A Mathematical Emergency.

Don Blazys · 01-13-2009

To: Modest,

Thanks.

Thanks a lot!

I was seriously considering hiring someone to show me how to post in LaTex,
so you probably just saved me some hard earned money.

Now, I'm off to practice.

Don.

Don Blazys · 01-15-2009

Due to numerous requests and "popular demand", I will now re-write the
summary of this topic/thread so that the equations are presented in "LaTex".

However, before I do so, I would like to make a few comment's.

The title of this topic/thread is in no way "inflated", "exagerated" or "overblown".

If students are being taught the wrong way to represent and eliminate common factors,
then that's a very serious matter,
and does indeed constitute a genuine "mathematical emergency"!

We simply can not allow nonsensical gibberish to be taught in our schools!

A lot of math departments of famous colleges and universities have known about
this result/discovery for years, but have yet to take any remedial action because
they are simply too embarrassed to admit that what they are teaching is wrong!

My advise for them is to "get over it", because truth is a very impartial thing.
It's bigger than all of us and doesn't care one iota about our "feelings".
Most importantly, it always wins in the end!

Now, here is what each and every student and teacher should know
about the proper representation and elimination of common factors.
__________________________________________________ __________________________________________________

Given the equation:

$Ta^x+Tb^y=Tc^z$ ,

where all the variables are positive integers, how do we eliminate the
common factor $T>1$ so that all three terms become "co-prime"?

Well, these days, students are being taught that we should divide each and every $T$ by $T$ ,
then "cross out" or "cancel out" the $T$ 's so that they "disappear". Doing so gives us:

$\left(\frac{T}{T}\right)a^x+\left(\frac{T}{T}\right)b^y=\left(\frac{T}{T}\right)c^z=a^x+b^y=c^z$ .

However, this is wrong because it falsely implies that $x, y$ and $z$ can all be greater than 2
when there is no common factor.

In reality, and as is stated in the "Beal Conjecture", when there is no common factor,
then we must have a "restriction" on either $x, y$ or $z$ so that either:

$x=$ {1, 2}, $y=$ {1, 2} or $z=$ {1, 2}

So, where did we go wrong?

Well, we never actually prevented $T=1$ .

You see, preventing $T=1$ is important, because a true or "non-trivial" common factor $T$
is defined as $T>1,$ and our equations must be consistent with our definitions
if we are to make any real progress and discover the truth.

Now, watch what happens if we refuse to prematurely "cross out" the cancelled $T$ 's,
and instead, re-write the co-prime equation:

$\left(\frac{T}{T}\right)a^x+\left(\frac{T}{T}\right)b^y=\left(\frac{T}{T}\right)c^z$ ,

so that it appears as either:

$\left(\frac{T}{T}\right)a^x+\left(\frac{T}{T}\right)b^y=T\left(\frac{c}{T}\right)^{\left(\frac{\frac{z\ln(c)}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}\right)}$ ,

or (in it's factored form),

$\left(\left(\frac{T}{T}\right)a^{(\frac{x}{2})}\right)^2 +\left(\left(\frac{T}{T}\right)b^{(\frac{y}{2})}\right)^2= \left(T \left(\frac{c}{T}\right)^{\left(\frac{\frac{(\frac{z}{2})\ln(c )}{\ln(T)}-1}{\frac{\ln(c)}{\ln(T)}-1}\right)}\right)^2$ .

Immediately, we find that by substituting just one "Blazys term",
we automatically eliminate any possibility that $T=1$ ,
and for the first time in the history of mathematics,
render our equations perfectly consistent with the definition of a non-trivial common factor.

Now, take a good close look at the last three equations.

