our findings properly,

which is to say in a way that is consistent with both common sense and logic.

 

Until recently, the basic (one exponentiation, one multiplication) algebraic term [math]Ta^x[/math]

has always been expressed simply as its "reflection" [math]Ta^x[/math], and we Hypographers

are the very first to write [math]Ta^x[/math] in ways other than [math]Ta^x[/math].

 

Personally, I think it is entirely possible for mankind to develop a new branch of mathematics

such that each and every independent variable is perfectly defined by its own unique domain,

while each and every operation results in some kind of change, and is therefore effectual and meaningful.

 

In my personal opinion (and I have given up trying to influence anyone else on this matter)

the notion of unity as the "identity element of multiplication" really doesn't qualify as an "axiom"

because to me, an axiom is supposed to be a "self evident" and "universal" truth

and thus can not possibly depend on the location or proximity of a 3 by 5 card

as the "Don Blazys experiment" so absolutely and clearly demonstrates!

 

By the way, what are the domains of the variables in all of those recently introduced terms?

Do they make any kind of logical sense at all?

 

Don.

[sanctus]

Do they make any kind of logical sense at all?

 

Yes, to show you that from an identity you cannot go outside the domain of validity of the given identity in order to conclude that T IN GENERAL cannot have a value outside the domain. Because following modest's and qfwfq's domains and then following your reasoning you get to the conclusion that T cannot be any integer n, etc...