Mathematical Consistency

[Ben]

the notion of 1 being the "identity element"

relative to multiplication is highly illogical.

No, it is merely a convention; you do not need to adopt if you don't want to.

 

So, suppose for example we ditch 1 as the multiplicative identity and choose, say, 2. So that 2x = x. This is OK, since there exists a one-to-one correspondence between the infinite sets, say, [math]\{1,2,3,...\}[/math] and [math]\{2,4,6,...\}[/math] and we have that [math] x = \frac{x}{2}[/math].

 

This is still OK. But the problems start here: Let's agree that [math] \ln(xy) = \ln(x) + \ln(y)[/math], so that [math]\ln(2y) = \ln (y)=\ln(2) + \ln(y)[/math].

 

So we are still left with the problem that we MUST have that [math]\ln(2) =0[/math], unless you are prepared to ditch the additive identity also.

 

Are you?

[Qfwfq]

Are you?

No! I prefer to call it apple, with any number times apple giving the same number and the logarithm of apple being 1.

 

:D

[Don Blazys]

Quoting Qfwfq

But exactly how is this question related to dividing by zero?

 

 

The question is...

 

Given the identity:

 

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

 

can we let [math]T=1[/math] if [math]x>1[/math] ?

 

 

 

Now, in order to see what the answer might be, let's let [math]T=1, a=2, x=3[/math] which gives us:

 

[math] 1*2^3= \left(1*2\right)^{\frac{\frac{3*\ln(2)}{0}+1}{\frac{\ln(2)}{0}+1}} [/math]

 

Thus we find that the answer is indeed "no" because the right hand side contains

two different "divisions by zero", both of which are strictly disallowed.

 

Don

[Qfwfq]

The question is...

 

Given the identity:

 

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

 

can we let [math]T=1[/math] if [math]x>1[/math] ?

 

 

 

Now, in order to see what the answer might be, let's let [math]T=1, a=2, x=3[/math] which gives us:

 

[math] 1*2^3= \left(1*2\right)^{\frac{\frac{3*\ln(2)}{0}+1}{\frac{\ln(2)}{0}+1}} [/math]

 

Thus we find that the answer is indeed "no" because the right hand side contains

two different "divisions by zero", both of which are strictly disallowed.

No Don, that is the answer to the question I had proposed, you are saying that in the [imath]T=0[/imath] case the rhs of the identity is undefined.

 

What is your point anyway?

[sanctus]

Qfwfq, no he is not ln(T)= 0 for T=1...

[Qfwfq]

no he is not ln(T)= 0 for T=1...

Err, uhm, ah... oh! You mean I wrote [imath]T=0[/imath] but I should have written [imath]T=1[/imath] :doh:

 

:D

[Ben]

What is your point anyway?

 

Something like this?

 

 

the notion of 1 being the "identity element" relative to multiplication is highly illogical.

 

Meanwhile, those so called "professors of mathematics" in our colleges and universities will continue to not only teach this gibberish and indoctrinate our young people with this rubbish, but will undoubtedly add insult to injury by demanding payment for their "services"!

 

You math majors who are paying thousands of dollars for an education are being taken for suckers.

 

You are being defrauded, duped and swindled!

 

Which I am sure all here will agree with.

 

Oh wait, what's that noise? Sounds like an axe being ground - maybe my hearing isn't all that it used to be......

[Don Blazys]

Quoting Myself:

The notion of 1 being the "identity element" relative to multiplication is highly illogical.

 

Quoting Ben:

Which I am sure all here will agree with.

 

At [math]T=1,[/math] the perfectly true, perfectly defined identity:

 

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

 

demonstrates that multiplying [math]a^x[/math] by [math]1[/math]

(so as to leave [math]a^x[/math] "unchanged")

would automatically result in divisions by 0,

and since divisions by 0 are strictly disallowed,

multiplication by 1 must also be strictly disallowed.

 

Everyone here agrees with that because

this identity is simple, incontrovertible,

and can easily be checked by anyone.

 

New discoveries in math will require math to change

and at this early stage, most folks are still afraid

because the necessary change is indeed somewhat radical.

 

However, in time, they will think it through,

and once they get used to the truth,

that fear will naturally just fade away.

