Don Blazys

To: Megabrain, In my opinion, you have made some very astute (and courageous) observations. Quoting Megabrain: The great mathematician Kronicker once remarked: "The integers (meaning positive) were created by God, all else is the work of man." The ancients simply dismissed "negative numbers" as "folly" and viewed them as merely a superfluous "mirror image" of positive numbers. Negative numbers became accepted and adopted only as a matter of convenience and "convention", and as such have become "necessary". However, I agree that there is something quite "unnatural" about them. In my opinion, they are simply not interesting, and I personally find them rather "ugly". There is a reason why concepts such as "prime numbers", "perfect numbers", "abundant numbers" etc. don't include negative integers! Don.

An Elementary Proof Of Both The Beal Conjecture And Fermat's Last Theorem. By: Don Blazys The Beal Conjecture can be stated as follows: For positive integers: [math]a, b, c, x, y, [/math] and [math]z[/math], if [math]a^x+b^y=c^z[/math], and [math]a, b[/math] and [math]c[/math] are co-prime, then[math]x, y[/math] and [math]z[/math] are not all greater than [math]2[/math]. Proof: Letting all variables herein represent positive integers, we form the equation: [math]c^z-b^y=a^x[/math].__________________________________________________________(1) Factoring (1) results in: [math]\left(c^\frac{z}{2}+b^\frac{y}{2}\right)\left(c^\frac{z}{2}-b^\frac{y}{2}\right)=a^x[/math].______________________________________________(2) Here, it will be assumed that the terms in (1) and (2) are co-prime, and that the only "common factor" they contain is the "trivial" unity, which can not be defined in terms of itself, and must therefore be defined as: [math]1=\left(\frac{T}{T}\right)[/math], where [math]T>1[/math]._________________________________________________(3) Re-stating (1) and (2) so that the "trivial common factor" [math]1=\left(\frac{T}{T}\right)[/math] and its newly discovered logarithmic consequences are represented, we now have both: [math]T\left(\frac{c}{T}\right)^{\left(\frac{\frac{z\ln©}{\ln(T)}-1}{\frac{\ln©}{\ln(T)}-1}\right)}-\left(\frac{T}{T}\right)b^y=\left(\frac{T}{T}\right)a^x[/math] __________________________________(4) and: [math]\left(T\left(\frac{c}{T}\right)^{\left(\frac{\frac{\frac{z}{2}\ln©}{\ln(T)}-1}{\frac{\ln©}{\ln(T)}-1}\right)}+\left(\frac{T}{T}\right)b^\frac{y}{2}\right)\left(\left(\frac{T}{T}\right)c^\frac{z}{2}-\left(\frac{T}{T}\right)b^\frac{y}{2}\right)=\left(\frac{T}{T}\right)a^x[/math].__________(5) At this point, we note that the definition of unity in (3) implies: [math]1=\left(\frac{T}{T}\right)=\left(\frac{c}{c}\right)[/math], which clearly means that [math]T=c[/math] must be allowable. We also note that the logarithms preventing [math]T=c[/math] "cancel out" and therefore cease to exist if and only if [math]z=1[/math] in (4), and [math]z=2[/math] in (5), which gives us both: [math]T\left(\frac{c}{T}\right)-\left(\frac{T}{T}\right)b^y=\left(\frac{T}{T}\right)a^x[/math] ___________________________________________(6) and: [math]\left(T\left(\frac{c}{T}\right)+\left(\frac{T}{T}\right)b^\frac{y}{2}\right)\left(T\left(\frac{c}{T}\right)-\left(\frac{T}{T}\right)b^\frac{y}{2}\right)=\left(\frac{T}{T}\right)a^x[/math].____________________(7) [math]T=c[/math] is now clearly allowable, and simplifying (6) and (7) shows that the original equations, as stated in (1) and (2), are now: [math]c-b^y=a^x[/math],__________________________________________________________(8) and: [math]\left(c+b^\frac{y}{2}\right)\left(c-b^\frac{y}{2}\right)=c^2-b^y=a^x[/math],_______________________________________(9) which proves not only the Beal Conjecture, but Fermat's Last Theorem (which is only the special case where [math]x=y=z[/math]) as well.

