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

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.

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**TheDude****Member**- Registered: 2007-10-23
- Posts: 361

Can you show the steps you took to get from

to

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

Wrap it in bacon

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**MrButterman****Member**- Registered: 2012-07-18
- Posts: 3

Don Blazys wrote:

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.

You are dividing by zero in your exponent, so there must be something wrong here

Your problem is at the third step. The fraction is equal to 0/0 and thus no longer equals the value in the second step.

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

Quoting MrButterman

You are dividing by zero in your exponent...

That is not true. I am not dividing by zero. In fact, I am doing exactly the opposite!

I am saying that:

wherewhich means that substituting

for is strictly disallowed, because division by zero is strictly disallowed.This is an extraordinarily serious issue because if the **symmetric axiom of equality** is flawed,

then the **substitution axiom of equality**, which states that we can "always" substitute

Quoting MrButterman:

Your problem is at the third step. The fraction is equal to 0/0 and thus no longer equals the value in the second step.

At

, your equations contain theIn this particular case, since the expression at

Thus, in this particular case, that indeterminate form , so at , your third step is clearly equal to your second step.

By contrast, at

, my equation has aQuoting MrButterman:

...so there must be something wrong here.

What's wrong here are those badly flawed axioms, which when taught to unsuspecting students,

actually hinders their ability to think rationally and thereby takes a lot of the fun out of math!

Don

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Could you reply to TheDude's question in post #2?

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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,636

Hi Don;

Here is my favorite argument.

In arithmetic operations as you are doing care must be taken not to use 0 / 0 as 1. There are simple methods to prove the danger, for instance

So,

dividing by 0?

replacing 0 / 0 by the assumed identity 0 / 0 = 1

so 1 = 2?

It is clear that the culprit is the asumption 0 / 0 = 1 in the arithmetic sense.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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

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.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Hi

This step is wrong:

Don Blazys wrote:

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

To: anonimnystefy,

Hi,

Quoting anonimnystefy:

This step is wrong:

All the steps are correct, including that one.

Don.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

No. that one is not correct. There is no explanation for subtracting ln(b)/ln(b).

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

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.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Hi

If we define x/x at 0 to be one, then the following would be true:

Which is not true! When we define division, we don't allow 0 to be the second argument.

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: 86,636

Hi;

I guess no one is reading post #6.

Because the limit of x / x as x approaches 0 is 1 that does not imply that 0 / 0 = 1. Those are two different concepts. As in post #6 the idea of using 0 / 0 as a number will lead to all kinds of contradictions. Isn't it just safer to eliminate 0 / 0 = 1 then to unravel all of mathematics?

It is illogical to assume 0 / 0 has different values when needed. We do not assume 3 / 3 has a different value according to where it is at. Post #6 is too simple in its workings. It is clear that assuming 0 / 0 = 1 is the mistake. Consider it a counter example to the concept 0 / 0 = 1.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

In the field axioms the multiplicative inverse axiom does not allow for zero to have a multiplicative inverse:

For each non-zero x in F there is a y in F such that x*y=1. Typically we write y as 1/x.

If x were zero then y would be "1/0" so x*(1/x) = 0*(1/0) = 0/0.

If zero had a multiplicative inverse then this would yield one.

On the other hand it has been shown many times that zero times any number is zero.

Thus we could conclude that 1=0. This makes any number equal to zero. Hence our

number system is reduced to a single number.

Such a system would be nice to work with, but not very useful.

The field axioms and the order axioms for the real number system do not involve limits.

Hence I have trouble "hooking up" with arguments about the real number system that

resort to limits from calculus. Calculus has lots of "strange things" due to pushing things

to infinity. But that's a totally different story.

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

Regarding the statement that

(b/b)*a^3 = b*(a/b)^[ ln(a^3/b)/ln(a/b) ] let the quantity in the brackets be named x.

Then we have (b/b)*a^3 = b*(a/b)^x = b*(a^x)/(b^x) = (a^x)/(b^(x-1))

Of course if a or b is zero then these have problems, so assume they are not zero.

If a=b=1 then all of these expressions produce 1.

Now suppose that a=b and they are not zero and not one.

Then the leftmost expression equals b*((a^3)/b) = a^3 whereas the rightmost expression

becomes (a^x)/(a^(x-1)) = a which cannot equal a^3 since a is not 0 or 1.

(b/b)*a^3 = a^3 and would be equal to (a^x)/(b^(x-1)) for very few pairs (a,b).

I see no problem with the log transformations of x into the various forms listed.

My problem is that I can't believe that the first equality as listed above is correct.

Logarithms are very TRICKY and easy things with which to have a problem (can't end

a sentence with a peposition!).

By the way, how do you get that math to display so pretty? Using LATEX?

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Hi noelevans

Yes, that is LaTeX. You can use the

`[math][/math]`

to display it.

*Last edited by anonimnystefy (2012-07-21 10:17:08)*

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

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

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

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.

Allowing 0 to have a multiplicative inverse would cause zero to be equal to one and would crater the system out since all numbers would then be equal to zero and hence equal to each other.

Complex analysis, calculus, etc. are wonderful and have their place and usefulness, but I can't see the relevance to problems involving only the field axioms and order axioms.

I did Google "removable singularity" and look at Wikipedia's site. The area I like to deal in most is foundations of arithmetic and algebra. I took complex analysis, but that was many years ago.

I complement you on your great knowledge of complex analysis and wish you well in your continued studies in mathematics. Be blessed!

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Don Blazys wrote:

In complex analysis, if we can't

asdefineat ,beingthen neither can we

asdefinebeing1at .Let's all Google the phrase "removable singularity" and find out!

Don

No one defined sin(x)/x to be 1. The limit of that expression is 1 when x approaches 0. Same for x/x. It is indeterminate and undefined at 0 but its limit as x approaches 0 is 1.

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

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

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.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Yes, we could define another function partially, so that it has no singularities, but that wouldn't be the same function we started with.

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.

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

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

stefy,

It's true that x/y <> (x-1)/(y-1) in general but in this particular instance it works out OK.

(3lna)/(lnb) - 1 (3lna-lnb)/(lnb) by writing 1 as (lnb)/(lnb) and combining terms.

----------------- = ------------------

(lna)/(lnb) - 1 (lna-lnb)/(lnb) by writing 1 as (lnb)/(lnb) and combining terms.

In the complex fraction the denominators lnb of numerator and denominator cancel leaving

(3lna-lnb)/(lna-lnb) which equals (ln(a^3/b))/ln(a/b).

I gotta learn LATEX! Actually I can do a nice job of writing this using Word Perfect, but I doubt

that I can apply it to this site.

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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

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

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Hi

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.

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

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.

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