John Gabriel's Thread on Mathematics.

[John Gabriel]

I do not welcome trolls on this thread. If you have something to say about the topic at hand, do so. Otherwise, please move on! I will not engage in any lengthy discussion, as I do not have the time. Many of the links contain old Knol articles I wrote which were highly rated during its curriculum vitae. Knol featured articles of mine that received thousands of hits every week. I won the top awards many times. You may not like my personality. That's fine because chances are I wouldn't like yours either! I ask that you refrain from exposing your stupidity on this thread as I cannot moderate comments.

1. Ask *one* question at a time or leave a *short* comment (I'd prefer not to read your comments, but please, if you must, a short comment only).
2. Do NOT write essays as I am not interested in your opinions.
3. Use proper English grammar and spelling.
4. Do NOT attack the character of a contributor.
5. Do NOT use profanity.
6. Do NOT spam the thread just because you can. It makes it hard for others who are serious and want to learn.
7. This is MY thread. If you don't like it, then please avoid reading it. More importantly, avoid defecating here before you move on to other threads.
8. If you must be an arsehole, please be kind enough to create your own thread.

Thanks in advance for your cooperation.

This thread is a labour of love for those who are earnest about mathematics.

Real mathematics educators, researchers and students will enjoy reading this thread.

Occasionally I will post new links to topics and comments in this thread.

The primary goal is to expose academic stupidity, and to stamp out ignorance. I also hope to spark enthusiasm for mathematics and encourage independent thought. My motto in life is summarised by the following lines:

All knowledge is questionable.
No knowledge is ever past further investigation.
All knowledge is subject to revision and correction.

My New Calculus website: http://thenewcalculus.weebly.com

I'll provide both old and new links here. This is the thread I will respond to mainly in sci.math.

Let's get started!

The New Calculus Auxiliary Equation:

http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-559.html#post27286

Differentials in the New Calculus:

http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-560.html#post27297

A comparison of Cauchy's Kludge and the New Calculus Derivative:

http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-560.html#post27304

The meaning of differentials in calculus:

http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-561.html#post27313

About integration:

http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-561.html#post27323

The New Calculus Integral:

http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-562.html#post27331

Partial derivative in the New Calculus:

http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-562.html#post27332

About Leibniz's contributions:

http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-562.html#post27335

Understanding Root Approximation Method & dispelling crap you were taught at school:

http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-563.html#post27336

[Spac...@hotmail.com]

one of my most recent interests, and an old one (although
it has nothing to do wight he regular tetragonal ("skware,
in particular [that is important to my method, as they say

> Understanding Root Approximation Method & dispelling crap you were taught at school:
>
>
>
> http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-563.html#post27336

I'll see if I can read it, later

[Spac...@hotmail.com]

ah, root of an equation, not just radicals. well,
how is a third power (qyubic equation, discontinuous at the origin?

(theirdpowering is not an especial property of the regular hexahedron,
but you don't have to bother with that

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in
news:b1b7cab9-af7d-4e00...@googlegroups.com:

> 6. Do NOT spam the thread just because you can. It makes it hard for
> others who are serious and want to learn. 7. This is MY thread. If you
> don't like it, then please avoid reading it. More importantly, avoid
> defecating here before you move on to other threads. 8. If you must be
> an arsehole, please be kind enough to create your own thread.

You don't OWN a thread. You may start it, but you don't own it.

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in
news:b1b7cab9-af7d-4e00...@googlegroups.com:

> http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-560.ht
> ml#post27297

Let f(x) = x^3

Using Cauchy's definition:

f'(x) = 3x^2 + 3xh + h^2

We can write that as

f'(x) = 3x^2 + Q(x,h) where Q(x,h) = 3xh + h^2

Set h = 0 and we get

f'(x) = 3x^2

Using New Calculus:

f'(x) = 3x^2 + 3x(n-m) + m^2 -mn + n^2

We can write that as

f'(x) = 3x^2 + Q(x,m,n) where Q(x,m,n) = 3x(n-m) + m^2 -mn + n^2

Set m = 0 and we get

f'(x) = 3x^2 + Q(x,0,n) where Q(x,0,n) = 3xn + n^2

Which is the same as Cauchy (just the change in pronumeral name from h to
n, which is insignificant)

Set n = 0 and we get

f'(x) = 3x^2

There is nothing new about this so-called 'New Calculus'. It adds
complexity, is defined in fewer placed and gives us nothing new or
interesting.

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:1d4b101f-734a-45a1...@googlegroups.com:

> On Sunday, 23 February 2014 08:59:57 UTC+2, Wizard-Of-Oz wrote:
>> John Gabriel <thenewc...@gmail.com> wrote in
>> news:b1b7cab9-af7d-4e00...@googlegroups.com:
>
>> http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-
560.h

>> tml#post27297 Let f(x) = x^3

>> Using Cauchy's definition:
>> f'(x) = 3x^2 + 3xh + h^2
>> We can write that as
>> f'(x) = 3x^2 + Q(x,h) where Q(x,h) = 3xh + h^2
>> Set h = 0 and we get
>> f'(x) = 3x^2
>> Using New Calculus:
>> f'(x) = 3x^2 + 3x(n-m) + m^2 -mn + n^2
>> We can write that as
>> f'(x) = 3x^2 + Q(x,m,n) where Q(x,m,n) = 3x(n-m) + m^2 -mn + n^2
>> Set m = 0 and we get
>> f'(x) = 3x^2 + Q(x,0,n) where Q(x,0,n) = 3xn + n^2
>
>> Which is the same as Cauchy
>

> That is not correct.

Yes it is. I note that you did not point out any errors in the above

> 1. The New Calculus does not use infinity, limits, infinitesimals or
> any other ill-formed concept.

Neither did I above. Yet you claim it is not correct.

> 2a) m, n and x have a special relationship in the new calculus and can
> be used to find the gradient of the tangent line.

As can old calculus. Nothing new there.

> 2b) h and x have no relationship and no bearing to the tangent line.

As with the new calculus, set what you call the Q function to zero and
you get the tangent. There is no difference there between old and new,
except you add an extra variable that gets set to zero.

> 3) The New Calculus does not contain any kludges such as division by 0
> in order to find the general derivative. Cauchy's definition is a
> kludge.

There is no difference in the 'kludges'. In both cases we rewrite the
function and then substitute h=0 or m=n=0 accordingly when there is no
division by h or m, n

>>(just the change in pronumeral name from h to n, which is
>>insignificant)
>

> Nonsense, nothing to do with pronumeral name/s.

That's what I said. So you claim what I said is nonsense and then agree
with it. Typical of a troll.

> h is in no way related
> to n.

It is identical when m = 0

> In fact Cauchy's definition cannot be translated to the New
> Calculus.

As I already said, the new calculus with m = 0 is just the old calculus.
And as you set m = 0 to get yout answer, there is effectively no
difference between them

> If you set h=m+n,

Noone said that you do. That's your own stupid idea.

> you get f'(x) = lim (m+n->0) [ f(x+m+n) - f(x)] /
> (m+n) which is hogwash and completely unrelated to the New Calculus.

I already showed setting m = 0 in yours give the identical results to
'old' calculus. There is nothing new of note in the new calculus

> One can however arrive at the New Calculus definition by using the
> Mean Value Theorem:
>
> If f'(x) =[ f(b)-f(a) ] / (b - a) then let b-a=m+n and b=x+n:
>
> So, f'(x)= [ f(x+n)-f(x-m) ] / (m+n)

So you just plagiarised the mean value theorum and claimed it as your
own. Nice of you to admit it.

>> Set n = 0 and we get
>> f'(x) = 3x^2
>> There is nothing new about this so-called 'New Calculus'. It adds
>> complexity, is defined in fewer placed and gives us nothing new or
>> interesting.
>

> That's just your ignorant opinion.

No, it is fact.

You do not allow a differential at a point of inflection, so there are
fewer places your method can supposedly work. And having both m and n
is more complex than just having h. You get the same answers that the
'old' calculus gives, so there is nothing new.

All facts.

> From your past comments, it is easy
> to see that for the most part, you keep getting things wrong.

I haven't so far.

[John Gabriel]

On Sunday, 23 February 2014 10:05:33 UTC+2, Moron-Of-Oz wrote:
> John Gabriel <thenewc...@gmail.com> wrote in
> news:1d4b101f-734a-45a1...@googlegroups.com:
> > On Sunday, 23 February 2014 08:59:57 UTC+2, Wizard-Of-Oz wrote:
> >> John Gabriel <thenewc...@gmail.com> wrote in
> >> news:b1b7cab9-af7d-4e00...@googlegroups.com:
> >> http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-560.html#post27297

> Yes it is. I note that you did not point out any errors in the above

I pointed out the errors, but you didn't understand them. I can't help you there.

> Neither did I above. Yet you claim it is not correct.

Of course you did. You divided by 0 when you are not allowed to divide by 0 in Cauchy's Kludge. One of the errors I pointed out to you. See error no. 3.

> As can old calculus. Nothing new there.

I showed it's not possible and you continue to be so obtuse.

> There is no difference in the 'kludges'. In both cases we rewrite the
> function and then substitute h=0 or m=n=0 accordingly when there is no
> division by h or m, n

You should pay more attention to detail.

> That's what I said. So you claim what I said is nonsense and then agree
> with it. Typical of a troll.

Are you sure? Go back and read what I said.

> It is identical when m = 0

m is never equal to 0 in the New Calculus difference quotient unless it has been simplified.

> In fact Cauchy's definition cannot be translated to the New
> Calculus.

> Noone said that you do. That's your own stupid idea.

You implied it. Troll!

> So you just plagiarised the mean value theorum and claimed it as your
> own. Nice of you to admit it.

Actually no. The mean value theorem is not the same as the secant theorem, but nice try!!

> You do not allow a differential at a point of inflection, so there are
> fewer places your method can supposedly work.

Of course not. The New Calculus is sound. There is no derivative (not differential idiot!) at an inflection point because no tangent line can be constructed there.

> And having both m and n is more complex than just having h. You get the same
> answers that the 'old' calculus gives, so there is nothing new.

For a simpleton like you, it makes no difference.

If you read my opening comment, you would have realised it stated that I do not welcome trolls.

For other simpletons, here is a detailed comparison of the transformations involved:

http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-563.html#post27341

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:47d13a88-203f-4fd8...@googlegroups.com:

> On Sunday, 23 February 2014 10:05:33 UTC+2, Moron-Of-Oz wrote:
>> John Gabriel <thenewc...@gmail.com> wrote in
>> news:1d4b101f-734a-45a1...@googlegroups.com:
>> > On Sunday, 23 February 2014 08:59:57 UTC+2, Wizard-Of-Oz wrote:
>> >> John Gabriel <thenewc...@gmail.com> wrote in
>> >> news:b1b7cab9-af7d-4e00...@googlegroups.com:
>> >> http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-
56
>> >> 0.html#post27297
>
>> Yes it is. I note that you did not point out any errors in the above
>
> I pointed out the errors, but you didn't understand them. I can't help
> you there.

