Mathematical Consistency

[Don Blazys]

Let all variables represent non-negative integers.

Then, 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] ?

 

 

Don

[sanctus]

I think when you derive this identity at some point you have an equality which only holds if T is different from 1. Or if you transform your identity (righ hyandside)

 

[math] (T\cdot a)^{\frac{x\ln(a)+\ln(T)}{\ln(a)+ln(T)}}[/math]

 

Then it is only valid as long not T=a=1...

 

So I think this identity is consistent under the proper conditions...coming from the derivation.

[Qfwfq]

can we let [math]T=1[/math] ?

Of course we can.

[Ben]

[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] ?

I agree with with Q.

 

If [math]T = 1[/math] then [math]Ta^x = a^x[/math], so that [math] x={\frac{\frac{x*ln(a)}{ln(T)}+1}{\frac{ln(a)}{ln(T)}+1}}= 1[/math] and therefore [math]Ta^x = a^1 = a[/math]

 

Proof: Given that [math]\ln 1 = 0[/math] then this implies that the numerator [math]\frac{x \ln a}{0} +1[/math] is crap. Likewise the denominator [math] \frac{ \ln a}{0}+1 =[/math] crap. And the quotient of crap by crap is quite clearly one. This completes the proof.

 

Hint: Division by zero is not legal under any jurisdiction outside of mental institutions

[Don Blazys]

Of course, I agree with all of you that we can let [math]T=1[/math],

but only in the special case where [math]x=1[/math], because arguably,

at [math]x=1[/math] the logarithms are not required and can be removed or

"cancelled out", leaving only [math]\left(T*a\right)[/math] where [math]T=1[/math] can then be viewed

as the multiplicand rather than the multiplier.

 

Now, let's include the term that Sanctus wrote and ask the slightly more specific question...

 

Given the identity:

 

[math] T*a^x=

(T\cdot a)^{\frac{x\ln(a)+\ln(T)}{\ln(a)+\ln(T)}}=

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

 

(which in theory includes all logically derived terms that form an identity with any of the above,)

 

can we let [math]T=1[/math] if [math]x>1[/math] ?

 

Don.

[CraigD]

Given the identity:

 

[math] T*a^x=

(T\cdot a)^{\frac{x\ln(a)+\ln(T)}{\ln(a)+\ln(T)}}=

\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] if [math]x>1[/math] ?

An expression of the form

[math]\mbox{expression}_A = \mbox{expression}_B = \mbox{expression}_C[/math]

 

isn’t really a well-formed truth expression, but rather a common shorthand for

 

[math]\mbox{expression}_A = \mbox{expression}_B[/math] AND [math]\mbox{expression}_A = \mbox{expression}_C[/math] AND [math]\mbox{expression}_B = \mbox{expression}_C[/math]

 

So Don’s question could be answered with a simple “no” for the whole expression, or for each equation separately: yes; no; no.

 

... (which in theory includes all logically derived terms that form an identity with any of the above,) ...

 

I have a hunch that this parenthetical states an idea of Don’s that a given identity is only true – or “mathematically consistent”, if every algebraic derivation of it is. This idea doesn’t make much sense to me. Because it’s always possible to make an expression undefined for some value of its variables (for example, by multiplying it by [math]\frac{a-k}{a-k}[/math], so that it’s not defined for [math]a=k[/math]), by this definition, no identity is “consistent”. A condition that’s never true isn’t very useful.

 

In a sense, this idea is the opposite of the more common mathematical goal of deriving expressions that eliminate undesired undefined domain values. (eg: “analytic continuation”)

[Don Blazys]

[math] \frac{a-k}{a-k} [/math] at [math] a=k [/math] results in the "indeterminate form" [math]\frac{0}{0}[/math].

 

Indeterminate forms are not strictly disallowed as are divisions by zero.

Here's the difference...