Notice that the first one tells us that $T=c$ is allowable while the next two tell us that
before we can allow $T=c$ , we must first let $z=1$ and $z=2$ respectively,
then immediately "cross out" or "cancel out" the logarithms themselves.
Thus, the last three equations must now appear as:

$\left(\frac{T}{T}\right)a^x+\left(\frac{T}{T}\right)b^y=\left(\frac{T}{T}\right)c^z$ ,

$\left(\frac{T}{T}\right)a^x+\left(\frac{T}{T}\right)b^y=T\left(\frac{c}{T}\right)$ ,

and

$\left(\left(\frac{T}{T}\right)a^{(\frac{x}{2})}\right)^2 +\left(\left(\frac{T}{T}\right)b^{(\frac{y}{2})}\right)^2= \left(T \left(\frac{c}{T}\right)\right)^2$ .

Now, and only now can we allow $T=c$ , or we can simply "cross out" or "cancel out"
the remaining $T$ 's, so that the above three equations appear as:

$a^x+b^y=c^z$ ,

$a^x=b^y=c$

and

$a^x+b^y=c^2$ .

Notice that the first of the above three equations is a lie because it falsely implies that
if we add together any two co-prime numbers $a^x, (x>2)$ and $b^y,(y>2)$ ,
then we might get a result $c^z$ , such that $z>2$ .

The other two equations tell us the truth, which is that
if we add together any two co-prime numbers $a^x, (x>2)$ and $b^y, (y>2)$ ,
then the exponent of $c$ must be either 1 or 2.

Try it yourself! Add together any two co-prime positive integers under the sun,
whose exponents are greater than 2,
and you will find that their sum will always have an exponent that is either 1 or 2.

Most importantly, notice that "indeterminate forms" such as 0/0 are never ever encountered
if we do the algebra correctly and "cross out" or "cancel out"
the expressions involving logarithms
the very moment that we let $z=1$ and $z=2$ .

Believe it or not, there are some mathematicians who don't think it's possible to "cross out"
or "cancel out" the expressions involving logarithms at $z=1$ and $z=2$ .

I think that they are mistaken.

I think that conjuring up "indeterminate forms" that don't even exist is just plain silly.

Not only is this the correct way to represent and eliminate common factors,
but it also shows that problems such as the "Beal Conjecture" and "Fermat's Last Theorem"
would never have been an issue and would therefore never even have existed
had mankind learned how to properly represent and eliminate common factors to begin with!

Thus, it is quite understandable that many in the "math community"
find this irrefutable result to be somewhat embarrassing.

It's the true and simple solution to those supposedly "hard" problems!

Don.

logy · 01-21-2009

Don:

I am beginning to see that you have a valid point there. I must admit that the math over there is beyond my skills, but I can see how, when doing a certain type of algebraic menipulation on a function that involves natural logarithms, and a function of the type $a^x+b^y=c^z$ something special (and beyond my understanding) happens.

BUT…what about the more obvious things?
And I will give a few examples
1. $\left(\frac{T}{T}\right)F(x)$ if not the same as $F(x)$ because T might be equal to zero, in which case $\frac{T}{T}$ is indeterminate !
2. When working with exponents $a^x$ is not always legal, since if a is a negative number the exponent cannot be a fraction since $(-1)^{0.5}$ has no real answer.
3. $\sqrt{x^2}$ is not the same as x, since if x is negative, it will become a positive !

this list can go on forever, i just came up with a couple examples from my fairly basic knowledge in math, but I believe that this kind of concepts are absolutely essential if you want to understand basic mathematics.

How many students are aware of the fundamentals? How many year 12 students who have taken calculus are capable doing simple differentiation using first principals?
How many students know where e, the natural logarithm comes from or are capable of writing and explaining for formula for the series that generates e?
(its $\lim_{n\to 0}(1+\frac{1}{n})^n$ by the way. just punch into a grahpics calculator $(1+(\frac{1}{(2^{96})}))^(2^{96}))$ :

(1+(1/(2^96)))^(2^96) = 2.7182818284590452353602874713355

e^1 = 2.7182818284590452353602874713527
(This was fun

)

I would even challenge the teachers here to see what percentage of their students know that 'ln' is simple $log_e$ (!)

If I where to use the word emergency, I would use it for this kind of thing. this is not something that concerns mathematicians, but also scientists and engineers (those are the epoeple that have made it possible for this forum to exist through microchips and magnetic drives and stuff like that). They use this level of math all the time and yet a large portion of them are clueless when it comes to actually understanding this level of mathematics.

