A Mathematical Emergency.

[Don Blazys]

To: Qfwfq.

I apologize for my continuous failure to post my equations using LaTex. My "computer skills" are rudimentary at best, so I beg you to remain both patient and curious and to bear with me as I am doing my best to improve those skills under somewhat difficult circumstances and conditions. There is nothing that I want more than to have other mathematicians understand what I am presenting, and I really, really appreciate it when good folks such as yourself and CraigD take the time to contribute to this discussion.

 

If you can find the time to visit my website (donblazys.com), then please do so. (You can also find it by doing a "Google search" on "cohesive terms".) It was put together by the computer teachers at the school where I work and contains a lot more information on the equation (identity) that we are discussing here.

 

Now, here's why this really is a "cool new discovery", why it does indeed have "profound implications" for mathematics, and why I named this thread "A Mathematical Emergency".

 

You see, at this very moment, students throughout the world are being taught that in order to render the terms in the equation:

 

Ta^x+Tb^y=Tc^z

 

relatively prime or "co-prime", we must first divide by T, then "cross out" the T's. However, doing so results in:

 

(T/T)a^x+(T/T)b^y=(T/T)c^z=a^x+b^y=c^z,

 

which is wrong and inconsistent with both the "Beal Conjecture", and all observed results, because it falsely implies that x, y, and z can all be greater than 2 after the largest possible common factor T has been cancelled and the terms are co-prime.

Now, the proper, correct and effective way to render the terms "co-prime" is as follows:

Instead of "crossing out" the T's, we actually use them to derive at least one "cohesive term" so that it can be firmly established that the cancelled common factor T>1. This gives us three possibilities, one of which is:

 

(T/T)a^x+(T/T)b^y=(T/T)c^z=

 

T(c/T)^((zln©/(ln(T))-1)/(ln©/(ln(T))-1)).

 

Factoring does not involve the cancelled variable T, and results in:

 

((T/T)a^(x/2))^2+((T/T)b^(y/2))^2=((T/T)c^(z/2))^2=

 

(T(c/T)^(((z/2)ln©/(ln(T))-1)/(ln©/(ln(T))-1)))^2,

 

where it is now clear that before we do anything else, we must first let z=1 in the unfactored case, and z=2 in the factored case, so that we can "cross out" and eliminate the logarithms that are preventing us from letting T=c, which common sense tells us must be allowable. This gives us both:

 

(T/T)a^x+(T/T)b^y=(T/T)c=T(c/T),

 

and

 

((T/T)a^(x/2))^2+((T/T)b^(y/2))^2=((T/T)c)^2=(T(c/T))^2,

 

where T=c is now clearly allowable. Simplifying, we note that our co-prime equations:

 

a^x+b^y=c,

 

and

 

(a^(x/2))^2+(b^(y/2))^2=c^2

 

are now perfectly consistent with both the "Beal Conjecture", and all observed results, because the implication that x, y and z can all be greater than 2 after the largest possible common factor T has been cancelled no longer exists.

Thus, by using a "cohesive term" to guarantee that the common factor T>1, we prevent the T's from being "crossed out" prematurely, and automatically prove both the "Beal Conjecture", and "Fermat's Last Theorem" (which involves only the "special case" where x=y=z).

Had mankind learned how to properly represent and eliminate common factors to begin with, (using at least one "cohesive term" to ensure that any common factor is "non-trivial") problems such as the BC and FLT would never have surfaced!

That, to me, is profound.

 

Don.

[CraigD]

(T/T)a^x+(T/T)b^y=(T/T)c^z=

 

T(c/T)^((zln©/(ln(T))-1)/(ln©/(ln(T))-1)).

 

Factoring does not involve the cancelled variable T, and results in:

 

((T/T)a^(x/2))^2+((T/T)b^(y/2))^2=((T/T)c^(z/2))^2=

 

(T(c/T)^(((z/2)ln©/(ln(T))-1)/(ln©/(ln(T))-1)))^2,

This step is algebraically incorrect. It should be

 

((T/T)a^(x/2))^2+((T/T)b^(y/2))^2=((T/T)c^(z/2))^2=

 

T^2 (c/T)^((((z/2)ln©/(ln(T))-1)/(ln©/(ln(T))-1)))2)

 

Rendered more readably

[math]\left(\frac{T}{T}a^\frac{x}2 \right)^2 +\left(\frac{T}{T}b^\frac{y}2 \right)^2 = \left(\frac{T}{T}c^\frac{z}2 \right)^2= T^2 \left(\frac{c}{T} \right)^{\frac{\frac{z}2\frac{\ln©}{\ln(T)}-1}{\frac{\ln©}{\ln(T)}-1}2}[/math]

 

This is because [math](a^b)^c = a^{bc}[/math]. [math]\left( a^b \right)^c \not= a^{\left(b^c\right)}[/math].

