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A Mathematical Emergency.


Don Blazys

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To: CraigD.

 

Here's the difference between a "division by zero", and an "indeterminate form":

 

6/3=2 is valid because: 2*3=6. However,

 

6/0= N is "undefined" because: N*0 does not equal 6, and

 

0/0= N is "indeterminate" because: any number N*0=0.

 

Thus, if 0/0= N, where N can be "any number", then the expressions in my proof clearly work out to:

 

(1)^(0/0)=1, therefore, c(1)^(0/0)= c, and (c(1)^(0/0))^2=c^2.

 

In other words, we need not know the value of the "indeterminate form" (0/0) in order to determine that

(1)^(0/0)=1 because 1 raised to any power, (including any "indeterminate power" such as 0/0), still equals 1.

 

Moreover, the all important doctrine of maintaining logical/mathematical consistency requires that our results have the exact same meaning regardless of whether we first let z=1 and z=2, then "cross out" the logarithms or whether we first let T=c, then let z=1 and z=2 so as to beget those absolutely benign "indeterminate forms".

 

In short, the "indeterminate forms" in no way derail my proof.

 

From my humble point of view, the proof works perfectly, and I expect that the editors and referees at the Journal Of The American Mathematical Society will see what I, along with many others, see. They have certainly had it long enough! (My website includes a few of the many letters of recomendation that I recieved, including one from a famous NASA/JPL planetary scientist that is directed to the AMS.)

 

"Cohesive terms" is just a name that I gave my discovery. Thus, at this time, you will not find that name or the construct to which it pertains anywhere but in my writings. (My innate humility prevented me from calling them "Blazys terms".) In my opinion, they are the most beautifull, elegant, and powerfull construct in all of mathematics. So much so that I no longer view the "Beal Conjecture" and "Fermat's Last Theorem" as "famous conjectures", but as "conjectures that would never have been made, had mankind learned to represent algebraic terms correctly to begin with".

 

I am very proud of both you, and Qfwfq. Despite my not being able to write out my mathematical expressions and equations in a format that you are used to, you fought your way through and have now come to a point where you nearly understand my proof! I am also very gratefull for all your help and advice.

 

Don.

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0/0= N is "indeterminate" because: any number N*0=0.

 

Thus, if 0/0= N, where N can be "any number", then the expressions in my proof clearly work out to:

 

(1)^(0/0)=1, therefore, c(1)^(0/0)= c, and (c(1)^(0/0))^2=c^2.

 

In other words, we need not know the value of the "indeterminate form" (0/0) in order to determine that

(1)^(0/0)=1 because 1 raised to any power, (including any "indeterminate power" such as 0/0), still equals 1.

Note Don that 0/0 is just as "undefined" as 6/0 is. The step from saying "any number N*0=0" to saying "0/0= N, where N can be 'any number'" is a paralogism; you should notice that the nexus involves dividing or multiplying by 0, so you draw your conclusions by an "as if" and quite circular argument.

 

The third assert is interesting because, while fraught with the formal woe of being an expression in which one term does not define a value, you justify the claim by the seemingly compelling argument that the whole expression is totally independent on the undefined term anyway, so it doesn't matter. Now, apart from the above difference between "can be any value" and "defines no value", even when loosely put in the former terms there remains the fact that "any value" can be [imath]\infty[/imath] too. Now this is a very subtle matter because the indepence can, at first glance, seem so intrinsic that one may be convinced by the argument. However one may use the rules of logarithms to write:

 

[math]1^a=e^{a\ln 1}=e^{0a}[/math]

 

which sheds a better light on the matter of independence; it is easier to interpret in the familiar terms of "any value" multiplied by zero. Now we know that [imath]0a=0[/imath] for any finite a and that it can be for infinite "values" of a too; notoriously [imath]0\cdot\infty[/imath] is also an indeterminate form, therefore so is [imath]0\frac00[/imath]. So, re-casting the matter as follows:

 

[math]1^{\frac00}=e^{0\frac00}[/math]

 

where it is manifest that the rhs is not independent on the exponent, one may understand that the exponential also isn't wholly independent on the blunt [imath]\frac00[/imath] form and there's no hook or crook in the equality because it relies only on the value of [imath]\ln 1[/imath] and of [imath]e^0[/imath], no whatsoever singularities. Therefore the rhs can't be equated to 1 as an apodictical certainty and the same goes for the lhs.

