Jump to content

The Most Important Equation Ever.


Don Blazys
 Share

Recommended Posts

  • Replies 59
  • Created
  • Last Reply

Top Posters In This Topic

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.

Link to comment
Share on other sites

Don't give up! You were right when you said that it shows the present method of eliminating common factors to be inadequate.

I never made any such claim; as I said before, cancellation is a deep property of the real numbers; if this equation contradicts it, which it does not, then the error must lie in the equation, not in the law of cancellation.

 

The real numbers, over which I will assume we are working, though we can allow the variables to take complex values with equal ease, form a field. Among the properties of a field is that it forms a group (an Abelian group in fact, but that is not relevant here) under multiplication. The definition of a group is thus: a set [math]G[/math] under an operation which I shall denote by simple juxtaposition forms a group if and only if all of the following conditions are met:

[math]\forall a,b\in G,ab\in G[/math] (closure; this is fairly trivial, and is often taken as implicit in the definition of the operation itself)

[math]\forall a,b,c\in G,(ab)c=a(bc)[/math] (associativity)

[math]\exists e\in G:\forall a\in G,ae=ea=a[/math] (identity)

[math]\forall a\in G,\exists b\in G:ab=ba=e[/math] (inverses, and we usually write [math]a^{-1}[/math] for [math]b[/math])

Note that these four conditions are taken as axiomatic; they are accepted as true by definition. Now, suppose that [math]a,b,c\in G[/math], and that:

1. [math]ab=ac[/math]

Now, proceeding from first principles, i.e. the above four axioms:

2. [math]a^{-1}(ab)=a^{-1}(ac)[/math] (inverses)

3. [math](a^{- 1}a)b=(a^{-1}a)c[/math] (associativity)

4. [math]eb=ec[/math] (inverses again)

5. [math]b=c[/math] (identity)

Thus the law of cancellation, one of the basic results of group theory, is derived. Your example actually corresponds to stage 2 above; stage 1 would feature the denominator of your [math]\frac{T}{T}[/math] appearing as part of the numerator of the RHS. Nevertheless, cancellation most certainly holds for it.

 

I have tried to keep this to the mathematics, and not to make it personal, but I would like to comment that I find your tone rather more patronising than I'm entirely comfortable with. This may well be accidental; online conversations are rather notorious for misapprehensions of emotional content.

Link to comment
Share on other sites

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.

 

This is wrong. Especially in that I personally have the derivation where I started with a^x on the right hand side. All the operations with the T's result in the same spot you started with. What's so unique about that?

 

What if I wrote "10*(5/5) = (3*5)-5." Which, then I can cancel the 5's on the LHS leaving: "10 = (3*5)-5". Does this mean I can also "deny that any operations with 5 ever took place"? I mean, just like you said above, I eliminated the 5's from the left hand side, they must be gone, right?

 

It isn't straightforward, but the influence of the T's cancel out. The T's don't matter. It is no different than writing f(x) = 5, where the x's clearly don't matter. Or writing f(x,y) = 2x+1^y, where the y clearly doesn't matter.

 

Yours is certainly more complicated looking, but still the same. It isn't that influential at all. It is just a simple "trick" of mathematics.

Link to comment
Share on other sites

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.

Edited by Don Blazys
multiple post merged
Link to comment
Share on other sites

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.

 

With all the viruses and spyware out there, I don't just click on unknown websites. Please just post things on here.

 

And, if you want to be completely 100% correct, the two sides actually aren't strictly equal. Both sides blow up at T=0, but the LHS has real values for all T!=0. The RHS for values of T<0 have to start using imaginary numbers.

 

I.e. the LHS using a=3, x=5, and T=-1 evaluates 243

But, the RHS evaluates to -0.089 + 0.038i

 

Not equal.

 

For it to be a true equality, both sides should be equal for the entire range of all variables.

 

Your is only true if and only if T>0.

 

And you have to include that restriction, otherwise is isn't a true statement.

 

So, in summary, the equation isn't strictly correct in that the limits of both sides aren't the same. And, if you'd post some of the why on this website, rather than my clicking to an unknown website (and supporting any ads or even seeming like I support what is borderline spam in my opinion), I'd much appreciate that.

Link to comment
Share on other sites

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.

Link to comment
Share on other sites

To: Dr.P. Whatever we do to one side of an equation, we should also do to the other.

 

 

But we arn't doing anything to either side. We arn't dividing through by T (other wise yes we would have to do the same to both sides) we are just canceling the Ts on the LHS. Sorry - so far I still don't get it (well, I do, you can cancel the Ts on the RHS without any problem) - I just don't see your reasoning for the RHS being effected by the cancellation.

 

OK lets do it your way of doing the same thing to each side:

(T/T) a^x = RHS... multiplying BOTH sides by T right?

 

=> T.a^x = T.RHS

 

Now - divide BOTH sides by T:

 

=> a^x = RHS just the same as cancelling the origional Ts on the LHS, but by doing what you asked for - the same thing to be done on both sides.