 

Just as:

 

[math]\frac{6}{3}=2[/math] implies the true statement [math] 2*3=6 [/math]

 

the indeterminate form:

 

[math]\frac{0}{0}=N[/math] implies the true statement [math] N*0=0[/math]

 

However, the division by zero:

 

[math]\frac{6}{0}=N[/math] implies the false statement [math]N*0=6[/math]

 

"True" and "false" are as different as "yes" and "no",

so your example in no way applies to this discussion,

which involves division by zero and a "no" answer

and does not involve an indeterminate form, and a "yes" answer!

 

 

In fact, indeterminate forms are so benign in this regard that if we asked...

 

Given the identity:

 

[math]a+k=\frac{a^2-k^2}{a-k}[/math]

 

can we let [math]a=7[/math] and [math]k=7[/math] ?

 

then the answer would be "yes" because it could be shown that

in this particular case, both sides are equal to [math]14[/math].

 

Quoting Craig D:

An expression of the form [math] \mbox{expression}_A = \mbox{expression}_B = \mbox{expression}_C [/math]

isn’t really a well-formed truth expression.

 

The identity:

 

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

 

is absolutely true.

 

I don't see how it can get any more true than "absolutely true"!

What would be a "better formed" expression and why?

 

Anyway, [math] \mbox{expression}_A = \mbox{expression}_C [/math] is:

 

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

 

where we both agree that the answer to the question "can we let [math]T=1[/math]" is no.

 

Now, can we let [math]T=1[/math] if we hide the right hand side under a rug, where no one could see it?

 

Please, just answer yes or no without any commentary.

 

Don

[CraigD]

[math]\frac{6}{3}=2[/math] implies the true statement [math] 2*3=6 [/math]

 

the indeterminate form:

 

[math]\frac{0}{0}=N[/math] implies the true statement [math] N*0=0[/math]

 

However, the division by zero:

 

[math]\frac{6}{0}=N[/math] implies the false statement [math]N*0=6[/math]

I understand your distinction between an indeterminate form and an undefined form that is not indeterminate (eg: [math]\frac{6}{0}[/math])

 

However, I think taking the concept of indeterminacy as implying that [math]\frac{0}{0}[/math] is permitted outside of the context of limits of continuous (eg: real valued) functions, is wrong. Indeterminacy is not a “magic” mathematical concept that can be applied to an integer-valued function to make division by zero defined. The concept is, in my experience, meaningful only in the context of limits of continuous functions, and then to signify that an expression has lost important information about a limit, and should be avoided, not that [math]\frac{0}{0}[/math] is permitted.

 

Division by zero can permitted by allowing a function’s range to include points of infinity – for example, to be elements of the extended reals.

I’ve encountered some off-beat schemes that involve infinite sets of infinite real numbers such that

[math]\frac{6}{0}=N=\infty_{\frac{6}{0}}[/math], so [math]N \dot 0 = 6[/math]

 

but only as novelties, not of much general usefulness.

 

The identity:

 

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

 

is absolutely true.

Guessing that you intend “absolutely true” to “not ‘relatively true’”, and that “relatively true” means “true for a domain smaller than all values of T, a, and x” these identities are not absolutely true, because [math] \mbox{expression}_A = \mbox{expression}_B [/math] and [math] \mbox{expression}_A = \mbox{expression}_C [/math] are not true for all values of T and a. [math] \mbox{Expression}_A[/math] has a domain of all values of integer triple (T, a, x). [math] \mbox{Expression}_B[/math]’s domain doesn’t contain any (1, 1, x).

[math] \mbox{Expression}_B[/math]’s domain doesn’t contain any (1, a, x).

 

So the 3 identities are only “relatively true”, for limited domains.

 

What would be a "better formed" expression and why?

[math] \mbox{expression}_A = \mbox{expression}_B[/math] AND [math] \mbox{expression}_A = \mbox{expression}_C[/math] AND [math] \mbox{expression}_B = \mbox{expression}_C[/math]

 

is well formed.

