A Mathematical Emergency.

Don Blazys · 01-29-2009

Hi Turtle,

And

.

Now, when you wrote:

Quoting Turtle:

Quote:

Substitute your "increase" with "make arrangements of "
and consider that you have actually piles of beans in the number that the
numerals represent. In this case, the 1 * 28 has the expression
" make 1 arrangement of 28" or "make 28 arrangements of 1",
or vice versa if you write 28*1. The product is of course
the sum of all arrangements then made.

you gave me an idea as to how the operation of multiplication can be
rigorously defined in terms of "beans"!

I like beans! They are good for my heart!
The more I eat, the more I'm

And although our lives may not be worth
a hill of beans in this crazy world,

this is our hill, and these are our beans!

There aren't enough "bean counters" here to warrant "bean shaped smilies"!

That's okay, I will use "O" instead. After all, it "kinda" looks like a bean.

Alrighty then!

In this this demonstration, we will use "repeated columns of beans"
to rigorously define multiplication as "repeated addition".
Now, if we call the number of beans in a column the "multiplicand",
and the number of columns the "multiplier",
then the multiplication $3*5$ can be shown as either:

3+3+3+3+3
O O O O O
O O O O O
O O O O O

or

5+5+5
O O O
O O O
O O O
O O O
O O O

Notice that the columns "repeat" in two ways,
as "five columns of three beans"
and as "three columns of five beans".

However, if we apply this definition
to the multiplication $1*5$ ,
then all we have is either:

1+1+1+1+1
O O O O O

or

5
O
O
O
O
O

where it is clear that
the columns "repeat" in only one way,
as "five columns of one bean".

The "one column of five beans" does not "repeat".

It occurs only once.

Thus, unity can be defined as a multiplicand, but not as a multiplier,
and "multiplication of unity" can be defined as the "repeated addition of unity"
while "multiplication by unity" can not be defined as the "repeated addition of five",
or, for that matter, as the "repeated addition" of anything!

Multiplication by unity is simply "undefined"!

Leave it to you

to inspire this most entertaining simplification!

Don.

Turtle · 01-29-2009

Quote:

Originally Posted by Don Blazys

Hi Turtle,
And

.
...
That's okay, I will use "O" instead. After all, it "kinda" looks like a bean.

In this this demonstration, we will use "repeated columns of beans"
in order to rigorously define multiplication as "repeated addition".
Now, if we call the number of beans in a column the "multiplicand",
and the number of times those columns occur the "multiplier",
then the multiplication $3*5$ can be shown as either:

3+3+3+3+3
O,O,O,O,O
O,O,O,O,O
O,O,O,O,O

or

5+5+5
O,O,O
O,O,O
O,O,O
O,O,O
O,O,O

However, if we apply this definition to the multiplication $1*5$ , all we have is either:

1+1+1+1+1
O,O,O,O,O

or

5
O
O
O
O
O

where it is clear that the $1$ column of $5$ beans does not "repeat" as do the $5$ columns of $1$ bean.

Thus, unity can be defined as a multiplicand, but not as a multiplier,
and "multiplication of unity" can be defined as "repeated addition of unity"
while "multiplication by unity" can not be defined as "repeated addition of five."

Leave it to you

to inspire this most entertaining simplification!

Don.

Roger the welcome. This is what the bean-counters pay me the big bucks for.

So to the worsmithery! The mistake here I think is that you have taken my generalized principle term "arrangement" and substituted a specific example term "column". You are correct that one arrangement is a column and the other not, but that is because the other arrangment of five beans is a row. Both arrangments contain a total of 5 beans and one might make a line of 5 separate beans on any compass heading with no effect on their quantity.

Imagine now that you have an endless supply of glass beakers in which to place arrangements of beans, which is to say amounts of beans as they will self-arrange physically in the beaker. Imagine also an endless supply of beans.

Let'c count out 5 beans and put them all in a single beaker. This is multiplication by unity. One beaker of 5 beans. Now we can proceed to take all but 1 of those beans from the beaker and set about seeing if we can put those beans taken out into other beakers such that all the beakers contain the same number of beans. If we need as many beakers as we have beans, the number of beans is Prime. So it is with 5. If we can have beakers with all the same numbers of beans & with more than 1 bean in each beaker, then the number of beans is Composite. For example with 28, we may have 4 beakers each with seven beans or seven beakers each with 4 beans, etcetera. See that the beakers are as variables and it makes no difference to the arrangements of the quantities what position the beakers have to one to another. At any point we may put all our beans in one beaker.