Link to comment
Share on other sites

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.

Link to comment
Share on other sites

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)).

 

Fair enough - if you want to derive the RHS from the left then yes. But once the equation has been set up and you have your

(T/T)a^x = RHS already then canceling the Ts would be fine for sure, and a bit tidier to give

a^x = RHS no? The RHS has T in it so you havn't lost the relationship between x and T by canceling.

 

Thanks for the explaination anyway - I do remember vaugely from school that there was a situation where our teacher advised us not to cancell so we could keep a variable in the equation to proove some derivation - I don't really remember alot about it though.

 

Also - sorry - but I think I should bail out of this thread, and let the real mathemeticians discuss this, as i think I'm getting a bit confused and maybe missing the point. :confused: Cheers anyway.

Link to comment
Share on other sites

Finally I think I see where you're coming from, and can formulate some kind of useful reply.

 

For one thing, as far as I'm aware, the Beal conjecture is a conjecture, and thus not proven; indeed, Andrew Beal has offered a prize of US$100 000 for a proof or counterexample. Even assuming it does hold, it deals with equations of the form [math]a^x+b^y=c^z[/math] with all of x, y and z greater than two. Considering your initial equation, [math]Ta^x+Tb^y=Tc^z[/math], which is equivalent to the above, certainly all three terms have a common factor of T; however, there is nothing indicating that they do not have additional common factors; that is, [math]a^x[/math], [math]b^y[/math] and [math]c^z[/math] are not necessarily coprime. If they are coprime, then the Beal conjecture is false, and the discoverer of the counterexample might be well advised to send it to Andrew Beal and claim hir prize money.

Link to comment
Share on other sites

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.

Link to comment
Share on other sites

  • 2 weeks later...
  • 3 weeks later...

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.

Link to comment
Share on other sites

I still don't really see the point in making things more complicated than is needed.

 

Don,

 

in response to your mini-article there, can you cite one example where

 

[math]Ta^x + Tb^y = Tc^z[/math] is not equivalent to [math]a^x + b^y = c^z[/math]?

 

I.e. what value of T makes the first equation untrue, when the second equation is true?

 

I am focusing on your line:

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

 

How is this "wrong"?

Link to comment
Share on other sites

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.

Edited by Don Blazys
Link to comment
Share on other sites

 

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.

 

 

 

But (T/T) is 1. It is a fundamental of the number system and operations as we know them. Anything (except 0) divided by itself is 1.

 

I don't see why multiplying anything by 1 implies that the exponents can be greater than 1.

 

[math]\frac{T}{T}\cdot 17 = 17 = {\sqrt{17}}^2 = {\sqrt[5]{17}}^5[/math]

 

Why does there have to be a restriction on the exponents? And if there is a restriction, why not just state it? When I write an equation for the temperature of a block of steel, for example, it is understood that T (representing temperature) is greater than or equal to zero. If there is a restriction on a value in an equation, then that restriction is a part of the equation.

 

If there is no stated restrictions on the variables, then they can take any value. Exponents less than 2 included.

 

Your terms using logarithms unnecessarily places restrictions on the variables, because with the logarithms, you are limited to positive values.

 

Use you first example just with negative numbers this time: [math](-6)^3 + (-3)^3 = (-3)^5[/math] If you do use your logarithm canceling, you will not get the same on the right and left hand sides, the logarithm of a negative number has an imaginary part. But, you can divide that equation by [math](-3)^3 = -27 [/math] and it is still correct. Where is the benefit from getting a wrong answer using your method? Where is the problem in diving the equation by -27? I remain thoroughly unconvinced of any problem at all. You said it yourself that "there is ... no value T which makes the first equation untrue when the second equation is true."

Link to comment
Share on other sites

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.

Edited by Don Blazys
Link to comment
Share on other sites

To: Bignose.

 

This topic is about the properties of non-negative integers.

 

 

This is NOT what this thread started talking about. (The word "integer" doesn't even appear until the 32rd post!!!) You asked if your equation was true, and called it "the most important equation ever". No talk about co-primes or limits on the exponents. Only talk about why canceling of terms is wrong (which it isn't). Check the first few pages of this thread.

 

NOW, this stuff about coprimes and whatnot. If that is what you wanted to talk about, why didn't you just state it up front? Why the deception? What was the point of asking about one thing then turning it into a discussion of something else, when that something else is obviously what you wanted to talk about in the first place?

 

Also, you never did "show" why canceling is wrong. Maybe you've showed that indeed when the common factors are divided out, one of the exponents has to be 1 or 2 (when limited to non negative integers). But that doesn't make canceling wrong. They are almost totally two different topics.