The concept of "nothing" or "no thing" is simply inconcievable. By definition, there is no frame of reference by which it can be communicated or even experienced. It simply doesn't exist, and that fact/truth, in and of itself, forces existence into being and sustains all other concepts and perhaps even the physical universe as well. Don.

To: Bignose, If I start a new thread on the BC, then can you post my proof in the very next post in LaTex so that other readers will find it easier to read? (I prefer versions #1 and #2) My website is perfectly safe to visit. It was created by the computer teachers at a Catholic high school. Don.

To:Bignose, It might be a few days or so before I can post again as I have to go in for some tests and scans to determine the progress of my Alzheimers. If this is my final post here, then thanks for remaining interested throughout. Don.

To: Bignose. You don't understand the difference detween meaningfull and trivial. Show me an example of a number other than a non-negative integer that does not require an "extra symbol built-in operation" and I will believe you! By your "logic", I can represent not just any number, but the entire universe using only the one symbol @. Utterly trivial. Don. To: D H The great mathematician Karl Gauss once remarked that: "Mathematics is largely a point of view." From my point of view, my "true colors", are "true mathematics". My topics, are both quite entertaining and therefore quite popular! I will soon be posting a straight forward formula that is even more astonishing than my "cohesive term": (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) which really is astonishing because it is the first and only elementary algebraic term in the history of mathematics that algebraically prevents us from "crossing out" the cancelled T's, and therefore engenders all kinds of interesting questions such as: "Which term is more relevant with regards to the properties of non-negative integers?", "Which term has the better defined variables?" and: "Does the above true equation show that multiplication/division by unity automatically results in division by zero?" Like I said before, letting T=1 sure seems to suggest that, doesn't it? From my humble point of view, these questions are truly profound and the only motivations for eliminating, moving or locking my topic would be fear and/or jealousy. Don. To: !Now. Thank you. It is real math, and people really are learning and asking questions! I'm new to posting in forums, and I'm not exactly sure what "OP" means, but from the context, I figure it means me. Which point of mine was proven wrong over and over? I'm really doing my best to bring only true (and therefore interesting) things to this forum so if I know where I made a mistake, then I will correct it immediately, I promise. Don. To: Captain Panic. I agree that Pi based number systems are both functional and widely used. They are also fun! I sometimes use them in my own research. Don.

To: Bignose. My logic is not wrong. Yours is! When you say that "Pi" can be represented as the one symbol "1" in the base Pi system, you are hillariously wrong! In the base "Pi" system, the value "Pi" is represented as the two symbols "10". Besides, symbolizing "Pi" as "10" does not constitute an evaluation of "Pi". Pi is transendental and therefore can't be evaluated in any base. Pi can only be approximated. Thus, to take any "approximate number" as a base is to render even the most simple numbers impossible to evaluate exactly. For instance, how do you represent the value "7" in the base Pi system? It takes an infinite number of symbols, doesn't it? Thus, all such "wierd" number system bases are rather contrived, aren't they? When I said that Pi requires an infinite number of symbols in order to be represented, I meant represented meaningfully. "10" is not a meaningfull representation of "Pi", any more than "#" is a meaningfull representation of the expression: 5^(1/2)+7. Thus, your argument that the expression: 5^(1/2)+7 can be "represented" using only the one symbol "#" is both trivial and meaningless! Most of us would prefer to represent: 5^(1/2)+7 meaningfully as 9.23606...., even though it does require an infinite number of symbols. Look at it this way. Every number that is not a non-negative integer requires an operation, (and therefore an extra symbol) in order to be properly, fully, and therefore meaningfully represented. Non-negative integers, on the other hand, are the only numbers that do not require "attached operations". Thus non-negative integers are the only "pure numbers", because they are not associated with any operation whatsoever. They are not "mixed" with symbols for operations. The imaginary number: i=(-1)^(1/2) can indeed be trivially represented as "i", but it's meaningfull representation: (-1)^(1/2) requires the two operations (and therefore the two symbols) "minus sign" and "radical". The non-negative integer: 2, on the other hand, does not have any operation anywhere near it. That's the difference between non-negative integers and all other numbers, and why I prefer to work with the former. I already promised that I will, from now on, state up front the restrictions on the variables that I am using. Nobody likes confusion, so that's one dead horse you can now leave alone. Don.

