A Mathematical Emergency.

[modest]

(1) Which term has the better defined variables?

 

Assuming you mean “better-defined domain”, the answer is neither. So long as you keep track of the domain of each variable while you’re doing arithmetic, the domain is always perfectly defined.

 

(2) Is it logical to proceed from perfectly defined variables to poorly defined variables

and call it a "simplification"?

 

You’re working backwards. Consider your identity:

[math]\frac{T}{T}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]

And this one, which I've just made up:

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

The domain of x on the left side includes all real numbers while the right side does not work for x=0. But, this obviously doesn’t imply that any time you take [math]a^x[/math], x cannot equal zero. The restrictions on the domain of the variables in your identity is a result of the arithmetic you’ve done on it. For example, your identity needs the restriction [math]T \neq a[/math], because that would result in division by zero. Yet, we can avoid this restriction by constructing the identity differently. Here’s another example,

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

Nothing here prevents T from equaling a in either side of the equation. There’s nothing unusual going on here. We’re just starting with the term [math](T/T)a^x[/math] and applying a bunch of logarithmic identities. One result of all the substitutions and identities is to restrict the domain of the variables. Consider,

[math]x = (x + x)/2[/math]

Nothing about this constrains x against being any real number. It certainly can be zero, or negative just as well as it can be a natural number. This extends to the following:

[math](xT) = ((xT) + (xT))/2[/math]

The following, however, does constrain the domain of x,

[math]x = 10^{\log_{10}(x)}[/math]

here x cannot be zero or negative. It’s domain is restricted because we’ve put it through the logarithmic ringer. There is no real number (positive, negative, or zero) to which you can raise 10 to in order to get a negative or zero answer. In other words, the restriction is a property of the right hand side of the equation and not the left. This extends to the following,

[math](xT) = 10^{\log_{10}(xT)}[/math]

Here the result of multiplying x and T cannot give a negative number nor can either x or T be zero for this identity to work. But, It is not the lhs operation of multiplying x and T that restricts their domain. It is, as before, the property of the logarithm on the right hand side. You do not keep this restricted domain when you’re not using the identity. For example, these are not improper,

[math]x=-4, T=3, \; (xT) = (-4)(3) = -12[/math]

[math]x=2, T=0, \; (xT) = (2)(0) = 0[/math]

[math]x=1, T=-3, \; (xT) = (1)(-3) = -3[/math]

You wouldn’t say these are wrong because there exists somewhere an identity that has a different domain for x and T. That would make no sense.

(3) Can the cancelled [math]T[/math]'s on both sides of the equation be "cancelled out" meaning "crossed out" and made to "disappear"?

Where you say "canceled out" or "crossed out", I prefer to think of it as replacing X/X with unity. If the term X/X is a product then it can be eliminated because N•1=N. When doing this, it's important to remember the variable cannot equal zero—but, the answer is yes. The T's can be "crossed out" when [math]T \neq 0[/math]. That's what I showed yesterday. The right hand side of your identity can be simplified algebraically so that it looks just like the left.

 

This is more apparent if you look at your identity in a slightly simpler form:

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

The exponent is a log with a base equal to the base of the exponent. It's taking the log of [math]a^x/T[/math] which coincides exactly with the term on the left. The logarithmic function is an inverse to the exponential function. So, this is very much like multiplying by a variable then dividing by the same variable or adding a variable then subtracting the same variable. You end up where you started. You've done something then done its inverse. I see no point in doing that. It simplifies to:

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

or,

[math]\frac{T}{T}a^x=\frac{T}{T}a^x[/math]

and without an exponent, of course, the T's can "cancel".

 

~modest

[Don Blazys]

To: Modest,

 

Quoting Modest:

Assuming you mean “better-defined domain”, the answer is neither.

So long as you keep track of the domain of each variable while you’re doing arithmetic,

the domain is always perfectly defined.

 

Sorry Modest. You are a very smart person, but this time, you assumed incorrectly.

I said absolutely nothing about "better defined domains".

The question was:

 

Quoting myself:

Which term has the better defined variables?

 

That means the symbols: [math]T, a[/math] and [math]x[/math].

 

You know, when I was a taxi driver :steering:, each and every taxi in the fleet

had its own unique number painted on its sides.

Thus, there was a "one to one corespondence" between the names of the drivers,

and the numbers on the taxis that they drove.

The dispatcher almost always referred to us by the numbers on our taxis,

but in principle, names and numbers were "interchangeable".

The same basic principle applies here.

 

You see, "Blazys terms" are the only basic (one exponentiation, one multiplication)

terms in mathematics that allow us to refer to each and every variable by it's domain.

(That's one of the things that makes them so absolutely incredible!)