I made no errors, you pointed none out

>> Neither did I above. Yet you claim it is not correct.
>
> Of course you did.

Nope

> You divided by 0 when you are not allowed to divide
> by 0 in Cauchy's Kludge. One of the errors I pointed out to you. See
> error no. 3.

I did not do that at all. Point out the line where there is a division
by zero. Funny, you've snipped all the lines where I show you wrong.

>> As can old calculus. Nothing new there.
>
> I showed it's not possible and you continue to be so obtuse.

No .. you're just fooling yourself, but not anyone else

>> There is no difference in the 'kludges'. In both cases we rewrite
>> the function and then substitute h=0 or m=n=0 accordingly when there
>> is no division by h or m, n
>
> You should pay more attention to detail.

I did

>> That's what I said. So you claim what I said is nonsense and then
>> agree with it. Typical of a troll.
>
> Are you sure? Go back and read what I said.

Yes. I said the only difference was the change in pronumeral names
which is insignificant. You said that was nonsense and the the
pronumeral names had nothing to do with it. So you argreed with what
you claimed was nonsense.

>> It is identical when m = 0
>
> m is never equal to 0 in the New Calculus difference quotient unless
> it has been simplified.

And then it is identical. Just like in the old calculus. Nothing new

>> In fact Cauchy's definition cannot be translated to the New
>> Calculus.
>
>> Noone said that you do. That's your own stupid idea.

You are dishonestly quoting me out of context. Typical troll behaviour

> You implied it.

No.

I did not at all imply that one should set h = (m+n), that was your
idea, not mine.

>Troll!

Indeed you are

>> So you just plagiarised the mean value theorum and claimed it as your
>> own. Nice of you to admit it.
>
> Actually no. The mean value theorem is not the same as the secant
> theorem, but nice try!!

You plagiarised both then. Regardless, you offer nothing new.

>> You do not allow a differential at a point of inflection, so there
>> are fewer places your method can supposedly work.
>
> Of course not. The New Calculus is sound.

It is not new

> There is no derivative (not
> differential idiot!) at an inflection point

There is a derivative. And your formula gives one, but you don't allow
it.

> because no tangent line
> can be constructed there.

One doe not need a tangent line for a derivative

>> And having both m and n is more complex than just having h. You get
>> the same answers that the 'old' calculus gives, so there is nothing
>> new.
>
> For a simpleton like you, it makes no difference.

There is nothing new

> If you read my opening comment, you would have realised it stated that
> I do not welcome trolls.

YOU are the troll, not me. As is evidenced be your dishonest snipping
and quoting out of context

> For other simpletons, here is a detailed comparison of the
> transformations involved:

> http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-
563.html#post27341

So you show that your new calculus is nothing more or less than the mean
value theorum. Nothing new at all. You've just stolen an old idea and
changed a few pronumerals (which, as we agree, makes no difference)

And I showed that setting m = 0 in your new calculus gives the old
calculus, and as the new calculus requires you setting m=n=0 effectively
there is no difference between them.

\

[John Gabriel]

You saying it ain't so, does not make it true. Therefore, unless you can prove me wrong, I suggest you move on. Again, this thread is not for trolls.

Your assertions are not facts. Furthermore, if you want a response in future, limit your questions to one. I have no interest in reading your rot. Others can see that I am correct and you are obtuse. That's all I care about. Discussion over. Please DO NOT comment here again, unless you obey the rules. Create your own thread if you like! Call it "John Gabriel's not new calculus" or whatever shit you please. But once again, I am asking you nicely: Please, DO NOT comment here again. I have reported every one of your comments as SPAM. I would have not even bothered responding to you if I could delete your comments. You are in one word, a dunce.

Thank you for your cooperation in advance!

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:c126aa16-1cac-44bf...@googlegroups.com:

The facts do

> Therefore, unless you
> can prove me wrong, I suggest you move on. Again, this thread is not
> for trolls.

Then stop posting in it

> Your assertions are not facts.

They are. Dmonstrable facts

> Furthermore, if you want a response in
> future, limit your questions to one. I have no interest in reading
> your rot.

You're the one with rot.

> Others can see that I am correct and you are obtuse.

No

> That's
> all I care about. Discussion over. Please DO NOT comment here again,

Not your call

> unless you obey the rules.

You don't get to make the rules

> Create your own thread if you like! Call it
> "John Gabriel's not new calculus" or whatever shit you please. But
> once again, I am asking you nicely: Please, DO NOT comment here again.

I ignore your request

> I have reported every one of your comments as SPAM.

Typical dishonesty on your part. Noone cares about what you report.

> I would have not
> even bothered responding to you if I could delete your comments. You
> are in one word, a dunce.

You are a fucking moron. Two words

> Thank you for your cooperation in advance!

Fuck you, in advance.

[Dirk Van de moortel]

John Gabriel <thenewc...@gmail.com> wrote:
> I do not welcome trolls on this thread.

Then you shouldn't have opened it.
This is sci.math.trolls

Dirk Vdm

[John Gabriel]

Now you have removed all doubt that you are a troll.

[John Gabriel]

Flawed and New Calculus:

http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-563.html#post27344


f(x)=x^2

Flawed:

f'(x)= 2x+h

At this stage we have the gradient of a non-parallel secant line. If we set h=0, then we don't have anything: neither a secant line nor a tangent line.

f(x)-f(x) / 0 is meaningless nonsense.

We only have a general derivative obtained through a KLUDGE.

New Calculus:

f'(x) = 2x + n - m

At this stage we have the gradient of a parallel secant line and also the gradient of the tangent line. If we set m=n=0, then we *still* have the gradient of the tangent line. Morover, any valid (m,n) pair will still work in the New Calculus gradient, that is, f'(x) = 2x + n - m, which gives the general derivative if m=n=0, and also the numeric derivative when x=c.

The New Calculus - the first and only rigorous formulation of calculus in human history.

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:ebb04bed-7ce5-42f1...@googlegroups.com:

> Flawed and New Calculus:
>
> http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-
563.ht
> ml#post27344
>
>
> f(x)=x^2

OK

> Flawed:
>
> f'(x)= 2x+h

No flaw there

> At this stage we have the gradient of a non-parallel secant line. If
> we set h=0, then we don't have anything: neither a secant line nor a
> tangent line.

Wrong .. we have the slope of the tangent line

> f(x)-f(x) / 0 is meaningless nonsense.

Just as it is in your new calculus when you set m = n = 0

> We only have a general derivative obtained through a KLUDGE.

Just as in your new calculus. Except that the old calculus says to take
a limit as h approaches zero. It is clear that as h is made smaller and
smaller, the formula 2x+h gets closer to 2x. 2x is the limit, and the
slope of the curve at x.

> New Calculus:
>
> f'(x) = 2x + n - m
>
> At this stage we have the gradient of a parallel secant line and also
> the gradient of the tangent line.

We also have gradients of all different values, for secants both
parallel and not parallel, between arbitrary points on the curve. Even
if you limit it to m >= 0 and n >= 0 you still have inifinitely many
secants, only some of which will be parallel to the tangent line.

That means f'(x) is indeterminate until you provide particular m and n
values or constraints upon them.

> If we set m=n=0, then we *still*
> have the gradient of the tangent line.

And the same division by zero that you complained about above. Using
limits avoids the actual division by zero, but you don't use limits, so
it is a real problem for you.

> Morover, any valid (m,n) pair

How do you determine a valid pair?

> will still work in the New Calculus gradient, that is, f'(x) = 2x + n
> - m, which gives the general derivative if m=n=0, and also the numeric
> derivative when x=c.

What c? There was no mention of a c there.

> The New Calculus - the first and only rigorous formulation of calculus
> in human history.

Not terribly rigorous when its main formula

f'(x) = (f(x-m)+f(x+n)) / (m+n)

gives different values for the supposed gradient or derivative or slope
at x depending on what m and n values you choose. You cannot justify
the equal sign there as a general formula, unless you add in the
condition that it is only true for certain values of m and n which need
to them be determined. You could add in a limit as m->0 and n->0, but
you don't have that.

In fact, for other than linear functions f, and for arbitrary m and n,
your formula does NOT give any unique value, so you can't say it is
equal to the value of f'(x).

Your kludge is to choose values for m and n that give the same answer as
when m = n = 0. But, of course, your formula has a division be zero in
that case, just as you were complaining about in the old calculus. So
you need the additional 'kludge' of simplifying the expression to remove
the division before one substitutes.

You've not removed any flaws at all, and you have a less rigorous and
more complicated approach that offers nothing new that wasn't already
covered by the 'old' calculus.

[Art Hopkins]

>"John Gabriel" wrote in message
>news:b1b7cab9-af7d-4e00...@googlegroups.com...

>I do not welcome trolls on this thread.

Shut up, idiot.

[John Gabriel]

Get off my thread troll!

[John Gabriel]

To those who are not trolls and happen to be reading my thread:

Be warned that a troll called Wizard-Of-Oz and his sidekicks will do all they can to sabotage this thread. As you can see from their comments, there is no mathematics, no refutation. All they can do is insult me, write long irrelevant rants and ignore the facts.

I encourage you to read my work and study it for yourselves. These trolls will try to influence you not to read my work. Ignore them and decide for yourselves.

If my work was so bad as they claim, then why do they go to such trouble to convince you not to read it? Only one reason: they want to control your minds.

I am here to free your minds from these fools and to guide you to the light. But I tell you NOT to believe anything I say. Prove it all for yourselves.

Go go the links provided in the comments and study the same. I will be ignoring these trolls because one cannot reason with an irrational mind. I tried but as you can see from their comments, there is no substance.

You can tell a troll when he is asked to leave a thread and continues to spew out his vile rot. I asked Moron-Of-Oz to leave this thread but he declined in the true nature of the troll that he is.

If my work is so bad, then ask yourself why it is the academic bourgeoisie are trying to shut me down... could it be that they KNOW by accepting my work, theu effectively sign a confirmation that they have been incompetent, ignorant and stupid the last few hundred years.

So, decide for yourselves. If you have any questions, post these here or at my New Calculus page and I will respond to you.

[John Gabriel]

I am a real mathematician, not just some guy with a PhD in mathematics.

Most of the trolls on this forum have never accomplished anything in their lives. As for me, I am the discoverer of the first and only rigorous formulation of calculus in human history.

http://thenewcalculus.weebly.com

The academic bourgeoisie has already shut down 3 of my previous sites. Ask yourself why they did this. Why is it so? If my ideas and work are genuinely crap, then people will eventually stop reading them.

But the opposite is happening. More and more people are studying my New Calculus and the number of visitors to my sites increase by the day.

Don't let others think for you. It's the greatest insult to anyone with dignity.

[Dirk Van de moortel]

John Gabriel <thenewc...@gmail.com> wrote:
> To those who are not trolls

If by now you haven't realised that we are all de-facto trolls,
then there is something severely wrong with your ability to
judge other people's mindset. There is a name for that
condition.