To: Hadron Collider. Thanks, you made my day! Don. To: Bignose. No, it's not "utterly ridiculous"! The truth is that it is easy to write the number 10 using only one symbol. In the "base 12" or "duodecimal system", "10" would be written simply as "A" and "11" would be representet by "B". Thus, given a sufficiently large "base N" system, any non-negative integer whatsoever could be written using only one symbol. The non-negative integers are the only numbers that have this property! No other numbers do, or can! 10/3, (or in the "duodecimal system", A/3) is a perfectly good number. However, as we can see, it does require at least three symbols in order to be represented. (A, /, 3) The square root of 2 requires at least two symbols, (radical, 2) Phi=(5^(1/2)+1)/2 requires at least six symbols, (radical, 5, +, 1, /, 2 ), Pi and e, are trancendental, and therefore require an infinite number of symbols in any number system. (Pi and e are merely "names", and tell us absolutely nothing about their actual values.) 6/2=3 requires only one symbol. (representing 3 as 6/2 or 3.00000... is not representing it in the "least possible number of symbols".) Again, I sure didn't mean to antagonize anyone. Since most of your expertise lies in applied math rather than elementary number theory, I can see how this thread resulted in some pretty frustrating moments for you. The main thing is that we are learning from each other. My next thread will be a lot closer to your fields of expertise, I promise. Thanks for remaining interested! Don.

To: Bignose. Given the symbol X, if all we are told is that it represents a number, then what kind of number does it represent? Here is my answer to this question, and I hope it sheds some light on at least some of the difficulties that we encountered in this thread: Since the only clue we have is that X constitutes but one symbol, the only logical conclusion is that X must represent a non-negative integer, because all other numbers require more than one symbol in order to be properly and fully represented. In other words, the "intrinsic" value of an independent variable must be some non-negative integer because given the possibility of arbitrarily large bases for number systems, non-negative integers are the only numbers that can possibly be properly and fully represented using only one symbol. Having established that X intrinsically represents a non-negative integer, we can then proceed to derive all other reasonable numbers by performing operations on X. Thus, all negative integers can be represented by the two symbols -X, all unit fractions by the three symbols 1/X, and so on. Given the above, and the fact that my website makes it sufficiently clear that "cohesive terms" apply primarily to non-negative integers, I felt that it was unnecessary to stipulate what I believed was already quite obvious. In hindsight, I can see that this was a "bad call". I am by no means perfect, and as it turns out, I really should have stipulated that this thread is about non-negative integers a lot earlier, and a lot more often. Thanks for the valid constructive critisism. I will add the "non-negative integers only" stipulation to my initial post for the sake of any new readers that might happen upon this thread. As for cancellation being "wrong", well, there is this matter of semantics that must first be cleared up before we can come to an understanding. You see, for many people (and it's not their fault), the word "cancel" means "cross out". But as we have seen, given the equation: (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)), the T's do indeed "cancel", but can not be "crossed out". Thus, "crossing out" the T's on either side is indeed wrong, because on the right, it is literally impossible, while on the left, we need to retain those T's so that both sides, (not just the right) represent the exact same extraction and neutralization of the factor T. In other words, the representation: (T/T)a^x implies primality or co-primality while the representation: a^x does not, and both sides need to imply the exact same thing. Please keep in mind that "cohesive terms" are only about a decade old. Thus, all their consequences, ramifications and implications have not yet been fully explored. My guess is that "crossing out" variables is okay if we are dealing with non-discrete quantities such as irrationals and transendentals, but not okay if we are dealing with discrete quantities such as fractions and non-negative integers. If you look deeply enough into the many questions that are engendered by the very existence of "cohesive terms", then you will be amazed at how much work needs to be done in order to render the rest of mathematics consistent with them, and vice versa. Also, let me assure you that I did not come to this forum to annoy anyone. Quite the contrary, I came here to make friends and to share some rather unique findings with other mathematicians. If at times, I come across as "arrogant" or "grandiose", then I'm sure that it's only a subconcious attempt to make those who I am communicating with appear more humble, and therefore, more virtuous than I. I will heed your advice and start a new thread in about a week. Thanks for remaining interested. Don. To: Severian. This thread may be somewhat "clumsy", especially on my part, but it is far from being "pointless". The equation: (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) presents the entire mathematical community with all kinds of incredible new possibilities and very, very serious questions such as: "Does multiplication and/or division by unity automatically result in division by zero?" Letting T=1 sure does seem to indicate that, doesn't it? Progress is being made, and lots of people are interested. I even have a "record breaking" math topic by the same name on the "Marilyn vos Savant" website and recently recieved an e-mail from the mathematics department at Stanford University to the effect that my work is being "processed" for possible publication. Don.