 

Now, without the above manipulations, this was only less obvious than the more common case of multiplication by a factor such as [imath]\frac{x}{x}[/imath] which, Don, is not the same as multiplying by 1; this argument of yours totally forgets the very matter of the indeterminate form. It's the multiplication by an expression which defines a value of 1 only [imath]\forall x\neq 0[/imath] and defines no value for [imath]x=0[/imath]. Even if you remember your own contention, that for [imath]x=0[/imath] the expression "can be any value", you would have to admit that this isn't the same thing as multiplying by 1, or is it?

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To: Qfwfq.

 

When you say that the "indeterminate form":

(0/0) is just as "undefined" as the "undefined operation":

(6/0), then you are dismissing the expression:

(0/0) as "meaningless", when in fact, there are plenty of cases, contexts and circumstances in which equations such as:

(0/0)=1 can be viewed as "meaningfull". (Especially in the context of "limits".)

 

There are, however, no conditions under which equations such as:

(6/0)= N can be construed as meaningfull, so the expressions:

(0/0), and:

(6/0) must be viewed as "fundamentally different" from each other.

 

(0/0) "exists" as a logical construct.

(6/0) "does not exist" as a logical construct.

 

Also, (0/0)= (infinity) if and only if (infinity)*0=0. (Again, the meaning/value of (0/0) depends entirely on the context in which it occured.)

 

I suppose that different mathematicians have different criteria for what constitutes a "valid result". For me (and most mathematicians my age), those criteria are "consistency" and "beauty".

 

My garage is the same (consistent) regardless of whether I enter it from the front door, or the back door.

 

In the same way, at x=1, the equation:

 

(T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1))=

 

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

 

so that letting T=a results in:

 

(a/a)a=a(a/a)^(0/0)=a(a/a)=

 

(1)a=a(1)^(0/0)=a(1),

 

where, for the sake of consistency, we must conclude that the expressions:

 

(1) and (1)^(0/0)

 

are both equal, and "equally trivial" and may therefore be disregarded so as to result in the identity:

 

a=a=a.

 

Look at it this way. If, in the above case, we were somehow able to demonstrate that:

 

(1) and (1)^(0/0)

 

do not have the exact same value, then we would have to make the further "assumption" that logarithms are inherently "flawed", and must therefore be eliminated from mathematics!

 

I would then be forever remembered as the "mathematician who demonstrated that the properties of logarithms are bogus".

 

Like I said before, the proof is merely an unavoidable consequence of "cohesive terms", which are, in turn, merely an extention of the properties of logarithms,... and those properties are definitely not bogus.

 

Don.

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...there are plenty of cases, contexts and circumstances in which equations such as:

(0/0)=1 can be viewed as "meaningfull". (Especially in the context of "limits".)

Gosh Don, I was talking about the blunt [imath]\frac00[/imath] form and I said so. I was distinguishing it from limits of any kind of expressions in the exponent.

 

In terms of limits, it is clear that the ratio of two infinitesimals (a distinct thing from the ratio of 0 and 0), can have a limit of any value (according to what the two infinitesimals are). It is also trivial to argue that any limit of:

 

[math]1^{f(x)}[/math]

 

exists and can only be equal to 1. By "any limit of" I mean the limit for x approaching any accumulation point of the domain of [imath]f[/imath].

 

[imath]\frac00[/imath] has no domain, no accumulation points, it cannot have any limit. It simply does not and cannot define any value (for any value of [imath]x[/imath] or any other variable).

 

Do you understand the distinction Don?

 

There are, however, no conditions under which equations such as:

(6/0)= N can be construed as meaningfull, so the expressions:

Obviously, if by N you mean a finite value! (or an infinitesimal one) You attempted a sleight of hand here!

 

Also, (0/0)= (infinity) if and only if (infinity)*0=0. (Again, the meaning/value of (0/0) depends entirely on the context in which it occured.)
This gets you into the same paralogism I had already pointed out.

 

I suppose that different mathematicians have different criteria for what constitutes a "valid result". For me (and most mathematicians my age), those criteria are "consistency" and "beauty".
This is a very dubious statement, self-consistence is a criterion of validity, but beauty is not.
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To:Qfwfq.