Don Blazys · 01-22-2009

To: Logy,

Mathematics does not have a "king", but it does have a "queen",
and that queen is "number theory".

Number theory is different from other branches of mathematics in that
it deals primarily with the properties of non-negative integers.

Most importantly, in number theory, concepts such as "the fundamental theorem of arithmetic",
"common factors", "prime numbers", "perfect numbers", "abundant numbers", etc.
all become utterly meaningless if we allow unity to be both a "multiplier" and a "multiplicand".

For example, in number theory, a "perfect number" can be defined as:
"A positive integer such that the sum of its positive proper factors is equal to itself".
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!

However, if we allow only "multiplication of unity" and strictly prohibit "multiplication by unity",
then $6$ must be factored as:

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

and the concept of a "perfect number" remains intact.

You see, the "queen of mathematics" herself is telling us that "multiplication by unity"
is a really dumb idea because it not only destroys important concepts in number theory,
but it also accomplishes absolutely nothing and shouldn't even be considered an "operation"!

Now, "coheseve terms" (also known as "Blazys terms") were developed for use in number theory.
They are superior to "conventional" algebraic terms in that they don't allow multiplication by unity,
They are also superior in other ways.
For instance, if we consider the actual meaning of the equation:

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

where all the variables represent non-negative integers,
then it is clear that the greatest possible factor $T$
was "extracted", then "eliminated" (meaning "cancelled"),
so that both sides must now represent the same "prime number".

However, prime numbers do not have exponents, so we are now stuck with
the seemingly impossible task of eliminating the exponent.

Again, the "Blazys term" on the right comes to the rescue by telling us that
before we can allow $T=a$ (which common sense tells us must be allowable),
we must first let $x=1$ , then immediately "cross out" or "cancel out"
the expressions involving logarithms, which leaves us with:

$\frac{T}{T}a=T\left(\frac{a}{T}\right)$ ,

which simplifies to:

$a=a$ .

So not only do "Blazys terms" provide us with a more meaningfull and accurate
representation of "common factors" (as we saw in post #92),
but they also give us a more meaningfull and accurate representation of factors....period!

To put it simply, without "Blazys terms", even our most simple equations will wrongly imply that
eliminating the greatest possible factor from a number does not necessarily result in a prime!

You know Logy, you are absolutely right in your assertion that a lot of "high level"
mathematicians don't really understand the underlying principles of the mathematics
that they use, depend on and teach every day.

They are like parrots, mynah birds and

who do everything by rote,
mindlessly regurgitate only what they have been taught,
and never ever bother to actually reason, or think things through for themselves.

They are absolutely blind to the fact that "Blazys terms" have the potential to unify
number theory (where multiplication by unity is prohibited), and set theory, algebra,
calculus, etc. (where multiplication by unity is allowed).

You are also right in that the "mathematical emergency" does indeed extend far beyond
the idiotic way in which factors and common factors are presented to students.

Now here is a question for you.

Since it has now been demonstrated that the way in which factors and common factors
are presented to students is every bit as unscientific as "creationism",
shouldn't the courts intervene and put an end to it, just as they did with
the "theory" of "intelligent design"?

Don.

logy · 01-22-2009

to: Don

the math is getting too time consuming for me. If one day i get into number theory, and am still active in this forum i will revisit this topic.

In response to your question, for all i know, you may well be correct. if someone where to come to me 1000 years ago and say that the earth was flat i would dismiss him, but not because his idea is rediculous, rather because it doesnt make sense, how come the people at the bottom don't fall off?

only once i see a good explanation would I consider a new idea. (gravity for the above exmple would exlain why people don't fall off) . I believe that this notion is shared among most scientifically oriented people.

modest · 01-22-2009

Quote:

Originally Posted by Don Blazys

For example, in number theory, a "perfect number" is defined as:
"A number such that the sum of its proper factors is equal to itself".
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!