To: Bignose. This topic is about the properties of non-negative integers. Concepts such as "co-prime" require non-negative integers. 17^(1/2) and 17^(1/5) are not even rational numbers, much less non-negative integers! Thus, there is no way to define them as "co-prime"! My "cohesive terms" were specifically designed for use with non-negative integer variables. They allow us to solve hitherto intractable problems in "number theory" which is that branch of mathematics where we mathematicians focus primarily on the properties of non-negative integers. Thus, the logarithmic preclusion of negative numbers is a most desirable property. Now, if we assume that our variables represent non-negative integers, then the equation: A^X+B^Y=C^Z does allow all the exponents to be greater than 2. However, if we write: (T/T)a^x+(T/T)b^y=(T/T)c^z, then it is logical to assume that the greatest common factor T was eliminated and that the terms are therefore co-prime. As I explained in my previous post and on my website, "cohesive terms" actually demonstrate (prove) that once the greatest common factor has been cancelled, we must have, as our "co-prime" representation, either: a+b^y=c^z, a^x+b=c^z, a^x+b^y=c, a^2+b^y=c^z, a^x+b^2=c^z or a^x+b^y=c^2. where one of the variables is clearly restricted to being either 1 or 2. Thus, the "co-prime" representation: a^x+b^y=c^z is clearly "inadequate" and therefore "wrong" because it does not have such a restriction. If you don't believe me, then try to find an equation of the form: a^x+b^y=c^z where all variables are positive integers, all terms are co-prime, and all exponents are greater than 2. Don.

To:Bignose. Given the two equations: Ta^x+Tb^y=Tc^z and a^x+b^y=c^z, there is, of course, no value T which makes the first equation untrue when the second equation is true. The reason why: (T/T)a^x+(T/T)b^y=(T/T)c^z=a^x+b^y=c^z is wrong, is because it implies that the exponents x,y and z can all be greater than 2 after the "common factor" T has been cancelled. Take the simple numerical example: 6^3+3^3=3^5, and note that all three exponents are greater than 2. Also note that this equation contains the "common factor" 3^3=27. Now, extract and eliminate that common factor by writing: (27)(2)^(3)+(27)(1)^(3)=(27)(3)^(2), and: (27/27)(2)^(3)+(27/27)(1)^(3)=(27/27)(3)^(2)= 2^3+1^3=3^2 and note that all three exponents are no longer greater than 2. Such will always be the case, but the representation a^x+b^y=c^z doesn't reflect that! There is absolutely no restriction on either x,y or z! Thats why it's wrong! Now, consider the general representation: ((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 eliminating the outermost parenthesis does not involve the cancelled variable T and therefore results in: (T/T)a^x+(T/T)b^y=(T/T)c^z= T(c/T)^((zln©/(ln(T))-1)/(ln©/(ln(T))-1)). Applying our numerical examples by letting: a=2, b=1, c=3, x=3, y=3 and z=2, we find that since both T and 27 cancel, it is obvious that we need not let T=27 and that we must be able to allow T=3, and that this, in turn, requires that we first "cross out" the logarithms by letting z=2 and z=1 respectively, which is perfectly consistent with our numerical example! Thus, by using "cohesive terms", we actually demonstrate that if the equation: a^x+b^y=c^z has no common factor, then x,y and z can not all be greater than 2. (Just like we saw in our numerical example.) So you see, when it comes to the proper representation and elimination of common factors, "cohesive" terms reflect reality, and "non-cohesive" terms don't. Don.

To: Paganinio. Thanks. It is important! Moreover, it's new, and not in any books yet! I'm certain that it will lead to many other discoveries, so I hope you study it, and tell your friends and teachers about it. You see, if you love math, then you must love to explore, and "cohesive terms" open up a lot of "uncharted territory", because they are one of the few basic and fundamental constructs in math whose concequences and ramifications have yet to be fully and adequately explored. What could be more exiting than that? Happy exploring! Don.