 

As mathematicians, we must distinguish between "reasonable implications" and "unreasonable implications".

 

Now, if N is a non-negative integer, then the "blunt form":

 

(0/0)=N

 

implies that any non-negative integer N multiplied by 0 equals 0.

 

That's a "reasonable implication" because N*0=0 is a "true statement".

 

However, the "blunt form":

 

(6/0)=N

 

implies that any non-negative integer N multiplied by 0 equals 6,

 

and that's an "unreasonable implication" because N*0=6 is a "false statement".

 

Thus, the result: (1)^(0/0)=(1)^N=1

 

is predicated on a "reasonable implication" and a "true statement", so we need not define the exact value of either (0/0) or N.

 

In fact, if it were somehow concievable that the "blunt form" (0/0)=N defined a "particular value", then that would be a most stupid and ugly thing because it would assign a completely unwarranted significance to that "particular value" and would thereby falsely imply that for the equation:

 

(1)^(0/0)= (1)^N=1,

 

some particular value of (0/0) or N is "more true" than the rest!

 

By contrast, the "result":

 

(1)^(6/0)=(1)^N

 

is predicated on an "unreasonable implication" and a "false statement" so we need not even consider it.

 

Please keep in mind that indeterminate forms such as (0/0) are often initially encountered in their "blunt forms", and that in some cases (such as my proof), they should remain that way, while in other cases, they should be evaluated by introducing more subtle concepts such as limits.

 

Speaking of limits, it would be easy to introduce the "limit as T approaches c" in my proof, and thereby show that:

 

z=1 and z=2 results in (1)^(1),

 

while z>2 results in (1)^(infinity)="indeterminate",

 

but it would also be quite unnecessary, and would, in my opinion, diminish some of it's generality, simplicity, power, and of course, beauty.

 

Don.

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As a mathematician, I distinguish between "conclusive arguments" and "non sequiturs". Don, I'm also beginning to see that you need to straighten out your grasp on the fundamentals of logic. Let's look at the fallacies in your argument:

Now, if N is a non-negative integer, then the "blunt form":

 

(0/0)=N

 

implies that any non-negative integer N multiplied by 0 equals 0.

 

That's a "reasonable implication" because N*0=0 is a "true statement".

Actually, it is a truism:

 

[math]\frac00=N\Rightarrow N\cdot0=0[/math]

 

is formally a true implication, but only because the consequent is true and not by force of any argument; you might as well put in any satement as the implicant (even a false one!). There is circularity in attempting to argue the implication; after multiplying both sides of [imath]\frac00=N[/imath] by zero and then applying the implicant to the resulting rhs, you could only conlclude [imath]N\cdot0=N\cdot0[/imath] as a consequent. At this point, in order to deduce [imath]N\cdot0=0[/imath] from that, one would need to use it (i. e. already know it is true).

 

The trouble is that, since the argument relies on multiplying both sides by zero, even if it could fully lead to the implication ([imath]\Rightarrow[/imath]), it certainly couldn't demonstrate a co-implication ([imath]\Leftrightarrow[/imath]). It therefore does not justify using the true statement [imath]N\cdot0=0[/imath] to justify [imath]\frac00=N[/imath] at all. I'm under the impression you need to avoid confusion between modus ponens and modus tollens, your argument demonstrates nothing about [imath]\frac00[/imath] defining any value; it is inconlusive.

 

The implication [imath]\frac{51}{3}=2\Rightarrow N\cdot0=0[/imath] is also formally a true one, what can you deduce from it?

 

because N*0=6 is a "false statement".
Of course it is a false satement and I already said so, so we need not even consider it. Friday I left out that a fair comparison would be one between:

 

[math]\frac00=N[/math] and [math]\frac60=\infty[/math]

 

where both equalities may be seen as possibly resulting from true limits, by replacement of infinitesimal terms with blunt zeroes. What I meant Friday by "sleight of hand" is that your comparison wasn't a fair one; it's blithering obvious that [imath]\frac60=N[/imath] can't be similarly obtained.

