On Sunday, 30 July 2023 at 19:34:28 UTC-4, markus moron klyver wrote:
> söndag 30 juli 2023 kl. 23:56:09 UTC+2 skrev Eram semper recta:
> > On Sunday, 30 July 2023 at 17:09:14 UTC-4, Eram semper recta wrote:
> > > On Sunday, 30 July 2023 at 17:05:40 UTC-4, Eram semper recta wrote:
> > > > On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
> > > > > You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
> > > > >
> > > > > https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
> > > > >
> > > > > Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
> > > > If queer troll studied Euclid's Elements, then he would know about Euclid's Division Lemma.
> > > >
> > > >
> > > > Consider the expression:
> > > > f(x+h) - f(x) = d*h + q
> > > And in fact that is how my identity is stated:
> > >
> > > (f(x+h)-f(x))/h = f'(x) + Q(x,h)
> > >
> > > Add a remainder term and we have:
> > >
> > > f(x+h)-f(x) = [f'(x) + Q(x,h)]*h + R(x,h)
> > >
> > > The result follows easily, unless you are a math graduate from Sweden or France.
> > > >
> > > > Where:
> > > >
> > > > h is the potential factor
> > > > d and q are some magnitudes
> > > > If q = 0, then f(x+h) - f(x) = d*h
> > > >
> > > > In this case, f(x+h) - f(x) can be written as h multiplied by the magnitude d.
> > > > There is no remainder when dividing f(x+h) - f(x) by the magnitude h.
> > > > Therefore, by the generalised Euclidean definition, h would be considered a factor of f(x+h) - f(x) when q = 0.
> > > > This aligns with Euclid's Division Lemma, where a number is said to be a factor if there is no remainder when dividing another magnitude by it..
> > > >
> > > > Like I said, you can't fix stupid.
> > > >
> > > > Both Jean Pierre Messager (Python, YBM. JPM) and shithead markus klyver are MORONS.
> > Even from the troll's initial premise, he claimed that (h*f(x))/h means "h is a factor of f(x)". The fucking moron doesn't realise that h is a factor of h*f(x), not of f(x). Yes. It really is this bad in Sweden and France..
> >
> > To any normal person, r = p x q means that p and q are factors of r regardless of what q and r are. They can be any magnitude.
> >
> > The vile crank will then in his confused brain recycle back to his drivel "x can be written as x=a*h, where a is a real number ("magnitude" in your non-standard terminology). This means that h "divides" x without a remainder."
> >
> > But that is not what is being said by h being a factor of f(x+h)-f(x) because f(x+h)-f(x) is some expression k multiplied by h, that is, kh/h.
> >
> > Thus, if k = kh/h, then h is a factor of kh, not k. <----- fucking vile faggot can't comprehend this. Unbelievable how dense this cunt is.
> You can re-wri <crap>

No, moron, no. Not even wrong.

On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
> You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
>
> https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
>
> Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.

If f(x+h) - f(x) = k for some magnitude k then k can be rewritten as k = hd, where d is some other magnitude. This is proved in my theorem:

https://www.academia.edu/62358358/My_historic_geometric_theorem_of_January_2020

Formal proof on slide 10.
Slide 25 gives a visual proof.

This means h "wholly measures k", since k is a multiple of h. Therefore, by the definition established, h is a factor of f(x+h) - f(x) and we can divide both sides by h to get:

(f(x+h) - f(x))/h = d

This provides a very clear, straightforward way to determine if h is a factor and what the quotient will be, without needing the formal complete/remainder definition which is too complex for stupid people.

On Sunday, 30 July 2023 at 20:45:51 UTC-4, Eram semper recta wrote:
> On Sunday, 30 July 2023 at 19:29:13 UTC-4, crank markus klyver wrote:
> > söndag 30 juli 2023 kl. 23:05:40 UTC+2 skrev Eram semper recta:
> > > On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
> > > > You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
> > > >
> > > > https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
> > > >
> > > > Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
> > > If queer troll studied Euclid's Elements, then he would know about Euclid's Division Lemma.
> > >
> > >
> > > Consider the expression:
> > > f(x+h) - f(x) = d*h + q
> > >
> > > Where:
> > >
> > > h is the potential factor
> > > d and q are some magnitudes
> > > If q = 0, then f(x+h) - f(x) = d*h
> > >
> > > In this case, f(x+h) - f(x) can be written as h multiplied by the magnitude d.
> > > There is no remainder when dividing f(x+h) - f(x) by the magnitude h.
> > > Therefore, by the generalised Euclidean definition, h would be considered a factor of f(x+h) - f(x) when q = 0.
> > > This aligns with Euclid's Division Lemma, where a number is said to be a factor if there is no remainder when dividing another magnitude by it.
> > >
> > > Like I said, you can't fix stupid.
> > >
> > > Both Jean Pierre Messager (Python, YBM. JPM) and shithead markus klyver are MORONS.
> > Well, let's use actual numbers then. If f(x+h)-f(x)=5 and h=2, then
> That is never the case you fucking idiot. Never. h is a factor of f(x+h)-f(x) meaning that if k = f(x+h)-f(x), we can rewrite k = hd in which case hd/h = d.

I bet the idiot Klyver was thinking something like this:

f(x)=x^2 with x= 1/4 and h=2 in which case f(x+h)-f(x)=5 and h=2. LMAO.

But this is not at all a problem! 5 = 2 x 2.5 and so 2x2.5/2 = 2.5.

It wouldn't surprise me at all.

Gosh, the troll is dense!

