A Mathematical Emergency.

Don Blazys · 02-03-2009

To: Modest,

Quoting Modest:

Quote:

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:

Quote:

Which term has the better defined variables?

That means the symbols: $T, a$ and $x$ .

You know, when I was a taxi driver

, 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":

$\left(T*a\right)^{\left(\frac{\frac{x\ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1}\right)}$

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

$T=\{2, 3, 4...\}$ , $a=\{1, 2, 3...\}$ and $x=\{0, 1, 2...\}$

then I can say "Consider the variable whose domain is $\{0, 1, 2...\}$ ,
and immediately, everybody knows that I am referring to the variable $x$ .

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

$Ta^x$

whose variables are intrinsically defined as:

$T=a=x=\{0, 1, 2...\}$

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 $\{0, 1, 2...\}$ ,
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,

and this would lead to both anger and frustration.

Then, quite possibly, a fight would ensue,

and somebody would get badly injured, or even killed!

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 $T>1$ .
"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 $T$ 's.

When you did that, you changed my logical statement that says:
"This term guarantees that $T>1$ and will not allow trivial common factors."

into your own illogical statement that says:
"This term does not guarantee that $T>1$ and will allow trivial common
factors."

Try eliminating the $T$ 's in the "Blazys term" equation:

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

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 $T$ 's on both sides of the equation be "cancelled out"
meaning "crossed out" and made to "disappear"?
Correct answer: "No".

Quoting Modest:

Quote:

Nothing about this constrains x against being any real number.

This tread is about non-negative integers only.

Quoting Modest:

Quote:

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 · 02-06-2009

Quote:

Originally Posted by Don Blazys

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 $1*5$ can be expressed as either:
...
Don.

Roger the Euclid & erhmmm...no.

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.

modest · 02-06-2009

Quote:

Originally Posted by Don Blazys

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...

Quote:

Originally Posted by Don Blazys

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:

$T=T\frac{\sqrt{T-1}}{\sqrt{T-1}}$

~modest

Don Blazys · 02-07-2009

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 $1*5=5$ ,
which can also be expressed as $5*1=5$ ,
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" $1$ ,
and end up with the "product" $5$ ,
then we can be absolutely certain that a multiplication by the "multiplier" $5$
did indeed occur, because the "product" $5$
is "different" from the "multiplicand" $1$ ,
and therefore qualifies as a "new value".

However, if we begin with the "multiplicand" $5$ ,
and end up with the "product" $5$ ,
then we can't be certain that a multiplication by the "multiplier" $1$
ever occured, because the "product" $5$
is exactly the same as the "multiplicand" $5$ ,
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!

Wow, I did it!

Don.

Don Blazys · 02-07-2009

To: Modest,

Quoting Modest:

Quote:

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:

Quote:

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 $\left(\frac{T}{T}\right)a^x$
for the "Blazys term" in order to get $a^x=a^x$ !

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:

$E=MC^2$ to simply $E=E$ ?

Why not then divide both sides by $E$ and further "reduce" it to $1=1$ ?

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:

Quote:

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:

$T=T\frac{\sqrt{T-1}}{\sqrt{T-1}}$

The "restrictions" in your term:

$T=T\frac{\sqrt{T-1}}{\sqrt{T-1}}$

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

Not so in an as is "Blazys term"!

Don.

Turtle · 02-07-2009

Quote:

Originally Posted by Don Blazys

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 · 02-07-2009

Quote:

Originally Posted by Don Blazys

Quote:

Originally Posted by 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.

Quote:

Originally Posted by Don Blazys

Quote:

Originally Posted by 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!

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,

$\frac{T}{T}a^x$

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

$T\left(\frac{a}{T}\right)^{\left(\frac{\frac{x\ln(a)}{\ln(T)}-1}{\frac{\ln(a)}{\ln(T)}-1}\right)}$

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

$\frac{T}{T}a^x$

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.

Quote:

Originally Posted by Don Blazys

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,

$\frac{T}{T}$

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

$\frac{e^{\ln(T)}}{T}$

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?

Quote:

Originally Posted by Don Blazys

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 $a \neq 0$ 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,

$x=e^{\ln{x}}$

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

Quote:

Originally Posted by Don Blazys

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 $\left(\frac{T}{T}\right)a^x$
for the "Blazys term" in order to get $a^x=a^x$ !

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

$T\left(\frac{a}{T}\right)^{\left(\frac{\frac{x\ln(a)}{\ln(T)}-1}{\frac{\ln(a)}{\ln(T)}-1}\right)}$

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

Quote:

Originally Posted by Don Blazys

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 $\left(\frac{T}{T}\right)a^x$ does not imply $T \not= a$ or $a \not= 0$ 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.

Quote:

Originally Posted by Don Blazys

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

$E=MC^2$ to simply $E=E$ ?

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 · 02-07-2009

To: Turtle,

Quoting Turtle:

Quote:

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 · 02-07-2009

To: Modest,

Quoting Modest:

Quote:

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":

$T\left(\frac{a}{T}\right)^{\left(\frac{\frac{x\ln(a)}{\ln(T)}-1}{\frac{\ln(a)}{\ln(T)}-1}\right)}$

Now, you come along and write:

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

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.

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 · 02-08-2009

Quote:

Originally Posted by Don Blazys

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:

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

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

$\frac{5}{5}5^2=5\left(\frac{5}{5}\right)^{\left(\frac{\frac{2\ln(5)}{\ln(5}-1}{\frac{\ln(5)}{\ln(5)}-1}\right)}$

$25=\textrm{unde}\textrm{fined}$

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.

Quote:

Originally Posted by Don Blazys

...Now, you come along and write:

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

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

A Mathematical Emergency.

Hypography?

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