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**Don Blazys****Member**- Registered: 2012-06-06
- Posts: 32

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.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,861

Just because my posts aren't in LaTeX implies tgat I am wrong? That is truly false logic.

I am not beating around the bush. I have directly stated in post #44. If you don't see it I can explain it in more detail or have somebody else explain it to you.

Neither of you have read my comment on cmowla's graphs.

*Last edited by anonimnystefy (2012-07-26 20:56:32)*

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**Don Blazys****Member**- Registered: 2012-06-06
- Posts: 32

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.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,861

No, it just implies that I am on my phone. If I was on my computer like I am now, I would be glad to LaTeX anything that needs LaTeXing.

The two graphs aren't the same. The second one has singularities along the line x=y.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**Don Blazys****Member**- Registered: 2012-06-06
- Posts: 32

Which of course proves my point!

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,861

No it doesn't. Did you look at post #44? Can you comprehend it correctly?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**Don Blazys****Member**- Registered: 2012-06-06
- Posts: 32

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

Don.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,861

anonimnystefy wrote:

Don Blazys wrote:This is the flawed step. He turned an exponential with base a^3/b to an exponential of the form 1^(log_1(...)) (which is "against the rules") and then turned that into a fraction using logarithm rules.

This is the post I am talking about. It shows that you are wrong.

Here lies the reader who will never open this book. He is forever dead.

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**Don Blazys****Member**- Registered: 2012-06-06
- Posts: 32

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

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,861

It is not an identity. It is an identity when a<>b.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,913

I am sorry fellows but at the beginning of a similar thread I warned everyone involved that name calling and fighting will not be tolerated. This thread is closed until tempers cool.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,913

The thread is reopened for posting.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**Don Blazys****Member**- Registered: 2012-06-06
- Posts: 32

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

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,861

It is flawed because you assumed in the begnning that a=b. The identity doesn't hold when a=b.

Here lies the reader who will never open this book. He is forever dead.

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**Stangerzv****Member**- Registered: 2012-01-30
- Posts: 173

I agree with anonimnystefy. You should put the condition at the end and not at the beginning because you know at the end the condition is not valid, implying it is not valid through out the derivation. In other words, it is wrong to put a=b in the first place. You should refer to the flaw proof of 2=1 when it is assumed at the beginning that a=b.

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**Don Blazys****Member**- Registered: 2012-06-06
- Posts: 32

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

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,861

The axioms of equality do not say we can do that. The axioms of equality never mention logarithms in the first place.

Here lies the reader who will never open this book. He is forever dead.

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**Don Blazys****Member**- Registered: 2012-06-06
- Posts: 32

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.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,861

The substitution cannot be done because the of the restrictions that the lofarithms pose. The "identity" works for a<>b, so substituting a=b is flawed.

Here lies the reader who will never open this book. He is forever dead.

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**Don Blazys****Member**- Registered: 2012-06-06
- Posts: 32

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

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,861

I never said you cannot substitute a/a with b/b. Your further steps are incorrect.

Here lies the reader who will never open this book. He is forever dead.

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**cmowla****Member**- Registered: 2012-06-14
- Posts: 57

Sorry Don, but even though

(where ),Your whole argument is redundant.

Let me explain.

**(Notice that it IS possible for **

Since we have

as well as, then your logarithmic function (if we choose not to simplify it to

) just illustrates a case where the condition holds.The 3D graph of your log function is different than the 3D graph of

, but that has nothing to do with the validity of your argument (it is neither for you nor against you) because theHere's a polynomial function which shows where the condition

must hold:This supports the other possibility that a can be equal to b (a must be equal to b, that is...just as in your log function where a must not be equal to b....and we are assuming that we don't simplify our functions completely to (b/b)a^3).

An even more trivial case (which you probably thought of but didn't post) for when a must not equal b is:

and so suppose we have:

Again, the graph of

and are different, but they do not disagree with the conditions. How could they? They are equal expressions once you simplify the more complicated one into the simpler one.*Last edited by cmowla (2012-08-04 12:40:08)*

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**Don Blazys****Member**- Registered: 2012-06-06
- Posts: 32

Given the identity:

can we substitute

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

Don

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,861

That is not an identity when a=b.

Here lies the reader who will never open this book. He is forever dead.

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**Don Blazys****Member**- Registered: 2012-06-06
- Posts: 32

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

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