 

Please keep in mind that indeterminate forms such as (0/0) are often initially encountered in their "blunt forms", and that in some cases (such as my proof), they should remain that way, while in other cases, they should be evaluated by introducing more subtle concepts such as limits.
I should keep what in mind? You appear not to have understood what I posted on Friday; I was talking about limits, wasn't I? My point was indeed that these things make sense only in the context of computing limits, with the zeroes actually being infinitesimals! Try not to miss my points if you reply to them.
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To: Qfwfq

 

There is no fallacy in my argument.

 

First of all, let's keep in mind that we are discussing the meanings of the expressions:

 

(1)^(0/0) and

 

(1)^(N/0), N>0

 

as they relate to my proof of the BC.

 

Thus, if all variables are positive integers and all terms assumed co-prime, then:

 

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

 

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

 

and:

 

((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 letting z=1 and z=2 respectively allows us to "cross out" the logarithms, which results in:

 

(T/T)c=T(c/T), and

 

((T/T)c)^2=(T(c/T))^2.

 

Now, my questions to you are:

 

(1.) Can we now let T=c? (I say yes.)

 

(2.) Can we "cross out" the logarithms and let T=c if z>2? (I say no.)

 

(3.) Did letting z=1 and z=2, then "crossing out" the logarithms eliminate the possibility of encountering those pesky "indeterminate forms"? (I say yes.)

 

(4.) Do those pesky "indeterminate forms" occur if and only if fail to "cross out" the logarithms at z=1 and z=2 before we let T=c? (I say yes.)

 

(5.) Are those pesky "indeterminate forms" trivial? (I say yes.)

 

(6.) Do those pesky "indeterminate forms" somehow prevent us from letting T=c at z=1 and z=2? (I say no.)

 

(7.) Does the possibility of "division by zero" prevent us from letting T=c at z>2? (I say yes.)

 

(8.) Can we re-define T as "any positive real number other than unity", then show that at T=1 and T=2, the "limit as T approaches c" is c and c^2 respectively? (I say yes.)

 

Please let me know what your answers to these questions are.

 

By the way, I disagree that:

 

(0/0)=N and (6/0)=(infinity)

 

is a "fair comparison", because while any number:

 

N*0=0,

 

there is no number called "infinity" such that:

 

(infinity)*0=6.

 

Clearly, if limits are not introduced, then the "blunt" equation:

 

(0/0)=N

 

still belongs in the realm of numbers, while:

 

(6/0)=(infinity)

 

remains entirely nonsensical.

 

I also can't see how you can possibly claim that the implication:

 

51/3=2 (implies) N*0=0

 

is "formally true". The equation 51/3=2 is false, while the equation N*0=0 is true. How can something false imply, "formally" or otherwise, something true?

 

Don.

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There is no fallacy in my argument.

 

First of all, let's keep in mind that we are discussing the meanings of the expressions:

 

(1)^(0/0) and

 

(1)^(N/0), N>0

 

as they relate to my proof of the BC.

This is no way to address an opponents points in debating mathematical or scientific topics.

 

Please let me know what your answers to these questions are.
As you have failed to properly address many of my points and reply to my queries, I'm under no obbligation to do so despite that I could. It would not be worth the waste of time and I'm posting during breaks at work and limited time elsewhere and it's time consuming to parse your notation, copy it onto paper more readably and make certain I don't get the parentheses botched up. Folks have other things to attend to, if you fail to be cooperative and more compliant.

 

By the way, I disagree that:

 

(0/0)=N and (6/0)=(infinity)

 

is a "fair comparison", because while any number:

 

N*0=0,

 

there is no number called "infinity" such that:

 

(infinity)*0=6.

Please note that [imath]0\cdot\infty[/imath] is also notoriously an indeterminate form; the comparison is a fair one. Wasn't that the reason why I said they are equally meaningless?

 

Clearly, if limits are not introduced, then the "blunt" equation:

 

(0/0)=N

 

still belongs in the realm of numbers, while:

 

(6/0)=(infinity)

 

remains entirely nonsensical.

No, they are both meaningless.

 

I also can't see how you can possibly claim that the implication:

 

51/3=2 (implies) N*0=0

 

is "formally true". The equation 51/3=2 is false, while the equation N*0=0 is true. How can something false imply, "formally" or otherwise, something true?