[Kharakov]

P.S. What kind of calculator do you have that is accurate to 18 decimal places, or is such accuracy available only in conjunction with a computer? Are they expensive?

If you want something cheap and simple to use, Window's powertoys has a calculator upgrade with 32 digit precision.

 

http://www.microsoft.com/windowsxp/downloads/powertoys/xppowertoys.mspx

 

Click on the powercalc.exe link.

 

Or you can just search microsoft.com for power calculator (or powertoys). Works nice. Just make sure to switch view to scientific (from standard) to get the enhanced features.

[Qfwfq]

I apologize for my continuous failure to post my equations using LaTex.

Put your mouse pointer over some of Craig's equations and you will see the code (when it isn't too long); you should be able to figure enough for your purposes here. To use it, just slap it iside MATH or IMATH tags as follows:

 

[MATH]\frac{x+y}{a+\ln b}[/MATH] or [iMATH]a^{x+y} + B_n[/iMATH]

 

If you can find the time to visit my website (donblazys.com), then please do so.

I already did, even printed the two .pdf files to look at them at home but, as I said Friday, I don't get why you say that

 

[MATH]\frac{T}{T}a^x+\frac{T}{T}b^y=\frac{T}{T}c^z=a^x+b^y=c^z[/MATH]

 

is wrong. :confused: The rest of your post is much like in the file.

 

This step is algebraically incorrect.

Er, I don't see the mistake in:

 

[math]\left(\frac{T}{T}a^\frac{x}2 \right)^2 +\left(\frac{T}{T}b^\frac{y}2 \right)^2 = \left(\frac{T}{T}c^\frac{z}2 \right)^2= T^2 \left(\frac{c}{T} \right)^{\frac{\frac{z}2\frac{\ln©}{\ln(T)}-1}{\frac{\ln©}{\ln(T)}-1}2}[/math]

 

As well as his famous identity, he isn't applying: [math]\left( a^b \right)^c = a^{\left(b\;^c\right)}[/math] and he is applying: [math]\left( a^b \right)^c = a^{bc}[/math].

[Don Blazys]

To:CraigD. In post number 13 of this thread, I wrote:

 

(T/T)a^x+(T/T)b^y=(T/T)c^z=

 

T(c/T)^((zln©/(ln(T))-1)/(ln©/(ln(T))-1)).

 

Factoring does not involve the cancelled variable T, and results in:

 

((T/T)a^(x/2))^2+((T/T)b^(y/2))^2=((T/T)c^(z/2))^2=

 

(T(c/T)^(((z/2)ln©/(ln(T))-1)/(ln©/(ln(T))-1)))^2.

 

This is algebraically correct. It is not a mistake.

Now when you wrote: "It should be:

 

((T/T)a^(x/2))^2+((T/T)b^(y/2))^2=((T/T)c^(z/2))^2=

 

T^2(c/T)^((((z/2)ln©/(ln(T))-1)/(ln©/(ln(T))-1)))2),"

 

you inadvertently put in an odd number of parenthesis, that is, thirteen ")", but only twelve "(".

 

I'm sure that this was just a "typo", and that what you really meant to write was either:

 

T^2(c/T)^(((z/2)ln©/(ln(T))-1)/(ln©/(ln(T))-1)2),

 

or:

 

T^2(c/T)^((zln©/(ln(T))-2)/(ln©/(ln(T))-1)),

 

which is exactly what we get when we remove the outermost parenthesis in my original post!

 

If your intent was indeed to eliminate the outermost parenthesis in my original version, then your LaTex rendering is correct, and the logarithmic exponent that you wrote, which is:

 

z ln©

------- -1

2 ln(T)

-------------- 2

ln©

------- -1

ln(T)

 

can also be written as:

 

zln©

------- -2

ln(T)

-------------

ln©

------- -1

ln(T)

 

I did not make an algebraic mistake.

 

Don.