For instance, if we write the "Blazys term":

 

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

 

whose variables are both intrinsically and perfectly defined by their own unique domains as:

 

[math]T=\{2, 3, 4...\}[/math], [math]a=\{1, 2, 3...\}[/math] and [math]x=\{0, 1, 2...\}[/math]

 

then I can say "Consider the variable whose domain is [math]\{0, 1, 2...\}[/math],

and immediately, everybody knows that I am referring to the variable [math]x[/math].

 

However, if we "simplify" the above term so that it appears as:

 

[math]Ta^x[/math]

 

whose variables are intrinsically defined as:

 

[math]T=a=x=\{0, 1, 2...\}[/math]

 

then we can no longer refer to each and every variable by it's domain!

Thus, if I were now to say "Consider the variable whose domain is [math]\{0, 1, 2...\}[/math],

then nobody would know which variable I am referring to because all three variables

are now abysmally defined by the same exact domain!

 

Confusion would abound,

:confused: :QuestionM :shrug: :huh: :Confused:

and this would lead to both anger and frustration.

:mad: :angry: :eshout: :cussing: :Angry: :censored:

Then, quite possibly, a fight would ensue,

:Bump2: :hammer:

:Tupac: :Guns: :pirate:

and somebody would get badly injured, or even killed!

:eouch: :edizzy: :Tupac: :dead:

 

Now, there are many ways to restrict the domains of variables.

The important thing however, is that we restrict them in a way that makes sense.

For instance, non-trivial common factors are, by definition, greater than unity,

so for problems that involve common factors,

it makes perfect sense to use "Blazys terms" rather than "terms with poorly defined variables"

because only "Blazys terms" have the restriction [math]T>1[/math].

"Terms with poorly defined variables" actually allow trivial common factors to creep in,

and are therefore completely inadequate for use in problems that involve common factors.

 

Most importantly, once a logical restriction is complete, we must not tamper with it

for it constitutes an important and irrevocable logical statement.

Notice how you had to increase the domains of the variables in the "Blazys term" before

you could "cancel out" (meaning "cross out" and "render invisible") the cancelled [math]T[/math]'s.

 

When you did that, you changed my logical statement that says:

"This term guarantees that [math]T>1[/math] and will not allow trivial common factors."

 

into your own illogical statement that says:

"This term does not guarantee that [math]T>1[/math] and will allow trivial common

factors."

 

Try eliminating the [math]T[/math]'s in the "Blazys term" equation:

 

[math]\frac{T}{T}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]

 

without increasing the domains of the variables!

If you do, then you will find that it's quite impossible!

 

Thus, the answers to the questions are as follows:

 

(1) Which term has the better defined variables?

Correct answer: "The Blazys term".

 

(2) Is it logical to proceed from perfectly defined variables to poorly defined variables

and call it a "simplification"?

Correct answer: "No".

 

(3) Can the cancelled [math]T[/math]'s on both sides of the equation be "cancelled out"

meaning "crossed out" and made to "disappear"?

Correct answer: "No".

 

Quoting Modest:

Nothing about this constrains x against being any real number.

 

This tread is about non-negative integers only.

 

Quoting Modest:

Like I said before, I’m not a mathematician.

 

Well, judging by the vigor and tenacity with which you engage in the pursuit of mathematical truth, you certainly have the heart of one! :)

 

Don.

[Turtle]

To: Turtle,

 

Thanks for the info on Euclid and why unity is not a prime.

...

Now, in order to introduce the all important element of "repetition" into this model,

let's equate "bean" to "multiplicand" and "beaker" to "multiplier".

Then, the multiplication [math]1*5[/math] can be expressed as either:

...

Don.

 

Roger the Euclid & erhmmm...no. :eek2: :hyper: I mean no, let's not equate the beaker to a multiplier. The beaker is at best a variable just like I introduced it, and as you recall I started the bean counting without beakers. Moreover I started with the beans to show why "repetition" "increase" is a misleading choice of words for your graphic illustration of multiplication, and I pointed out "arrangement" is a better term. This is because you can have an arrangement that doesn't repeat, but all repetition is a special case of an arrangment. Capiche?

 

Edit: made some strike-throughs and additions as I mentioned the wrong term. :eek2: :naughty: :hyper:

[modest]

Thus, the answers to the questions are as follows:...

 

(3) Can the cancelled 's on both sides of the equation be "cancelled out"

meaning "crossed out" and made to "disappear"?

Correct answer: "No".

 

I now see this claim has a history...

 

For instance, one college student bet six of his friends and classmates ten dollars each that he could re-write the term:

 

(T/T)a^x

 

so that it is algebraically impossible to "cross out" the T's. At first they didn't take the bet because they thought that it must be some kind of "silly trick" (such as writing the term on a piece of paper, then flushing it down the toilet before anyone else can get their hands on it), so he assured them that the result would be legitimate and invited them to search the internet for a way to make "crossing out" the T's algebraically impossible. The next day, four of them did take the bet, (two of them being math majors who, after doing an exhautive internet search, were now convinced that such a result couldn't possibly exist). He then wrote:

 

(T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)),

 

and even though it took a couple of weeks for them all to agree that the identity is true, finally got his money.