>
> <bullshit>

måndag 31 juli 2023 kl. 02:46:56 UTC+2 skrev Eram semper recta:
> On Sunday, 30 July 2023 at 19:34:28 UTC-4, markus moron klyver wrote:
> > söndag 30 juli 2023 kl. 23:56:09 UTC+2 skrev Eram semper recta:
> > > On Sunday, 30 July 2023 at 17:09:14 UTC-4, Eram semper recta wrote:
> > > > On Sunday, 30 July 2023 at 17:05:40 UTC-4, Eram semper recta wrote:
> > > > > On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
> > > > > > You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
> > > > > >
> > > > > > https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
> > > > > >
> > > > > > Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
> > > > > If queer troll studied Euclid's Elements, then he would know about Euclid's Division Lemma.
> > > > >
> > > > >
> > > > > Consider the expression:
> > > > > f(x+h) - f(x) = d*h + q
> > > > And in fact that is how my identity is stated:
> > > >
> > > > (f(x+h)-f(x))/h = f'(x) + Q(x,h)
> > > >
> > > > Add a remainder term and we have:
> > > >
> > > > f(x+h)-f(x) = [f'(x) + Q(x,h)]*h + R(x,h)
> > > >
> > > > The result follows easily, unless you are a math graduate from Sweden or France.
> > > > >
> > > > > Where:
> > > > >
> > > > > h is the potential factor
> > > > > d and q are some magnitudes
> > > > > If q = 0, then f(x+h) - f(x) = d*h
> > > > >
> > > > > In this case, f(x+h) - f(x) can be written as h multiplied by the magnitude d.
> > > > > There is no remainder when dividing f(x+h) - f(x) by the magnitude h.
> > > > > Therefore, by the generalised Euclidean definition, h would be considered a factor of f(x+h) - f(x) when q = 0.
> > > > > This aligns with Euclid's Division Lemma, where a number is said to be a factor if there is no remainder when dividing another magnitude by it.
> > > > >
> > > > > Like I said, you can't fix stupid.
> > > > >
> > > > > Both Jean Pierre Messager (Python, YBM. JPM) and shithead markus klyver are MORONS.
> > > Even from the troll's initial premise, he claimed that (h*f(x))/h means "h is a factor of f(x)". The fucking moron doesn't realise that h is a factor of h*f(x), not of f(x). Yes. It really is this bad in Sweden and France.
> > >
> > > To any normal person, r = p x q means that p and q are factors of r regardless of what q and r are. They can be any magnitude.
> > >
> > > The vile crank will then in his confused brain recycle back to his drivel "x can be written as x=a*h, where a is a real number ("magnitude" in your non-standard terminology). This means that h "divides" x without a remainder."
> > >
> > > But that is not what is being said by h being a factor of f(x+h)-f(x) because f(x+h)-f(x) is some expression k multiplied by h, that is, kh/h.
> > >
> > > Thus, if k = kh/h, then h is a factor of kh, not k. <----- fucking vile faggot can't comprehend this. Unbelievable how dense this cunt is.
> > You can re-wri <crap>
>
> No, moron, no. Not even wrong.
Of course you can. f(x)= (f(x)/h)*h holds true by definition.

On Monday, 31 July 2023 at 04:39:51 UTC-4, markus wrote:
> måndag 31 juli 2023 kl. 02:46:56 UTC+2 skrev Eram semper recta:
> > On Sunday, 30 July 2023 at 19:34:28 UTC-4, markus moron klyver wrote:
> > > söndag 30 juli 2023 kl. 23:56:09 UTC+2 skrev Eram semper recta:
> > > > On Sunday, 30 July 2023 at 17:09:14 UTC-4, Eram semper recta wrote:
> > > > > On Sunday, 30 July 2023 at 17:05:40 UTC-4, Eram semper recta wrote:
> > > > > > On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
> > > > > > > You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
> > > > > > >
> > > > > > > https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
> > > > > > >
> > > > > > > Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
> > > > > > If queer troll studied Euclid's Elements, then he would know about Euclid's Division Lemma.
> > > > > >
> > > > > >
> > > > > > Consider the expression:
> > > > > > f(x+h) - f(x) = d*h + q
> > > > > And in fact that is how my identity is stated:
> > > > >
> > > > > (f(x+h)-f(x))/h = f'(x) + Q(x,h)
> > > > >
> > > > > Add a remainder term and we have:
> > > > >
> > > > > f(x+h)-f(x) = [f'(x) + Q(x,h)]*h + R(x,h)
> > > > >
> > > > > The result follows easily, unless you are a math graduate from Sweden or France.
> > > > > >
> > > > > > Where:
> > > > > >
> > > > > > h is the potential factor
> > > > > > d and q are some magnitudes
> > > > > > If q = 0, then f(x+h) - f(x) = d*h
> > > > > >
> > > > > > In this case, f(x+h) - f(x) can be written as h multiplied by the magnitude d.
> > > > > > There is no remainder when dividing f(x+h) - f(x) by the magnitude h.
> > > > > > Therefore, by the generalised Euclidean definition, h would be considered a factor of f(x+h) - f(x) when q = 0.
> > > > > > This aligns with Euclid's Division Lemma, where a number is said to be a factor if there is no remainder when dividing another magnitude by it.
> > > > > >
> > > > > > Like I said, you can't fix stupid.
> > > > > >
> > > > > > Both Jean Pierre Messager (Python, YBM. JPM) and shithead markus klyver are MORONS.
> > > > Even from the troll's initial premise, he claimed that (h*f(x))/h means "h is a factor of f(x)". The fucking moron doesn't realise that h is a factor of h*f(x), not of f(x). Yes. It really is this bad in Sweden and France.
> > > >
> > > > To any normal person, r = p x q means that p and q are factors of r regardless of what q and r are. They can be any magnitude.
> > > >
> > > > The vile crank will then in his confused brain recycle back to his drivel "x can be written as x=a*h, where a is a real number ("magnitude" in your non-standard terminology). This means that h "divides" x without a remainder."
> > > >
> > > > But that is not what is being said by h being a factor of f(x+h)-f(x) because f(x+h)-f(x) is some expression k multiplied by h, that is, kh/h.
> > > >
> > > > Thus, if k = kh/h, then h is a factor of kh, not k. <----- fucking vile faggot can't comprehend this. Unbelievable how dense this cunt is.
> > > You can re-wri <crap>
> >
> > No, moron, no. Not even wrong.
> Of course you can. f(x)= (f(x)/h)*h holds true by definition.

Straw man argument again. I did not say you can't, moron! I am saying as I have been all along that you are confused and you have no argument.

måndag 31 juli 2023 kl. 02:45:51 UTC+2 skrev Eram semper recta:
> On Sunday, 30 July 2023 at 19:29:13 UTC-4, crank markus klyver wrote:
> > söndag 30 juli 2023 kl. 23:05:40 UTC+2 skrev Eram semper recta:
> > > On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
> > > > You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
> > > >
> > > > https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
> > > >
> > > > Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
> > > If queer troll studied Euclid's Elements, then he would know about Euclid's Division Lemma.
> > >
> > >
> > > Consider the expression:
> > > f(x+h) - f(x) = d*h + q
> > >
> > > Where:
> > >
> > > h is the potential factor
> > > d and q are some magnitudes
> > > If q = 0, then f(x+h) - f(x) = d*h
> > >
> > > In this case, f(x+h) - f(x) can be written as h multiplied by the magnitude d.
> > > There is no remainder when dividing f(x+h) - f(x) by the magnitude h.
> > > Therefore, by the generalised Euclidean definition, h would be considered a factor of f(x+h) - f(x) when q = 0.
> > > This aligns with Euclid's Division Lemma, where a number is said to be a factor if there is no remainder when dividing another magnitude by it.
> > >
> > > Like I said, you can't fix stupid.
> > >
> > > Both Jean Pierre Messager (Python, YBM. JPM) and shithead markus klyver are MORONS.
> > Well, let's use actual numbers then. If f(x+h)-f(x)=5 and h=2, then
> That is never the case you fucking idiot. Never. h is a factor of f(x+h)-f(x) meaning that if k = f(x+h)-f(x), we can rewrite k = hd in which case hd/h = d.
>
> <bullshit>
The point is that you can always do that. Whatever f(x+h)-f(x) is, you can always factorize out h as a real number.

Ii have pointed out or before, and it seems like you want to see x and g as indeterminates. But then you're derivative definition only works for polynomials. In which case the real definition using limits is clearly superior.