 

Don.

Ask a logician. My only remedy to this shall be to suggest you learn the basics of logic; you have fully confirmed your lack of grasp on them; my reply of yesterday to your point:
That's a "reasonable implication" because N*0=0 is a "true statement".
was overly cautious because it wasn't clear whether or not you meant the very fact that the consequent being identically true is sufficient condition for the implication being true. Apparently this isn't so, you are therefore quite convinced of the argument which I detailed for you in order to expose the fallacies.
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To:Qfwfq.

 

In post#57, you ended your first paragraph with: "Let's look at the fallacies in your argument".

 

Well, I already know that there are no fallacies in my argument, so the only possible reply from my point of view was and is: "There are no fallacies in my argument."

 

I suppose that I could also have mentioned that "my grasp of fundamental logic" is quite good, and that from my point of view, it's your grasp of fundamental logic that needs to be "straightened out", but I prefer not to indulge in such childish gibberish.

 

Neither of us would bother to reply to the other if either of us was truly inept.

 

Moreover, I never ever considered you to be my "opponent".

 

I thought that we were exploring my proof together, as friends.

 

Now, here's where we differ:

 

I hold that the "blunt" equation:

 

(1)^(0/0)=1

 

is both true and correct, while you continue to assert that the expression

 

(1)^(0/0)

 

is both "undefined" and "meaningless".

 

Now, for positive integer variables, and co-prime terms, given the equations:

 

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

 

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

 

and

 

((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, ________________(2)

 

if we let z=1 and z=2 respectively, then the above becomes:

 

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

 

T(c/T)^((ln©/(ln(T))-1)/(ln©/(ln(T))-1)) ________________________(3)

 

and

 

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

 

(T(c/T)^((ln©/(ln(T))-1)/(ln©/(ln(T))-1)))^2 _____________________(4)

 

where the logarithms can now be "crossed out", so that all we have left is:

 

(T/T)a^x+(T/T)b^y=(T/T)c=T(c/T) ______________________________(5)

 

and

 

((T/T)a^(x/2))^2+((T/T)b^(y/2))^2=((T/T)c)^2=(T(c/T))^2 __________(6)

 

letting T=c in all of the above equations now results in:

 

(c/c)a^x+(c/c)b^y=(c/c)c=c(1)^(0/0), ___________________________(7)

 

 

((c/c)a^(x/2))^2+((c/c)b^(y/2))^2=((c/c)c)^2=(C(1)^(0/0))^2, ______(8)

 

 

(c/c)a^x+(c/c)b^y=(c/c)c=c(1) __________________________________(9)

 

and

 

((c/c)a^(x/2))^2+((c/c)b^(y/2))^2=((c/c)c)^2=(c(1))^2, ____________(10)

 

where it is now obvious that the equation:

 

(1)^(0/0)=1

 

is perfectly consistent with the cancellation of the logarithms.

 

If the expression:

 

(1)^(0/0)

 

was "undefined", as you claim it is, then both (7) and (8) would become:

 

a^x+b^y=c= "undefined"

 

and

 

a^x+b^y=c^2= "undefined",

 

and that would make no sense whatsoever, because clearly, defined co-prime equations do indeed exist!

 

The properties of logarithms can not render all co-prime equations "undefined"!

 

Thus, it must be the case that:

 

(1)^(0/0)=1,

 

and

 

(1)^(0/0) is not "undefined".

 

The proof works!

 

Don.

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In post#57, you ended your first paragraph with: "Let's look at the fallacies in your argument".

 

Well, I already know that there are no fallacies in my argument, so the only possible reply from my point of view was and is: "There are no fallacies in my argument."

Please note that these forums have Rules which include something against this type of reply. The onus of proving your point when challenged is on you and I provided a clear argument against your claim about "reasonable implication" which you had not supported at all.

 

I suppose that I could also have mentioned that "my grasp of fundamental logic" is quite good, and that from my point of view, it's your grasp of fundamental logic that needs to be "straightened out", but I prefer not to indulge in such childish gibberish.
Which childish gibberish?