 

Regardless if you've formerly believed this is true, I've now shown (post #117) that it is not. Your "Blazys term" does reduce to your non-Blazys term through pure force of algebra.

 

Also, if you're interested in restricting the domain of a variable it can be accomplished with other simpler methods. For example, T > 1 is true for:

[math]T=T\frac{\sqrt{T-1}}{\sqrt{T-1}}[/math]

 

~modest

[Don Blazys]

To:Turtle,

 

Wikipedia defines the word "operation" as follows:

 

"In it's simplest meaning in mathematics and logic,

an action or procedure which produces a new value

from one or more input values."

 

Wikipedia also says this about the word "multiplication":

 

"Multiplication is defined for whole numbers in terms of repeated addition".

 

Also, an "operand" is defined as "an initially considered or introduced quantity

on which an operation is then performed".

 

Now, let's consider the multiplication [math]1*5=5[/math],

which can also be expressed as [math]5*1=5[/math],

and for the purpose of this illustration,

let's call the number on the far left the "multiplicand",

and the number on the far right the "product".

 

Now, if we begin with the "multiplicand" [math]1[/math],

and end up with the "product" [math]5[/math],

then we can be absolutely certain that a multiplication by the "multiplier" [math]5[/math]

did indeed occur, because the "product" [math]5[/math]

is "different" from the "multiplicand" [math]1[/math],

and therefore qualifies as a "new value".

 

However, if we begin with the "multiplicand" [math]5[/math],

and end up with the "product" [math]5[/math],

then we can't be certain that a multiplication by the "multiplier" [math]1[/math]

ever occured, because the "product" [math]5[/math]

is exactly the same as the "multiplicand" [math]5[/math],

and therefore, does not qualify as a "new value".

 

I happen to prefer a "model of multiplication" that makes a clear distinction between

unity as a "multiplicand" and unity as a "multiplier", and subscribe to the belief that

unity can only be a "multiplicand", because as a "multiplier",

it (unity) is both ineffectual and superfluous.

 

However, since everyone but me seems to view

"multiplication by unity" as a "meaningfull concept",

I will not press the issue and will concede that

I am completely alone in my point of view,

which for the life of me, I can't change.

 

Math may be important, but for me,

it's not nearly as important as being a kind and helpfull person.

 

I will now proceed to multiply each and every number ever recorded in the entire

universe by unity!

 

:dust::dust::dust::dust:

:dust::dust::dust::dust:

:dust::dust::dust::dust:

:dust::dust::dust::dust:

:dust::dust::dust::dust:

 

Wow, I did it!

 

Don.

[Don Blazys]

To: Modest,

 

Quoting Modest:

I now see this claim has a history...

 

It has a ten year history and from my point of view,

that "claim" is really just a "self evident and irrefutable fact".

 

Quoting Modest:

Regardless if you've formerly believed this is true, I've now shown (post #117) that it is not.

Your "Blazys term" does reduce to your non-Blazys term through pure force of algebra.

 

You did nothing of the kind!

Essentially, all you did in (post #117) was take the derivation for

"cohesive terms" or "Blazys terms" that I have in my "Proof of the Beal Conjecture"

which is on my website (donblazys.com) and state it backwards!

In other words, in my derivation, I systematically restricted the domains of the variables

until it was possible to refer to each and every variable by its particular domain.

Then you stated my derivation in reverse and increased the domains of the variables

until it was impossible to refer to each and every variable by it's particular domain!

If you really want to "reduce" my "Blazys terms", then you must do so without changing

the domains of my variables! Otherwise, it's not the "force of algebra"

but the "farce of algebra" that you are employing.

 

You know, you would have arrived at the exact same vaccuous "conclusion" had you spared us

all the machinations of (post #117) and simply substituted [math]\left(\frac{T}{T}\right)a^x[/math]

for the "Blazys term" in order to get [math]a^x=a^x[/math]!

 

Thus, according to your "line of reasoning" we shouldn't even bother with "restrictions" at all

because according to you, they could always be "undone" by "reversing the derivation"!

If that were the case, I would have simply "undone" my "Blazys terms" ten years ago

and "deemed" them to have "exactly the same properties as those terms whose variables

can't be referred to by their domains"!

How ridiculous would that be?

What would then be the point of mathematics?

Why not then just use "the force of algebra" to "reduce" Einstiens famous equation:

 

[math]E=MC^2[/math] to simply [math]E=E[/math] ?

 

Why not then divide both sides by [math]E[/math] and further "reduce" it to [math]1=1 [/math] ?

 

You see, as mathematicians, we must develop a sense of what is meaningfull and what is not.

Math is both a science and an art.

 

Here's the "bottom line" which we may henceforth refer to as the "Blazys Theorem".