On Monday, 31 July 2023 at 04:46:09 UTC-4, markus...:
> måndag 31 juli 2023 kl. 02:45:51 UTC+2 skrev Eram semper recta:
> > On Sunday, 30 July 2023 at 19:29:13 UTC-4, crank markus klyver wrote:
> > > söndag 30 juli 2023 kl. 23:05:40 UTC+2 skrev Eram semper recta:
> > > > On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
> > > > > You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
> > > > >
> > > > > https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
> > > > >
> > > > > Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
> > > > If queer troll studied Euclid's Elements, then he would know about Euclid's Division Lemma.
> > > >
> > > >
> > > > Consider the expression:
> > > > f(x+h) - f(x) = d*h + q
> > > >
> > > > Where:
> > > >
> > > > h is the potential factor
> > > > d and q are some magnitudes
> > > > If q = 0, then f(x+h) - f(x) = d*h
> > > >
> > > > In this case, f(x+h) - f(x) can be written as h multiplied by the magnitude d.
> > > > There is no remainder when dividing f(x+h) - f(x) by the magnitude h.
> > > > Therefore, by the generalised Euclidean definition, h would be considered a factor of f(x+h) - f(x) when q = 0.
> > > > This aligns with Euclid's Division Lemma, where a number is said to be a factor if there is no remainder when dividing another magnitude by it..
> > > >
> > > > Like I said, you can't fix stupid.
> > > >
> > > > Both Jean Pierre Messager (Python, YBM. JPM) and shithead markus klyver are MORONS.
> > > Well, let's use actual numbers then. If f(x+h)-f(x)=5 and h=2, then
> > That is never the case you fucking idiot. Never. h is a factor of f(x+h)-f(x) meaning that if k = f(x+h)-f(x), we can rewrite k = hd in which case hd/h = d.
> >
> > <bullshit>
> The point is that you can always do that.

That is not the point, you idiot. It is known that you can always do that.

> Whatever f(x+h)-f(x) is, you can always factorize out h as a real number.
>
> Ii have pointed out or before, and it seems like you want to see x and g as indeterminates.

They are not indeterminates, you monkey. In fact, it doesn't matter at all what is h. You are just too dense.

> But then you're derivative definition only works for polynomials.

That is an argument by erroneous conclusion because I proved that it works for all smooth functions.

> In which case the real definition using limits is clearly superior.

Your meds are obviously not helping you. Did you know that now you can get assisted suicide in Switzerland also for your mental condition?

Look into it!

måndag 31 juli 2023 kl. 03:12:33 UTC+2 skrev Eram semper recta:
> On Sunday, 30 July 2023 at 20:45:51 UTC-4, Eram semper recta wrote:
> > On Sunday, 30 July 2023 at 19:29:13 UTC-4, crank markus klyver wrote:
> > > söndag 30 juli 2023 kl. 23:05:40 UTC+2 skrev Eram semper recta:
> > > > On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
> > > > > You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
> > > > >
> > > > > https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
> > > > >
> > > > > Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
> > > > If queer troll studied Euclid's Elements, then he would know about Euclid's Division Lemma.
> > > >
> > > >
> > > > Consider the expression:
> > > > f(x+h) - f(x) = d*h + q
> > > >
> > > > Where:
> > > >
> > > > h is the potential factor
> > > > d and q are some magnitudes
> > > > If q = 0, then f(x+h) - f(x) = d*h
> > > >
> > > > In this case, f(x+h) - f(x) can be written as h multiplied by the magnitude d.
> > > > There is no remainder when dividing f(x+h) - f(x) by the magnitude h.
> > > > Therefore, by the generalised Euclidean definition, h would be considered a factor of f(x+h) - f(x) when q = 0.
> > > > This aligns with Euclid's Division Lemma, where a number is said to be a factor if there is no remainder when dividing another magnitude by it..
> > > >
> > > > Like I said, you can't fix stupid.
> > > >
> > > > Both Jean Pierre Messager (Python, YBM. JPM) and shithead markus klyver are MORONS.
> > > Well, let's use actual numbers then. If f(x+h)-f(x)=5 and h=2, then
> > That is never the case you fucking idiot. Never. h is a factor of f(x+h)-f(x) meaning that if k = f(x+h)-f(x), we can rewrite k = hd in which case hd/h = d.
> I bet the idiot Klyver was thinking something like this:
>
> f(x)=x^2 with x= 1/4 and h=2 in which case f(x+h)-f(x)=5 and h=2. LMAO.
>
> But this is not at all a problem! 5 = 2 x 2.5 and so 2x2.5/2 = 2.5.
>
> It wouldn't surprise me at all.
>
> Gosh, the troll is dense!
>
> >
> > <bullshit>
You even came up with an example, and also proved that as rational numbers that 2 and 2.5 are factors of 5. You don't want to allow that, so that's why you need a better (and working) definition of what a factor is

måndag 31 juli 2023 kl. 10:49:55 UTC+2 skrev Eram semper recta:
> On Monday, 31 July 2023 at 04:46:09 UTC-4, markus...:
> > måndag 31 juli 2023 kl. 02:45:51 UTC+2 skrev Eram semper recta:
> > > On Sunday, 30 July 2023 at 19:29:13 UTC-4, crank markus klyver wrote:
> > > > söndag 30 juli 2023 kl. 23:05:40 UTC+2 skrev Eram semper recta:
> > > > > On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
> > > > > > You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
> > > > > >
> > > > > > https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
> > > > > >
> > > > > > Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
> > > > > If queer troll studied Euclid's Elements, then he would know about Euclid's Division Lemma.
> > > > >
> > > > >
> > > > > Consider the expression:
> > > > > f(x+h) - f(x) = d*h + q
> > > > >
> > > > > Where:
> > > > >
> > > > > h is the potential factor
> > > > > d and q are some magnitudes
> > > > > If q = 0, then f(x+h) - f(x) = d*h
> > > > >
> > > > > In this case, f(x+h) - f(x) can be written as h multiplied by the magnitude d.
> > > > > There is no remainder when dividing f(x+h) - f(x) by the magnitude h.
> > > > > Therefore, by the generalised Euclidean definition, h would be considered a factor of f(x+h) - f(x) when q = 0.
> > > > > This aligns with Euclid's Division Lemma, where a number is said to be a factor if there is no remainder when dividing another magnitude by it.
> > > > >
> > > > > Like I said, you can't fix stupid.
> > > > >
> > > > > Both Jean Pierre Messager (Python, YBM. JPM) and shithead markus klyver are MORONS.
> > > > Well, let's use actual numbers then. If f(x+h)-f(x)=5 and h=2, then
> > > That is never the case you fucking idiot. Never. h is a factor of f(x+h)-f(x) meaning that if k = f(x+h)-f(x), we can rewrite k = hd in which case hd/h = d.
> > >
> > > <bullshit>
> > The point is that you can always do that.
> That is not the point, you idiot. It is known that you can always do that..
> > Whatever f(x+h)-f(x) is, you can always factorize out h as a real number.
> >
> > Ii have pointed out or before, and it seems like you want to see x and g as indeterminates.
> They are not indeterminates, you monkey. In fact, it doesn't matter at all what is h. You are just too dense.
> > But then you're derivative definition only works for polynomials.
> That is an argument by erroneous conclusion because I proved that it works for all smooth functions.
> > In which case the real definition using limits is clearly superior.
> Your meds are obviously not helping you. Did you know that now you can get assisted suicide in Switzerland also for your mental condition?
>
> Look into it!
And we are back to you wishing me dead because I'm telling you that your definitions aren't working. 🤷‍♂️

No, it doesn't work for all "smooth functions" (and you have your own definition of what "smooth" means anyway, so no one will be able to understand anything you say because you re-define words as it suits you). If you use the indeterminates approach and want to extend it to a broader class of functions you need limits. Which is the very thing you want to avoid.