Why would the number 1 want to be included in the divisors twice? The divisors of 6 are {-6,-3,-2,-1,1,2,3,6}. A perfect number is “a positive integer which is the sum of its proper positive divisors”—proper meaning to exclude the number itself. The proper positive divisors of 6 are then {1,2,3}.

I think including 1 twice would assume the divisors are being multiplied, which isn't the case. That 1 • 2 • 3 = 6 (or 1 • 1 • 2 • 3 = 6) is coincidental as far as perfect numbers go. The next perfect number, 28, sums perfectly (28 = 1 + 2 + 4 + 7 + 14), but doesn't multiply perfectly (28 $\neq$ 1 • 2 • 4 • 7 • 14).

So, I'm confused what you did there. And, also in the case of the fundamental theorem of arithmetic, an example such as $432 = 2^4 \cdot 3^3$ doesn't include the factor 1 because 1 isn't prime. My only guess is that you're thinking $432 = 1 \cdot 2^4 \cdot 3^3$ wouldn't be unique if $432 = 1^2 \cdot 2^4 \cdot 3^3$ were allowed. Is that the case? I'm probably way off base.

In any case, the multiplicative identity follows as second order logic from the Peano axioms of the natural numbers. So, I don't see how it is at all incompatible with number theory. The wikipedia page on Peano axioms under the arithmetic section has:

Quote:

It is easy to see that 1 is the multiplicative identity:
a • 1 = a • (S(0)) = a + (a • 0) = a + 0 = a

and using more common logic syntax, axiom #7 under Equivalent axiomatizations:

Quote:

7. $\forall x \in \mathbb{N}. \ x \cdot 1 = x$ i.e., one is the identity element for multiplication.

My understanding is that one is included in the natural numbers, so the above explicitly allows 1 x 1 = 1... which I can't reconcile with your statement about number theory not allowing "multiplication by unity and multiplication of unity".

~modest

Don Blazys · 01-23-2009

To: Modest,

Please allow me to both clarify and elaborate.

If we are to be consistent in our our use of nomenclature,
then the words "proper factor" must mean the exact same thing as "proper divisor",
because in the case of "proper divisor",
the word "proper" is clearly taken to mean "other than itself".

Also, as mathematicians, if there is one thing that we absolutely must agree on,
it's that we must always remain consistent in our mathematical definitions.

However, in both the "math community" and on the internet,
there is indeed a whole lot of inconsistency, disagreement and confusion
on what the words "proper factor" and "proper divisor" should mean,
which only adds to the "urgency of the mathematical emergency".

Definitions should be both simple and easy to understand,
and since multiplication is "easier" and "more simple" than division,
let us henceforth define "perfect numbers" as I did in post #94,
using "proper factors" rather than "proper divisors".

Now, according to the only logically consistent definition of a "proper factor",
the proper factors of $28$ are:

$1, 2, 4, 7$ and $14$ ,

and $28$ is called a "perfect number" because:

$1+2+4+7+14=28=(1*2*4*7*14)^{\left(\frac{1}{2}\right)}$ .

Note that the proper factors of $6$ "multiply perfectly" as a "first root",
while the proper factors of $28$ "multiply perfectly" as a "second root".
If we continue in this manner, then we will find that the proper factors of
the next perfect number, $496$ , also "multiply perfectly", but as a "fourth root",
and that in fact, all perfect numbers "multiply perfectly" as some "Nth root".
(It's one of the "deeper" reasons why these numbers are called "perfect".)
Thus, the argument that I presented in post #94 is not
"coincidental as far as perfect numbers go", but applies throughout.

Now, let's take another look at the above equation and note that
the numbers being added together on the far left are an "exact match"
with the numbers being multiplied together on the far right.

Clearly, we have a "one to one correspondence" and a "symmetry",
that we dare not break, lest we destroy the foundations of number theory itself.

Therefore, as serious seekers of both truth and beauty,
we are now compelled to come up with some reasonable model, paradigm,
or explanation, as to why unity should occur as a factor... only once.