Any term that precludes trivial common factors and whose variables can be individually

refered to by their domains will not allow cancelled common factors to be "crossed out".

 

Quoting Modest:

Also, if you're interested in restricting the domain of a variable

it can be accomplished with other simpler methods. For example, T > 1 is true for:

 

[math]T=T\frac{\sqrt{T-1}}{\sqrt{T-1}}[/math]

 

The "restrictions" in your term:

 

[math]T=T\frac{\sqrt{T-1}}{\sqrt{T-1}}[/math]

 

as is can be "crossed out" and are thus "ineffectual".

 

Not so in an as is "Blazys term"!

 

Don.

[Turtle]

To:Turtle,

 

I happen to prefer a "model of multiplication" that makes a clear distinction between

unity as a "multiplicand" and unity as a "multiplier", and subscribe to the belief that

unity can only be a "multiplicand", because as a "multiplier",

it (unity) is both ineffectual and superfluous.

...

Don.

 

I don't have a problem with models, but so far I fail to see how yours is an emergency. For whatever failings you ascribe to the standard model, we have done pretty well by it over the centuries in the products of its application. What in the Calculus used in engineering a bridge for example, would be "wrong" under your model?

[modest]

I now see this claim has a history...

 

It has a ten year history and from my point of view,

that "claim" is really just a "self evident and irrefutable fact".

 

I understand. I do not mean to attack your work of the last 10 years. My interest is in understanding and exploring mathematics. While your conclusions may not make sense to me, I do respect the work you've done in making them.

 

Regardless if you've formerly believed this is true, I've now shown (post #117) that it is not.

Your "Blazys term" does reduce to your non-Blazys term through pure force of algebra.

 

You did nothing of the kind!

Essentially, all you did in (post #117) was take the derivation for

"cohesive terms" or "Blazys terms" that I have in my "Proof of the Beal Conjecture"

which is on my website (donblazys.com) and state it backwards!

 

I don’t have time to sort through the papers on your site, but I suspect I have indeed done exactly what you did but in reverse :) You started with the term,

[math]\frac{T}{T}a^x[/math]

applied a bit of algebra and algebraic identities until you got,

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

I then applied (no doubt an equal amount of) algebra and algebraic identities to get,

[math]\frac{T}{T}a^x[/math]

You thread it with a needle... I’ve done your doing in reverse. Why this would mean I've not done what I claimed to do is beyond me.

 

In other words, in my derivation, I systematically restricted the domains of the variables until it was possible to refer to each and every variable by its particular domain. Then you stated my derivation in reverse and increased the domains of the variables until it was impossible to refer to each and every variable by it's particular domain!

 

It seems again that you might be moving the goalpost. You went from “crossing out the T's is algebraically impossible” to “crossing out the T's is algebraically impossible without changing the domain of the variables”. Perhaps I've simply misunderstood the story you told above, but I'm having a hard time inserting anything about domain restrictions in it. Perhaps you can clarify...

 

If I bet somebody money that I can rewrite the term,

[math]\frac{T}{T}[/math]

so that it’s “algebraically impossible” to cross out the T’s, then I wrote:

[math]\frac{e^{\ln(T)}}{T}[/math]

Is it now “algebraically impossible” to cross out the T’s because the domain would change? I would win that bet? Is that what you were saying with that story? Really?

 

If you really want to "reduce" my "Blazys terms", then you must do so without changing the domains of my variables! Otherwise, it's not the "force of algebra" but the "farce of algebra" that you are employing.

 

That's not normal algebra. If I want to substitute 1 for a/a then I can do that in normal algebra. I will retain the restriction [math]a \neq 0[/math] as I go, but the restriction does NOT need to be implicit in the equation! It's completely normal to retain the restriction explicitly.

 

By your method if I were on step 73 of a proof and I have,

[math]x=e^{\ln{x}}[/math]

then there would be no step 74. No further algebra possible. The proof would be unattainable.

 

You know, you would have arrived at the exact same vaccuous "conclusion" had you spared us

all the machinations of (post #117) and simply substituted [math]\left(\frac{T}{T}\right)a^x[/math]

for the "Blazys term" in order to get [math]a^x=a^x[/math]!

 

Don't forget [math]T \not= a[/math], [math]T \not= 0[/math], [math]T \not= 1[/math], and [math]a \not= 0[/math]. But yes, knowing this is a proven identity, if I were confronted 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]

I would feel absolutely confident using a^x (with the domain restrictions I just mentioned).

 

Thus, according to your "line of reasoning" we shouldn't even bother with "restrictions" at all because according to you, they could always be "undone" by "reversing the derivation"!

 

No Don. No. What I'm saying is that finding the native term [math]\left(\frac{T}{T}\right)a^x[/math] does not imply [math]T \not= a[/math] or [math]a \not= 0[/math] just because some identity exists with those particular restrictions. You are trying to say multiplication by unity results in division by zero and other such conclusions based on your identity. You are wanting to use the domain restrictions apart from the identity. I do not agree with that.