 

By definition, an implication is a true assert if there are no cases in which the consequent is false and the implicant true. Without this being the valid requistite, there would be no point in arguing out any implication. It is trivial that it holds if at least one of the following can be said about a given implication:

 

  • there are no cases in which the consequent is false
  • there are no cases in which the implicant is true

 

Now Don, I agree that at if least one of the above holds the implication is useless. Indeed this is the reason I don't see the point of your argument, based on [imath]0\cdot\infty=0[/imath] being true. In the second case above, it is impossible to apply modus ponens and superfluous to apply modus tollens; vice versa in the first case. Where am I needing to be straightened out?

 

If the expression:

 

(1)^(0/0)

 

was "undefined", as you claim it is, then both (7) and (8) would become:

 

a^x+b^y=c= "undefined"

 

and

 

a^x=b^y=c^2= "undefined",

 

and that would make no sense whatsoever, because clearly, defined co-prime equations do indeed exist!

 

The properties of logarithms can not render all co-prime equations "undefined"!

 

Thus, it must be the case that:

 

(1)^(0/0)=1,

 

and

 

(1)^(0/0) is not "undefined".

Gosh I'm finally beginning to see that your argument is along the lines of:

 

[math]\frac{x}{x}=1[/math] is an identity [math]\forall x[/math] (rather than [math]\forall x\neq0[/math] as I say)

 

therefore, since 1 is defined even when [imath]x=0[/imath] then so must [imath]\frac{x}{x}[/imath] be. This appears to be the main source of disagreement.

 

Mathematics, as Gauss meant to imply in the sentence you quoted here weeks ago, is all a matter of: "Define, construct and work out the consequences!" Now Don, if you say the above is an identity [imath]\forall x[/imath] rather than [imath]\forall x\neq0[/imath] you imply that the expression [imath]\frac{x}{x}[/imath] defines a function that's equal to 1 and continuous even for the value [imath]x=0[/imath] and therefore, in order to adopt a non-standard definition of the terms function and continuous then you must state exactly what your definition of these terms is. If you can make the whole thing self-consistent it's an Alternative Theory, otherwise it's a Strange Claim and this case includes if you insist on your claims about implications but fail to support them.

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Let me first address a simple question Don asked about formal logic.

The implication [imath]\frac{51}{3}=2\Rightarrow N\cdot0=0[/imath] is also formally a true one, what can you deduce from it?
I also can't see how you can possibly claim that the implication:

 

51/3=2 (implies) N*0=0

 

is "formally true". The equation 51/3=2 is false, while the equation N*0=0 is true. How can something false imply, "formally" or otherwise, something true?

Qfwfq is correct, because, in formal logic, “something false can imply something true”.

 

Like all formalism, formal logic is simply a collection of rules exactly defining the meaning of the symbols - operators and term - used in it. Unlike more well-known formalisms such as arithmetic, statements in ordinary formal logic evaluate to one of only two values: True and False.

 

The implication operator, [imath]\Rightarrow[/imath], is defined as follows:

[math]\begin{array}{|c|c||c|} a & b & a \Rightarrow b \\

\hline

False&False&True\\

False&True&True\\

True&False&False\\

True&True&True\\

\end{array}[/math]

 

So the statement [imath]\mbox{False} \Rightarrow \mbox{True}[/imath] evaluates to True.

 

It's common to mix arithmetic statements with logical ones. Such a statement,

[math](1=2) \Rightarrow (1=1)[/math]

while intuitively silly-looking, is formally true.

 

On the less simple question of the Don's claim to have proven Beal's conjecture, I can offer only vague comment and advice.

 

I don't believe Don's proof is correct

 

Proofs are considered correct , categorizing roughly, for either formally or “sociologically”.

 

A formal proof is correct when each statement of a proof is “mechanically” produced according to a formal operation postulated to be correct for the formal system in question.

 

A “sociological” proof is “correct” when the intended audience of the proof believe it to be so. The importance of a sociological proof depends strongly on its audience. For example, for mathematical proofs, the acceptance as true by many professional academic mathematicians is more important than the acceptance, of, say, a gathering of friends.

 

By neither of these standards are Don's proofs correct.

 

My advice to Don is to take the various unusual phrases he's used in his proofs, and give them exact, formal definitions. “Cohesive term”, I believe, would be the best one to start with. State precisely the rule for transforming one expression into another given by the phrase. Don't attempt to prove anything, and avoid vague statements such as “cohesive terms are a consequence of the properties of logarithms”, and simply define the concept.