 

 

Quoting Ben:

Oh wait, what's that noise? Sounds like an axe being ground.

 

Do not ask for whom that axe grinds...

it grinds for thee! ;)

 

Quoting Ben:

Maybe my hearing isn't all that it used to be......

 

Your hearing is just fine.

Indeed, you have good ears.

It's what's between those ears

that may be in question here!

 

Don.

[Don Blazys]

To: CraigD,

 

Quoting CraigD:

I recall that Don posted a derivation the identity in post #1 some time ago,

either at hypography or on his own webpages.

Perhaps he’ll take pity on me and my poor memory and algebra and repost them here.

 

The derivation is similar to the one in the "notes"

at the end of my Proof of Beal's Conjecture.

 

You can find it here:

 

http://donblazys.com/02.pdf

 

Speaking of which, are you any good at writing stuff out in

the language of "predicate calculus" or "first order logic"?

 

I need some help in translating my proof into that language

so that it can be verified by computer.

 

Don.

[Ben]

It's what's between those ears

that may be in question here

Now, now, there is no need to be rude.

 

First this: you realize, of course that it is NOT true in general that [math]Ta^x = (Ta)^x[/math]? I know you never asserted as much, just want to clear the decks, as it were. Let's quote your monster so-called "identity" so we can keep our bearings. Here it is:

 

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

 

I claim this is false when [math]x = 0[/math]. For,

 

[math] Ta^0= (Ta)^0 =T,\,\,\, \frac{0\ln(a)}{\ln(T)}+1= 1[/math] in the numerator, and [math]T[/math] can be whatever we want. Agreed?

 

So let's assume, for a contradiction, that [math]T \ne T'[/math] and set [math]a =1[/math]

 

So that [math]\frac{0}{ \ln T'} +1 =1[/math] in the denominator which implies that [math]T = T'[/math], contrary to our datum

[sanctus]

Don, referring to your post #25. I see no problem with the identity element there, even if T=1 is disallowed on the rh-side. It simply means that this identity is only valid for T[math]\neq[/math]1 and then you don't have any contradiction anymore ;-).

[Qfwfq]

Now, now, there is no need to be rude.

Absolutely right.

 

I claim this is false when [math]x = 0[/math].

No Ben, it isn't false when [math]x = 0[/math], it simply reduces to:

 

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

 

which is a mighty fine identity with the exceptions of [imath]T=1[/imath] and [imath]Ta=1[/imath].

[modest]

Don, this is how I would derive the identity:

 

(1)... [math]Ta^x = Ta^x[/math]

 

using the logarithmic identity [math]b^{\log_{b}(z)}= z[/math] where [math]b = Ta[/math] and [math]z = Ta^x[/math]

 

(2)... [math]Ta^x = (Ta)^{\left(\log_{Ta}\left(Ta^x\right)\right)} [/math]

 

using [math]\log_b z = {\log_d z \over \log_d b}[/math]

 

(3)... [math]Ta^x = (Ta)^{\left( \ln(Ta^x) \over \ln(Ta) \right)} [/math]

 

using [math] \log_b(xy) = \log_b(x) + \log_b(y) [/math]

 

(4)... [math]Ta^x = (Ta)^{\left( \ln(a^x) + \ln(T) \over \ln(a) + \ln(T) \right)} [/math]

 

using [math] \log_{b} (x^y) = y \log_{b}(x) [/math] for [math]\ln(a^x)[/math]

 

(5)... [math]Ta^x = (Ta)^{\left( x \ln(a) + \ln(T) \over \ln(a) + \ln(T) \right)} [/math]

 

using [math]z \cdot 1 = z[/math] where z is the exponent,

 

(6)... [math]Ta^x = (Ta)^{\left(x \ln(a) + \ln(T) \over \ln(a) + \ln(T) \right) \left( 1 \right) } [/math]

 

using [math]1 = z/z[/math] where z = [math]1 / \ln(T)[/math] and distributing,

 

(7)... [math]Ta^x = (Ta)^{\left( \dfrac{\frac{x \ln(a)}{\ln(T)} + \frac{\ln(T)}{\ln(T)}}{\frac{\ln(a)}{\ln(T)} + \frac{\ln(T)}{\ln(T)}} \right)} [/math]

 

using a/a=1 where [math]a = \ln(T)[/math]

 

(8)... [math]Ta^x = (Ta)^{\left( \frac{\frac{x*\ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1} \right)}[/math]

 

 

Step 2 requires [math]Ta \neq 1[/math] because the identity [math]b^{\log_{b}(z)}= z[/math] requires [math]b \neq 1[/math] and also step 7 and 8 require [math]\ln(T) \neq 0[/math] and therefore [math]T \neq 1 [/math].