After all, considering the incredible depth
of the relationships that abound in number theory,
it would not be unreasonable to suppose
that a reasonable explanation does indeed exist,
and that it is both significant and profound.

Do we have any clues? Can we compile some evidence?

You betcha and yes we can!

Consider the following:

In number theory, multiplication is strictly and stringently defined as "repeated addition".
Thus, the multiplication: $3*5$ is viewed as either:

$3+3+3+3+3$ or $5+5+5$ ,

where the number itself is called the "multiplicand",
and the number of times that it occurs is called the "multiplier".

However, applying this definition to the multiplication: $1*5$ yields either:

$1+1+1+1+1$ or $5$ ,

and it is now obvious that unity can be defined as a multiplicand,
but not as a multiplier, because clearly, the representation: $5$
qualifies as neither a "repeated addition", nor a "repeated" anything!

This is absolutely indisputable, utterly irrefutable evidence that
"multiplication and/or division by unity can not be defined.

It is so simple that it can be understood by virtually all preschoolers,
yet it is ignored by virtually the entire mathematical community!

Thus, the "queen of mathematics" has spoken both clearly and decisively. She said:
"Let this not go in one ear and out the other. Multiplication by unity is undefined!"

By contrast, the "Peano" notion of unity being a "multiplicative identity"
or "identity element" is entirely artificial and without foundation, because in truth,
it does nothing more than present us with a rather vaccuous statement
that unity as a factor is "essentially superfluous".
It completely ignores the above direct evidence, which is, in fact, incontravertible,
and in no way addresses the underlying significance of what actually occurs.

However, the most compelling and convincing evidence that
"multiplication and or division by unity is undefined" is the "Blazys term" itself,
for at $T=1$ , it actually demonstrates, and clearly shows,
that multiplication and/or division by unity results in division by zero!

Thus, given the following four unquestionable facts:

(Fact #1): The properties of logarithms exist and are valid.

(Fact #2): "Blazys terms" are a logical and therefore unavoidable consequence
of the properties of logarithms.

(Fact #3): At $T=1$ , "Blazys terms" demonstrate that
multiplication and/or division by unity results in division by zero.

(Fact #4):"Blazys terms" demonstrate that multiplication is non-commutative
with respect to unity, and that unity can exist only as a multiplicand.

It logically follows that:

(1): The properties of logarithms demonstrate
that multiplication and/or division by unity
is exactly as undefined as "division by zero",
and that unity can exist only as a multiplicand,
so that multiplication must be non-commutative
with respect to unity.

and

(2): The Peano axiom regarding "multiplicative identities"
is inconsistent with and contradicted by the properties of logarithms.

Don.

modest · 01-25-2009

Quote:

Originally Posted by Don Blazys

To: Modest,

Please allow me to both clarify and elaborate.

Quote:

Originally Posted by Don Blazys

If we are to be consistent in our our use of nomenclature,
then the words "proper factor" must mean the exact same thing as "proper divisor",
because in the case of "proper divisor",
the word "proper" is clearly taken to mean "other than itself".

Also, as mathematicians, if there is one thing that we absolutely must agree on,
it's that we must always remain consistent in our mathematical definitions.

However, in both the "math community" and on the internet,
there is indeed a whole lot of inconsistency, disagreement and confusion
on what the words "proper factor" and "proper divisor" should mean,
which only adds to the "urgency of the mathematical emergency".

Definitions should be both simple and easy to understand,
and since multiplication is "easier" and "more simple" than division,
let us henceforth define "perfect numbers" as I did in post #94,
using "proper factors" rather than "proper divisors".

Well, that's a rather large preamble... and I'm not exactly sure I see the point. Perfect numbers have been defined for over 2,000 years. Factors, for longer. If there's some online confusion and inconsistency... I'm completely out of the loop on that.

Quote:

Originally Posted by Don Blazys

Now, according to the only logically consistent definition of a "proper factor",
the proper factors of $28$ are:

$1, 2, 4, 7$ and $14$ ,

and $28$ is called a "perfect number" because:

$1+2+4+7+14=28=(1*2*4*7*14)^{\left(\frac{1}{2}\right)}$ .