Dirk Vdm

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:5552742f-acc8-480e...@googlegroups.com:

> As I said, your assertions and inability to understand are not my
> problem. Fuck off from here!!! You are an idiot and no one cares what
> you have to say.

Translation: Wizard has proved Gabriel is wrong, and Gabriel is trying to
discredit Wizard because Gabriel cannot prove Wizard is wrong; Gabriel
wishes Wizard would stop proving him wrong and he hopes that noone else
notices.

[John Gabriel]

We've realised from the start who are the trolls. Thank you. Now would you please refrain from defecating any further on this thread. I would have banned you if I could, but unfortunately I cannot moderate on this forum.

[John Gabriel]

He thinks that because I am not playing his game, you will think that perhaps his non-arguments are valid. But I'll let the mathematics speak for me, not long useless rants. Again, this troll is unwelcome on this thread, but do you think that will stop him?

[John Gabriel]

I want to bring to your attention the following link:

http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-116.html

That was the first comment of my 0.999... debate on STATU. Thousands of comments later I have exposed many false notions about mathematics. I also uncover the ignorance and stupidity of mathematicians the last few hundred years.

If you have the patience to read all my comments, you will no doubt learn more than you have ever learned about mathematics on your life's journey.

As a bonus, you will learn the New Calculus, the greatest accomplishment of any human mind. :-)

Here are some of my favourite quotes:

Whatever I imagine is real, because whatever I imagine is well defined.

A good mathematician is like a fine artist: the objects arising from concepts in the mind of a mathematician are only as appealing as they are well defined.

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:5552742f-acc8-480e...@googlegroups.com:

> As I said, your assertions and inability to understand are not my
> problem. Fuck off from here!!! You are an idiot and no one cares what
> you have to say.

Funny, that's what everyone thinks of you.

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:75ccc1d8-a0e1-4898...@googlegroups.com:

> On Sunday, 23 February 2014 23:51:32 UTC+2, Wizard-Of-Oz wrote:
>> John Gabriel <thenewc...@gmail.com> wrote in
>>
>> news:5552742f-acc8-480e...@googlegroups.com:
>>
>>
>>
>> > As I said, your assertions and inability to understand are not my
>>
>> > problem. Fuck off from here!!! You are an idiot and no one cares
>> > what
>>
>> > you have to say.
>>
>>
>>
>> Translation: Wizard has proved Gabriel is wrong, and Gabriel is
>> trying to
>
>>
>> discredit Wizard because Gabriel cannot prove Wizard is wrong;
>> Gabriel
>>
>> wishes Wizard would stop proving him wrong and he hopes that noone
>> else
>
>>
>> notices.
>
> He thinks that because I am not playing his game,

You think it is a game? That explains a lot.

> you will think that
> perhaps his non-arguments are valid. But I'll let the mathematics
> speak for me, not long useless rants. Again, this troll is unwelcome
> on this thread, but do you think that will stop him?

The mathematics has spoken .. it shows that you're a moron.

[John Gabriel]

On Monday, 24 February 2014 07:59:48 UTC+2, Wizard-Of-Oz wrote:

Troll: The mathematics has spoken .. it shows that you're a moron.

JG: I have never noticed any mathematics or semblance of a sound argument in your comments. Can you tell me where to find these? :-)

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:12834dec-9b6f-4f7e...@googlegroups.com:

They are the bits you snip.

Your claims are refuted.

Your mathematical skills are around the level of a primary schoool student
.. still playing with blocks.

You just don't have the mental capacity for abstract thinking and
mathematics. But your ego is so over-inflated that your can't see it.

Its truly pathetic

[John Gabriel]

On Monday, 24 February 2014 13:25:19 UTC+2, Wizard-Of-Oz wrote:

> > JG: I have never noticed any mathematics or semblance of a sound argument in your comments. Can you tell me where to find these? :-)

Troll: Your mathematical skills are around the level of a primary schoool student ... still playing with blocks.

JG: Um, no Mr. Troll. That would not be an answer. :-) Again, I ask you:
Can you tell me where to find these? I do not see any mathematics in any of your comments, only troll rants.

Troll: You just don't have the mental capacity for abstract thinking and mathematics. But your ego is so over-inflated that your can't see it.

JG: His teachers and professors have told him this so many times, he now simply repeats it. Ha, ha!

Troll: Its truly pathetic

JG: It truly is pathetic. But I don't think he gets it, as pathetic as it is!

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:bbe8ee98-8170-462d...@googlegroups.com:

> On Monday, 24 February 2014 13:25:19 UTC+2, Wizard-Of-Oz wrote:
>
>> > JG: I have never noticed any mathematics or semblance of a sound
>> > argument in your comments. Can you tell me where to find these?
>> > :-)
>
>
> Troll: Your mathematical skills are around the level of a primary
> schoool student ... still playing with blocks.
>
> JG: Um, no Mr. Troll. That would not be an answer. :-) Again, I ask
> you: Can you tell me where to find these? I do not see any
> mathematics in any of your comments, only troll rants.

Again, you are too stupid to recognise them when you see them

> Troll: You just don't have the mental capacity for abstract thinking
> and mathematics. But your ego is so over-inflated that your can't
> see it.
>
> JG: His teachers and professors have told him this so many times, he
> now simply repeats it. Ha, ha!

Not at all

> Troll: Its truly pathetic
>
> JG: It truly is pathetic. But I don't think he gets it, as pathetic as
> it is!

I'm glad you realise you are pathetic.

[John Gabriel]

Another troll comment!

[John Gabriel]

These links are re-posted here as evidence that I have refuted everyone of the non-arguments by Moron-Of-Oz. I shall keep re-posting it every time he comments here.

Yes, my New Calculus is really NEW:

http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-563.html#post27341

Old and New - from definition:

http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-563.html#post27344

My New Calculus: http://thenewcalculus.weebly.com

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:be88e735-2b11-468b...@googlegroups.com:

No matter how many times you post it, it is still utter garbage. Post it
all you like. You'll just have more people seeing what a deluded moron you
are.

[Giddy Armstrong]

Silence your repetitious blitherings, jerk.

[John Gabriel]

[John Gabriel]

I will not allow trolls to bury the real content in this thread. Take your shit and post it elsewhere.

[Spac...@hotmail.com]

this excerpt shows that you Rock cosmoS, and John is his own worst trolL

[John Gabriel]

Let the endpoints of a parallel secant line on a function f be [c-m, f(c-m)] and [c+n, f(c+n)].

The gradient of any such parallel secant line is given by:

f'(c)= (f(c+n)-f(c-m))/(m+n) [A]

Now, m+n divides f(c+n)-f(c-m) exactly, that is,

f'(c) = gradient(c) + Q(c,m,n) [B]

where gradient(c) is the expression of the gradient in terms of c only.
Q(c,m,n)=0 in [B]. However, any (m,n) pair of any parallel secant line will satisfy Q(c,m,n)=0. Moreover, the same (m,n) will produce the gradient regardless of whether [A] or [B] is used.

Let's do an example:

f(x)=x^3.

Suppose we want to find the derivative at x=3.

f'(x) = { 3x^2(m+n) + 3x(n-m)(m+n)+(m^2 - mn + n^2)(m+n) } / (m+n) [A]

f'(x) = 3x^2 + 3x(n-m)+m^2 - mn + n^2 [B]

Or

f'(x) = gradient(x) + Q(x,m,n) [C]

where gradient(x)=3x^2 and Q(x,m,n)= 3x(n-m)+m^2 - mn + n^2

To find any valid pair, we use x=3 with either of m or n. So let's choose m=1.

Then,

Q(3,1,n)=3(3)(n-1)+(1)^2 - (1)n + n^2 = 9n-9+1-n+n^2=n^2+8n-8=0

So, n=-4+2sqrt(6) or n=-4-2sqrt(6)

Since -4+2sqrt(6) lies to the right of x=3, the (m,n) pair we require is (1,-4+2sqrt(6)).

Substituting this into [A], we have f'(3)=27 as expected.

We could also have used [B] with the pair (1,-4+2sqrt(6)):

f'(3)=3(3)^2+3(3)(-4+2sqrt(6)-1)+(1)^2-(1)(-4+2sqrt(6))+(-4+2sqrt(6))^2 [B]

f'(3)=27+9(-5+2sqrt(6))+1-(-4+2sqrt(6))+16-16sqrt(6)+24

f'(3)=27-45+18sqrt(6)+5-2sqrt(6)+16-16sqrt(6)+24

f'(3)=27-45+18sqrt(6)+5-2sqrt(6)+16-16sqrt(6)+24

f'(3)=27 as expected.

Of course you do not need to go through all this because

f'(x) = gradient(x) + Q(x,m,n) = gradient(x)

So, that's how you can find any valid (m,n) pair you like.

[John Gabriel]

On Monday, 24 February 2014 20:41:47 UTC+2, Spac...@hotmail.com wrote:

You can't do this with Newton's flawed calculus. There is no auxiliary equation because Cauchy's definition is a kludge.

The New Calculus is the first and only rigorous formulation of calculus in human history.

You should not capitalise the last letter of certain words - it makes your comment hard to read. But then being a troll that you are, I suppose it should not be too surprising. Why do you do it? I have always wondered what makes trolls do the things they do. Please do not comment here again.

Go and post your crap elsewhere. Thanks in advance.

[Spac...@hotmail.com]

why do you have no interest in either book seven or
fourteen of _The elementS?... so, Whatm if
the fourteenth book is "deuterocanonicos?

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:bca901e4-e31e-46c1...@googlegroups.com:

> On Monday, 24 February 2014 20:41:47 UTC+2, Spac...@hotmail.com
> wrote:
>> this excerpt shows that you Rock cosmoS, and John is his own worst
>> trolL
>>
>>
>>
>> > And the same division by zero that you complained about above.
>> > Using
>>
>> >
>>
>> > limits avoids the actual division by zero, but you don't use
>> > limits, so
>>
>> >
>>
>> > it is a real problem for you.
>>
>> >
>>
>> >
>>
>> >
>>
>> > > Morover, any valid (m,n) pair
>>
>> >
>>
>> >
>>
>> >
>>
>> > How do you determine a valid pair?
>>
>> >
>>
>> >
>>
>> >
>>
>> > > will still work in the New Calculus gradient, that is, f'(x) = 2x
>> > > + n
>>
>> >
>>
>> > > - m, which gives the general derivative if m=n=0, and also the
>> > > numeric
>>
>> >
>>
>> > > derivative when x=c.
>>
>> >
>>
>> >
>>
>> >
>>
>> > What c? There was no mention of a c there.
>
> Let the endpoints of a parallel secant line on a function f be [c-m,
> f(c-m)] and [c+n, f(c+n)].
>
> The gradient of any such parallel secant line is given by:
>
> f'(c)= (f(c+n)-f(c-m))/(m+n) [A]

That is also the forlmula for a non-parallel secant line

There is nothing in that formula that forces it to be parallel.

f'(c) is then not the derivative of f at c, it is the slow of some
arbitrary secant at c.