I do hope this has bean more helpful than knot.

Don Blazys · 01-30-2009

To: Modest,

When I wrote:

Quote:

Originally Posted by Don Blazys
Thus, if we allow both "multiplication by unity" and "multiplication of unity",
then $6$ can be factored as:

$6=3*2*1*1$ , where $3+2+1+1=7$

and the entire concept of a "perfect number" simply collapses!

all I meant to convey was that mathematics, in it's present form,
does absolutely nothing to preclude or prevent "indiscriminant" multiplication by unity,
which is, as we have already shown in this thread, "undefined".

Present day mathematics relies on us being polite, considerate and friendly enough
to express a simple number such as "seven", by writing $7$ .

It (present day mathematics) does not take into account the possibility that
mathematicians who are impolite, inconsiderate and unfriendly can,
at any time, choose to annoy the rest of us by expressing "seven" as:

$((1*1)*((1*1)^7)*(1*1^7)*(((1*1*1*7)/(1^7))/1))$

and then claim that it is "merely redundant", but not wrong, or impossible,
or inconsistent with number theory.

"Blazys terms", on the other hand, would never allow such shenanegans
because they simply don't allow "multiplication by unity".

Moreover, "Blazys terms" demonstrate that "multiplication by unity" is
inconsistent with the properties of logarithms and actually results in division by zero!

You see, when I wrote the above quote, I was merely pointing out the fact that
indescriminant multiplications by unity, when performed on numbers that are part of
some legitimate factorization, compromize the integrity of that factorization.

In other words, all I did was take the legitimate proper factorization of $6$
and showed a particular one to one correspondence by writing:

$6=3*2*1$ where $3+2+1=6$ ,

Then, to point out the sheer stupidity of multiplication by unity,
I "indiscriminantly" multiplied the $1$ on the left by $1$ so as to obtain:

$6=3*2*1*1$ where $3+2+1=6$ ,

where the equation on the left now "appears" as a "factorization" but is,
in fact, nothing more than a vaccuous and therefore idiotic statement.

Then, to illustrate the utter absurdity of any attempt to now regain,
under this condition, a one to one correspondence,
I added $1$ to both sides of the equation on the right so as to show:

$6=3*2*1*1$ where $3+2+1+1=7$ .

That's all there is to it!

I was simply illustrating how "redundant but harmless" multiplications by unity
don't belong in number theory.

None of this involved some "new way of factoring", or anything of that sort!

Don.

Turtle · 01-30-2009

Quote:

Originally Posted by Don Blazys

all I meant to convey was that mathematics, in it's present form,
does absolutely nothing to preclude or prevent "indiscriminant" multiplication by unity,
which is, as we have already shown in this thread, "undefined".

Present day mathematics relies on us being polite, considerate and friendly enough
to express a simple number such as "seven", by writing $7$ .

It (present day mathematics) does not take into account the possibility that
mathematicians who are impolite, inconsiderate and unfriendly can,
at any time, choose to annoy the rest of us by expressing "seven" as:

$((1*1)*((1*1)^7)*(1*1^7)*(((1*1*1*7)/(1^7))/1))$

and then claim that it is "merely redundant", but not wrong, or impossible,
or inconsistent with number theory.

...
Don.

All those ones prompted me to recall a discussion of why 1 is not Prime. Four answers are given here, but it's #2 that appears most relevant to your expression here. Boldenation mine.

Why is the number one not a prime?

Quote:

...Answer Two: Because of the purpose of primes.
The formal notion of primes was introduced by Euclid in his study of perfect numbers (in his "geometry" classic The Elements). Euclid needed to know when an integer n factored into a product of smaller integers (a nontrivially factorization), hence he was interested in those numbers which did not factor. Using the definition above he proved:
The Fundamental Theorem of Arithmetic
Every positive integer greater than one can be written uniquely as a product of primes, with the prime factors in the product written in order of nondecreasing size.
Here we find the most important use of primes: they are the unique building blocks of the multiplicative group of integers. In discussion of warfare you often hear the phrase "divide and conquer." The same principle holds in mathematics. Many of the properties of an integer can be traced back to the properties of its prime divisors, allowing us to divide the problem (literally) into smaller problems. The number one is useless in this regard because a = 1*a = 1*1*a = ... That is, divisibility by one fails to provide us any information about a. ...

