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There is already way too much "bad blood" on this God forsaken planet,

and I refuse to contribute any more. Please close my threads.

Don

I apologize if post #75 was in any way viewed as "rebuking the poster for post #74".

I am being as polite as I know how, but if that is insufficient for this forum,

then please close this thread, along with my other thread here and I will politely go away.

Don.

The question in post #73 involves a paradigm shift.

http://en.wikipedia.org/wiki/Paradigm_shift

I know it's a "hard question", but I believe that the human mind

is capable of answering it with a simple **yes** or **no **.

Don

It's not a demand.

It's simply a polite request.

Those who are unable to answer with a simple **yes** or **no**

are free to post whatever they want, like or desire...

as long as they are having fun.

Don

Please just answer **yes** or **no** without any commentary whatsoever.

Given the identity:

can we substitute

for ?Please just answer yes or no without any commentary.

Don

Quoting anonimnystefy:

The substitution cannot be done because the of the restrictions that the lofarithms pose.

You see folks, that so called "**substitution axiom of equality**",

which states that we can * always* and

has been

We can not allow that utterly ridiculous "axiom" to be shoved down our children's throats!

Don

Quoting anonimnystefy:

The axioms of equality do not say we can do that.

Of course they do! Take for instance the **substitution axiom of equality** which states that

if two quantities are equal, then one can be replaced by the other in any equality or expression.

Well, the two quantities

and are indeed equal,but if we try to replace with in the identity ,

then we quickly find that it's quite impossible, because clearly, the above identity has thoroughly and irrevocably * negated* the closely related

Now, this is not the first time that a faulty axiom has been negated by a perfectly logical mathematical construct.

We must not forget that the negation of Euclids fifth axiom (parallel postulate) ultimately resulted in many vastly

superior "non-Euclidean" geometries, one of which even allowed Einstein to formulate his theories of relativity!

Don.

Quoting anonimnystefy:

The identity doesn't hold when a=b.

That's what I said in the opening post of this thread!

So, do we now agree that given the identity:

we can never let or substitute for

even though those so called "axioms of equality" say we can?

Don

Quoting anonimnystefy:

This is the flawed step.

Quoting anonimnystefy (one post later):

It is an identity when a<>b.

Well, if it's an identity, then there can be no "flawed step".

Don

That's an identity, pure and simple. You are wrong.

Yes, I comprehend it perfectly. It shows that you are wrong.

Don.

Which of course proves my point!

That your posts aren't even in LATEX shows that you can't

even take the time or trouble to articulate yourself clearly!

In other words, it implies that you are **not serious**.

cmowla's graphs clearly show that my equations are both true and correct.

Why you can't handle that, I don't know.

Don.

Your posts aren't even in LATEX. cmowla's "Mathematica" results are clear. You are simply wrong!

I wish there was a kinder way to say that, but... this issue is just too important for us to "beat around the bush."

Don.

To: cmowla,

Thanks for the extended derivation and for articulating the stipulation **"AND a and b are non zero"**.

Most of all, thanks for making it abundantly clear that the equations are both true and correct.

It should now be obvious to at least some of our more advanced readers that the axioms of mathematics

are indeed badly flawed and need to be revised immediately.

Math is supposed to be fun, but how can students have fun when their reasoning power has been compromised

by their having been indoctrinated into a system of logic that has, as its foundation, "axioms" that are badly flawed?

Indoctrinated minds can't think. They can only "parrot" what they have been taught.

* You* have seen first hand the damage that has been done. Just look at all the hassle that

in order to get this argument even

Why do you and I see the gist of this issue so clearly, while others are still struggling with it ?

I will venture to say that it's because you and I had teachers who did not force us to believe in

a bunch of gibberish and actually encouraged us to **think for ourselves!**

Now, the only question that remains is **what are we going to do about it?**

In another month from now, most students will be back in school, and many more will be indoctrinated into those

same badly flawed "axioms".

Are we going to let that happen?

Don.

So clearly, we all agree that division by zero is strictly disallowed.