 

[math] \mbox{expression}_A = \mbox{expression}_B = \mbox{expression}_C [/math]

 

is not well formed, because:

 

The “=” equivalence operator returns a value in the range {true, false}, and is binary, so describing its order of operations explicitly, the whole expression is

 

[math]( \mbox{expression}_A = \mbox{expression}_B ) = \mbox{expression}_C[/math]

 

Which we can rewrite generically as

 

[math]{ V: V \in {\mbox{true, false}} } = { B : B \in \mathbb{Z} }[/math]

 

Here, we can say the expression is not well-formed, because [math]B[/math] can’t have a value of true or false.

 

We can allow {true, false} to be a subset of [math]\mathbb{Z} [/math], {1, 0}. In this case, the expression is well formed, but false, because the range of expression_C doesn’t include 1 or 0.

 

 

Anyway, [math] \mbox{expression}_A = \mbox{expression}_C [/math] is:

 

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

 

where we both agree that the answer to the question "can we let [math]T=1[/math]" is no.

 

Now, can we let [math]T=1[/math] if we hide the right hand side under a rug, where no one could see it?

 

Please, just answer yes or no without any commentary.

Yes, if by “hiding the right hand side under a rug”, we mean replacing the right hand side of the equation with an expression derived from the left hand side that’s defined for [math]T=1[/math], such as:

 

[math] (T\cdot a)^{\frac{x\ln(a)+\ln(T)}{\ln(a)+\ln(T)}} [/math]

[Ben]

 

Given the identity:

 

[math]a+k=\frac{a^2-k^2}{a-k}[/math]

 

can we let [math]a=7[/math] and [math]k=7[/math] ?

 

then the answer would be "yes" because it could be shown that

in this particular case, both sides are equal to [math]14[/math].

Then show it. In my ignorance I get that [math]7+7 = 14[/math] and yet [math]49-49=0,\,\,\, 7-7=0 ,\,\, \frac{0}{0} \ne 14[/math]. Or anything else, as far as I can see

 

 

The identity:

 

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

 

is absolutely true.

In fact it is absolutely false! Have you any idea how to do natural logarithmic arithmetic?

 

 

Now, can we let [math]T=1[/math] if we hide the right hand side under a rug, where no one could see it?

 

Please, just answer yes or no without any commentary.

 

Yes, but it's not interesting; [math] Ta^x = a^x[/math] when [math]T=1[/math]. Let's write home to mother about this Earth-shattering discovery.

 

But you still have a problem with your RHS - it is complete and utter nonsense. Can you not see why?

[CraigD]

The identity:

 

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

 

is absolutely true.

In fact it is absolutely false! ...

Recheck your work, Ben, or just evaluate for several values. Though your calculator likely will have small errors due to rounding transcendental numbers (you can avoid this by using an integer rather than [imath]e[/imath]-base logarithms), you can quickly (assuming you’ve got a user-friendly calculator) convince yourself the identities are correct for all but T=a=1 and T=1, respectively.

 

I recall that Don posted a derivation the identity in post #1 some time ago, either at hypography or on his own webpages. Perhaps he’ll take pity on me and my poor memory and algebra and repost them here.

[Don Blazys]

Here's an experiment that we all can do. I call it "the Don Blazys experiment".

 

Get two 3 by 5 cards.

 

On one of the cards, write: [math] T*a^x [/math]

 

and on the other card write: [math] = \left(T*a\right)^{\frac{\frac{x*\ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1}} [/math]

 

Now, lay the two cards side by side so that what you see is: [math] T*a^x = \left(T*a\right)^{\frac{\frac{x*\ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1}} [/math]

 

and ask a mathematician "Can we let [math]T=1[/math] ?" (The correct answer is "no".)

 

Then, put the card with the term involving logartthms in your pocket where no one can see it,

point to the remaining card, ask the same exact question and enjoy the enormously entertaining

"song and dance", complete with all kinds of "harrumphing and hand waving" that is sure to ensue

when you then explain that this experiment demonstrates that...

 

the possibility of allowing the "unit coefficient" [math]T=1[/math]

depends entirely upon where the card with

the term involving logarithms is located!