[Don Blazys]

To: Kharakov. Thanks for the information. Much obliged.

 

Don.

[Don Blazys]

To: Qfwfq.

 

(T/T)a^x+(T/T)b^y=(T/T)c^z=a^x+b^y=c^z

 

is wrong because it is incomplete.

 

You see, the above suggests that once the greatest common factor T has been cancelled and the terms are co-prime, then it is still possible to have:

x>2, y>2 and z>2, all at the same time.

 

It's not!

 

In reality, when the terms are co-prime, we must have either:

x={1,2}, y={1,2} or z={1,2}. In other words, at least one of the three exponents must be either 1 or 2 after the greatest common factor T has been cancelled.

 

To put it bluntly, the above representation is lying, because it says that there is no restriction on the value of either x,y or z when in reality, there is!

 

It is also "lame" because it allows T=1 when in reality, 1 is not a "common factor"!

 

Thus, the truth comes to light only after we substitute either:

 

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

 

T(b/T)^((yln(:)/(ln(T))-1)/(ln(:doh:/(ln(T))-1))

 

or

 

T(c/T)^((zln©/(ln(T))-1)/(ln©/(ln(T))-1))

 

for one of the terms in the equation:

 

(T/T)a^x+(T/T)b^y=(T/T)c^z.

 

Note that the terms involving logarithms make a lot more sense because they actually require that the cancelled common factor T>1.

 

This is a very serious matter, because our present methods of representing and eliminating common factors is clearly inadequate, and should not be taught in our schools.

 

Don.

[Qfwfq]

First, Don, sorry if I couldn't show you better yesterday how to use LaTeX; the example meant to be:

 

[MATH]\frac{x+y}{a+\ln b}[/MATH] or [iMATH]a^{x+y} + B_n[/iMATH] to render as:

 

[MATH]\frac{x+y}{a+\ln b}[/MATH] or [iMATH]a^{x+y} + B_n[/iMATH]

 

The imath tag is slightly better for small things in lines of text.

 

That said, and before discussing what you mean by a few of the things you say, I'd like you to answer a couple of quick questions. First: How would you set about calculating the following limit?

 

[MATH]\lim_{x\rightarrow 1}\frac{2x^6-2x^5+x^4+4x^3-12x^2+11x-4}{3x^5+2x^4-5x^3+4x^2-12x+8}[/MATH]

 

Second: Is the following an identiy, or would one strictly have to specify something?

 

[MATH]a=e^{\ln a}[/MATH]

 

Actually, I would further ask you whether the above has any implication about the trivial identity:

 

[MATH]a=a[/MATH]

[CraigD]

Put your mouse pointer over some of Craig's equations and you will see the code (when it isn't too long); you should be able to figure enough for your purposes here.

Even easier, rather than clicking on the [New Reply] button, Click on the

button at the bottom of a post containing a rendered equation. You can then copy (in most browsers, highlight the [math]…LaTeX stuff…[/math], rightclick and chose Copy) the equation. LaTeX is pretty intuitive – though little if at any easier to read before it’s rendered than plain text with lots of ()s.

 

The quote button also provides a nice alternative to posting text like "In post #n, xxx said:"

This step is algebraically incorrect.

Er, I don't see the mistake in:

 

[math]\left(\frac{T}{T}a^\frac{x}2 \right)^2 +\left(\frac{T}{T}b^\frac{y}2 \right)^2 = \left(\frac{T}{T}c^\frac{z}2 \right)^2= T^2 \left(\frac{c}{T} \right)^{\frac{\frac{z}2\frac{\ln©}{\ln(T)}-1}{\frac{\ln©}{\ln(T)}-1}2}[/math]

This is algebraically correct. It is not a mistake.

Don and QfwfQ are correct – once again, I’m so disoriented by all the nested ()s in it, I’m not able to effective read Don’s math, and think it’s wrong when it’s not :doh:

 

:) I like the note in post #22, which elaborates prettily into

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

 

However, I don’t’ see how these identities provide or give any direction toward a proof or disproof of Fermat’s last ([math]a^n +b^n \not= c^n, a,b,c,n \in \mathbb{Z}, a,b,c>0, n>2 [/math] or Beal’s conjectures ([math]a^x +b^y = c^z, a,b,c,n \in \mathbb{Z}, A,B,C>0, x,y,z>2[/math] only if [math]A, B[/math] and [math]C[/math] have a common prime factor).