No, you can't have a unique decomposition f'(x)+Q(h) just from "factors". You need to be able to say something more about Q which makes it unique. There are infinitely many ways to write f'(x)+h*W(h). So that means neither f'(x) nor Q(h)=h*W(h) are uniquely defined. What you want to say, and I'm stressing this, is that Q(h) goes to 0 as h goes to 0. This would be the correct definition and the one we are using in mainstream mathematics.

On Monday, 31 July 2023 at 05:06:33 UTC-4, markus... wrote:
> måndag 31 juli 2023 kl. 10:49:55 UTC+2 skrev Eram semper recta:
> > On Monday, 31 July 2023 at 04:46:09 UTC-4, markus...:
> > > måndag 31 juli 2023 kl. 02:45:51 UTC+2 skrev Eram semper recta:
> > > > On Sunday, 30 July 2023 at 19:29:13 UTC-4, crank markus klyver wrote:
> > > > > söndag 30 juli 2023 kl. 23:05:40 UTC+2 skrev Eram semper recta:
> > > > > > On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
> > > > > > > You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
> > > > > > >
> > > > > > > https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
> > > > > > >
> > > > > > > Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
> > > > > > If queer troll studied Euclid's Elements, then he would know about Euclid's Division Lemma.
> > > > > >
> > > > > >
> > > > > > Consider the expression:
> > > > > > f(x+h) - f(x) = d*h + q
> > > > > >
> > > > > > Where:
> > > > > >
> > > > > > h is the potential factor
> > > > > > d and q are some magnitudes
> > > > > > If q = 0, then f(x+h) - f(x) = d*h
> > > > > >
> > > > > > In this case, f(x+h) - f(x) can be written as h multiplied by the magnitude d.
> > > > > > There is no remainder when dividing f(x+h) - f(x) by the magnitude h.
> > > > > > Therefore, by the generalised Euclidean definition, h would be considered a factor of f(x+h) - f(x) when q = 0.
> > > > > > This aligns with Euclid's Division Lemma, where a number is said to be a factor if there is no remainder when dividing another magnitude by it.
> > > > > >
> > > > > > Like I said, you can't fix stupid.
> > > > > >
> > > > > > Both Jean Pierre Messager (Python, YBM. JPM) and shithead markus klyver are MORONS.
> > > > > Well, let's use actual numbers then. If f(x+h)-f(x)=5 and h=2, then
> > > > That is never the case you fucking idiot. Never. h is a factor of f(x+h)-f(x) meaning that if k = f(x+h)-f(x), we can rewrite k = hd in which case hd/h = d.
> > > >
> > > > <bullshit>
> > > The point is that you can always do that.
> > That is not the point, you idiot. It is known that you can always do that.
> > > Whatever f(x+h)-f(x) is, you can always factorize out h as a real number.
> > >
> > > Ii have pointed out or before, and it seems like you want to see x and g as indeterminates.
> > They are not indeterminates, you monkey. In fact, it doesn't matter at all what is h. You are just too dense.
> > > But then you're derivative definition only works for polynomials.
> > That is an argument by erroneous conclusion because I proved that it works for all smooth functions.
> > > In which case the real definition using limits is clearly superior.
> > Your meds are obviously not helping you. Did you know that now you can get assisted suicide in Switzerland also for your mental condition?
> >
> > Look into it!
> And we are back to you wishing me dead because I'm telling you that your definitions aren't working. 🤷‍♂️

They're working just fine. Your brain has never worked, you fucking moron.

>
> No, it doesn't work for all "smooth functions" <shit>

YES, IT DOES WORK FOR ALL SMOOTH FUNCTIONS, FUCKING TOSSER!!!

Your failure to understand does not mean shit. All you have ever had are straw man arguments and insults. Not to mention that you are a class one troll.

f(x+h)-f(x) = h* tan ( pi/2 - atan (h/ (f(x+h)-f(x)))) where atan is the inverse tan function and all angles are expressed in radians.

This means that h is a factor of f(x+h)-f(x).

I can't help it that the poor annoying moron and his fellow idiot Jean Pierre Messager (they feed off each other) has a fear of trigonometry.

Nothing about any complicated measure theory for bird brains in the above statement. E.O.D.

This happens to be on slide 25 of my article which proves the theorem:

https://www.academia.edu/62358358/My_historic_geometric_theorem_of_January_2020

Your medication isn't helping you.

DIE QUICKLY!

On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
> You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
>
> https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
>
> Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.

I the great John Gabriel say that h is a factor of f(x+h)-f(x) because

f(x+h)-f(x) = h* tan ( pi/2 - atan (h/ (f(x+h)-f(x)))) where atan is the inverse tan function and all angles are expressed in radians.

This means that h is a factor of f(x+h)-f(x).

Don't believe me, you stupid cunts?

Download this applet and try it with ANY smooth function you like:

https://drive.google.com/file/d/1BHrD7WlSxQowFRqBF-2Rj7UQL9JeYSJ-/

On Monday, 31 July 2023 at 05:06:33 UTC-4, markus...@gmail.com wrote:
> måndag 31 juli 2023 kl. 10:49:55 UTC+2 skrev Eram semper recta:
> > On Monday, 31 July 2023 at 04:46:09 UTC-4, markus...:
> > > måndag 31 juli 2023 kl. 02:45:51 UTC+2 skrev Eram semper recta:
> > > > On Sunday, 30 July 2023 at 19:29:13 UTC-4, crank markus klyver wrote:
> > > > > söndag 30 juli 2023 kl. 23:05:40 UTC+2 skrev Eram semper recta:
> > > > > > On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
> > > > > > > You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
> > > > > > >
> > > > > > > https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
> > > > > > >
> > > > > > > Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
> > > > > > If queer troll studied Euclid's Elements, then he would know about Euclid's Division Lemma.
> > > > > >
> > > > > >
> > > > > > Consider the expression:
> > > > > > f(x+h) - f(x) = d*h + q
> > > > > >
> > > > > > Where:
> > > > > >
> > > > > > h is the potential factor
> > > > > > d and q are some magnitudes
> > > > > > If q = 0, then f(x+h) - f(x) = d*h
> > > > > >
> > > > > > In this case, f(x+h) - f(x) can be written as h multiplied by the magnitude d.
> > > > > > There is no remainder when dividing f(x+h) - f(x) by the magnitude h.
> > > > > > Therefore, by the generalised Euclidean definition, h would be considered a factor of f(x+h) - f(x) when q = 0.
> > > > > > This aligns with Euclid's Division Lemma, where a number is said to be a factor if there is no remainder when dividing another magnitude by it.
> > > > > >
> > > > > > Like I said, you can't fix stupid.
> > > > > >
> > > > > > Both Jean Pierre Messager (Python, YBM. JPM) and shithead markus klyver are MORONS.
> > > > > Well, let's use actual numbers then. If f(x+h)-f(x)=5 and h=2, then
> > > > That is never the case you fucking idiot. Never. h is a factor of f(x+h)-f(x) meaning that if k = f(x+h)-f(x), we can rewrite k = hd in which case hd/h = d.
> > > >
> > > > <bullshit>
> > > The point is that you can always do that.
> > That is not the point, you idiot. It is known that you can always do that.
> > > Whatever f(x+h)-f(x) is, you can always factorize out h as a real number.
> > >
> > > Ii have pointed out or before, and it seems like you want to see x and g as indeterminates.
> > They are not indeterminates, you monkey. In fact, it doesn't matter at all what is h. You are just too dense.
> > > But then you're derivative definition only works for polynomials.
> > That is an argument by erroneous conclusion because I proved that it works for all smooth functions.
> > > In which case the real definition using limits is clearly superior.
> > Your meds are obviously not helping you. Did you know that now you can get assisted suicide in Switzerland also for your mental condition?
> >
> > Look into it!
> And we are back to you wishing me dead because I'm telling you that your definitions aren't working. 🤷‍♂️
>
> No, it doesn't work for all "smooth functions" (and you have your own definition of what "smooth" means anyway, so no one will be able to understand anything you say because you re-define words as it suits you). If you use the indeterminates approach and want to extend it to a broader class of functions you need limits. Which is the very thing you want to avoid.
>
> No, you can't have a unique decomposition f'(x)+Q(h) just from "factors". You need to be able to say something more about Q which makes it unique. There are infinitely many ways to write f'(x)+h*W(h). So that means neither f'(x) nor Q(h)=h*W(h) are uniquely defined. What you want to say, and I'm stressing this, is that Q(h) goes to 0 as h goes to 0. This would be the correct definition and the one we are using in mainstream mathematics.

You get a special mention here, you vile little queer cunt!

https://www.youtube.com/watch?v=6uI74Fx8Yec

Cfank John Gabriel aka Eram semper recta wrote:
> On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
>> You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
>>
>> https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
>>
>> Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
>
> I the great John Gabriel say that h is a factor of f(x+h)-f(x) because
>
> f(x+h)-f(x) = h* tan ( pi/2 - atan (h/ (f(x+h)-f(x)))) where atan is the inverse tan function and all angles are expressed in radians.
>
> This means that h is a factor of f(x+h)-f(x).

Any non-zero number is a factor of any other, dumbass...

>
> Don't believe me, you stupid cunts?

Your "theorem" is void of any content, moreover we have provided
several counter-examples.

Find another hobby, math is not your thing John.

Le 31/07/2023 à 15:56, Eram semper recta a écrit :
> On Monday, 31 July 2023 at 05:06:33 UTC-4, markus...@gmail.com wrote:
>> måndag 31 juli 2023 kl. 10:49:55 UTC+2 skrev Eram semper recta:
>>> On Monday, 31 July 2023 at 04:46:09 UTC-4, markus...:
>>>> måndag 31 juli 2023 kl. 02:45:51 UTC+2 skrev Eram semper recta:
>>>>> On Sunday, 30 July 2023 at 19:29:13 UTC-4, crank markus klyver wrote:
>>>>>> söndag 30 juli 2023 kl. 23:05:40 UTC+2 skrev Eram semper recta:
>>>>>>> On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
>>>>>>>> You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
>>>>>>>>
>>>>>>>> https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
>>>>>>>>
>>>>>>>> Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
>>>>>>> If queer troll studied Euclid's Elements, then he would know about Euclid's Division Lemma.
>>>>>>>
>>>>>>>
>>>>>>> Consider the expression:
>>>>>>> f(x+h) - f(x) = d*h + q
>>>>>>>
>>>>>>> Where:
>>>>>>>
>>>>>>> h is the potential factor
>>>>>>> d and q are some magnitudes
>>>>>>> If q = 0, then f(x+h) - f(x) = d*h
>>>>>>>
>>>>>>> In this case, f(x+h) - f(x) can be written as h multiplied by the magnitude d.
>>>>>>> There is no remainder when dividing f(x+h) - f(x) by the magnitude h.
>>>>>>> Therefore, by the generalised Euclidean definition, h would be considered a factor of f(x+h) - f(x) when q = 0.
>>>>>>> This aligns with Euclid's Division Lemma, where a number is said to be a factor if there is no remainder when dividing another magnitude by it.
>>>>>>>
>>>>>>> Like I said, you can't fix stupid.
>>>>>>>
>>>>>>> Both Jean Pierre Messager (Python, YBM. JPM) and shithead markus klyver are MORONS.
>>>>>> Well, let's use actual numbers then. If f(x+h)-f(x)=5 and h=2, then
>>>>> That is never the case you fucking idiot. Never. h is a factor of f(x+h)-f(x) meaning that if k = f(x+h)-f(x), we can rewrite k = hd in which case hd/h = d.
>>>>>
>>>>> <bullshit>
>>>> The point is that you can always do that.
>>> That is not the point, you idiot. It is known that you can always do that.
>>>> Whatever f(x+h)-f(x) is, you can always factorize out h as a real number.
>>>>
>>>> Ii have pointed out or before, and it seems like you want to see x and g as indeterminates.
>>> They are not indeterminates, you monkey. In fact, it doesn't matter at all what is h. You are just too dense.
>>>> But then you're derivative definition only works for polynomials.
>>> That is an argument by erroneous conclusion because I proved that it works for all smooth functions.
>>>> In which case the real definition using limits is clearly superior.
>>> Your meds are obviously not helping you. Did you know that now you can get assisted suicide in Switzerland also for your mental condition?
>>>
>>> Look into it!
>> And we are back to you wishing me dead because I'm telling you that your definitions aren't working. 🤷‍♂️
>>
>> No, it doesn't work for all "smooth functions" (and you have your own definition of what "smooth" means anyway, so no one will be able to understand anything you say because you re-define words as it suits you). If you use the indeterminates approach and want to extend it to a broader class of functions you need limits. Which is the very thing you want to avoid.
>>
>> No, you can't have a unique decomposition f'(x)+Q(h) just from "factors". You need to be able to say something more about Q which makes it unique. There are infinitely many ways to write f'(x)+h*W(h). So that means neither f'(x) nor Q(h)=h*W(h) are uniquely defined. What you want to say, and I'm stressing this, is that Q(h) goes to 0 as h goes to 0. This would be the correct definition and the one we are using in mainstream mathematics.
>
> You get a special mention here, you vile little queer cunt!
>
> https://www.youtube.com/watch?v=6uI74Fx8Yec

This idiotic and abusive video will be removed.