 

A final bit of advice: When people don't agree with you, don't simply state that you know you're right. Quoting from the site rules

Do not endlessly show us that *your* theory is the *only* truth. And don't follow this up by making people look stupid for pointing out that there are other answers, especially if they provide links and resources.
It will get you banned!

In short, when encountering disagreement, back up steps in your argument until you encounter agreement, then, at the first step at which you encounter disagreement, support your claim with links and references to accepted texts, and discuss, being prepared to acknowledge that you may be wrong.

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To:Qfwfq and Craig D,

 

You know, in my previous post, I made it absolutely clear that those pesky "indeterminate forms" don't even exist if we do the algebra correctly and "cross out" the logarithms at z=1 and z=2.

 

In other words, the beingness of the pesky expression: (1)^(0/0) is "zip", "zero", "nada" at z=1 and z=2.

 

Now, if they don't exist, then they are a non-issue and we are really going off topic in discussing them!

 

The proof works precisely because z=1 and z=2 are the only possible values of z that allow us to "cross out" the logarithms preventing T=c.

 

That's all there is to it. No point in discussing "indeterminate forms" that either can, or must be "crossed out" before they even exist and are therefore perfectly consistent with the result:

 

(1)^(0/0)=1,

 

just as the "counterintuitive" expression: (p)^(0), is perfectly consistent with the result:

 

((p)^(q))/((p)^(q))=(p)^(q-q)=1.

 

Take another look at my post #60 and consider this fact: If it were not the case that the expression:

 

(1)^(0/0)=1,

 

then logarithms themselves would have to be viewed as an "inconsistent construct" because their properties would not allow T=c in (3) and (4), but would allow T=c in (5) and (6).

 

We would then have to banish logarithms from mathematics altogether!

 

Don't you see, we simply can't eliminate logarithms from mathematics, so we absolutely must conclude that:

 

(1)^(0/0)=1,

 

because it's not just the only logical and consistent conclusion, but the only possible conclusion as well! I simply can't accept your view that the properties of logarithms are somehow "bogus".

 

We must keep in mind that new discoveries often show us new things and therefore bring about new points of view.

 

I discovered "cohesive terms" about a decade ago, and showed how they are derived and defined apart from "non-cohesive terms" throughout my website. As their inventor, I had to give them a name, and chose "cohesive term" because they are the first and only algebraic terms in the history of mathematics that actually prevent cancelled factors or cancelled common factors from "falling off" and getting "lost".

 

I introduced them here in this forum because that property alone is nothing less than astonishing!

 

Now, let's get back on topic and discuss whether or not the equation:

 

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

 

is telling us that, in principle, multiplication and/or division by unity automatically results in division by zero.

 

Letting T=1 sure seems to indicate that it does!

 

Is this a "strange claim"?

 

Perhaps, but that's a good thing because in this case, it is obvious that the truth is indeed stranger than fiction!

 

Don

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  • 1 month later...
(T/T)a^x=T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1))

 

is telling us that, in principle, multiplication and/or division by unity automatically results in division by zero.

 

Don, have you studied ring theory? Field theory? Abstract algebra of any sort? Quite honestly, what this implies is that the way mathematics has been done is a sham. I really don't even have to read the previous posts to know that if you claim that your argument implies this, then it is false. Stranger than fiction? I think not. If you knew how about constructions and completions of fields, it would help to understand that this statement is simply wrong. If this were the case, mathematics would be reduced to trivial statements. Mathematicians may have pondered such questions as these at one point (perhaps even ones that would seem more obvious) but, I stress, multiplication by unit does NOT result in division by zero. Do you need the entire mathematical community to ensure you about this?

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To: Nootropic,

 

I'm almost 60 years old, so I have had plenty of time to familiarize myself with all those particular branches of mathematics... and then some. How about you?

 

I never said, or even implied that "the way mathematics has been done is a sham". Those are your words, not mine.

 

Mathematics must adapt to and be consistent with new discoveries if it is to remain both both vital and dynamic. Therefore, new discoveries must be disseminated rather than "swept under a rug" for fear of change.

 

Courage is required!