 

So, I agree with the other members that your identity is true for positive integers where [math]Ta \neq 1[/math], [math]T \neq 1[/math].

 

At [math]T=1,[/math] the perfectly true, perfectly defined identity:

 

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

 

demonstrates...

 

I don't believe the identity is defined for T=1. You made steps in the derivation that are true and defined only if T is not equal to one.

 

~modest

[Don Blazys]

Quoting Modest:

I don't believe the identity is defined for T=1.

You made steps in the derivation that are true and defined only if T is not equal to one.

 

Not only is the identity "undefined" at [math]T=1[/math],

it is also "nonsensical" at [math]T=1[/math].

 

That is a good thing because we want the variables to be perfectly

and meaningfully defined by their own unique domains as follows:

 

[math]T\in\{2,3,4,5...\}[/math], [math]a\in\{1,2,3,4...\}[/math], [math]x\in\{0,1,2,3...\}[/math].

 

If the term on the right "didn't exist", then all we would have is the term [math]T*a^x[/math] where

the variables are abysmally and redundantly "defined" by the same domain as follows:

 

[math]T\in\{0,1,2,3...\}[/math], [math]a\in\{0,1,2,3...\}[/math], [math]x\in\{0,1,2,3...\}[/math].

 

Now, here's the question...

 

If we write:

 

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

 

then [math]T\neq1[/math] in the term on the left,

 

and if we move the term on the right down a few an inches as follows:

 

[math] Ta^x[/math]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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

 

then it is still the case that [math]T\neq1[/math] in the term on the left.

 

So... how far down must we move the term on the right before we can let [math]T=1[/math] in the term on the left?

 

Don

[modest]

Not only is the identity "undefined" at [math]T=1[/math],

it is also "nonsensical" at [math]T=1[/math].

 

I couldn't agree more.

 

Now, here's the question...

 

...So... how far down must we move the term on the right before we can let [math]T=1[/math] in the term on the left?

 

Twenty eight and one half inches. I've found, in my experience, that is the distance something must be moved in order to be ignored. It also, coincidentally, happens to be the distance from the center of my desk to the trash can.

 

In all honesty, an identity has a left side and a right side. The domain of your identity does not include T=1 no matter how one obfuscates that it is an identity. The term Ta^x itself has no such domain restriction. It would be the same with the identity a=1. The domain of a is one. The term a, itself, has no such restriction on the domain of a (because, you know, a=1 is not the same as a).

 

~modest

[Qfwfq]

If you do it in Reykjavik it's enough to bring it down below the ground, an elf will immediately grab it and obliterate it from existence, so then [math]Ta^x[/math] can live happily ever after. Otherwise, it depends on the mass of the interaction boson; if it's a [imath]\pi[/imath] meson then the radius of a proton is sufficient, but if it has zero mass like a photon you're totally screwed, only an elf can help you.

So... how far down must we move the term on the right before we can let [math]T=1[/math] in the term on the left?

We can always let [math]T=1[/math] in the term on the left but it simply doesn't make sense in this case to say it equals the term on the right when the term on the right doesn't make sense. For [math]T=1[/math] one term makes sense and the other doeasn't, so they can't be equal. It doesn't matter where you write them or whether you write them or not.

[sanctus]

Don, it is just like in the post where modest derived your identity, there are 2 steps wehere there are requirements for this identity to valid. If you then look at a case where the identity is not valid and use it as an argument against the identity element, then you have NO argument ;-).

 

It is same like saying:

step1) x*x=x^2

 

step2) x*sin(asin(x))=x^2

 

Step 2 is only valid if x[math]\leq[/math]1. But what you seem to do in all this thread is to ignore the validity domain of an identity and say that outside this domain something doesn't work. In my example what you seem to do is like saying x*x=x^2 does not work, because step 2 is not defined for x=2, for instance.