måndag 31 juli 2023 kl. 14:45:13 UTC+2 skrev Eram semper recta:
> On Monday, 31 July 2023 at 05:06:33 UTC-4, markus... wrote:
> > måndag 31 juli 2023 kl. 10:49:55 UTC+2 skrev Eram semper recta:
> > > On Monday, 31 July 2023 at 04:46:09 UTC-4, markus...:
> > > > måndag 31 juli 2023 kl. 02:45:51 UTC+2 skrev Eram semper recta:
> > > > > On Sunday, 30 July 2023 at 19:29:13 UTC-4, crank markus klyver wrote:
> > > > > > söndag 30 juli 2023 kl. 23:05:40 UTC+2 skrev Eram semper recta:
> > > > > > > On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
> > > > > > > > You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
> > > > > > > >
> > > > > > > > https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
> > > > > > > >
> > > > > > > > Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
> > > > > > > If queer troll studied Euclid's Elements, then he would know about Euclid's Division Lemma.
> > > > > > >
> > > > > > >
> > > > > > > Consider the expression:
> > > > > > > f(x+h) - f(x) = d*h + q
> > > > > > >
> > > > > > > Where:
> > > > > > >
> > > > > > > h is the potential factor
> > > > > > > d and q are some magnitudes
> > > > > > > If q = 0, then f(x+h) - f(x) = d*h
> > > > > > >
> > > > > > > In this case, f(x+h) - f(x) can be written as h multiplied by the magnitude d.
> > > > > > > There is no remainder when dividing f(x+h) - f(x) by the magnitude h.
> > > > > > > Therefore, by the generalised Euclidean definition, h would be considered a factor of f(x+h) - f(x) when q = 0.
> > > > > > > This aligns with Euclid's Division Lemma, where a number is said to be a factor if there is no remainder when dividing another magnitude by it.
> > > > > > >
> > > > > > > Like I said, you can't fix stupid.
> > > > > > >
> > > > > > > Both Jean Pierre Messager (Python, YBM. JPM) and shithead markus klyver are MORONS.
> > > > > > Well, let's use actual numbers then. If f(x+h)-f(x)=5 and h=2, then
> > > > > That is never the case you fucking idiot. Never. h is a factor of f(x+h)-f(x) meaning that if k = f(x+h)-f(x), we can rewrite k = hd in which case hd/h = d.
> > > > >
> > > > > <bullshit>
> > > > The point is that you can always do that.
> > > That is not the point, you idiot. It is known that you can always do that.
> > > > Whatever f(x+h)-f(x) is, you can always factorize out h as a real number.
> > > >
> > > > Ii have pointed out or before, and it seems like you want to see x and g as indeterminates.
> > > They are not indeterminates, you monkey. In fact, it doesn't matter at all what is h. You are just too dense.
> > > > But then you're derivative definition only works for polynomials.
> > > That is an argument by erroneous conclusion because I proved that it works for all smooth functions.
> > > > In which case the real definition using limits is clearly superior.
> > > Your meds are obviously not helping you. Did you know that now you can get assisted suicide in Switzerland also for your mental condition?
> > >
> > > Look into it!
> > And we are back to you wishing me dead because I'm telling you that your definitions aren't working. 🤷‍♂️
> They're working just fine. Your brain has never worked, you fucking moron..
>
> >
> > No, it doesn't work for all "smooth functions" <shit>
>
> YES, IT DOES WORK FOR ALL SMOOTH FUNCTIONS, FUCKING TOSSER!!!
>
> Your failure to understand does not mean shit. All you have ever had are straw man arguments and insults. Not to mention that you are a class one troll.
>
> f(x+h)-f(x) = h* tan ( pi/2 - atan (h/ (f(x+h)-f(x)))) where atan is the inverse tan function and all angles are expressed in radians.
>
> This means that h is a factor of f(x+h)-f(x).
>
> I can't help it that the poor annoying moron and his fellow idiot Jean Pierre Messager (they feed off each other) has a fear of trigonometry.
>
> Nothing about any complicated measure theory for bird brains in the above statement. E.O.D.
>
> This happens to be on slide 25 of my article which proves the theorem:
>
> https://www.academia.edu/62358358/My_historic_geometric_theorem_of_January_2020
>
> Your medication isn't helping you.
>
> DIE QUICKLY!
The point is, as I have said multiple times now, you can factorize h from every number. You haven't explained what a factor is our how Q is defined. That means that f' remains undefined too.

måndag 31 juli 2023 kl. 15:56:27 UTC+2 skrev Eram semper recta:
> On Monday, 31 July 2023 at 05:06:33 UTC-4, markus...@gmail.com wrote:
> > måndag 31 juli 2023 kl. 10:49:55 UTC+2 skrev Eram semper recta:
> > > On Monday, 31 July 2023 at 04:46:09 UTC-4, markus...:
> > > > måndag 31 juli 2023 kl. 02:45:51 UTC+2 skrev Eram semper recta:
> > > > > On Sunday, 30 July 2023 at 19:29:13 UTC-4, crank markus klyver wrote:
> > > > > > söndag 30 juli 2023 kl. 23:05:40 UTC+2 skrev Eram semper recta:
> > > > > > > On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
> > > > > > > > You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
> > > > > > > >
> > > > > > > > https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
> > > > > > > >
> > > > > > > > Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
> > > > > > > If queer troll studied Euclid's Elements, then he would know about Euclid's Division Lemma.
> > > > > > >
> > > > > > >
> > > > > > > Consider the expression:
> > > > > > > f(x+h) - f(x) = d*h + q
> > > > > > >
> > > > > > > Where:
> > > > > > >
> > > > > > > h is the potential factor
> > > > > > > d and q are some magnitudes
> > > > > > > If q = 0, then f(x+h) - f(x) = d*h
> > > > > > >
> > > > > > > In this case, f(x+h) - f(x) can be written as h multiplied by the magnitude d.
> > > > > > > There is no remainder when dividing f(x+h) - f(x) by the magnitude h.
> > > > > > > Therefore, by the generalised Euclidean definition, h would be considered a factor of f(x+h) - f(x) when q = 0.
> > > > > > > This aligns with Euclid's Division Lemma, where a number is said to be a factor if there is no remainder when dividing another magnitude by it.
> > > > > > >
> > > > > > > Like I said, you can't fix stupid.
> > > > > > >
> > > > > > > Both Jean Pierre Messager (Python, YBM. JPM) and shithead markus klyver are MORONS.
> > > > > > Well, let's use actual numbers then. If f(x+h)-f(x)=5 and h=2, then
> > > > > That is never the case you fucking idiot. Never. h is a factor of f(x+h)-f(x) meaning that if k = f(x+h)-f(x), we can rewrite k = hd in which case hd/h = d.
> > > > >
> > > > > <bullshit>
> > > > The point is that you can always do that.
> > > That is not the point, you idiot. It is known that you can always do that.
> > > > Whatever f(x+h)-f(x) is, you can always factorize out h as a real number.
> > > >
> > > > Ii have pointed out or before, and it seems like you want to see x and g as indeterminates.
> > > They are not indeterminates, you monkey. In fact, it doesn't matter at all what is h. You are just too dense.
> > > > But then you're derivative definition only works for polynomials.
> > > That is an argument by erroneous conclusion because I proved that it works for all smooth functions.
> > > > In which case the real definition using limits is clearly superior.
> > > Your meds are obviously not helping you. Did you know that now you can get assisted suicide in Switzerland also for your mental condition?
> > >
> > > Look into it!
> > And we are back to you wishing me dead because I'm telling you that your definitions aren't working. 🤷‍♂️
> >
> > No, it doesn't work for all "smooth functions" (and you have your own definition of what "smooth" means anyway, so no one will be able to understand anything you say because you re-define words as it suits you). If you use the indeterminates approach and want to extend it to a broader class of functions you need limits. Which is the very thing you want to avoid.
> >
> > No, you can't have a unique decomposition f'(x)+Q(h) just from "factors". You need to be able to say something more about Q which makes it unique. There are infinitely many ways to write f'(x)+h*W(h). So that means neither f'(x) nor Q(h)=h*W(h) are uniquely defined. What you want to say, and I'm stressing this, is that Q(h) goes to 0 as h goes to 0. This would be the correct definition and the one we are using in mainstream mathematics.
> You get a special mention here, you vile little queer cunt!
>
> https://www.youtube.com/watch?v=6uI74Fx8Yec
Now I'm queer too?