 

These conjectures are challenging because they are restricted to the integers ([math]\mathbb{Z}[/math]), not the reals ([math]\mathbb{R}[/math]). If [math]a,b,c \in \mathbb{R}[/math], it’s trivial to disprove either conjecture by constructed example (eg: [math]1^3 +1^3 = (\sqrt[3]{2})^3[/math])

 

Like any logarithm function,

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

Is true of for [math]a,x,T \in \mathbb{R}[/math], requiring only [math]T \not=a,1[/math] and [math]T,a>0[/math]. I don’t see how it can be restricted to the intergers in some way that makes it useful for proofs of Fermat’s last, Beals, or similar conjectures.

 

Don seems to me to be hinting that [math]\frac{a}{a} \not=1[/math] and/or [math]1 \cdot a \not= a[/math], negating a couple of the most fundamental postulates of arithmetic, which makes no sense to me at all. Also, I can’t make sense of statements like

Had mankind learned how to properly represent and eliminate common factors to begin with, (using at least one "cohesive term" to ensure that any common factor is "non-trivial") problems such as the BC and FLT would never have surfaced!

Fermat’s and Beale’s conjectures are either true or false (for the handful of people who actually understand the proof, Fermat’s last conjecture actually is true). Even if proven, how could these and similar conjectures not “surface”?

 

Either I’m missing something, or Don’s hints really are leading nowhere. :confused:

[Don Blazys]

To: Qfwfq.

 

At x=1,

 

2x^6-2x^5+x^4+4x^3-12x^2+11x-4

-------------------------------------

3x^5+2x^4-5x^3+4x^2-12x+8

 

= 0/0, which is what we mathematicians refer to as an "indeterminate form". We can, however, show that in this particular case, the "indeterminate form" 0/0=1.25.

 

All we have to do is simply divide the derivative of the numerator by the derivative of the denominator and again let x=1. Thus we have:

 

12x^5-10x^4+4x^3+12x^2-24x+11

-------------------------------------

15x^4+8x^3-15x^2+8x-12

 

which at x=1 gives us 1.25.

 

This handy little technique is known as "L'Hopitals rule", and is quicker and more accurate than trying to "approach" x=1 as a "limit".

 

As for the equation:

a=e^(lna),

it is indeed an identity for all "natural numbers":

a={1,2,3...},

whereas the trivial:

a=a

is an identity for all "non-negative integers":

a={0,1,2...}.

 

Don.

[Don Blazys]

To: CraigD.

Any particular independent variable constitutes essentially one symbol, as does any particular non-negative integer. Therefore, independent variables must "naturally" assume non-negative integer values only, and allowing them to assume other values (as is often done) is really just an unnatural "shortcut" that in certain cases may even have the potential to undermine the integrity of a higher order mathematical construct.

Thus, if we want our math to be "perfectly rigorous", then we absolutely must insist that our independent variables represent non-negative integers only, and that all other "reasonable" numbers (negatives, fractions, irrationals, transendentals, imaginaries, complex, and so on) then be derived by performing operations on them, and them alone.

With this in mind, please check out my proof of the Beal Conjecture on my website (donblazys.com).

Also, thanks for the added info on LaTex.

 

Don.

[Qfwfq]

This handy little technique is known as "L'Hopitals rule", and is quicker and more accurate than trying to "approach" x=1 as a "limit".

This is a statement that we mathematicians consider totally inaccurate. Don, de L'Hopital's theorem is about limits. How in the world can it be an alternative to "trying" to approach [imath]x=1[/imath] as a limit? :xx:

 

In any case, it is one way of calculating the limit but there is also an algebric method since numerator and denominator are both plain polynomials; the fact that they have a common zero at [imath]x=1[/imath] implies them having a common factor equal to [imath]x-1[/imath], therefore division (see Ruffini's method) leaves no remainder:

 

[math]\frac{2x^6-2x^5+x^4+4x^3-12x^2+11x-4}{3x^5+2x^4-5x^3+4x^2-12x+8}=\frac{(x-1)(2x^5+x^3+5x^2-7x+4)}{(x-1)(3x^4+5x^3+4x-8)}[/math]

 

As you can see, for all [imath]x\neq 1[/imath] and for which it is defined, the fraction I gave has the same value as the one in here:

 

[math]\forall x\neq 1:\,\frac{2x^6-2x^5+x^4+4x^3-12x^2+11x-4}{3x^5+2x^4-5x^3+4x^2-12x+8}=\frac{2x^5+x^3+5x^2-7x+4}{3x^4+5x^3+4x-8}[/math]

 

The fact is that the rhs does have a value at [imath]x=1[/imath], while the lhs is discontinuous there and so only has a limit but not a value. So, strictly, the above identity is a correct one only under the restriction [imath]x\neq 1[/imath] specified.