Don Blazys · 01-31-2009

To: Turtle,

Thanks for the info on Euclid and why unity is not a prime. When you wrote:

Quoting Turtle:

Quote:

Many of the properties of an integer can be traced back to the properties of its prime divisors,
allowing us to divide the problem (literally) into smaller problems.
The number one is useless in this regard because a = 1*a = 1*1*a = ...
That is, divisibility by one fails to provide us any information about a. ...

you were not just "whistling Dixie"!

Not only does "multiplication by unity" fail to provide us any information about
the properties of integers, it doesn't even provide us with any information as to
what operation is being performed!

"Multiplication by unity" is absolutely indistinguishable from "division by unity",
so both must be undefined.

$a*1=a/1$ ,

so multiplying and/or dividing by unity renders the
multiplication and division symbols utterly meaningless. Moreover,

$1*(a+b)=1*a+b=a+1*b=1*a+1*b$ ,

so there is no way to know if "multiplication and/or division by unity" is "distributive".

Multiplying and/or dividing by unity is like peddling a bicycle with a broken chain! It's a joke!
Sometimes I think that it has the one redeeming quality of being funny

,
but other than that, I can't think of even thing that it is good for. Now that "Blazys terms"
are making the rounds, how otherwise astute and talented mathematicians can continue
to subscribe to "multiplication and/or division by unity" is beyond me.

One of the really cool things about "Blazys terms" is that they actually demonstrate that
unity can be a multiplicand but not a multiplier, and this, in turn, allows us to define
"prime number" as:

"Any positive integer that can be both a multiplicand and a multiplier
whose only factors are unity and itself".

(Note that unity is excluded because it can't be both a multiplicand and a multiplier!)

Now, let's get back to "wordsmithing".
"Wordsmithing" is important!
Had mathematicians not found Euclids fifth postulate to be

"too wordy"
and therefore

"not worthy",
they would not have conducted further investigations

,
and would not have discovered "non-Euclidean geometry".

So, let us now contemplate the definition or "word equation":

multiplication=repeated addition.

Notice that without the word "repeated", we would simply have:

multiplication=addition,

and that, of course, would be just plain silly

and therefore "unacceptable".
Clearly, the word "repeated", which means: "having occured more than once",
or, in mathematical language: " $>1$ " is crucial to the definition of multiplication,
so we must now investigate

into how it applies in your

"bean and beaker" model.

Quoting Turtle:

Quote:

Imagine now that you have an endless supply of glass beakers
in which to place arrangements of beans, which is to say amounts of beans as they will
self-arrange physically in the beaker. Imagine also an endless supply of beans.

Let's count out 5 beans and put them all in a single beaker.
This is multiplication by unity. One beaker of 5 beans.
Now we can proceed to take all but 1of those beans from the beaker and
set about seeing if we can put those beans taken out into other beakers
such that all the beakers contain the same number of beans.
If we need as many beakers as we have beans, the number of beans is Prime.
So it is with 5.
If we can have beakers with all the same numbers of beans
& with more than 1 bean in each beaker, then the number of beans is Composite.
For example with 28, we may have 4 beakers each with seven beans or
seven beakers each with 4 beans, etcetera. See that the beakers are as variables
and it makes no difference to the arrangements of the quantities what position
the beakers have to one to another. At any point we may put all our beans in one beaker.

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:

" $5$ beans evenly distributed in $5$ beakers",

which can be viewed as "multiplication of unity" where unity is the "multiplicand", or

" $5$ beans in $1$ beaker",

which, as you pointed out, can be viewed as "multiplication by unity", where unity is
the "multiplier".

Now, it is clear that in the first case, both beans and beakers are "repeated",
as there are $5$ of each, and $5>1$ .
However, in the second case, the beaker is not "repeated" as there is only $1$ .
Thus, unity can not be a multiplier (beaker) but only a multiplicand (bean).
Beans always repeat, but beakers don't.

Your "bean and beaker" model has made all this more clear than ever.

Thanks

Don.

Turtle · 01-31-2009

Quote:

Originally Posted by Don Blazys

To: Turtle,

Now, let's get back to "wordsmithing".
"Wordsmithing" is important!
...
Now, it is clear that in the first case, both beans and beakers are "repeated",
as there are $5$ of each, and $5>1$ .
However, in the second case, the beaker is not "repeated" as there is only $1$ .
Thus, unity can not be a multiplier (beaker) but only a multiplicand (bean).
Beans always repeat, but beakers don't.

Your "bean and beaker" model has made all this more clear than ever.

Thanks

Don.