 

 

Why not then just use "the force of algebra" to "reduce" Einstiens famous equation:

 

[math]E=MC^2[/math] to simply [math]E=E[/math] ?

 

I can cancel the T's in your term using algebra. I mention this only to correct your persistent claims that it would be impossible. In no way does that imply this strawman that you're going on and on about.

 

~modest

[Don Blazys]

To: Turtle,

 

Quoting Turtle:

I don't have a problem with models,

but so far I fail to see how yours is an emergency.

For whatever failings you ascribe to the standard model,

we have done pretty well by it over the centuries in the products of its application.

What in the Calculus used in engineering a bridge for example,

would be "wrong" under your model?

 

As with most "emergencys", there are many aspects and issues to consider.

Let's begin with the fact that students are not being taught the proper way to

represent and eliminate "common factors". This is very well explained in post #92

of this thread, but in brief, students are being taught to use the wrong term when

representing and eliminating common factors. The terms they are taught to use

allow trivial common factors to creep in and are therefore woefully inadequate for

use in problems that involve common factors. "Blazys terms" are the only terms

powerfull enough to handle common factors, but I seem to be the only one pointing this out.

 

"Blazys terms" also correct a lot of logical inconsistencies in the calculus.

I have an article called "Cohesive Terms and the Calculus" on my website (donblazys.com).

It's a somewhat "hard read", but it will give you some idea of the "loose ends" that

"cohesive terms" or "Blazys terms" tie up in a most surprising and wonderfull way!

 

"Blazys terms" are also the only terms powerfull enough to solve hitherto intractable

problems such as the "Beal Conjecture" and "Fermat's Last Theorem",

yet students are being denied the amazing opportunity to solve those celebrated problems

for themselves because many in the math community are simply too embarrased to admit that the method for eliminating common factors that they have been teaching is simply wrong!

 

Well, I don't care one iota about their fragile overinflated egos or their "feelings".

We simply can't allow student's to continue being taught things that are wrong!

That's why I am posting on this issue in this and other forums.

I am documenting my efforts to further the cause of mathematical truth while

allowing my detractors to document their ineptness and/or irresponsibility.

 

Don.

[Don Blazys]

To: Modest,

 

Quoting Modest:

I can cancel the T's in your term using algebra.

I mention this only to correct your persistent claims that it would be impossible.

 

Okay, here is my "Blazys term":

 

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

 

Now, you come along and write:

 

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

 

But that doesn't allow you to "cross out" the [math]T[/math]'s on the left,

or, on the right, because the cancelled [math]T[/math]'s are representative

of the notion that the greatest factor [math]T[/math] was extracted,

then cancelled, so that both terms must now represent the same prime number.

 

Moreover, since what we really have here is an "identity",

the domain's of all the variables on both sides, are exactly the same.

 

Now, what I don't get is how you take the term on the right out of this identity and

thereby expand the domains of it's variables?

 

From my point of view, once you establish an identity, you can't "renege" on it.

 

Don.

[modest]

Moreover, since what we really have here is an "identity", the domain's of all the variables on both sides, are exactly the same.

 

You bring up an excellent point. What you have here:

[math]\frac{T}{T}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]

is not (in the strictest sense) an identity. For T=5, a=5, x=2, you get,

[math]\frac{5}{5}5^2=5\left(\frac{5}{5}\right)^{\left(\frac{\frac{2\ln(5)}{\ln(5}-1}{\frac{\ln(5)}{\ln(5)}-1}\right)}[/math]

[math]25=\textrm{unde}\textrm{fined}[/math]

The equality is untrue with all values where T = a and also with negative a’s and T’s. When no domain restriction is specifically stated then it’s assumed that the variables of an identity can represent any real value. You need to explicitly state the domain of your variables otherwise your identity is not true.

...Now, you come along and write:

 

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

 

But that doesn't allow you to "cross out" the T's on the left,

or, on the right, because the cancelled T's are representative

of the notion that the greatest factor T was extracted,

then cancelled, so that both terms must now represent the same prime number.

 

Yes, you are allowed to cancel the T's on the right. It is basic arithmetic. At most you are offering a reason to choose not to cancel the T's which is a valid choice. But, your statements about the "algebraic impossibility" of canceling the T's in your term or canceling (T/T) in the lhs of your identity are wrong. Your attitude and tone of demanding whoever you're conversing with (and, indeed, all of mathematics) bend to your way of thinking here comes off as arrogant and demeaning.

 

~modest

[Don Blazys]

To: Modest,

 

Quoting Modest:

Your attitude and tone of demanding whoever you're conversing with

(and, indeed, all of mathematics) bend to your way of thinking here comes off as arrogant and demeaning.

 

Thanks for the constructive criticism.

I sincerely apologize for any of my posts that come off as "arrogant" or "demeaning".