Your labelling is dishonest and deliberately misleading

> Now, m+n divides f(c+n)-f(c-m) exactly, that is,

Define "exactly". Prove your assertion

> f'(c) = gradient(c) + Q(c,m,n) [B]

So it is the gradient plus some other number. So you are saying that
the derivative at c is not the same as the gradient at c.

> where gradient(c) is the expression of the gradient in terms of c
> only. Q(c,m,n)=0 in [B].

How do you know? It will depend on the value of m and n

So your method then is to try to divide and simplify (f(c+n)-f(c-m))/
(m+n) and then hope that all the terms in m and n can be grouped
together. And that those terms in m and n will end up as zero for some
m and n. And that what is left will be the gradient.

Where is your rigorous justification for such a simplification to always
be possible?

Where is your rigorous justification for the Q function to have 0 as a
possible value.

Where is your rigorous justification for the terms other than the Q
function to be the gradient at c.

Where is your rigorous justification for there to be no divisions by
zero when you subsitute values for c, m and n into Q.

> However, any (m,n) pair of any parallel
> secant line will satisfy Q(c,m,n)=0.

Where is your rigorous justification for Q being zero for secants
parallel to the gradient (slope or derivative) of f at c?.

> Moreover, the same (m,n) will
> produce the gradient regardless of whether [A] or [B] is used.

Where is your rigious justification for [A] to be able to be expressed
in form [B] for every function f?

> Let's do an example:
>
> f(x)=x^3.
>
> Suppose we want to find the derivative at x=3.
>
> f'(x) = { 3x^2(m+n) + 3x(n-m)(m+n)+(m^2 - mn + n^2)(m+n) } / (m+n)
> [A]

Only for SOME values of m and n

> f'(x) = 3x^2 + 3x(n-m)+m^2 - mn + n^2 [B]

Only for SOME values of m and n

> Or
>
> f'(x) = gradient(x) + Q(x,m,n) [C]

Only for SOME values of m and n

Where is your rigious justifications that the terms you get after you
groups terms in m and n as Q(x,m,n) will be the gradient?

> where gradient(x)=3x^2 and Q(x,m,n)= 3x(n-m)+m^2 - mn + n^2
>
> To find any valid pair, we use x=3 with either of m or n. So let's
> choose m=1.
>
> Then,
>
> Q(3,1,n)=3(3)(n-1)+(1)^2 - (1)n + n^2 = 9n-9+1-n+n^2=n^2+8n-8=0
>
> So, n=-4+2sqrt(6) or n=-4-2sqrt(6)

So you find m and n pairs by assuming the Q(x,m,n) will be zer, and then
assuming that the resultant secant will have the same slope as the f(x)
curve at x

> Since -4+2sqrt(6) lies to the right of x=3, the (m,n) pair we require
> is (1,-4+2sqrt(6)).
>
> Substituting this into [A], we have f'(3)=27 as expected.
>
> We could also have used [B] with the pair (1,-4+2sqrt(6)):
>
> f'(3)=3(3)^2+3(3)(-4+2sqrt(6)-1)+(1)^2-(1)(-4+2sqrt(6))+(-4+2sqrt(6))^
2
> [B]
>
> f'(3)=27+9(-5+2sqrt(6))+1-(-4+2sqrt(6))+16-16sqrt(6)+24
>
> f'(3)=27-45+18sqrt(6)+5-2sqrt(6)+16-16sqrt(6)+24
>
> f'(3)=27-45+18sqrt(6)+5-2sqrt(6)+16-16sqrt(6)+24
>
> f'(3)=27 as expected.
>
> Of course you do not need to go through all this because
>
> f'(x) = gradient(x) + Q(x,m,n) = gradient(x)
>
> So, that's how you can find any valid (m,n) pair you like.

There is no point in finding m and n, if you are simply deliberately
chosing m and n that make Q(x,m,n) = 0 and then substituting those m and
n into Q(x,m,n) to get 0.

So you method is, get a formula for the slope of any secant, and express
its endpoints as offsets from c (m and n). You simplify the expression
and hope that you end up with a term not involving the offset and
another term that do involved the offsets. Then you hope that the first
term is actually the gradient and that the second term can give a value
of zero for some values or m and n. And you hope that there is no
division be zero in either term.

And all that is supposed to be rigorous? Where is your rigorous proofs
that each of the assumption or hopes in the above is valid?

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:e86b387c-af56-4063...@googlegroups.com:

> JG: f'(c)= (f(c+n)-f(c-m))/(m+n) [A]
>
> Idiot: That is also the formula for a non-parallel secant line.

> There is nothing in that formula that forces it to be parallel.
>

> JG: It's not the formula for a non-parallel secant line, because I
> stipulate that it's the slope of a parallel secant line. It would be
> meaningless nonsense otherwise.
>
> Idiot: f'(c) is then not the derivative of f at c, it is the slope of

> some arbitrary secant at c.
>

> JG: It is *exactly* the derivative of f at c. If not, then the mean
> value theorem is false and everyone knows that can't be so.
>
> JG: Now, m+n divides f(c+n)-f(c-m) exactly, that is,
>
> Idiot: Define "exactly". Prove your assertion
>
> JG: It was proven in the New Calculus course lessons which had you
> bothered to read, you would have known.
>
> JG: f'(c) = gradient(c) + Q(c,m,n) [B]
>
> Idiot: So it is the gradient plus some other number.
>
> JG: It is the gradient plus ZERO. Q(c,m,n) is always ZERO.
>
> Idiot: So you are saying that the derivative at c is not the same as
> the gradient at c.
>
> JG: It's *exactly* the same.

>
> where gradient(c) is the expression of the gradient in terms of c
> only. Q(c,m,n)=0 in [B].
>

> Idiot: How do you know? It will depend on the value of m and n
>
> JG: No. It never depends on the value of m or n. These do not affect
> the slope of any straight line. If you studied the New Calculus course
> lessons, you would know all these facts.
>
> Idiot: So your method then is to try to divide and simplify

> (f(c+n)-f(c-m))/ (m+n) and then hope that all the terms in m and n can
> be grouped together.
>

> JG: Grrr! There is no chance in the New Calculus. (f(c+n)-f(c-m))/
> (m+n) will produce the derivative exactly. No hoping or wishing.
>
> Idiot: And that those terms in m and n will end up as zero for some m
> and n.
>
> JG: Q(x,m,n) is exactly ZERO. No "ending up" or any other rot.
>
> Idiot: And that what is left will be the gradient.

>
> Where is your rigorous justification for such a simplification to
> always be possible?
>
> Where is your rigorous justification for the Q function to have 0 as a
> possible value.
>
> Where is your rigorous justification for the terms other than the Q
> function to be the gradient at c.
>
> Where is your rigorous justification for there to be no divisions by
> zero when you subsitute values for c, m and n into Q.
>

> JG: Study the New Calculus course lessons!!!!
>
> JG: However, any (m,n) pair of any parallel secant line will satisfy
> Q(c,m,n)=0.
>
> Idiot: Where is your rigorous justification for Q being zero for

> secants parallel to the gradient (slope or derivative) of f at c?.
>

> JG: Study the New Calculus course lessons.
>
> JG: Moreover, the same (m,n) will produce the gradient regardless of

> whether [A] or [B] is used.
>

> Idiot: Where is your rigious justification for [A] to be able to be

> expressed in form [B] for every function f?
>

> JG: Study the New Calculus course lessons.
>
> JG: Let's do an example:

>
> f(x)=x^3.
>
> Suppose we want to find the derivative at x=3.
>
> f'(x) = { 3x^2(m+n) + 3x(n-m)(m+n)+(m^2 - mn + n^2)(m+n) } / (m+n)
> [A]
>

> Idiot: Only for SOME values of m and n
>
> JG: No. Once we have any (m,n) pair, then we know that any other pair
> (p,q) such that c-m<c-p and c+p<c+n will work.

>
> f'(x) = 3x^2 + 3x(n-m)+m^2 - mn + n^2 [B]
>

> Idiot: Only for SOME values of m and n
>
> JG: Answered.
>
> JG: Or f'(x) = gradient(x) + Q(x,m,n) [C]
>
> Idiot: Only for SOME values of m and n
>
> JG: Answered.
>
> Idiot: Where is your rigorous justifications that the terms you get

> after you groups terms in m and n as Q(x,m,n) will be the gradient?
>

> JG: Study the New Calculus course lessons.
>
> JG: where gradient(x)=3x^2 and Q(x,m,n)= 3x(n-m)+m^2 - mn + n^2

>
> To find any valid pair, we use x=3 with either of m or n. So let's
> choose m=1.
>
> Then,
>
> Q(3,1,n)=3(3)(n-1)+(1)^2 - (1)n + n^2 = 9n-9+1-n+n^2=n^2+8n-8=0
>
> So, n=-4+2sqrt(6) or n=-4-2sqrt(6)
>

> Idiot: So you find m and n pairs by assuming the Q(x,m,n) will be
> zero, and then assuming that the resultant secant will have the same

> slope as the f(x) curve at x
>

> JG: No, I don't *assume* anything. I know Q(x,m,n) will be zero
> because that's how I designed it. Nothing is left to chance in the New
> Calculus. There are no kludges as in Cauchy's definition.
>
> JG: Since -4+2sqrt(6) lies to the right of x=3, the (m,n) pair we

> require is (1,-4+2sqrt(6)).
>
> Substituting this into [A], we have f'(3)=27 as expected.
>
> We could also have used [B] with the pair (1,-4+2sqrt(6)):
>

> So, that's how you can find any valid (m,n) pair you like.
>

> Idiot: There is no point in finding m and n, if you are simply
> deliberately choosing m and n that make Q(x,m,n) = 0 and then

> substituting those m and n into Q(x,m,n) to get 0.
>

> JG: You don't have to find m and n at all. But if you had bothered
> studying the New Calculus lessons, you would have known this.
>
> Idiot: So you method is, get a formula for the slope of any secant,

> and express its endpoints as offsets from c (m and n).
>

> JG: Not any secant line. Only a parallel secant line.
>
> Idiot: You simplify the expression and hope that you end up with a

> term not involving the offset and another term that do involved the
> offsets.
>

> JG: Moron.
>
> Idiot: Then you hope that the first term is actually the gradient and

> that the second term can give a value of zero for some values or m and
> n. And you hope that there is no division be zero in either term.
>

> JG: There is no need for hope or wishing or anything of the sort.
> There is no chance in the New Calculus. It is the first and only

> rigorous formulation of calculus in human history.
>

> Idiot: And all that is supposed to be rigorous? Where is your

> rigorous proofs that each of the assumption or hopes in the above is
> valid?
>

> JG: It is all 100% rigorous. Your inability to grasp these things is
> the problem.
>
> This is my final response to you. If you have any more questions, go
> and ask them at my FaceBook page or at Space Time and the Universe on
> the 0.999... thread.
>

Looks like you lied about not posting here again.

And you also haven't addressed any of my questions other than flat out
lying and claiming its in your course (which it isn't).