On Monday, 31 July 2023 at 12:55:09 UTC-4, Python wrote:
> Le 31/07/2023 à 15:56, Eram semper recta a écrit :
> > On Monday, 31 July 2023 at 05:06:33 UTC-4, markus. shithead wrote:
> >> måndag 31 juli 2023 kl. 10:49:55 UTC+2 skrev Eram semper recta:
> >>> On Monday, 31 July 2023 at 04:46:09 UTC-4, markus...:
> >>>> måndag 31 juli 2023 kl. 02:45:51 UTC+2 skrev Eram semper recta:
> >>>>> On Sunday, 30 July 2023 at 19:29:13 UTC-4, crank markus klyver wrote:
> >>>>>> söndag 30 juli 2023 kl. 23:05:40 UTC+2 skrev Eram semper recta:
> >>>>>>> On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
> >>>>>>>> You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
> >>>>>>>>
> >>>>>>>> https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
> >>>>>>>>
> >>>>>>>> Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
> >>>>>>> If queer troll studied Euclid's Elements, then he would know about Euclid's Division Lemma.
> >>>>>>>
> >>>>>>>
> >>>>>>> Consider the expression:
> >>>>>>> f(x+h) - f(x) = d*h + q
> >>>>>>>
> >>>>>>> Where:
> >>>>>>>
> >>>>>>> h is the potential factor
> >>>>>>> d and q are some magnitudes
> >>>>>>> If q = 0, then f(x+h) - f(x) = d*h
> >>>>>>>
> >>>>>>> In this case, f(x+h) - f(x) can be written as h multiplied by the magnitude d.
> >>>>>>> There is no remainder when dividing f(x+h) - f(x) by the magnitude h.
> >>>>>>> Therefore, by the generalised Euclidean definition, h would be considered a factor of f(x+h) - f(x) when q = 0.
> >>>>>>> This aligns with Euclid's Division Lemma, where a number is said to be a factor if there is no remainder when dividing another magnitude by it.
> >>>>>>>
> >>>>>>> Like I said, you can't fix stupid.
> >>>>>>>
> >>>>>>> Both Jean Pierre Messager (Python, YBM. JPM) and shithead markus klyver are MORONS.
> >>>>>> Well, let's use actual numbers then. If f(x+h)-f(x)=5 and h=2, then
> >>>>> That is never the case you fucking idiot. Never. h is a factor of f(x+h)-f(x) meaning that if k = f(x+h)-f(x), we can rewrite k = hd in which case hd/h = d.
> >>>>>
> >>>>> <bullshit>
> >>>> The point is that you can always do that.
> >>> That is not the point, you idiot. It is known that you can always do that.
> >>>> Whatever f(x+h)-f(x) is, you can always factorize out h as a real number.
> >>>>
> >>>> Ii have pointed out or before, and it seems like you want to see x and g as indeterminates.
> >>> They are not indeterminates, you monkey. In fact, it doesn't matter at all what is h. You are just too dense.
> >>>> But then you're derivative definition only works for polynomials.
> >>> That is an argument by erroneous conclusion because I proved that it works for all smooth functions.
> >>>> In which case the real definition using limits is clearly superior.
> >>> Your meds are obviously not helping you. Did you know that now you can get assisted suicide in Switzerland also for your mental condition?
> >>>
> >>> Look into it!
> >> And we are back to you wishing me dead because I'm telling you that your definitions aren't working. 🤷‍♂️
> >>
> >> No, it doesn't work for all "smooth functions" (and you have your own definition of what "smooth" means anyway, so no one will be able to understand anything you say because you re-define words as it suits you). If you use the indeterminates approach and want to extend it to a broader class of functions you need limits. Which is the very thing you want to avoid.
> >>
> >> No, you can't have a unique decomposition f'(x)+Q(h) just from "factors". You need to be able to say something more about Q which makes it unique.. There are infinitely many ways to write f'(x)+h*W(h). So that means neither f'(x) nor Q(h)=h*W(h) are uniquely defined. What you want to say, and I'm stressing this, is that Q(h) goes to 0 as h goes to 0. This would be the correct definition and the one we are using in mainstream mathematics.
> >
> > You get a special mention here, you vile little queer cunt!
> >
> > https://www.youtube.com/watch?v=6uI74Fx8Yec
> This idiotic and abusive video will be removed.

Not a chance, fuck face! sci.math has your details and I am not doxxing you..

You miserable little French Jew Fuck. Gosh, I would love to have just one minute with you and your fuckbuddy Giberlt Strang. I would wring both your necks!

Eat my SHIT and die, you miserable, vile JEW FUCK!!!!!!!!!

Oh, and I am still waiting for FBI. LMFAO!!!!!!!!!!!!!!