In my heart of hearts, I know that I am nothing but a tired old

fuddy duddy with a cantankerous disposition and a fading memory,

which at the very least qualifies me as being "fundamentally humble".

:bow::bow::bow::bow::bow::bow::bow::bow::bow:

(Perhaps not quite to the point of being "Modest" :hihi::):lol:, but humble nevertheless.)

Anyway, I am well aware of the fact that I do get a little "too blunt" or "abrasive"

while defending my arguments and positions, but I still prefer to leave those parts in,

if only to make for a more lively and entertaining read.

(Believe it or not, there are some people out there who think that math is "boring"!)

Truly, it's all in fun, and if I really believed that those who I communicate with are

somehow not able to reason at my level or above, I would simply stop communicating.

The important thing here is that I will always admit when I am wrong,

(for me, it's part of being a man) which is, unfortunately,

more than I can say for many others in the "math community".

(If you want to get an idea of just how bumbling, yet pompous

many in the math community really are, just read some of the letters that they wrote to

Marilyn vos Savant after she gave the correct answer to the "Monty Hall Problem".)

 

Quoting Modest:

Yes, you are allowed to cancel the T's on the right.

It is basic arithmetic. At most you are offering a reason to choose

not to cancel the T's which is a valid choice. But, your statements about

the "algebraic impossibility" of canceling the T's in your term or canceling

(T/T) in the lhs of your identity are wrong.

 

"Crossing out" the [math]T[/math]'s on both sides of the equation:

 

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

 

would result in:

 

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

 

Now, if you or anyone else can explain what on Earth the "expression":

 

[math]\ln()[/math]

 

is supposed to mean, then I will concede that I am wrong in my assertion that it is

algebraically impossible to "cross out" or "cancell out" the [math]T[/math]'s on the left hand side.

(By the way, an alternate view of "crossing out" or "cancelling out"

is to "divide each and every [math]T[/math] by [math]T[/math],

but that doesn't work either because it results in "division by zero".)

 

Now, as for the term on the right, if we cross out those [math]T[/math]'s only,

then we will be left with:

 

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

 

which is "problematic" in several respects. For one thing, the term without the [math]T[/math]'s

shows no evidence that the greatest possible factor [math]T[/math] was extracted, then cancelled,

which means that it does not necessarily represent a prime as does the term on the left.

 

Also, there is now no evidence that the term on the left was derived from the term on the right,

which means that the equation can no longer be viewed as an "identity".

 

Most importantly however, the [math]T[/math]'s on the right tell us that it must be possible

to let [math]T=a[/math], which litterally forces us to let [math]x=1[/math]

which in turn allows us to "cross out" the expressions involving logarithms,

so that all we are left with is:

 

[math]T\left(\frac{a}{T}\right)=\left(\frac{T}{T}\right)a[/math]

 

where we may now let [math]T=a[/math], or simply "cross out" the remaining [math]T[/math]'s so as to have:

 

[math]a=a[/math]

 

which, if you think about it, makes perfect sense, because if the greatest possible factor [math]T[/math]

was extracted, then cancelled, then the terms must represent the same prime number,

and prime numbers do not have exponents!

 

In other words, if it weren't for the "Blazys term" on the left, then any extraction and

cancellation of the greatest possible factor [math]T[/math] would be utterly meaningless,

because there would be no logical argument that would result in the elimination of the exponent [math]x[/math].

 

The "Blazys term" represents a "higher order of logic", and is thus a "template" for all other

terms.

 

Thus, if we are confronted with the term:

 

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

 

we can simply "invoke" the "Blazys term" to show that

 

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

 

Your example:

 

[math]\frac{5}{5}5^2=5^2[/math]

 

still contains [math]5[/math] as a factor, and is thus an incomplete

(and therefore "undefined") extraction and cancellation of the factor T.

 

eliminating it gives us:

 

[math]\frac{5}{5}5=T\left(\frac{5}{T}\right)^{\left(\frac{\frac{\ln(5)}{\ln(T)}-1}{\frac{\ln(5)}{\ln(T)}-1}\right)}=T\left(\frac{5}{T}\right)=5\left(\frac{5}{5}\right)=5[/math]

 

Notice how we let [math]x=1[/math] and "cancelled out" or "crossed out"

the logarithms before we let [math]T=5[/math].

Most importantly, notice how "Blazys terms" correctly show that

any incomplete extraction and cancellation of factors

is indeed "undefined"! Isn't that awesome?

 

You see Modest, all I am doing is correcting a "weakness"

in how factors and common factors are both represented and eliminated.

I should think that the math community would be overjoyed at the

prospect of correcting something that is not only meaningless,

but embarassing as well!

 

Don.

[CraigD]

"Crossing out" the [math]T[/math]'s on both sides of the equation:

 

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

 

would result in:

 

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

If by “crossing out the Ts”, one means “removing every occurrence of the character “T”, this is true. However, this typographical is not a valid algebraic transformation, any more than “crossing out the ‘+’s” is a valid transformation of

“1+2+3=6” into “123=6”.