Troll like you can't help but lie, though.

[John Gabriel]

You are one dishonest, vile and despicable excuse for a human being. Don't you realise that anyone who reads my response to you will know immediately that you are the liar and the troll? Did it even occur to you? You can't fool people.

I am back here to stay. Get used to it!

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:6e79b08c-c84d-4394...@googlegroups.com:

> You are one dishonest, vile and despicable excuse for a human being.

Not at all. You have no evidence to support your faulty allegations.

I have plenty of evidence that they apply to you, of course. Almost every
post you make adds to the evidence. Not that you matter .. you're an
insignificant nobody .. just an annoying little mosquito that the rest of
us swat.

> Don't you realise that anyone who reads my response to you will know
> immediately that you are the liar and the troll?

No .. they will see that you are

> Did it even occur to
> you? You can't fool people.

I'm not .. and neither are you. We all know you're a moron

> I am back here to stay. Get used to it!

I couldn't give a fuck. You're a nobody.

And I've shown your 'new calculus' is nonsense .. it is not at ALL
rigorous, and it is nothing more than MVT and secant rule, applies in fewer
places than old calculaus and adds unnecessary complexity. Its a total
waste, just like you.

[John Gabriel]

So one of the local morons (Wizard of Oz) claimed that in the New Calculus difference quotient, m+n is not guaranteed to divide f(x+n)-f(x-m).

I want to expose this fucking idiot's stupidity in this comment.

If I had given him the link https://www.filesanywhere.com/fs/v.aspx?v=8b6c6889606075ada9 , it would have taken better brains than his a few hours to understand.

So, instead I gave the monkey a simple proof, which is published at my New Calculus course website (Lesson 2 at: http://johngabrie1.wix.com/newcalculus#!new-calculus-course/cahx).

He didn't get it then and he never will.

The sorry bastard immediately accused me of being a liar and dishonest. See for yourself who is being dishonest!

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:f4851205-f015-4709...@googlegroups.com:

> So one of the local morons (Wizard of Oz) claimed that in the New
> Calculus difference quotient, m+n is not guaranteed to divide
> f(x+n)-f(x-m).

You didn't define what you mean by 'divides exactly'.

Usually it means that both numbers are integers and the result is an
integer.

If you expect it to mean something else, then you need to define it

> I want to expose this fucking idiot's stupidity in this comment.

Really .. more likely just be further evidence of your own

> If I had given him the link
> https://www.filesanywhere.com/fs/v.aspx?v=8b6c6889606075ada9

It doesn't address your claims

> , it
> would have taken better brains than his a few hours to understand.

Only a couple of minutes

> So, instead I gave the monkey a simple proof, which is published at my
> New Calculus course website (Lesson 2 at:
> http://johngabrie1.wix.com/newcalculus#!new-calculus-course/cahx).

You gave me no proof. You just asserted

When you're working with reals (or rationals) every number divides every
number. Saying something exactly divides means the result has to be an
integer eg 3.5 exactly divides 7 (because it gives 2)

> He didn't get it then and he never will.

Because you're just bleating nonsense

> The sorry bastard immediately accused me of being a liar and
> dishonest. See for yourself who is being dishonest!

You are. And you have just been again.

[John Gabriel]

On Wednesday, 26 February 2014 11:34:26 UTC+2, Wizard-Of-Oz wrote:

> You didn't define what you mean by 'divides exactly'.

And you claim to be a "mathematician" ? Haaaaar, haaar, harr!!! :-)

Fucking moron! Any real mathematician will know what divides exactly means.

Fuckwad!!!!

> Usually it means that both numbers are integers and the result is an
> integer.

No, you moron. It means there is NO remainder. Stupid fuck. It applies to those objects you and your fucking moron colleagues think are *real numbers*.

Grrrr, you fucking dimwit. I am going to shit and piss all over the lot of you. Anyone who knows you personally, will have no doubt you are retards by the time I am done with you!!! :-))) Ha, ha!

> If you expect it to mean something else, then you need to define it

I defined it very well. That you don't have what it takes to understand is not surprising. Most Americans are retards. You are one of them.

> > https://www.filesanywhere.com/fs/v.aspx?v=8b6c6889606075ada9
> It doesn't address your claims

Hee, hee. Middle finger to you pedophile!!! Anyone who reads it will know you are a blustering fool.

> Only a couple of minutes

It only takes 3 seconds to realise you are stupid.

The fact that all you fuckwads continue to respond to my comments proves I have something. Keep going!!!! :-)

> > So, instead I gave the monkey a simple proof, which is published at my
> > New Calculus course website (Lesson 2 at:
> > http://johngabrie1.wix.com/newcalculus#!new-calculus-course/cahx).

> You gave me no proof. You just asserted

That doesn't fly. Sorry, just trying to appear erudite and claiming assertions won't help your cause. Muah!!!!

> When you're working with reals (or rationals) every number divides every
> number. Saying something exactly divides means the result has to be an
> integer eg 3.5 exactly divides 7 (because it gives 2)

Yes fuckwad. Now, when you calculate a gradient, what do you get? That's right!!!!! You get what you think is a *real number*, don't you? Sure!!! Are there any remainders there O queerie? :-)

You need a few days to understand even the simplest concepts.

> Because you're just bleating nonsense

Meeeh. Stupid. People will read it and know you are a liar. Don't you see that?
They will begin to ignore you because you are DELUDED.

> You are. And you have just been again.

Now that's an assertion. And I shit and piss on you and your fellow bumchums - Robin, Dick van der Moron and all you spineless scum.

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:1119bce1-e065-443b...@googlegroups.com:

> On Wednesday, 26 February 2014 11:34:26 UTC+2, Wizard-Of-Oz wrote:
>
>> You didn't define what you mean by 'divides exactly'.
>
> And you claim to be a "mathematician" ?

I am

> Haaaaar, haaar, harr!!! :-)

Laughing like a loon I see

> Fucking moron! Any real mathematician will know what divides exactly
> means.

But I want to know what YOU mean by it.

> Fuckwad!!!!

No math from you

>> Usually it means that both numbers are integers and the result is an
>> integer.
>
> No, you moron.

Yes, you moron. That's what it usually means.

> It means there is NO remainder.

Which is only applicable when we express the result as an integer
quotient and some (possibly zero) remainder

I gave an example .. 7 divided by 3.5 gives 2 with no remainder, so 3.5
exactly divides 7. Conversely, 7 divided 4.5 gives 1 with a remainder of
2.5, so 4.5 does not exactly divide 7.

It is meaningless to talk about rational or real numbers exactly
dividing each other otherwise.

> Stupid fuck. It applies
> to those objects you and your fucking moron colleagues think are *real
> numbers*.
>
> Grrrr, you fucking dimwit. I am going to shit and piss all over the
> lot of you. Anyone who knows you personally, will have no doubt you
> are retards by the time I am done with you!!! :-))) Ha, ha!

Noone will care what a moron like you says. Shit and piss all you like.

>> If you expect it to mean something else, then you need to define it
>
> I defined it very well.

You didn't at all.

It seems now you are saying that you take it to mean the the numerator
divided by the denomintaor gives a whole number with no remainder.

> That you don't have what it takes to
> understand is not surprising. Most Americans are retards. You are one
> of them.

Really?

>> > https://www.filesanywhere.com/fs/v.aspx?v=8b6c6889606075ada9
>> It doesn't address your claims
>
> Hee, hee. Middle finger to you pedophile!!! Anyone who reads it will
> know you are a blustering fool.

No, they won't. But they will see that it does NOT prove what you
claim: that in your formula for the secant gradient the demoniator (m+n)
exactly divides the numerator.

>> Only a couple of minutes
>
> It only takes 3 seconds to realise you are stupid.

Nonsense .. it takes a couple of minutes to read, but is easily
understood along the way

> The fact that all you fuckwads continue to respond to my comments
> proves I have something. Keep going!!!! :-)

Not when we continually prove that you don't

>> > So, instead I gave the monkey a simple proof, which is published at
>> > my New Calculus course website (Lesson 2 at:
>> > http://johngabrie1.wix.com/newcalculus#!new-calculus-course/cahx).
>
>> You gave me no proof. You just asserted
>
> That doesn't fly. Sorry, just trying to appear erudite and claiming
> assertions won't help your cause. Muah!!!!

You gave no proof. You just asserted.

>> When you're working with reals (or rationals) every number divides
>> every number. Saying something exactly divides means the result has
>> to be an integer eg 3.5 exactly divides 7 (because it gives 2)
>
> Yes fuckwad.

So again you say that it means there is an integer quotient and no
remainder

> Now, when you calculate a gradient, what do you get?
> That's right!!!!! You get what you think is a *real number*, don't
> you? Sure!!! Are there any remainders there O queerie? :-)

There may well be when you express it as an integer quotient and
(possibly zero) remainder

> You need a few days to understand even the simplest concepts.

Nope

>> Because you're just bleating nonsense
>
> Meeeh. Stupid.

Yes .. you stupid

> People will read it and know you are a liar.

They will see that I post the truth

> Don't you
> see that?

Just more of your lies

> They will begin to ignore you because you are DELUDED.

Projecting your own faults onto me doesn't make them go away.

>> You are. And you have just been again.
>
> Now that's an assertion.

Nope

> And I shit and piss on you and your fellow
> bumchums - Robin, Dick van der Moron and all you spineless scum.

I'm sure you do .. its what is typical of vile disgusting creatures like
you. Now .. how about you just go away like you promised before .. h,
that's right .. you can't be trusted because you are a proven liar.

[John Gabriel]

The line p is at an angle of theta to the horizontal.

In dimwit American parlance, the rise of theta is sqrt(3) and the run is 2.

The gradient of the line is therefore according to Newtoni, gradient=sqrt(3)/2, which is a *real* number. :-)

Fuckwad Moron-of-Oz claims that gradient has some sort of remainder which invalidates it. :-) lol.

[John Gabriel]

On Wednesday, 26 February 2014 12:59:33 UTC+2, Wizard-Of-Oz wrote:
> I am

BULLSHIT. What is your name? No anonymous fuckwad is ever taken seriously. In your case, I don't think it will help you much providing your name because you make errors every time you think!

> Laughing like a loon I see

You bet! You are laughable! If you really are a mathematician the likes of fuckwad Robin Chapman and others who frequent these forums, why do you continue to respond to me?!!! Do you like being proved ignorant? Do you like the verbal abuse? I shall continue to expose your ignorance and also abuse you because you can't communicate in a civil manner. You are a disgusting, vile and sick human being who is not worthy of the air it breathes.

> But I want to know what YOU mean by it.

I mean exactly what every other mathematician means by it. NO REMAINDER when dividing by (m+n)!!!!

FUCKING MORON.

> No math from you

Hee, hee.

> > It means there is NO remainder.

That's right. There is no remainder when dividing the new calculus difference quotient by (m+n). You get the gradient.

For the fucking 15th time:

f(x+n)-f(x-m) / (m+n) = k

k is the gradient.