Note that the proper factors of $6$ "multiply perfectly" as a "first root",
while the proper factors of $28$ "multiply perfectly" as a "second root".
If we continue in this manner, then we will find that the proper factors of
the next perfect number, $496$ , also "multiply perfectly", but as a "fourth root",
and that in fact, all perfect numbers "multiply perfectly" as some "Nth root".
(It's one of the "deeper" reasons why these numbers are called "perfect".)
Thus, the argument that I presented in post #94 is not
"coincidental as far as perfect numbers go", but applies throughout.

First, let me say, that's really cool, Don.

It appears to be true (or, at least, the first few have no counterexample). I've written a little algorithm to test it up to the 7th perfect number which was successful. The roots were 1,2,4,6,12,16,18. My brain is not working well-enough this Sunday morning to figure why that works... any opinion on that would be much-appreciated.

Second, I should say... what you have there is not a definition of a perfect number. It appears I was incorrect using the term "coincidental", but it is, nevertheless, superfluous. A perfect number is an "integer which is the sum of its proper divisors". It's really that simple. You could say the integer (n) is half its divisor function ( $2n = \sigma(n)$ ), or equal to its restricted divisor function ( $n = s(n)$ ).

Perfect Number -- from Wolfram MathWorld

But, you cannot say it is "equal to some integer root of the product of its factors". The integer 30 would qualify under that definition as would many others.

$(1\cdot2\cdot3\cdot5\cdot6\cdot10\cdot15)^{1/3} = 30 \neq (1+2+3+5+6+10+15)$

Moreover, I see nothing about the following equality that prohibits multiplication by unity.

$(1+2+4+7+14) = (1\cdot2\cdot2\cdot4\cdot7\cdot14)^{1/2}$

I think you want to imply that multiplying the rhs by one would necessitate adding one to the lhs. If that's the case, I don't follow. One is a factor of 28 once.

Quote:

Originally Posted by Don Blazys

Therefore, as serious seekers of both truth and beauty,
we are now compelled to come up with some reasonable model, paradigm,
or explanation, as to why unity should occur as a factor... only once.

Because $1\cdot1\neq28$ . One is a factor of 28 because (1 • 28) = 28. You could alternatively write (28 • 1) = 28, but that doesn't make 1 a factor twice any more than it doubles up any of the other factors.

Either I'm missing something big or there's nothing about number theory that precludes multiplication by unity. Since multiplication by unity is included in the axioms of natural number arithmetic, I'm pretty confident that it isn't precluded.

~modest

Turtle · 01-25-2009

Hey Don. Hey Modest.

Just thought I'd pop in with some tid-bits I'm somewhat familiar with.

When I saw the summing of Perfect factors here, I recalled that the sum of the reciprocals of the proper factors of a Perfect number equals 2 (including the reciprocal of the number itself). Do you guys recall that? Does it shed any light on the discussion?

e.g. >> $\frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{7}+\frac{1}{14}+\frac{1}{28} = 2$

When I saw Modest mention 30 satisfying some of Don's criteria, I recalled that it is one of my "Strange" numbers, which is to say it is abundant by 12. Again, I don't know if any of that forwards this conversation.

So I'm not too up on the logarithms but I have been reading all the posts here and maybe some of it will sink in.

If we need them, we have threads on Perfect numbers and the Strange numbers.

Perfect Numbers: An Aural Dissertation

Strange Numbers

modest · 01-25-2009

Ok, my brain has had time to warm up. What Don points out is not a property of perfect numbers and it's not "one of the "deeper" reasons why these numbers are called "perfect".". If fact, every number with an even number of factors will follow Don's little rule. I pulled 30 out of my hat, but consider its factors:
(1 • 30) = 30
(2 • 15) = 30
(3 • 10) = 30
(5 • 6) = 30
Each of these multiply to the number in question. So, the result of multiplying all the factors must be some multiple of the number being factored

A perfect number can't be a perfect square so it must have an even number of factors. Really Don?

~modest

A Mathematical Emergency.

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