Therefore, we must all agree that given the equations in this post:

Quoting myself from post #1:

The "foundations of mathematics" are its

axioms, which are defined as "self evident truths".

So, let's have some fun with them. Let's "shake" those foundations a little and see what happens!Consider the "symmetric axiom of equality" which states that "if

, then .Well, if

where ,and the properties of logarithms allow

where ,then clearly, that so called "symmetric axiom of equality" is neither self evident, nor always true!

Don.

it's not always true that "if

then ",and we can't always substitute for .

Don

To: noelevans,

Quoting noelevans:

We can certainly define a function f(x)=x/x by f(x)=1 if x<>0 and f(x)=a (a any real number) if x=0. And it is certainly nice to define this as 1 since this is the limit of the function as x approaches 0.

But this is not to say that the actual number 0 divided by itself (0/0) is one. That would be equivalent to saying that zero has a multiplicative inverse, which is precluded in the field axioms.

I agree.

We can not assign a specific value to the indeterminate form

without knowing the circumstances under which it occured.

In and of itself,

can bejust as implies the

implies the

while implies the

From my point of view, since

implies athe symbol can be used to make other

can indeed be construed as meaning

"one raised to any power equals one".

Don.

Hi,

Quoting anonimnystefy:

There is no logarithmic law that allows you to transform (3*log(a)/log(b))/(log(a)/log(b)) into (3*log(a)/log(b) -1)/(log(a)/log(b) -1).

Two functions are same if and only if they map the same domain into the same range in exactly the same way. When you remove the singularity at 0 you change the domain of the function sin(x)/x from R\{0} to R.

Can you post the above in LATEX? I'm sure that our readers will appreciate it!

Thanks,

Don.

Quoting anonimnystefy:

Your "identity" isn't correct. You cannot subtract 1 from the denominator and the numerator

of a fraction and say it is the same fraction. x/y<>(x-1)(y-1) in the general case.

Please look carefully.

We are * not* "subtracting 1 from the denominator and the numerator of a fraction".

We are

Quoting anonimnystefy:

Yes, we could define another function partially, so that it has no singularities,

but that wouldn't be the same function we started with.

That's kind of like saying that after somebody "pops a zit", they don't have the same face they started with.

I tend to view it as the * same* function but in the light of a higher order of logic.

The important thing is that my identity has a **non-removable singularity** at

while MrButtermans equations have a

Therefore my identity presents a much stronger argument for eliminating the symmetric and substitution axioms of equality.

However, if you want to join my crusade to eliminate those shoddy axioms using MrButtermans much weaker equations,

then I still welcome your support because really, those axioms have got to go.

Don

To:anonimnystefy,

Quoting anonimnystefy:

No one defined sin(x)/x to be 1.

Quoting the article "Removable singularity" from Wikipedia:

...the function

has a singularity at .

This singularity can be removed by defining ,

which is the limit of as tends to .

Please note the phrase "removed by * defining*".

Quoting anonimnystefy:

But, either way, you didn't derive the formula correctly.

It's not a formula. It's an identity, and it's correct.

To see that it's correct, apply the property: ln(a/b)=ln a - ln b * before* the change of base. It works out the same.

Don.

In complex analysis, if we can't **define**

then neither can we **define**

Let's all Google the phrase "removable singularity" and find out!

Don

To: bobbym,

Quoting bobbym:

In arithmetic operations as you are doing care must be taken not to use 0 / 0 as 1.

I agree. Care must be taken and we can't just * let* the

We must * first* know the details of how it occured in order to give it a specific value.

For instance, in the expression

Thus, we can

Don.

To: anonimnystefy,

Hi,

Quoting anonimnystefy:

This step is wrong:

All the steps are correct, including that one.

Don.

To: TheDude,

Quoting TheDude:

Can you show the steps you took to get from

to

It's not obvious to me how you do that.

The "Blazys identity" is derived as follows:

Note that it is * not possible* to derive this identity if

the coefficient of the first term is either

Don.