To: Josy. The Beal Conjecture says that if (T/T)a^x+(T/T)b^y=(T/T)c^z are co-prime, then x,y and z can not all be greater than 2. Now, the above equation must be co-prime because the greatest possible common factor T was cancelled, and we know that T must be the greatest possible common factor because if other common factors such as Q, R and S did exist, then they could all be multiplied together to form the product T= Q*R*S. I know that it's hard to believe, but had mathematicians learned how to represent and eliminate common factors properly to begin with, then problems such as the Beal Conjecture and FLT would never have surfaced! In other words, the proper representation and elimination of common factors (using "cohesive terms") automatically solves those problems! Don. To: Dr.P. If we "cross out" the T's on the left, then a^x can be construed as (1)a^x=(N/N)a^x where N does not equal T, and although the equation would remain true, we would be either ambiguous (in the case of 1) or inconsistent (in the case of N/N) with regards to the value of the cancelled common factor. Besides, if we "cross out" the T's on the left, then most people would not be able to tell at a glance that the T's do indeed cancel. Don.

On The Proper Representation And Elimination Of Common Factors. By: Don Blazys. Our students 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 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 term that employs logarithms in order to firmly establish that the 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 common factor T has been cancelled no longer exists. Thus, by using logarithms 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). This method of representing and eliminating common factors is obviously more consistent and effective than simply "crossing out" the T's. What I find both amusing and confusing is why it is not yet being taught in our schools. I discovered it a decade ago! The above article is a lot easier to read on my website (donblazys.com). There is also another version of it on the Marilyn vos Savant Home Page Forum along with a lot more spirited discussion. Don. To: Dr.P All I am saying is that if we are confronted with the term: (T/T)a^x, then we should not "cross out" the T's because if we do, then all we are left with is a^x, and if all we have left is a^x, then we can't derive the more powerfull term: T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)). "Crossing out" the T's isn't mandatory. There is no mathematical "law" that says the T's must be "crossed out". Students should be taught that they have a choice...that they can choose to re-write (T/T)a^x as T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) where it must be the case that any factor or common factor T>1, and where it is algebraically impossible to "cross out" the cancelled T's. My invention allows us to retain cancelled variables rather than lose them and doesn't allow the "trivial common factor" (1) to rear it's ugly head! It is therefore more truthfull and accurate than the term on the left. I even gave algebraic terms that have these properties a name. I call them "cohesive terms". Don.

To: Bignose. Had you gone to my website, then you would have found that the variables in my equation represent non-negative integers. Any particular independent variable constitutes but one symbol and since non-negative integers are the only numbers that can possibly be represented using only one symbol, they are the most logical candidates for representation by independent variables. Thus, in order to be "perfectly rigorous", independent variables should represent non-negative integers only. All other numbers can then be derived by performing operations on those variables or the terms containing them. (By the way, my website is perfectly safe. It was designed by the computer teachers at the school where I work and in the few months that it has been up, it has recieved thousands upon thousands of visitors. I get e-mails about it all the time and no one has ever complained about it messing up their computer. However, for the sake of convenience, I will type out the article "On The Proper Representation And Elimination Of Common Factors" in my next post.) Now, the identity: (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) is true for all non-negative integers T={2,3,4...}, a={1,2,3...} and x={0,1,2...}. Note that it contains the expression: ln(a)/(ln(T))= log T a, (T being the base of the logarithm). Now since the base of a logarithm is commonly defined as "any positive real number other than unity", why you would choose to "evaluate" what happens at T=(-1) is beyond me. Don.

To: The Tree. Well, the fella who "corrected" you (Bignose) has since corrected himself and now agrees that (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) is indeed a true equation. What do you say now? True or false? Don. To: Josy. In your previous post, you said "The point is that if the equation in question held in general, it would prove that cancellation of common factors doesn't work". I agree with that. I also agree that the equation: (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) does not contradict cancellation because clearly, the T's do indeed cancel on both sides. (I'm glad to see that you have changed your mind and have finally come to the realization that the above equation is true.) Now, the question is, can we "cross out" the cancelled T's and make them "dissapear" the way we were taught in school? Don. To: Bignose. You were wrong when you said that the equation (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) is false and now you are wrong when you say that the cancelled T's "don't matter". The examples that you give in your last post, namely 10*(5/5)=(3*5)-5, f(x)=5 and f(x,y)=2x+1^y where the expressions (5/5), x and y don't matter are truly trivial and have absolutely nothing to do with my equation, where the cancelled T's really do matter. If you want to see how and why they matter, then go to my website (donblazys.com), and read the article "On The Proper Representation And Elimination Of Common Factors". There you will find that unlike the "examples" that you provided, my equation actually contains a mechanism that allows us to eliminate common factors in a way that is truly consistent with both logic, and all observed results. Don.