Now, k(m+n) = f(x+n)-f(x-m)

Do you see (m+n) on the RHS? f(x+n)-f(x-m) is EXACT. There is no remainder. Stupid!!!!

> Noone will care what a moron like you says. Shit and piss all you like.

And is that why you CARE so much that you continue to respond? Weasel!!! You care a lot!!!!

> It seems now you are saying that you take it to mean the the numerator
> divided by the denomintaor gives a whole number with no remainder.

BULLSHIT. It means that you get the gradient exactly.

> No, they won't. But they will see that it does NOT prove what you
> claim: that in your formula for the secant gradient the demoniator (m+n)
> exactly divides the numerator.

That's EXACTLY what they will see if they study it. That the denominator (m+n) EXACTLY divides the numerator. Do you see any part of the gradient that has (m+n) as a denominator anywhere? MORONNNNN!!!!!!

> Nonsense .. it takes a couple of minutes to read, but is easily
> understood along the way

I know that's not true in your case because of what you continue to write.
In fact, that document is not at all necessary in understanding the facts. Lesson 2 of my calculus course proves this simply and elegantly with only high school algebra.

STUPID!!!!

> Not when we continually prove that you don't

Ha, ha! I am the one proving you are wrong!


That is a very sound proof.

> I'm sure you do .. its what is typical of vile disgusting creatures like
> you. Now .. how about you just go away like you promised before .. h,
> that's right .. you can't be trusted because you are a proven liar.

Hee, hee. YOU WILL GO AWAY. I shall remain and publish the truth. I shall also expose the vile rot that you and others are spewing out on this and other forums. GET USED TO ME!

:-) Haaaar, haaaar, haaar.

[YBM]

Le 26/02/2014 12:23, John Gabriel a écrit :
> [nonsensea and rant] :-) Haaaar, haaaar, haaar.

Complaints-To: groups...@google.com

Lot of people got their google account suspended for far less than
that.

[John Gabriel]

What you don't know is that I have been reporting everyone of your comments as SPAM. :-)

[John Gabriel]

On Wednesday, 26 February 2014 12:59:33 UTC+2, Wizard-Of-Oz wrote:

I am still waiting for you to give me even just ONE counterexample which proves that (m+n) does not divide f(x+n)-f(x-m).

:-) Haaar, haaar.

You can't because there isn't any.

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:41413c0f-b13a-465d...@googlegroups.com:

You are the one who talks about exact divisors (as you are) which means you
need to talk about an integer quotient and a (possibly zero) remainder. If
the remainder is zero, it is an exact division. If your answer for the
division is sqrt(3)/2, then that is NOT an integer quotient with a zero
remainder at all. So there is no exact division going on as you claim.

I said nothing about a remainder invalidating anything, other than it
invalidating your claim of exact division.

But you do also claim that f'(x) = gradient(x) + Q(x,m,n) .. so that means
the derivative is the gradient plus some value that depends on your choice
of m and n. You claim it always gives zero, but that is only the case for
SOME value of m and n. Your formula for f'(x) can give different values
depending on the m and n you chose, unless you stipulate some condition on
them. All you do is the equivalent of "pick the values for m and n that
give the answer you want, hoping that there will be an m and n that give
that answer, and then when you use them you get the answer you want".
Rather circular logic there. Your justifications for there being m and n
values are the MVT and secant theorum from the old calculus that you say is
flawed. If the old calculus is flawed and not rigorous, and you use it to
justify your new calculus, then your new calculus is not rigorous either.

[John Gabriel]

I am not playing your moronic game. What you wrote is shit and YOU know it.

Here are your options:

1. Give me a counter example that shows m+n does not divide f(x+n)-f(x-m)

OR

2. Admit you are wrong.

No more nonsense.

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:74a24d01-1b2c-48e8...@googlegroups.com:

> On Wednesday, 26 February 2014 12:59:33 UTC+2, Wizard-Of-Oz wrote:
>
>
> I am still waiting for you to give me even just ONE counterexample
> which proves that (m+n) does not divide f(x+n)-f(x-m).

I never said you cannot divide f(x+n)-f(x-m) by (m+n). Unless m+n is zero,
of course.

Every rational number can be divided by every non-zero rational number

Every real number can be divided by every non-zero real number

But YOU claimed it always divided *exactly*

To say a divides b exactly means the result of dividing b by a is an
integer quotient (with a zero remainder). This is written a | b, and
usually read as just "a divides b" (it is implicit that the result of the
division is an integer) .. one doesn't even really need the addition of
'exactly'. a || b is sometimes used for "a exactly divides b" and is a
more strict relationship. We don't need to go into that here.

>:-) Haaar, haaar.
>
> You can't because there isn't any.

Every answer for the division that is not an integer value is an counter
example of your claim for exact division.

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:ae2bc465-8f11-48ce...@googlegroups.com:

> I am not playing your moronic game. What you wrote is shit and YOU
> know it.

Your admission of defeat is duly noted

[John Gabriel]

On Wednesday, 26 February 2014 15:12:03 UTC+2, Wizard-Of-Oz wrote:

> But YOU claimed it always divided *exactly*

I stated very clearly that m+n always divides f(x+n)-f(x-m) EXACTLY.

Still waiting for you to show me a counterexample where this is not true.

In order for you to do this, you must show me an example where you simplify the quotient and you have a term with m+n in the denominator. You can't. Moron!

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:f4b56b21-8e1a-4fd6...@googlegroups.com:

> On Wednesday, 26 February 2014 15:12:03 UTC+2, Wizard-Of-Oz wrote:
>
>> But YOU claimed it always divided *exactly*
>
> I stated very clearly that m+n always divides f(x+n)-f(x-m) EXACTLY.

That's what I said you claimed.

That means the result is an integer quotient with a zero remainder

You agreed that that is what divides exactly means

> Still waiting for you to show me a counterexample where this is not
> true.

I have

> In order for you to do this, you must show me an example where you
> simplify the quotient and you have a term with m+n in the denominator.

So now you are saying that that is what you mean by 'divides exactly'. I
did ask you what you meant, and you did't say.

> You can't. Moron!

YOU need to proof that noone can.

You claim what you say is rigorous, so you need to PROVE that for any
function f(x), you can simplify (f(x+n)-f(x-m))/(m+n) and that
simplification will not have any term involving (m+n) in the denominator

Try f(x) = sin(x)

[Roland Franzius]

Am 26.02.2014 15:20, schrieb Wizard-Of-Oz:
> John Gabriel <thenewc...@gmail.com> wrote in
> news:f4b56b21-8e1a-4fd6...@googlegroups.com:
>
>> On Wednesday, 26 February 2014 15:12:03 UTC+2, Wizard-Of-Oz wrote:
>>
>>> But YOU claimed it always divided *exactly*
>>
>> I stated very clearly that m+n always divides f(x+n)-f(x-m) EXACTLY.
>
> That's what I said you claimed.
>
> That means the result is an integer quotient with a zero remainder

No, its just highschool calculus where the little child begins to
calculte slopes following Newton

(x^n-y^n)/(x-y) -> sum_(k=0=^n-1 x(n-1)y^k -> n x^(n-1) if y=x

Nobody needs to go back to integer division to define calculus limits on
real and complex polynomials.

--

Roland Franzius

[Wizard-Of-Oz]

Roland Franzius <roland....@uos.de> wrote in news:lel04u$6as$1
@newsserver.rrzn.uni-hannover.de:

He didn't say anything about polynomial division. Saying one value
exactly divides another can have different meanings depending on the
context.

That's why I asked for clarification of what he meant. And for a proof
of the exact division.

Especially as there is no specification the f(x) is a polynomial.

===
> Now, m+n divides f(c+n)-f(c-m) exactly

Define "exactly". Prove your assertion

===

Rather than defining what he mean and proving it, he launched into a
tirade of insults and big-noting himself, as appears usual for him.

So I suggested a possible meaning for it. More insults ensued.

He still hasn't provided a proof that for any arbitrary function f(x)
that m+n divides f(c+n)-f(c-m) exactly. Perhaps his method should be
constrained to functions where that is true, but then though it may be
easy to show how a particular given function works by simply do the
division and simplifyiing and examining the result, but that is not a
general proof or condition.

He may be able to show example, or even prove it for certain classes of
function. But for his method to be as rigorous as he claims, it can't
rely on assertion for important steps. And if he claims the old
calculus is flawed and not rigorous, it means he cannot use anything
that is in the old calculus. He has to prove it from scratch.

[John Gabriel]

On Wednesday, 26 February 2014 16:20:20 UTC+2, Wizard-Of-Oz wrote:

> That means the result is an integer quotient with a zero remainder

Irrelevant bullshit. I explained to you precisely what I meant, more than once. You are an argumentative arsehole.

> I have

No you have not.

> So now you are saying that that is what you mean by 'divides exactly'. I
> did ask you what you meant, and you did't say.

My grade 10 students get this and English is their second language. Aren't you ashamed of yourself? You act like a 2 year old.

> YOU need to proof that noone can.

Lesson 2 of the New Calculus. Go read it!!!!!

> You claim what you say is rigorous, so you need to PROVE that for any
> function f(x), you can simplify (f(x+n)-f(x-m))/(m+n) and that
> simplification will not have any term involving (m+n) in the denominator
> Try f(x) = sin(x)

It works for all differentiable functions, including sin(x).

YOU HAVE YET TO SHOW ME A COUNTER-EXAMPLE!

[John Gabriel]

Thank you. Finally someone with common sense. :-)

[John Gabriel]

On Wednesday, 26 February 2014 17:38:23 UTC+2, Wizard-Of-Oz wrote:

> > Now, m+n divides f(c+n)-f(c-m) exactly
> Define "exactly". Prove your assertion

I proved it a few times but you are as thick as a brick wall.

Baby proof: Take any straight line you wish. Any straight line. Say f(x)=px+q.
We can do this because the parallel secant line is always a STRAIGHT LINE.

So, f'(x) = p*(x+n)+q - p*(x-m) - q / (m+n)
f'(x) = px+pn+q - px+pm - q / (m+n) = p(m+n) / (m+n) = p.

See moron? It really does not matter what f(x) is, because we choose the endpoints of a parallel secant line which lie on f(x).

That was the baby proof. Then I gave you the hard proof which is one I did first decades ago. Here it is again: https://www.filesanywhere.com/fs/v.aspx?v=8b6c6889606075ada9

> Rather than defining what he mean and proving it, he launched into a
> tirade of insults and big-noting himself, as appears usual for him.

If you asked nicely and stopped behaving like an asshole, I would not have insulted you. You should not come on here pretending to be a mathematician. I am a real mathematician and you can't fool me. I don't care how many degrees you have, you don't get to be a mathematician without significant accomplishments and recognition. You are very immature and are constantly provoking me. You have only yourself to blame for the insults. Add to that the fact that you are incredibly obtuse. Grrrr.