 

In algebra, the correct meaning “crossing out” like terms is “summing the exponents of like terms”. So

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

is rewritten

[math]T^1 a^b T^{-b}[/math]

summing the exponents of like terms gives

[math]T^{1-b} a^b[/math]

which may be rewritten

[math]\frac{a^b}{T^{b-1}}[/math]

 

I don’t doubt that there are many “mathematical crises” in the form of schoolteachers who understand Math so poorly they can’t distinguish between typographical and algebraic transformations. The solution, however, is IMHO better education of teachers and more careful selection of teachers by schools to prevent people who don’t understand fundamental math from teaching it, not new, informal, and unsupported ideas about algebra, like “Blayzs terms”.

In my heart of hearts, I know that I am nothing but a tired old fuddy duddy with a cantankerous disposition and a fading memory, which at the very least qualifies me as being "fundamentally humble".

Ah, but you’re a very fun tired, cantankerous old fuddy duddy, and a valued member of hypography. :)

 

Humility, in my experience, is an easier character trait to claim than to attain. It’s easy to express humility by acknowledging ones shortcomings. It’s harder to express humility by recognizing that criticism of a cherished idea is correct, and the idea is wrong.

[modest]

Thanks for the constructive criticism.

I sincerely apologize for any of my posts that come off as "arrogant" or "demeaning".

 

Thank you, no problem, and I did indeed mean it as constructive criticism and it’s wonderful you took it as such. A respected member of our forums here sums up nicely where I was coming from in this post. It’s excellent advice. In fact, I'll quote it:

When introducing revolutionary breakthroughs, probably the worst approach is to say "everything you know is wrong, and I'm about to prove it to you." It may sound more sneaky, but you'll find that you have a much easier time if you introduce what you're talking about as a "slight improvement" on existing theory--something that you almost do up until you hit the excerpted quote!

 

Let someone *else* characterize your astounding breakthrough as such: when you say it yourself, it pretty much guarantees that it will be dismissed out of hand, even by those who should know better.

 

Successful Physics requires good Marketing, :cheer:

Buffy

 

Yes, you are allowed to cancel the T's on the right. It is basic arithmetic. At most you are offering a reason to choose not to cancel the T's which is a valid choice. But, your statements about the "algebraic impossibility" of canceling the T's in your term or canceling (T/T) in the lhs of your identity are wrong.

 

"Crossing out" the [math]T[/math]'s on both sides of the equation:

 

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

 

would result in:

 

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

 

Nice :hihi: We can call this the whiteout method of canceling common factors... you just erase any mention of ‘em :)

 

Ok... seriously though... I think you’re probably making a play on the words “crossing out”, but you’re the only one using that phrase so there’s really no point in exaggerating its meaning. I’m saying “cancel” which is the appropriate mathematical term for eliminating a product that equals one. It is not algebraically impossible to cancel (T/T) in the RHS of this identity:

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

[math]\forall \ a,x,T \in \Bbb{R} \ \colon T>1, a>0, T \neq a[/math]

Every reason you give for leaving (T/T) is a reason to choose to leave it and not a reason to prohibit the operation. By the way, I believe the line following the identity properly sets your domain which is necessary to include for reasons I gave in my last post.

 

As far as the other thing I said “canceling the T's in your term”. I’m referring to this term:

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

In particular—the story you told about some student winning a bet because T could not be eliminated from this term using algebra. I contend (and I think I’ve shown) that T can be eliminated from the term so long at T isn’t zero. I realize that doing so goes against what you’re using this term for, but that does not make it algebraically impossible. It is possible. As I said before, it’s just as possible as canceling the T’s in this term:

[math]\frac{e^{\ln(T)}}{T} \ a[/math]

I’m not saying you would do this to your identity. That would defeat any purpose of having the identity. I’m saying by normal rules of algebra, it can be done.

 

 

Now, as for the term on the right, if we cross out those [math]T[/math]'s only,

then we will be left with:

 

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

 

which is "problematic" in several respects. For one thing, the term without the [math]T[/math]'s

shows no evidence that the greatest possible factor [math]T[/math] was extracted, then cancelled,

which means that it does not necessarily represent a prime as does the term on the left.

Possibly a fine reason to *choose* to leave (T/T).

 

Also, there is now no evidence that the term on the left was derived from the term on the right,

which means that the equation can no longer be viewed as an "identity".

 

No. Notice the Pythagorean trigonometric identity,

[math]\sin^2 x + \cos^2 x = 1[/math]

The right side of this identity shows no evidence that it was derived from the left yet it keeps its title “identity”. Examples abound.