 

______________________WARNING!!!___________________________

 

If you are a student, then please do not try this experiment on your teacher

as their "harrumphing and hand waving song and dance" may go on long after

class has been dismissed!

[Ben]

Recheck your work, Ben, or just evaluate for several values. Though your calculator likely will have small errors

Oddly enough I don't posses a calculator so I cannot accept your invitation.

the identities are correct for all but T=a=1 and T=1, respectively.

So is it your contention that an identity can be absolutely true mod some exceptions? This is the thrust of Don's claim. My logic says otherwise; maybe "conditionally true" might be better? Hardly interesting though.

 

PS I confess I earlier overstated my case: "not absolutely true" is not equivalent to "absolutely false"

[sanctus]

Doesn't it depend on how one defines "absolutely true"? I mean it makes perfect sense to say [math]\forall T \neq 1[/math] one has the following identity (for whatever identity). Then this identity is absolutely true [math]\forall T \neq 1[/math].

 

But ok, you can this is not "absolutely true", but only true for all T satisfying the condition...as long as we don't have a definition of what "absoultely true" means (either for all real numbers or all numbers in a given domain...), this sounds more a bit like semantics.

 

So let's define "absolutely true" ;-)

[Don Blazys]

You see folks, 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]

 

is both embarassing and humiliating to the "math community"

because it conclusively and unequivocally demonstrates

that the concept of a "unit coefficient" is badly flawed

and that the notion of 1 being the "identity element"

relative to multiplication is highly illogical.

 

The consequences of this are enormous, because the "unit coefficient"

is often applied in calculations and 1 being the "identity element"

for multiplication is universally accepted and cited as an "axiom" !

 

Who would have thought that ultimately, they must both go the way of

the "luminiferous aether" and "phlogiston".

 

Meanwhile, those so called "professors of mathematics" in our colleges and universities

will continue to not only teach this gibberish and indoctrinate our young people with this rubbish,

but will undoubtedly add insult to injury by demanding payment for their "services"!

 

You math majors who are paying thousands of dollars for an education are being taken for suckers.

You are being defrauded, duped and swindled! You are getting fleeced

and the wool is being pulled over your eyes! You should at least get your money back!!!

 

Don.

[Qfwfq]

Get two 3 by 5 cards.

 

On one of the cards, write: [math] T*a^x [/math]

 

and on the other card write: [math] = \left(T*a\right)^{\frac{\frac{x*\ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1}} [/math]

 

Now, lay the two cards side by side so that what you see is: [math] T*a^x = \left(T*a\right)^{\frac{\frac{x*\ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1}} [/math]

 

and ask a mathematician "Can we let [math]T=1[/math] ?" (The correct answer is "no".)

 

Then, put the card with the term involving logartthms in your pocket where no one can see it,

point to the remaining card, ask the same exact question and enjoy the enormously entertaining

"song and dance", complete with all kinds of "harrumphing and hand waving" that is sure to ensue

when you then explain that this experiment demonstrates that...

 

the possibility of allowing the "unit coefficient" [math]T=1[/math]

depends entirely upon where the card with

the term involving logarithms is located!

 

______________________WARNING!!!___________________________

 

If you are a student, then please do not try this experiment on your teacher

as their "harrumphing and hand waving song and dance" may go on long after

class has been dismissed!

Don, you're asking the mathematician the wrong question.

 

The answer to the question you are asking is yes in any case.

 

If instead you write that equality and ask "In the [math]T=1[/math] case, is this equality defined and true?" then the correct answer is: No, because the expression on the rhs is undefined for [math]T=1[/math].

 

You see, mathematicians are a very odd type of beast. If you don't ask them the right question, they don't give you the right answer

[Don Blazys]

Quoting Qfwfq:

 

The answer to the question you are asking is yes in any case.

 

I disagree.

 

The answer to the question "Can we let T=1?" is "no" because we can't divide by zero.

 

Don.

[Qfwfq]

The answer to the question "Can we let T=1?" is "no" because we can't divide by zero.

But exactly how is this question related to dividing by zero?