> He still hasn't provided a proof that for any arbitrary function f(x)
> that m+n divides f(c+n)-f(c-m) exactly. Perhaps his method should be
> constrained to functions where that is true, but then though it may be
> easy to show how a particular given function works by simply do the
> division and simplifyiing and examining the result, but that is not a
> general proof or condition.

Ha, ha, ha! If I hadn't done this, the New Calculus would be worthless you dimwit!!!!!

> He may be able to show example, or even prove it for certain classes of
> function. But for his method to be as rigorous as he claims, it can't
> rely on assertion for important steps.

Go and read the materials. :-)

> And if he claims the old calculus is flawed and not rigorous, it means he
> cannot use anything that is in the old calculus. He has to prove it from scratch.

And I do prove everything from scratch. But you don't know because you didn't bother to read it.

Next time you ask a question, include the document name also.

[Spac...@hotmail.com]

this seems to be quite picquayune, or merely uncotrovertible:
A little known fact is that arc length (itself an average) is the core component of line integrals and every other integral.

[Soap Research]

I'll give it a try. Let f(x) = sin(x)

f(x+n)-f(x-m) = sin(x+n) - sin(x-m)
= sin(x) cos(n) + cox(x) sin(n)
- (sin(x) cos(m) - cos(x) sin(m))
= sin(x) (cos(n) - cos(m)) + cos(x) (sin(n) + sin(m))
= sin(x) * 2 cos((n+m)/2) cos((n-m)/2)
+ cos(x) * 2 sin((n+m)/2) cos((n-m)/2)

Now, Mr. Gabriel, would you please help me and indicate how I can simplify (m+n) in the previous example? How can I reach f'(x) = cos(x) + Q(x, m, n) from here? BTW, you can't use anything from "flawed old calculus", only your new calculus.

Thanks.

[John Gabriel]

It's wrong from the first line. Tsk, tsk.

Here you go:

http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-442.html#post25653

[Soap Research]

So, correct me if I'm wrong, but the derivative cannot be done if sin(x) gives non-terminating values. Saying that sin(Pi/3) = sqrt(3)/2 is meaningless because saying that sqrt(3)/2 = 0.8660254... is a non-sense in your new calculus. Right?

[konyberg]

Hi.
Is n and m both integers in your definition of divisible? So if some number T exactly divisible of (m+n) is an integer timed (m+n)? Or do you mean that the division leads to a rational number, or what?
KON

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:133bc1a0-1ca8-49a3...@googlegroups.com:

> It works for all differentiable functions, including sin(x).

Prove that it does. For all differentiable functions. Oh, and you need to
define what YOU mean by "differentiable functions" in your 'new calculus'.
Is it just a circular defintition that a differentiable function is one
where you get the exact division .. that would be cheating, of course.

> YOU HAVE YET TO SHOW ME A COUNTER-EXAMPLE!

No I don't. You have to prove that it does. Without using 'old calculus;
which you claim is flawed and not rigorous. You claim to have rigorous
proof of your method.

You are the one with a claimed new rigorous calculus .. the first ever.
I'm not going to do your work for you.

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:2fec528b-f819-405f...@googlegroups.com:

> On Wednesday, 26 February 2014 17:38:23 UTC+2, Wizard-Of-Oz wrote:
>
>> > Now, m+n divides f(c+n)-f(c-m) exactly
>> Define "exactly". Prove your assertion
>
> I proved it a few times but you are as thick as a brick wall.

Not once. Showing examples where it divides does NOT prove that it
divides for all functions.

> Baby proof: Take any straight line you wish. Any straight line. Say
> f(x)=px+q. We can do this because the parallel secant line is always a
> STRAIGHT LINE.
>
> So, f'(x) = p*(x+n)+q - p*(x-m) - q / (m+n)
> f'(x) = px+pn+q - px+pm - q / (m+n) = p(m+n) / (m+n) = p.
>
> See moron? It really does not matter what f(x) is, because we choose
> the endpoints of a parallel secant line which lie on f(x).

That's not a proof at all, just an example for a particular class of
function for which ther eis an exact divisor.

Now prove it for all function

> That was the baby proof. Then I gave you the hard proof which is one I
> did first decades ago. Here it is again:

Still not a proof, and not even relevant. You show that (m+m) |
(m^p+n^p) and (m+n)|(m^p-n^p)

Try again

>> Rather than defining what he mean and proving it, he launched into a
>> tirade of insults and big-noting himself, as appears usual for him.
>
> If you asked nicely

I did

> and stopped behaving like an asshole,

I didn't behave like you at all

> I would not
> have insulted you.

You seem to insult anyone who is smart enough to disagree with you

> You should not come on here pretending to be a
> mathematician. I am a real mathematician and you can't fool me.

You're a deluded moron so convinced of his own self worth that he can't
see his own flaws

> I
> don't care how many degrees you have, you don't get to be a
> mathematician without significant accomplishments and recognition.

So you aren't one then

> You
> are very immature and are constantly provoking me.

You're so easily provoked because of your over inflated ego and
delusions of mathematical grandeur. And your high-school level
mathemtacis

> You have only
> yourself to blame for the insults. Add to that the fact that you are
> incredibly obtuse. Grrrr.

Not at all.

>> He still hasn't provided a proof that for any arbitrary function f(x)
>> that m+n divides f(c+n)-f(c-m) exactly. Perhaps his method should be
>> constrained to functions where that is true, but then though it may
>> be easy to show how a particular given function works by simply do
>> the division and simplifyiing and examining the result, but that is
>> not a general proof or condition.
>
> Ha, ha, ha! If I hadn't done this, the New Calculus would be worthless
> you dimwit!!!!!

It *is* worthless because you haven't. I asked for you to show your
rigor (your 'lessons' and just vague whitewash and overview), and
nothing was forthcoming.

It's also worthless because it is nothing new, is less applicable, and
more complex.

>> He may be able to show example, or even prove it for certain classes
>> of
>
>> function. But for his method to be as rigorous as he claims, it
>> can't rely on assertion for important steps.
>
> Go and read the materials. :-)

I did

>> And if he claims the old calculus is flawed and not rigorous, it
>> means he
>
>> cannot use anything that is in the old calculus. He has to prove it
>> from
> scratch.
>
> And I do prove everything from scratch. But you don't know because you
> didn't bother to read it.

I read it

> Next time you ask a question, include the document name also.

Where is the proof that (m+n) divides exactly? .. which one of your
'lessons' shows that proof. It was only evident as an assertion and an
example. One needs a proof.

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:54a6c874-13f0-4999...@googlegroups.com:

> Here you go:
>
> http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-442.ht
> ml#post25653

Hilarious .. the hypocrisy just oozes from that.

You say that series like that do not terminate and do not represent
numbers. You even claim that 0.9999 = 9/10+9/100+9/1000+9/1000+... is NOT
the same as one. You reject the whole notion of non-terminating infinite
serier and sequences. And yet you use them in your example for sin(x).
You would also need to prove that there is a similar series for every
function (from scratch) AND that for every such series you will be alwaus
be able to exactly divide by (m+n)

[konyberg]

What you haven't told is this: Is 2/sqrt(2) =sqrt(2) or 1/0.5 = 2 an exact division.What do you mean? In my mathematics an division into numbers is between the natural nunbers.
KON

[John Gabriel]

We can take any straight line we wish. Say f(x)=px+q. We can do this because the parallel secant line is always a STRAIGHT LINE.

So, f'(x) = p*(x+n)+q - p*(x-m) - q / (m+n)
f'(x) = px+pn+q - px+pm - q / (m+n) = p(m+n) / (m+n) = p.

It really does not matter what f(x) is, because we choose the endpoints of a parallel secant line which lie on f(x). f(x+n)-f(x-m) / m+n is ALWAYS the gradient of a STRAIGHT LINE. By the above proof, m+n ALWAYS divides f(x+n)-f(x-m).

Q.E.D.

Unlike Taylor Series which is always an approximation, the Gabriel Polynomial is exact. (On page 5, there is the Gabriel polynomial).

https://www.filesanywhere.com/fs/v.aspx?v=8b6c688a615e75aa71a1

And of course the sine series can be derived without calculus whatsoever:

https://www.filesanywhere.com/fs/v.aspx?v=8b6c688a615e76bcaa69

[Spac...@hotmail.com]

this is just a common error, of "equating
a nonterminating rational number with an integer;
0.9999... is equal to 1.0000..., not to 1 <period;
this sole ambiguity is detailed by Stevin in 15th cce,
but his notation was really bad (at least,
in the first edition of his huge, best-selling, multi-translated
from teh Latin [Stevin is a pen name

[John Gabriel]

No. m and n are any "real" numbers. The New Calculus is extremely simple and elegant, at the same time it is far more powerful than Newton's junk. You do not have an auxiliary equation in Newton's calculus. You do not have well-defined differentials or integrals.

Why don't you go through the New Calculus lessons? There are 9 short and simple lessons at the link: http://johngabrie1.wix.com/newcalculus#!new-calculus-course/cahx

I'll be glad to answer any questions you have. I can also tell you that I have taught the New Calculus to 10th graders in Asia in 2 weeks. That's quite amazing given that English is their second language.

There are no ill-defined concepts in the New Calculus: no limits, no infinity, no infinitesimals.

It is the first and only rigorous formulation of calculus in history. You can post questions at my FaceBook page: facebook.com/thenewcalculus

[John Gabriel]

The auxiliary equation is a new and powerful feature in the New Calculus. It is not available or even possible in Newton's flawed formulation. Check out the following GeoGebra applet:

https://www.filesanywhere.com/fs/v.aspx?v=8b6c688a61626f7d6b9b

There are no viruses or malicious software at any of my sites.

If Cauchy's definition of derivative were correct, Newton's root approximation method would never have worked:

https://www.filesanywhere.com/fs/v.aspx?v=8b6c688a6162707bb399

There are many more interesting applets at this link:

https://www.filesanywhere.com/fs/v.aspx?v=8b6c688a616272a66d9d

Also check out over 65 articles at the following link:

https://www.filesanywhere.com/fs/v.aspx?v=8b6c688a61627578b196

[John Gabriel]

There is an article available at this site which discusses the New Calculus at a high school level:

http://download.math10.com/en/university-math/calculus/NewCalcForDummies.pdf

Also therein are new definitions, basic proofs of differentiation formulas and some simple examples.

[John Gabriel]

I realise many of you are extremely dull. So here is yet another way to understand that (m+n) always divides f(c+n)-f(c-m).

The equation of any secant line is given by s(x)=kx+p.

If the endpoints of the parallel secant line (to a tangent at c) are [(c-m,f(c-m)); (c+n,f(c+n))], then its gradient is:

f'(c) = { f(c+n)-f(c-m) }/(m+n)

So, s(x)=f'(c)x+p

Therefore, f'(c) = k

Proof: f'(c)= { s(c+n)-s(c-m) } / (m+n)
= { k(c+n)+p-[k(c-m)+p] } / (m+n)
= { kc + kn +p - kc + km -p } / (m+n)
= k(m+n)/(m+n) = k

[John Gabriel]

So, what's in the greatest unpublished work in mathematics?

