Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,860

I am not able to at the moment. I will be able in a few hours, though.

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

Offline

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

Don,

I've viewed some of the axioms as "shoddy" myself over the years. For example the axioms of commutativity state that for each x and y in field F, x+y=y+x and xy=yx. To me this is not as much a property of addition and multiplication as it is a property of the way we write these expressions.

We write in a linear left to right manner. So we have to choose one of the variables to write first and the other is written second. The axiom states that it doesn't matter which way we write it. The result is supposed to be the same. But suppose we superimpose 1 and 3 instead of writing 1+3 or 3+1. If someone doesn't see us write it, they wouldn't know which we wrote first and which second or whether we wrote part of each and then the rest of each, etc. Hence it is more of a concession to our method of writing sums than it is to the sum itself.

We could perhaps view the symmetry axiom similarly. If we wish to indicate the equality of a and b then we can write a=b or b=a. But our linear language makes us do one or the other. So the axiom says that it doesn't matter.

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.

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,860

noelevans wrote:

Don,

I've viewed some of the axioms as "shoddy" myself over the years. For example the axioms of commutativity state that for each x and y in field F, x+y=y+x and xy=yx. To me this is not as much a property of addition and multiplication as it is a property of the way we write these expressions.

We write in a linear left to right manner. So we have to choose one of the variables to write first and the other is written second. The axiom states that it doesn't matter which way we write it. The result is supposed to be the same. But suppose we superimpose 1 and 3 instead of writing 1+3 or 3+1. If someone doesn't see us write it, they wouldn't know which we wrote first and which second or whether we wrote part of each and then the rest of each, etc. Hence it is more of a concession to our method of writing sums than it is to the sum itself.We could perhaps view the symmetry axiom similarly. If we wish to indicate the equality of a and b then we can write a=b or b=a. But our linear language makes us do one or the other. So the axiom says that it doesn't matter.

Commutativity is not just the property that we n write it any way we want. It is needed to be stated because all operations are defined on ordered tuples of numbers.

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

Offline

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

Don Blazys wrote:

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

Whoops, my bad haha

Don Blazys wrote:

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 theremovable singularitywhich is so utterly trivial that we mathematicians refer to it as "cosmetic".

In this particular case, since the expression at.by definition

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 anon-removable singularitywhich demonstrates that some axioms are not always true!

The third step (and beyond) is only valid as long as a = b. But you are applying it to cases where a =/= b!

Offline

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

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.

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,860

Hi Don Blazys

That is not how it works. You cannot change the value of 0/0 whenever it doesn't fit.

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

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,724

I agree, and first stated that with a little false proof that should be well known in post #6 +.

**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.**

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,860

I saw it. But apparently Don Blazys didn't.

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

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,724

I saw it. But apparently Don Blazys didn't.

That is harsh and argumentative.

The reason is because a bobbym post is a strange sort of entity. It is defined like this:

**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.**

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,860

So it isn't a number, obviously.

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

Offline

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

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

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,860

We can substitute a/a with b/b ! You have illegal steps in your work!

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

Offline

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

anonimnystefy wrote:

We can substitute a/a with b/b ! You have illegal steps in your work!

I'm not sure how he made any changes to the function in his work. You have been stating that he has made an error by subtracting 1 from the numerator and the denominator when clearly he followed the properties of logs.

Here are all of the steps (flawless):

Of course, those steps can only be true if he would have specified that a and b are non zero in addition to the

part.where ,AND a and b are non zeroand the properties of logarithms allow

where ,AND a and b are non zero

His overall argument (if he would have put in those extra restrictions which avoids 0/0) reminds me of set theory.

This makes sense in a way why Don Blazys' argument doesn't work for a = 0 and thus b = 0, because the empty set does not have the possibility of being a proper subset of itself:

.But then again, what about the next natural question that follows:

Is

(because he claims that ).Certainly this cannot be true, but what happens if we exclude the empty set from set theory? The definition of a set would change...

So maybe that's where this argument might be headed...

Very interesting argument for non-zero a (and thus non-zero b), which is a big chuck of the reals if you ask me!

I know very little about set theory, so maybe someone could add on to this/point out some things I said which might be incorrect about set theory.

*Last edited by cmowla (2012-07-25 18:19:00)*

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,860

Hi cmowla

is a set relation, not a number relation.Hi Don Blazys

This step is flawed:

Don Blazys wrote:

You cannot have a 0 in the denominator of a fraction. ln(a^3/b) and ln(a/b) are both 0,so that is the flawed step. You turned a 3 into 0/0.

*Last edited by anonimnystefy (2012-07-25 20:56:20)*

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

Offline

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

Hi anonimnystefy,

anonimnystefy wrote:

Hi cmowla

is a set relation, not a number relation.