 

As for the equation:

a=e^(lna),

it is indeed an identity for all "natural numbers":

a={1,2,3...},

whereas the trivial:

a=a

is an identity for all "non-negative integers":

a={0,1,2...}.

Don, we mathematicians don't get why you are restricting to integers and naturals, neither why you restrict to non-negative even in the case of [imath]a=a[/imath]. Surely you aren't proposing to teach students in this manner?

 

The identity [imath]a=e^{\ln a}[/imath] is correct [imath]\forall x > 0,\, x\in\mathbb{R}[/imath].

 

The identity [imath]a=a[/imath] is correct [imath]\forall x\in\mathbb{R}[/imath].

 

Let's forget about the subtleties of complex values! You failed to address the last thing I had asked: does the identity [imath]a=e^{\ln a}[/imath] have any bearing on [imath]a=a[/imath]? As you can see, the domain of validity of the first does not restrict the second one. Likewise can be said about the two polynomial fractions, the discontinuity of the one, wherein it has no value, does not by any means imply the other not being valid for [imath]x=1[/imath].

[Don Blazys]

To:Qfwfq.

 

When I wrote that applying L'Hopitals rule is quicker and more accurate than trying to "approach" x=1 as a "limit", I simply meant that it is easier to evaluate the expression involving derivatives at x=1 than it is to evaluate the original expression at x=1.00000000001 or x=.99999999999 or some such approximation to 1.

 

I certainly didn't mean to imply that applying L'Hopitals rule automatically allows us to dispose of the notion of limits. (I might be crazy, but I'm not that crazy!)

 

Now, the question is, at x=1, does 0/0=1.25?

 

I say yes, because I don't subscribe to the view that at x=1, the indeterminate form 0/0 necessarily constitutes a "discontinuity". For me, the result 0/0 is absolutely consistent with the result 1.25, so the two results can, in that particular case, be set equal to each other.

In other words, from my point of view, if one method of evaluation at x=1 yields 0/0, while another, equally valid method of evaluation at x=1 yields 1.25, then maintaining consistency actually requires that the two results be regarded as "interchangeable" in that particular case.

 

As for why I restrict the variable "a" in the identity a=a to non-negative integers, well, I am of the opinion that the restriction is both "built in" and "natural".

You see, non-negative integers are the only numbers that can possibly be represented using essentially one symbol such as "a". (All other numbers require some operation as part of their representation).

Now, since every imaginable and reasonable number can be derived by performing operations on non-negative integers, if we reserve the variable

"a" to represent non-negative integers only, then we can derive all other numbers by performing operations on "a".

Thus, if we wish to represent "negative integers", then we should do so by writing "(-a)" rather than just "a".

By the same token, if we want to represent "unit fractions", then we should write "1/a" instead of just "a".

"Negative unit fractions" are better represented as "(-1/a)" than simply by "a", and so on.

It's just a matter of being concise, precise and rigorous, not to mention logical and artistic.

I am well aware of the fact that this amount of rigor is quite "impractical" and "unnecessary" for most everyday problems in mathematics.

Therefore, I would not present the above facts to students who have no interest in my research.

 

Don.

[Qfwfq]

I simply meant that it is easier to evaluate the expression involving derivatives at x=1 than it is to evaluate the original expression at x=1.00000000001 or x=.99999999999 or some such approximation to 1.

You might notice I did not do such a pointless thing, I did something else instead. Which one is quicker is a matter quite beside my point and depends much on specific case.

 

Now, the question is, at x=1, does 0/0=1.25?

No. It does not determine a value. Any value multiplied by 0 gives 0; this goes for 1.25 but also for countless others. Therefore it makes no sense to assert the = sign.

 

I say yes,

You are even going against your previous contention.

 

In other words, from my point of view, if one method of evaluation at x=1 yields 0/0, while another, equally valid method of evaluation at x=1 yields 1.25, then maintaining consistency actually requires that the two results be regarded as "interchangeable" in that particular case.