The work consists of approximately 2000 pages.

1. A complete rewrite of Euclid's Elements. The rewrite is how Euclid should have written the Elements.
2. Selected chapters from The Works Of Archimedes. These are a massive improvement on Thomas Heith's work. There is much that Heath missed, which is explained in a simple yet surprisingly ingenious way. If you have ever looked at the Works of Archimedes, you will know that they are formidable till this day.
3. Over 800 pages are devoted to the New Calculus.
4. The last few chapters (also the most exciting) are a new kind of mathematics based entirely on tangent objects. There is also a new multi-variable calculus that is far less complex and easier to learn in these chapters.

Sorry, but I am not prepared to reveal anything more until I get an offer to publish. If I don't, you will never get to read this exciting book. In fact, it will probably take another 50-100 years for even the brightest among you to even get close to where I am now. :-)

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:b5e4e5ba-de73-4f0c...@googlegroups.com:

For dummies, and written BY an even bigger one. Intelligent and
knowledgeable people, if they even both reading it given the disgusting
nature of its author, see it for what it is and dismiss it as useless.

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:d040b6c4-975a-4a21...@googlegroups.com:

Deluded moron. Noone cares. And your reputation means you have zero
credibility, noone will belief this book exists. Just more lies and
delusion from the biggest moron of mathematics.

[John Gabriel]

Well, I'm afraid that you wouldn't even understand that one, because it's for dummies, not down syndrome learners. I don't know that I'll be writing one for down syndrome learners anytime soon. But keep following me, you might be pleasantly surprised some day. :-)

[John Gabriel]

Coming from you that is a very big compliment. I wouldn't have it any other way. :-) Now if someone smart and respectable said that of me, well maybe then would be the time for me to be concerned. But from a nobody who can't grasp even the simplest concepts and proofs, it's really nothing for me to worry about.

Has it occurred to you that when you say you don't care and you continue to comment on these threads, others are laughing at you. I'll let you try to figure this one out. Ha, ha. Tsk, tsk. Stupidicus maximus!!!

[John Gabriel]

The auxiliary equation is a new and powerful feature in the New Calculus. It is not available, or even possible in Newton's flawed formulation. Check out the following GeoGebra applet:

https://www.filesanywhere.com/fs/v.aspx?v=8b6c688a61626f7d6b9b

NOTE: There are no viruses or malicious software at any of my sites.

If Cauchy's definition of derivative were correct, Newton's root approximation method would never have worked:

https://www.filesanywhere.com/fs/v.aspx?v=8b6c688a6162707bb399

There are many more interesting applets at this link:

https://www.filesanywhere.com/fs/v.aspx?v=8b6c688a616272a66d9d

Also check out over 65 articles at the following link:

https://www.filesanywhere.com/fs/v.aspx?v=8b6c688a61627578b196

I encourage you to post questions and comments at: facebook.com/thenewcalculus

The New Calculus is the first and only rigorous formulation of calculus in human history. It can be learned in just 2 weeks by anyone with just a high school knowledge of mathematics.

It is based on well-defined concepts. There are no flaws or contradictions as one finds in existing mathematics. The derivative and integral are redefined without use of infinity, limits or infinitesimals.

The New Calculus is the way Newton and Leibniz would have liked to have discovered calculus.

Since the age of 14, when I first taught myself Newton's calculus, I have been working on arriving at this wonderful new knowledge. I am certain that you will be just as excited as I am about it, once you study the materials at my website: http://johngabrie1.wix.com/newcalculus#!new-calculus-course/cahx

Don't miss out on the greatest accomplishment of the human mind!

For centuries, mathematicians and non-mathematicians were unable to accomplish what I have done in the New Calculus.

And although I have not shared everything I know, because I some day hope to reap a reward for all my labours, I can tell you with the utmost conviction and truth, that the greatest mathematics is yet to be discovered.

It will not be based on the flawed mathematics of today. So, to get a headstart , learn the new calulus! :-)

[dull...@sprynet.com]

How do you differentiate f(x) = sin(x) ???

[John Gabriel]

Knisley (http://faculty.etsu.edu/knisleyj/calculus/Crisis.htm) is one of the academics in charge of the calculus reform project in the United States. He is not a mathematician from what I know, but he has a PhD in mathematics. This is what Knisley has to say:

"Clearly, our calculus course does not prepare scientists in other fields to recognize, understand, and utilize the calculus that many of their fields are based upon. Thus, when it comes to calculus, we don't get it the first time around, our colleagues don't get it, and our students are still not getting it. It's no wonder that one of the most common occurrences in higher education is that of a non-mathematics faculty member discovering that something they were doing is calculus. And at the very least, we feel justified in asserting that there still is a crisis in calculus instruction." - Jeff Knisley

While I don't endorse all of Knisley's views, I think his statement says a lot about the state of calculus education, but more than that, it says something most academics don't like to hear: The calculus they have been using is flawed. In fact, the foundations of mathematics as currently taught are also flawed.

Note he admits they don't get it. They never did get it. It's one thing to learn how to use theorems and results, but entirely a different matter when it comes to understanding. How can there be real understanding if the foundations are flawed? How can anyone really understand calculus when it is based on ill-formed concepts and definitions?

Well, the answer is no one can. One of the powerful features of the New Calculus, is that it is easy to understand. It does not require years of study to master. Moreover, it is both simple and elegant at the same time. There is much one can do with the New Calculus that is not possible using the existing flawed formulation. How can you know for sure? Study it! :-)

[John Gabriel]

Like this:

http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-442.html#post25653

[Ross Donglemeier]

Shut the fuck up, you reposting moron.

"John Gabriel" wrote in message
news:b5e4e5ba-de73-4f0c...@googlegroups.com...

There is an article available at this site which discusses the New Calculus

at a high school level, but no-one give a flying fuck!

[John Gabriel]

We know your IQ is 43. No need to tell us again.

[John Gabriel]

John Gabriel's construction of rational numbers from nothing in 5 easy steps:

1. A magnitude is the idea of size of extent. We can either tell that two magnitudes are equal or not. If we can tell they are not equal, then we know which is smaller or bigger, but we can't tell how much bigger or smaller. This is called qualitative measurement (without numbers).

2. We can form ratios of magnitudes. AB : CD where AB and CD are line segments. The expression AB : CD means the comparison of magnitudes AB and CD.

3. A ratio of equal magnitudes, say AB : AB or CD : CD allows us to use either as the standard of measurement, that is, the unit. The unit is a ratio of equal magnitudes.

4. The unit enables us now to compare AB and CD if both are exact multiples of the unit that measures both. We can now perform quantitative measurement, because we can tell how much greater AB is than CD or how much less AB is than CD.

5. Finally, if a magnitude is only part of a unit, then we arrive at a ratio of numbers, say AB : CD where AB and CD are multiples of the unit. AB : CD now means the comparison of numbers AB and CD. When we write AB/CD, it is called a fraction.

So, in five steps I have derived the concept of number for you. There is one thing left - what happens when you can't measure a magnitude that is not a multiple of a unit and can't be expressed exactly using any part of a unit? This is called an incommensurable magnitude and the best you can do is provide an approximation such as 3.14159... or 1.414..., etc.

Now that we have defined number, we state the axioms of arithmetic:

John Gabriel's axioms of arithmetic:

1. The difference (or subtraction) of two numbers is that number which describes how much the larger exceeds the smaller.

2. The difference of equal numbers is zero (same as no difference when compared).

3. The sum (or addition) of two numbers is that number whose difference with either of the two numbers is either of the two numbers.

4. The quotient (or division) of two numbers is that number that measures either number in terms of the other.

5. If a unit is divided by a number into parts, then each of these parts of a unit, is called the reciprocal of that number.

6. Division by zero is undefined.

7. The product (or multiplication) of two numbers is the quotient of either number with the reciprocal of the other.

8. The difference of any number and zero is the number.

Observe that all the basic arithmetic operations are defined in terms of difference, which is the most primitive operator.

[fom]

He actually provided that elsewhere. It does depend
upon an expansion. I do not know exactly how
constructive mathematics deals with such
things. But, under the premise that a repetitive
operation need not be assumed as fully expanded
if it is feasible to expand it to any given length,
he proceeds with the division. That general premise
would be abstracted from what I have read in Markov.
But, I do not know specific implementations within
calculations.

Ultimately, the problem is that the expansion
is not proven. In constructive mathematics
like Bishop and Bridges, there is a development
from a definition of number and so there are
identity criteria available for proving such
things.

Mr. Gabriel does mathematics from the standpoint
of "ideas of magnitudes". There is no "idea of
identity" for an "idea of magnitude" that I
could find.

[John Gabriel]

On Thursday, 27 February 2014 21:45:27 UTC+2, fom wrote:
> On 2/27/2014 8:55 AM, dull...@sprynet.com wrote:


> He actually provided that elsewhere. It does depend
> upon an expansion.

Of course it relies on expansion. The sine series as found by Newton did not require any calculus at all.

> Ultimately, the problem is that the expansion
> is not proven.

It has been proven - without calculus.

> Mr. Gabriel does mathematics from the standpoint
> of "ideas of magnitudes". There is no "idea of
> identity" for an "idea of magnitude" that I
> could find.

You have no idea how I do things!

[John Gabriel]

What really amuses me is how the brainless fools on this forum are distraught, because they can no longer bullshit without being exposed as the idiots they really are.

Poor little John Gabriel. He is so insignificant, yet every academic trembles at his name. They must silence and destroy him because his success means their exposure as ignorant, incompetent and stupid. To acknowledge John Gabriel is equivalent to academic suicide.

How can the world know how stupid they've been? They just can't have that. lol!

[Wizard-Of-Oz]

John Gabriel <thenewc...@gmail.com> wrote in

news:32040599-4cb2-41ed...@googlegroups.com:

Your definition of a unit is self-contradictory

And by your axioms you get 2 - 3 = 1, 2 + 3 = 6, 2 / 6 = 3, 0 - 3 = 3

You need to work on your axiom and number construction as they are
rather sloppy and ambiguous.

[Ross Donglemeier]

Shut up, moron.

[Spac...@hotmail.com]

RotF, it's just what i do

[John Gabriel]

On Friday, 28 February 2014 02:09:59 UTC+2, Wizard-Of-Oz wrote:

> Your definition of a unit is self-contradictory

You are a contradiction!

> And by your axioms you get 2 - 3 = 1, 2 + 3 = 6, 2 / 6 = 3, 0 - 3 = 3

As the axiom says: the smaller is subtracted from the larger.

You should really give up studying math. Try something that does not require much logic. Ask your teacher to help you!

[John Gabriel]

On Friday, 28 February 2014 03:26:58 UTC+2, Ross Donglemeier wrote:
> Shut up, moron.

Dingleberry, you mumbling something?

[Jay Wheeler]

shut up imbecile

[John Gabriel]

On Friday, 28 February 2014 05:59:57 UTC+2, Jay Wheeler wrote:
> shut up imbecile

Greetings Retard!!! And you hail from? :-)