To: Dr.P. Whatever we do to one side of an equation, we should also do to the other. Also, we must always keep in mind what a particular equation actually means. Now, the equation (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) is an "identity", which means that it is true for all values of the variables. The right hand side tells us that if T>1, then it is algebraically impossible to eliminate the T's in the "usual manner" by "crossing them out" and rendering them invisible. This must also hold for the left hand side because both sides tell us that the factor T was extracted and then used as a divisor. Thus, to "cross out" the T's on the left is to deny that any operations involving T ever took place! In other words, "crossing out" the T's on the left "destroys" the identity because it leaves us no way to derive the term on the right from the term on the left. Don.

To: Bignose. Congratulations! You got it! Sorry about not using "LaTex" but I never even touched a personal computer until a few months ago and my "computer skills" are, at this time, limited to checking and sending e-mails, performing "Google searches" and posting in forums. I'm almost 60 years old and "early onset alzheimers" makes it difficult to learn and retain new skills. You know, when I was in high school back in the 60's, we didn't have personal computers, calculators, or any of that stuff. I still use a "slide rule", pencil and paper... and for most mathematicians my age, parenthesis make equations easier, rather than more difficult to understand. In other words, I wasn't trying to make things difficult for you. Perhaps you would be kind enough to post the formal derivation of my equation in "LaTex" in your next post on this topic. That way, this new generation of mathematicians that relies on the "high tech" that my generation invented will find my work easier to understand. You can find that derivation on my website (donblazys.com). There, you will also find the point of this topic. Don. To: Josy. (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) is a proven identity, but it is very very "counterintuitive" because even though the T's cancel (are neutralized) on both sides, they can't be "crossed out" on either side because if we "cross out" the T's on the left, then we have no way to logically derive the term on the right, where the T's can"t be "crossed out". Don't give up! You were right when you said that it shows the present method of eliminating common factors to be inadequate. By the way, you can find more information about this equation on my website, and on the Marilyn Vos Savant forum. Thanks for your interest. Don. To: The Tree. I predict that you will change your mind yet again. Good luck! Don.

To: Everyone who reads this. It's election time, so let's have a vote. Is the equation: (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) true or false? No commentary please, just vote true or false over the weekend. I will be back on Monday to count the votes. Don. I will cast the first vote. I vote TRUE. Don.

To: Dr. Jekyll and Bignose. (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) is a proven identity. Therefore, it must be true for all possible values of the variables. The special case T=a requires x=1 because if Ta^x is a composite number and T is the greatest possible factor, then (T/T)a^x must represent a prime, and primes can't be represented as having exponents. Don. To: Tree. First you said that the equation is true, and now you are saying that it's false. You were right the first time! Don.

To: Dr.P. The exponent that I wrote is ((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)). Don't confuse it for x((ln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)). Don To: Dave, Bignose and Tree. The point of this thread is that if the equation: (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) is true, then even though the T's cancel, they can not be "crossed out" and rendered "invisible" in the way that is taught in our schools. Moreover, letting T=1 demonstrates that multiplication and/or division by unity automatically results in division by zero! Don. To: Dave and the Tree. What does "lulz" mean?

To: Bignose. The equation is true. The expressions involving logarithms "cancel out" if and only if x=1. Otherwise, they don't. Don.

You are correct, (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) is indeed a true equation (identity). Obviously, it would remain true even if we somehow transported the right hand side all the way to the other side of the universe. Now, can we "cross out" the T's and make them "disappear" the way kids are being taught to do in school? Don.

Since unity can be defined as 1=(T/T) where T>1, the answer for T is not indeterminate. This is a simple true/false, yes/no question. Don.

Is the equation: (T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) true or false? If true, then can we "cross out" the T's and make them "disappear" the way we were taught in school? Don.