On the boldened I have to ask, "what beaker?". Why the beaker I already mentioned, i.e. the 1 you repeated when you mentioned it. As soon as you refer to any "it", you have repeated "it". As I say though, I don't see what you're onto with the logrithms (when they mentioned it was upcoming in math class I thought we were going to drum on trees so I skipped.

), so I'll keep reading along for awhile. If by any chance you can use your Blazy constant to help me root out some Strange Anomolies, I'd be forever in your debt.

modest · 01-31-2009

Quote:

Originally Posted by Don Blazys

It (present day mathematics) does not take into account the possibility that
mathematicians who are impolite, inconsiderate and unfriendly can,
at any time, choose to annoy the rest of us by expressing "seven" as:
$((1*1)*((1*1)^7)*(1*1^7)*(((1*1*1*7)/(1^7))/1))$

But, if we’re allowed to use the normal axioms of the natural numbers where N•1=N, then the unwieldy thing above can be simplified.

Quote:

Originally Posted by Don Blazys

Then, to point out the sheer stupidity of multiplication by unity,
I "indiscriminantly" multiplied the 1 on the left by 1 so as to obtain:

6=3*2*1*1 where 3+2+1=6,

where the equation on the left now "appears" as a "factorization" but is,
in fact, nothing more than a vaccuous and therefore idiotic statement.

Ok, I think we’re on the same page. Multiplying a factor by unity does not imply adding one to its divisors.

Quote:

Originally Posted by Don Blazys

Moreover, "Blazys terms" demonstrate that "multiplication by unity" is
inconsistent with the properties of logarithms and actually results in division by zero!

I’m not too sure what a Blazys term is, I joined this thread only recently. What I’ve seen is the identity:

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

To me, this doesn’t imply anything about multiplication by unity. As long as $T \neq 0$ , both sides simplify algebraically to $a^x$ . Is this identity the source of “Blazys terms”?

Looking at the rhs of your identity,

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

And, for ease of readability, let’s look only at the exponent,

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

Substituting 1 = ln(t)/ln(t)

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

Subtracting fractions with like bases and rearranging,

$\frac{x\ln(a)-\ln(T)}{\ln(a)-\ln(T)}$

substituting $x\ln(a) = \ln(a^x)$ using $\log_{b} (x^y) = y \log_{b}(x)$ (source),

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

using $\log_{b}(x)-\log_{b}(y) = \log_{b}(x/y)$ ,

$\cfrac{\left(\ln\Big(\cfrac{a^x}{T}\Big)\right)}{\left(\ln\Big(\cfrac{a}{T}\Big)\right)}$

using $\log_{a}(b)=\frac{\log_{c}(b)}{\log_{c}(a)}$ ,

$\log_{a/T}\left(\frac{a^x}{T}\right)$

This is log base (a/T) of (a^x/T). The right hand side of your identity is now,

$T\left(\frac{a}{T}\right)^{\left(\log_{a/T}\left(a^x/T\right)\right)}$

using $b^{\log_{b}(x)}=x$ ,

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

The whole identity then,

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

Where $T\neq0$ your identity simplifies to:

$a^x=a^x$

Like I said before, I’m not a mathematician. I don’t know what implications this has (and I’d be very appreciative of any opinion offered on that), but I can’t imagine it holds any deep consequences against multiplication by unity. Unless (0/0) is considered unity, I don’t see how the above is applicable at all.

~modest

Don Blazys · 01-31-2009

To: Turtle

Quoting Turtle:

Quote:

On the boldened I have to ask, "what beaker?".
Why the beaker I already mentioned, i.e. the 1 you repeated when you mentioned it.
As soon as you refer to any "it", you have repeated "it".

The really cool thing about your "bean and beaker" model is that
the definition:

multiplication = repeated addition

as it applies to unity can be demonstrated using physical objects.
We need not think of the word "repeated" as "mentioned" or "refered"
but simply as "more than one bean or beaker",
and as we will soon find out, "more than one sound"!
So, let's actually perform this demonstration as a "laboritory experiment",
using real beans and beakers! Got your beans and beakers ready? Okay, here we go!

Quoting Turtle:

Quote:

Imagine now that you have an endless supply of
glass beakers in which to place arrangements of beans,
which is to say amounts of beans as they will
self-arrange physically in the beaker.
Imagine also an endless supply of beans.

Let's count out 5 beans and put them all in a single beaker.
This is multiplication by unity. One beaker of 5 beans.

Now, pick up each and every beaker, and shake it once.
All you will hear is "one sound" from the beaker containing the 5 beans.
The rest of the beakers will not produce a sound because they are empty.
Most importantly, note that the sound was not repeated. It occured only once.