 

Most importantly however, the [math]T[/math]'s on the right tell us that it must be possible

to let [math]T=a[/math], which litterally forces us to let [math]x=1[/math]

which in turn allows us to "cross out" the expressions involving logarithms,

so that all we are left with is:

 

[math]T\left(\frac{a}{T}\right)=\left(\frac{T}{T}\right)a[/math]

 

where we may now let [math]T=a[/math], or simply "cross out" the remaining [math]T[/math]'s so as to have:

 

[math]a=a[/math]

 

But T cannot equal a in your identity—it results in division by zero (if x=1 or not). So, I’m not sure where you’re coming from with that. Similarly the rest of your post goes more into that topic which I’ll have to think about to a greater extent.

 

~modest

[Don Blazys]

To: Craig D,

 

Quoting Craig D:

Ah, but you’re a very fun tired,

cantankerous old fuddy duddy, and a valued member of hypography.

 

Thanks. :shrug: Hypography is an incredible science forum,

and being a valued member means a lot to me.

If I collect the "Beal prize", then I will have enough money to

be a generous sponsor. :lol:

 

Quoting Craig D:

Humility, in my experience,

is an easier character trait to claim than to attain.

It’s easy to express humility by acknowledging ones shortcomings.

It’s harder to express humility by recognizing that criticism of

a cherished idea is correct, and the idea is wrong.

 

I agree.

 

Now, why so many in the math community continue to cherish the demonstrably wrong,

totally ineffectual present day method of representing and eliminating common factors

using terms that idiotically allow trivial common factors to creep in is beyond me.

It doesn't match obsereved results, or even common sense and clearly needs fixing!

 

A worldwide show of humility on this matter is definitely in order!

 

As for me, well, I already proved my humility over ten years ago by admitting

that the method for representing and eliminating common factors that I was taught

and had gotten so accustomed to using, was not only wrong,

but was so blatantly wrong as to constitute a joke!

I also had the courage to do something about it,

so I developed a stronger, more logical algebraic term that

doesn't allow trivial common factors to creep in.

 

I still feel that algebraic terms that don't allow trivial common factors,

and whose variables are so well defined that they can actually be

refered to by their domains, are probably the only hope we've got

to make the representation and elimination of common factors meaningfull.

 

So let's explore this possible solution together.

If it turns out that my "Blazys terms" are also inadequate for

representing and eliminating common factors, then I will be the first to admit it.

(Only a fool would stand by a result that is demonstrably wrong.)

However, if that turns out to be the case, then the problem of

meaningfully representing and eliminating common factors will still be with us

and a worldwide show of humility on this matter will still be in order!

 

Don.

[Don Blazys]

To: Modest,

 

Quoting Modest:

As I said before,

it’s just as possible as canceling the T’s in this term:

 

[math]\frac{e^{\ln(T)}}{T}\ * a[/math]

 

The above term, as is, allows us to divide each and every [math]T[/math] by [math]T[/math],

which results in:

 

[math]\frac{e^{\ln(T/T)}}{T/T}\ * a=a[/math]

 

However, the "Blazys term" as is does not allow us to divide each and every

[math]T[/math] by [math]T[/math]. That's a big difference!

 

Quoting Modest:

[math]\sin^2x+\cos^2x=1[/math]

 

The right side of this identity shows no evidence that it was derived from the left

yet it keeps its title “identity”. Examples abound

.

 

[math]\sin^2x=1-\cos^2x[/math]

 

is how I prefer to view it.

 

Quoting Modest:

But T cannot equal a in your identity—

it results in division by zero (if x=1 or not).

So, I’m not sure where you’re coming from with that.

Similarly the rest of your post goes more into that topic

which I’ll have to think about to a greater extent.

 

If:

 

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

 

then letting [math]x=1[/math] results in:

 

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

 

where the entire exponent on the left "cancels out" using the "white out" method!

Thus, at [math]x=1[/math], all we are left with is:

 

[math]T\left(\frac{a}{T}\right)=\left(\frac{T}{T}\right)a[/math]

 

Now, and only now can we let [math]T=a[/math].

 

You see, if [math]T\neq a[/math] were indeed true for all values of [math]x[/math],

then we would have the greatest mystery ever on our hands because

we would then be required to explain exactly why the properties of logarithms

should insanely preclude [math]T=a[/math] when both the term on the right

and common sense tell us that [math]T=a[/math] must be possible

under certain conditions such as [math]x=1[/math]

 

For me, it's a pretty safe bet that the properties of logarithms are not insane!

 

In short, if there is absolutely no condition under which my "Blazys terms" will allow [math]T=a[/math],

then I have discovered a huge flaw in the properties of logarithms because clearly,

[math]T=a[/math] must be allowable!

After all Mr. Napier was a great mathematician. He was not a "nut" , and his invention,

when used properly, could not possibly result in "nutty" preclusions!

 

So you see, it absolutely must be the case that

[math]T=a[/math] is indeed possible if and only if [math]x=1[/math].

 

Don.