 

Most importantly however, new discoveries must be studied so that all of their consequences and ramifications are properly, which is to say logically, interpreted.

 

Initially, there may be disagreements with regards to various possible interpretations, (such as the disagreement between Newton and Liebnitz on the subject of limits) but in the end, the truth will prevail, and mathematics will grow, thrive and flourish rather than stagnate.

 

Your fear that my equation would "reduce mathematics to trivial statements" is without foundation. My equation happens to be true, and the truth can never render the robust and sublime body of knowledge called mathematics "trivial".

 

Now, let's muster up all of our courage and ask ourselves the following questions:

 

Given the true equation:

 

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

 

if our variables are to represent non-negative integers, then:

 

(1) Which term has the better defined variables?

 

(2) Keeping in mind that unity is not an actual common factor but a "trivial" or "degenerate" common factor, which term is more suitable for representing actual common factors?

 

(3) Keeping in mind that what we do to one side of an identity, we must also do to the other, can we "cross out" the cancelled T's the way we were taught in school?

 

(4) What occurs at T=1 and what does it imply?

 

When answering these questions, please remember that the above equation is absolutely new to mathematics ( I discovered it only a decade ago) so there is, as of yet, no "general consensus" on what it actually means.

 

It's proper interpretation is perhaps the greatest challenge that the mathematical community has ever faced! Try answering those four simple questions and see for yourself!

 

Don.

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Do you not realize that if multiplication by 1, in, say the real numbers, results in division by zero then let x be a nonzero real number (note this is is also finite because we are not working in the "extended real number system" [ie. the closure of the real numbers]) x = 1*x = x/0 (which is again, not defined). And then by the density of the rationals in the real numbers, we have some sequence of rational numbers converging to x. So basically what this is saying is that NO sequence of rationals converges to any real number and density can be proven use elementary properties of the real numbers without resorting to topology. So basically this apparent "discovery" contradicts the entire field of real analysis. Now what is correct: Some "amazing" discovery or a field built upon almost 200 years of solid foundation?

 

If we are talking about division rings with an identity (unity), then how exactly do we deal with division by zero? This would make division rings (and hence fields) virtually unusable. Personally, I find that you have an extremely disturbed view of what the term "variable" means, how mathematics is done, and mathematics in general.

 

You really don't realize that such a discovery attempts to invalidate numerous fields of study. It basically makes the real analysis classes undergraduates and graduates take each year useless, because apparently the entire field of real numbers is unusable. One sign of crackpot mathematics is its implications and you fail to realize the implications your apparent discovery imposes.

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To: Nootropic,

 

We are, at present, discussing algebraic terms as they apply to common factors.

 

You are conjuring up constructs involving "extended reals" that have absolutely nothing to do with my discovery and are invoking a completely different branch of mathematics that has no bearing on this discussion whatsoever!

 

Also, please back up your claims. If you think that my equation is "crackpot mathematics" then show us where the error lies.

 

If you can neither point to a mistake in my equation, nor answer those four simple questions, then all you are really doing is avoiding the issue.

 

Please stay on topic and answer the questions!

 

Don.

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You are conjuring up constructs involving "extended reals" that have absolutely nothing to do with my discovery and are invoking a completely different branch of mathematics that has no bearing on this discussion whatsoever!

 

I am not "conjuring up any constructs". Read a book on analysis, preferably Rudin's Principles of Mathematical Analysis. I am arguing on the basis that this apparent discovery implies things we know to be false (ie. multiplication by one results in division by zero). If someone came up with a proof of the Riemann Hypothesis that implied 2 = 1 (in the integers of course), then what are we going to accept? 2 = 1 or this proof? It doesn't take a great amount of intellect to realize what's right and wrong. CraigD has already pointed out that it is an identity for an exponential function, but it does not imply multiplication by 1 results in division by zero.

 

I was not working in the extended real number system. What I'm saying is that the "x/0" is not defined. Basically what you are saying is that a sequence that converges to a real number does, in fact, not converge. I can pick any concrete example, multiply it by one, and then still find a sequence that converges to it. I don't know why you are so insistent on this ridiculous statement.

 

We are, at present, discussing algebraic terms as they apply to common factors

 

Yes, and the real numbers are an algebraic object--a field. This is basic field theory arithmetic.

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