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Old 01-15-2009   #92 (permalink)
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Don Blazys
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Re: A Mathematical Emergency.

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.

Last edited by Don Blazys; 01-20-2009 at 07:11 PM..
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