 

I really feel like you've wasted a lot of people's time here, because you weren't upfront about your intentions. You never said anywhere at the beginning that your variables in you equation was limited to non-negative integers. Had you done so, it may have saved a lot of trouble. How were we supposed to know that? Just read you mind? You have to state these restrictions. And what your main point is -- we did ask you what your point was several times. All you've really done now is 1) people probably aren't going to want to discuss your work in this thread because it is so filled with things that you really didn't want to discuss in the 1st place and 2) if you should start another thread, people are going to wonder if the first post in that will actually be about what you want to talk about.

 

Don, this forum is a good place for discussions like about whether your proof is complete or not. But, you have to be upfront and honest about your intentions. This end around nonsense just annoys people. And, whether intentional or not, as was mentioned above, your tone comes of very condescending and arrogant. If you want to discuss your proof, then I suggest you start a new thread -- with a less grandiose and arrogant and more on-topic title -- and discuss the idea there.

Edited by Bignose
Link to comment
Share on other sites

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.

Link to comment
Share on other sites

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.

 

Utterly ridiculous.

 

X is a variable. In fact, in mathematics, it isn't even formally correct to just leave X hanging out there by itself.

 

You should write something like [math]X \in \mathbb{Z}^{+}[/math] where [math]\mathbb{Z}^{+}[/math] would represent the non-negative integers. There are other symbols used other than Z, but the point is to define what the variable is limited to.

 

In general, the assumption is if there is a variable without explicit limits on what it's domain is, it is completely open. I.e. [math] X \in \mathbb{C} [/math] where [math]\mathbb{C}[/math] is the set of complex numbers. Negative, positive, zero, imaginary, and reals are all in play.

 

A big part of mathematics is being as explicitly clear about what your symbols mean as possible. Just because you interpret a symbol one way doesn't mean that everyone else is going to interpret it that same way. Read any mathematics journal article (if you are planning on trying to publish your work, you should read a lot of them to get a feel for how it is going to be expected to be written, anyway), and they will state up front exact what every variable means, and what its values are limited to. Even many of the engineering and physics journals have a table at the begining or end of the article listing what every variable means (and as such explicitly or implicitly what their values are limited to. I.e. if an equation is written for total kinetic energy, k, it is going to be a positive real).

 

Had you stated this limitation up front, it would have saved much, much consternation.

 

===========

 

Your logic "told that it represents a number" is kind of illogical to me anyway. Why is 10/3 not as good of a number as 10? Both are "numbers". Aren't [math]\sqrt{2}, \pi, 14.101001000100001..., \frac{17}{13} [/math] numbers, too?

 

Furthermore, your logic also limits X to just 0,1,2,3,4,5,6,7,8, and 9. Because anything else would require "more than one symbol" to write it out. 10 requires two symbols. I just sure hope you don't ever work in binary, because, then you'll only be limited 0 and 1! But, [math]\pi[/math] or [math]e[/math] or [math]\phi[/math] are numbers that only require one symbol to write out -- are they supposed to be included too? [math]\frac{6}{2}[/math] is a number, too. It happens to be equal to 3, but it requires more than one symbol to write out. You probably didn't want to exclude it from consideration, did you?

 

The only time I've ever seen a variable be limited to the number of characters it takes to write something out is on those "letter" math puzzles:

 

somthing like

 

M A T H

+ F U N

----------

C O O L

 

and the trick is to figure out what digit each letter stands for. That doesn't mean that there isn't math out there like that, I just have never personally encountered a variable interpreted in that way before.

 

====

 

Just be more forthright about your intentions up front, and everyone will be happy. I actually don't know anything about "co-primes" and "cohesive terms" and what not, but know a lot of applied math (i.e. solving differential equations, integro-differential equations, statistics & stochastic calculus, etc., stuff as it applies to physics and engineering). I don't know a whole lot about the more fundamental stuff.

 

But, when someone shows up telling us that canceling is wrong and saying nothing else, it is going to rub people (i.e me) the wrong way. Had you stated up front that your variables were all going to be limited to non-negative integers, well, then I think that the reception you would have gotten would have been a lot friendlier.

Edited by Bignose
Link to comment
Share on other sites

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.

Link to comment
Share on other sites

Don,

 

I hate to seem like I am beating a dead horse here, but there are other examples where I can show your logic wrong.

 

Consider a number in base [math]\pi[/math]. In such a base [math]\pi_{10} = 1_{\pi}[/math] where the subscript denotes the base. That only requires one symbol, "1", if you are working in base [math]\pi[/math].

 

(And, yes, there has been a base [math]\pi[/math] proposed, see Bergman, G. "A Number System with an Irrational Base." Math. Mag. 31, 98-110, 1957/58. as one example)

 

There have also been transcendental bases proposed, even negative number bases. If you are going to include that in your considerations, then again, I think that every single number can be written in "1 character". So, once again, every single number is a good as any other, real, rational, irrational, etc.

 

The bigger point is that you understand the need to state upfront the restrictions on the variables, don't you? Otherwise, pretty much everyone assumes that any variable can take on any value: integer, negative, real, or complex.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
 Share


×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.