The other(s), which yield(s) 1.25, is not equally valid; it is a different thing. It is a computation of the limit but not of any value.

 

Therefore, I would not present the above facts to students who have no interest in my research.

Thank goodness!

 

Once [imath]\mathbb{Z}[/imath] has been constructed from [imath]\mathbb{N}[/imath] and then [imath]\mathbb{Q}[/imath] and then [imath]\mathbb{R}[/imath] and [imath]\mathbb{C}[/imath], I really don't see why we should only give a single-symbol name to natural numbers such as 823. Is it really a single symbol before we choose to call it a? :shrug:

[Don Blazys]

To:Qfwfq.

 

Well, I guess that we will just have to agree to disagree that the left hand side of your equation in post #29 is "discontinuous" at x=1.

 

I will maintain that it is "continuous" at x=1.

 

By the way, I'm not alone in my view.

 

If you Google search:

 

(limits and continuity "we discuss a number of functions"),

 

then you will find a discussion (example 4.2, #4) about the similar expression:

 

(x^2-a^2)/(x-a)

 

and why it is reasonable to view it as "continuous" at x=a.

 

I will also maintain my point of view that any non-negative integer such as 823 constitutes essentially "one symbol" because it is "purely numerical", while a number such as -823 constitutes essentially "two symbols" because it is comprised of both an operation (-), and a pure number (823),... and that since any independent variable such as "x" and any non-negative integer such as 823 share the property of constituting essentially "one symbol", independent variables are "naturally suitable" for representing non-negative integers only.

 

You know, even though we mathematicians disagree on these particular issues, there are other issues that we do agree on.

 

For instance, we both agree that students who are not interested in my research should not be forced to learn the mathematics that it involves. However, thanks to good folks such as yourself who are interested enough in my work to post on my topics, my topics are now recieving a lot of attention in this and in other forums. (I even have a "record breaking" math topic on the Marilyn vos Savant forum!) As a result, many students (and teachers) are becoming very interested in my research, my website is getting many thousands of visitors, and I am now getting more e-mails from around the world than I can possibly answer.

 

Some of those e-mails are quite interesting and entertaining.

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. More importantly however, he then decided to change his major from political science to math!

 

So you see, my research is worth it.

 

Don.

[Qfwfq]

then you will find a discussion (example 4.2, #4) about the similar expression:

 

(x^2-a^2)/(x-a)

 

and why it is reasonable to view it as "continuous" at x=a.

Well, it seems you meant #3, anyway notice that [imath]f(x)[/imath] is defined in such a way as to be continuous.

 

While the expression [imath]\frac{x^2-a^2}{x-a}[/imath] does not define any value int the [imath]x=a[/imath] case, the expression [imath]2a[/imath] does. The judicious reader may easily see that [imath]f(x)[/imath] is defined by one or the other expression according to the value of [imath]x[/imath]. It's as simple as that, it's the reason why discontinuity of third species is (loosely) called eliminatable; strictly it isn't the same function with or without the discontinuity.

 

I will also maintain my point of view that any non-negative integer such as 823 constitutes essentially "one symbol" because it is "purely numerical", while a number such as -823 constitutes essentially "two symbols" because it is comprised of both an operation (-), and a pure number (823),...

Would you also maintain a point of view that 823 not shorthand for [imath]8\cdot 10^2+2\cdot 10+3[/imath]?

 

You might choose to give all natural numbers an arbitrary name such as klurbfampheryqusdeckpghanmegg, not composed of tokens like eighthundredandtwentythree is. Couldn't the same be done for minuseighthundredandtwentythree as well? No law would dictate it would have to be minusklurbfampheryqusdeckpghanmegg rather than pakdangleflvsaquourpgoose or something...:shrug:

 

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.

Define what exactly you mean by "it is algebraically impossible to 'cross out' the T's" because I'm not sure exactly. Would you say the same in a case such as:

 

[math]\frac{e^{1-\ln 0}+T-1}{1+T+e^{i\pi}}[/math]

 

for example?

[Turtle]

I have no mathematical opinion on Don's expression, but I do endorse the mathematical exploration that led to it. :)

 

On the matter of 823 and a single symbol, one only has to change the base to achieve that. Any base over eight-hundred-twenty-three necessarily assigns a single symbol to every value less than that. The numeral is not the number. :)