Quoting Turtle:

Quote:

Now we can proceed to take all but 1of those beans
from the beaker and set about seeing if we can put those beans taken out
into other beakers such that all the beakers contain the same number of beans.

This gives us 5 beakers containing one bean each.
Again, pick up each and every beaker, one at a time, and shake it once.
You will now hear 5 sounds from the five different beakers containing one bean each.
Most importantly, note that the sound was repeated. It occured five times.

Our "laboritory experiment" is now over.
We can now take our pens and clipboards and record our observations and results as follows:

First observation:

The sound that was not repeated occured when the "beakermultiplier" count was " $1$ "
and the "beanmultiplicand" count was " $5$ in a beaker".
Conclusion: If multiplication = repeated addition = repeated sound,
then $1$ is not a multiplier because $1$ "beakermultiplier" resulted in
a "sound" or "addition" that was not repeated.

Second observation:

The sound that was repeated occured when the "beakermultiplier" count was $5$ ,
and the "beanmultiplicand" count was " $1$ in each beaker".
Conclusion: If multiplication = repeated addition = repeated sound,
then $1$ is a multiplicand because $1$ "beanmultiplicand" in each beaker
resulted in a "sound" or "addition" that was repeated.

Quote Turtle:

Quote:

As I say though, I don't see what you're onto with the logrithms
(when they mentioned it was upcoming in math class I thought
we were going to drum on trees so I skipped. ), so I'll keep reading along for awhile.
If by any chance you can use your Blazys constant to help me root out
some Strange Anomolies, I'd be forever in your debt.

We both share a deep interest in mathematical anomolies. Here's one:
The derivatives of "Blazys terms" can be modified to result in "prime counting functions"
that are more accurate than Li(x) to as far as Pi(x) has been calculated!
What makes them particularly special is that they are contained in either one or two terms!
Are you talking about stuff like that?

Don

Don Blazys · 02-01-2009

To: Modest,

Let's assume that all variables herein represent non-negative integers.

Quoting modest:

Quote:

I’m not too sure what a Blazys term is, I joined this thread only recently. What I’ve seen is the identity:

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

The term:

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

is indeed a "Blazys term".

It's also known as a "cohesive term", and the properties that set it apart from
"non-cohesive" terms are in my "Proof of the Beal Conjecture" that you can find
on my website, (donblazys.com).

One of it's hallmarks is that intrinsically, it's variables are perfectly defined
by their own unique domains as:

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

Compare this to the non-cohesive term:

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

whose variables are intrinsically defined by their domains as:

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

Now, before we move on to how all this effects "multiplication by unity",
we must first answer a few questions. (Perhaps you or someone else will help me.)
The questions are as follows:

(1) Which term has the better defined variables?

(2) Is it logical to proceed from perfectly defined variables to poorly defined variables
and call it a "simplification"?

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

Don.

modest · 02-01-2009

Quote:

Originally Posted by Don Blazys

(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.

Quote:

Originally Posted by Don Blazys

(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:

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

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

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

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 $a^x$ , 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 $T \neq a$ , because that would result in division by zero. Yet, we can avoid this restriction by constructing the identity differently. Here’s another example,

$\frac{T}{T}a^x=\left(\frac{Ta}{x}\right )^{\left(\cfrac{x\ln(a)-\ln(T)+\ln(Ta)-\ln(a)}{ln(a)-\ln(x)+\ln(Tx)-\ln(x)}\right)}$

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 $(T/T)a^x$ and applying a bunch of logarithmic identities. One result of all the substitutions and identities is to restrict the domain of the variables. Consider,

$x = (x + x)/2$

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:

$(xT) = ((xT) + (xT))/2$

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

$x = 10^{\log_{10}(x)}$

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,

$(xT) = 10^{\log_{10}(xT)}$

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,

$x=-4, T=3, \; (xT) = (-4)(3) = -12$

$x=2, T=0, \; (xT) = (2)(0) = 0$

$x=1, T=-3, \; (xT) = (1)(-3) = -3$

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.

Quote:

Originally Posted by Don Blazys

(3) Can the cancelled $T$ '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 $T \neq 0$ . 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:

$\frac{T}{T}a^x=T\left(\frac{a}{T}\right)^{\left(\log_{a/T}\left(a^x/T\right)\right)}$

The exponent is a log with a base equal to the base of the exponent. It's taking the log of $a^x/T$ 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: