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

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

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:

To: anonimnystefy,

Hi,

Quoting anonimnystefy:

This step is wrong:

All the steps are correct, including that one.

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 indeterminate form

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.

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?

In complex analysis, if we can't define

Don Blazys wrote:

In complex analysis, if we can't define

as being

at

,

then neither can we define

as being 1 at

.

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.

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.

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 applying the properties of logarithms.

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

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.