I know. I said "It reminds me of" set theory, not that it "is". I was giving a comparison and trying to brainstorm a way to try to interpret how this result can be explained. Set theory is not completely solid (paradox), so I thought I could make a comparison (not that they are equal).

anonimnystefy wrote:

Hi Don Blazys

This step is flawed:

Don Blazys wrote:You cannot have a 0 in the denominator of a fraction. ln(a^3/b) and ln(a/b) are both 0,so that is the flawed step. You turned a 3 into 0/0.

How is either zero?

unless unless (which he clearly gave as a restriction).If

then that just makes the entire exponent zero. That is,because

.And since he wrote

, then a and b cannot be zero...but that's the only restriction.Also, look at this:

So between this and what I said earlier in this post, I cannot see where the errors are.

Please don't say (0/0) again unless you can show how that happens, GIVEN THE RESTRICTIONS I gave in this post.

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,860

Hi cmowla

I am not looking at your restrictions. Blazys started with the assumption that a=b and ended up with the conclusion that a=/=b. But I think I have finallfound the flawed step! It is the step I quoted in my last bost, but for different reasons. The step showed there can be brought out only if a=/=b, which contradicts the starting assumption.

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

Offline

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

Hi, anonimnystefy,

anonimnystefy wrote:

Hi cmowla

I am not looking at your restrictions. Blazys started with the assumption that a=b and ended up with the conclusion that a=/=b. But I think I have finallfound the flawed step! It is the step I quoted in my last bost, but for different reasons.

I ask kindly that you show your reasons (in fact, I asked in my last post for you to explain yourself).

anonimnystefy wrote:

The step showed there can be brought out only if a=/=b, which contradicts the starting assumption.

The starting assumption was:

, where a = b.This equality is true whether a = b or a does not equal b. As long as a and b are non-zero, it's a true statement.

And yes, there is a contradiction when he wrote

(and, as I have said several times already, for non-zero a and b)....because he cleverly made one which works for non-zero a and b. That contradiction is the core of his argument.

So, unless I have not fully identified what you found as incorrect yet, the only portion of his argument that was incorrect was that he didn't mention that a and b must be non-zero (which was pretty obvious to me, even though he didn't state it).

*Last edited by cmowla (2012-07-26 05:28:40)*

Offline

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

Also, I forgot to comment on what you were actually implying when you wrote:

anonimnystefy wrote:

This step is flawed:

Don Blazys wrote:You cannot have a 0 in the denominator of a fraction. ln(a^3/b) and ln(a/b) are both 0,so that is the flawed step. You turned a 3 into 0/0.

A 3 into (0/0)? The exponent should be a 3? Nope.

It's a/b raised to the exponent. Not

.It's not obvious why he chose to start with b(a/b)^x = a^3, where we can algebraically solve for x (as I did in my last post), but it works out nicely.

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,860

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.

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

Offline

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

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.

Can you explain why it's against the rules? If it means anything, (a/b)^(Log[a^3/b]/Log[a/b]) = a^3/b with Mathematica.

The 3D Graphs of the two are a little different though. Is that what you are talking about?

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,860

Don't you see the difference in the graphs? There is a weird (x=y) line on the second graph.

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

Offline

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

anonimnystefy wrote:

Don't you see the difference in the graphs? There is a weird (x=y) line on the second graph.

Obviously you don't read everything in my posts...

cmowla wrote:

The 3D Graphs of the two are a little different though. Is that what you are talking about?

You didn't comment at all about the result Mathematica gave?

. b can equal a (so the step he did wasn't invalid like you say it is), just not as a 3D graph.And because x cannot equal y in the 3D graph, this just explains his restriction that a cannot equal b. But the work from Mathematica proves that he did not make an error when expressing a^3/b with logs.

Offline

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

I think nobody could shake the fundamental of mathematics. The reason why all seem working is because of the division by zero is ignored. The concept of proving 2=1 also ignores the division by zero. This is why when you have statement like a=b and the derivation of the equation consist of expression that leads to the division by zero then the statement a=b must be not true in the first step. Consider this, a=b, multiplying both sides by a we get a^2=ab and minus both sides by b^2 yields, a^2-b^2=ab-b^2=> (a+b)(a-b)=b(a-b)=>a+b=b and a=b, thus 2b=b=>2=1. Anything leads to the division by zero through a-b or ln(a/b) must not be true in the beginning when it is stated a=b.

Offline

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

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.

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,860

Don Blazys wrote:

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.

have seen first hand the damage that has been done. Just look at all the hassle thatYouhad to go throughyou

in order to get this argument evenunderstood....... if that!partiallyWhy 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 tothink 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.

The equations are not both true!!! Have you read my post above. I have clearly shown why it is not true and you still go on and on about how mathematics is flawed. You have just done something that is bot allowed and that resulted in something that is frobidden in mathematics. This also shows clearly that you have no intention of discusing this matter properly as you didnt even think about reading and replying to my last post explaining why your argument is flawed and not math.

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

Offline