On Monday, 31 July 2023 at 13:27:02 UTC-4, markus...@gmail.com wrote:
> måndag 31 juli 2023 kl. 14:45:13 UTC+2 skrev Eram semper recta:
> > On Monday, 31 July 2023 at 05:06:33 UTC-4, markus... wrote:
> > > måndag 31 juli 2023 kl. 10:49:55 UTC+2 skrev Eram semper recta:
> > > > On Monday, 31 July 2023 at 04:46:09 UTC-4, markus...:
> > > > > måndag 31 juli 2023 kl. 02:45:51 UTC+2 skrev Eram semper recta:
> > > > > > On Sunday, 30 July 2023 at 19:29:13 UTC-4, crank markus klyver wrote:
> > > > > > > söndag 30 juli 2023 kl. 23:05:40 UTC+2 skrev Eram semper recta:
> > > > > > > > On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
> > > > > > > > > You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
> > > > > > > > >
> > > > > > > > > https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
> > > > > > > > >
> > > > > > > > > Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
> > > > > > > > If queer troll studied Euclid's Elements, then he would know about Euclid's Division Lemma.
> > > > > > > >
> > > > > > > >
> > > > > > > > Consider the expression:
> > > > > > > > f(x+h) - f(x) = d*h + q
> > > > > > > >
> > > > > > > > Where:
> > > > > > > >
> > > > > > > > h is the potential factor
> > > > > > > > d and q are some magnitudes
> > > > > > > > If q = 0, then f(x+h) - f(x) = d*h
> > > > > > > >
> > > > > > > > In this case, f(x+h) - f(x) can be written as h multiplied by the magnitude d.
> > > > > > > > There is no remainder when dividing f(x+h) - f(x) by the magnitude h.
> > > > > > > > Therefore, by the generalised Euclidean definition, h would be considered a factor of f(x+h) - f(x) when q = 0.
> > > > > > > > This aligns with Euclid's Division Lemma, where a number is said to be a factor if there is no remainder when dividing another magnitude by it.
> > > > > > > >
> > > > > > > > Like I said, you can't fix stupid.
> > > > > > > >
> > > > > > > > Both Jean Pierre Messager (Python, YBM. JPM) and shithead markus klyver are MORONS.
> > > > > > > Well, let's use actual numbers then. If f(x+h)-f(x)=5 and h=2, then
> > > > > > That is never the case you fucking idiot. Never. h is a factor of f(x+h)-f(x) meaning that if k = f(x+h)-f(x), we can rewrite k = hd in which case hd/h = d.
> > > > > >
> > > > > > <bullshit>
> > > > > The point is that you can always do that.
> > > > That is not the point, you idiot. It is known that you can always do that.
> > > > > Whatever f(x+h)-f(x) is, you can always factorize out h as a real number.
> > > > >
> > > > > Ii have pointed out or before, and it seems like you want to see x and g as indeterminates.
> > > > They are not indeterminates, you monkey. In fact, it doesn't matter at all what is h. You are just too dense.
> > > > > But then you're derivative definition only works for polynomials.
> > > > That is an argument by erroneous conclusion because I proved that it works for all smooth functions.
> > > > > In which case the real definition using limits is clearly superior.
> > > > Your meds are obviously not helping you. Did you know that now you can get assisted suicide in Switzerland also for your mental condition?
> > > >
> > > > Look into it!
> > > And we are back to you wishing me dead because I'm telling you that your definitions aren't working. 🤷‍♂️
> > They're working just fine. Your brain has never worked, you fucking moron.
> >
> > >
> > > No, it doesn't work for all "smooth functions" <shit>
> >
> > YES, IT DOES WORK FOR ALL SMOOTH FUNCTIONS, FUCKING TOSSER!!!
> >
> > Your failure to understand does not mean shit. All you have ever had are straw man arguments and insults. Not to mention that you are a class one troll.
> >
> > f(x+h)-f(x) = h* tan ( pi/2 - atan (h/ (f(x+h)-f(x)))) where atan is the inverse tan function and all angles are expressed in radians.
> >
> > This means that h is a factor of f(x+h)-f(x).
> >
> > I can't help it that the poor annoying moron and his fellow idiot Jean Pierre Messager (they feed off each other) has a fear of trigonometry.
> >
> > Nothing about any complicated measure theory for bird brains in the above statement. E.O.D.
> >
> > This happens to be on slide 25 of my article which proves the theorem:
> >
> > https://www.academia.edu/62358358/My_historic_geometric_theorem_of_January_2020
> >
> > Your medication isn't helping you.
> >
> > DIE QUICKLY!
> The point is, ...

YOU NEVER HAD A POINT, you dipstick SHITHEAD!

Eat my shit and DIE!!!!!!!!!!!!!!!!

On Monday, 31 July 2023 at 12:55:09 UTC-4, Python wrote:
> Le 31/07/2023 à 15:56, Eram semper recta a écrit :
> > On Monday, 31 July 2023 at 05:06:33 UTC-4, markus.. wrote:
> >> måndag 31 juli 2023 kl. 10:49:55 UTC+2 skrev Eram semper recta:
> >>> On Monday, 31 July 2023 at 04:46:09 UTC-4, markus...:
> >>>> måndag 31 juli 2023 kl. 02:45:51 UTC+2 skrev Eram semper recta:
> >>>>> On Sunday, 30 July 2023 at 19:29:13 UTC-4, crank markus klyver wrote:
> >>>>>> söndag 30 juli 2023 kl. 23:05:40 UTC+2 skrev Eram semper recta:
> >>>>>>> On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
> >>>>>>>> You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
> >>>>>>>>
> >>>>>>>> https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
> >>>>>>>>
> >>>>>>>> Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
> >>>>>>> If queer troll studied Euclid's Elements, then he would know about Euclid's Division Lemma.
> >>>>>>>
> >>>>>>>
> >>>>>>> Consider the expression:
> >>>>>>> f(x+h) - f(x) = d*h + q
> >>>>>>>
> >>>>>>> Where:
> >>>>>>>
> >>>>>>> h is the potential factor
> >>>>>>> d and q are some magnitudes
> >>>>>>> If q = 0, then f(x+h) - f(x) = d*h
> >>>>>>>
> >>>>>>> In this case, f(x+h) - f(x) can be written as h multiplied by the magnitude d.
> >>>>>>> There is no remainder when dividing f(x+h) - f(x) by the magnitude h.
> >>>>>>> Therefore, by the generalised Euclidean definition, h would be considered a factor of f(x+h) - f(x) when q = 0.
> >>>>>>> This aligns with Euclid's Division Lemma, where a number is said to be a factor if there is no remainder when dividing another magnitude by it.
> >>>>>>>
> >>>>>>> Like I said, you can't fix stupid.
> >>>>>>>
> >>>>>>> Both Jean Pierre Messager (Python, YBM. JPM) and shithead markus klyver are MORONS.
> >>>>>> Well, let's use actual numbers then. If f(x+h)-f(x)=5 and h=2, then
> >>>>> That is never the case you fucking idiot. Never. h is a factor of f(x+h)-f(x) meaning that if k = f(x+h)-f(x), we can rewrite k = hd in which case hd/h = d.
> >>>>>
> >>>>> <bullshit>
> >>>> The point is that you can always do that.
> >>> That is not the point, you idiot. It is known that you can always do that.
> >>>> Whatever f(x+h)-f(x) is, you can always factorize out h as a real number.
> >>>>
> >>>> Ii have pointed out or before, and it seems like you want to see x and g as indeterminates.
> >>> They are not indeterminates, you monkey. In fact, it doesn't matter at all what is h. You are just too dense.
> >>>> But then you're derivative definition only works for polynomials.
> >>> That is an argument by erroneous conclusion because I proved that it works for all smooth functions.
> >>>> In which case the real definition using limits is clearly superior.
> >>> Your meds are obviously not helping you. Did you know that now you can get assisted suicide in Switzerland also for your mental condition?
> >>>
> >>> Look into it!
> >> And we are back to you wishing me dead because I'm telling you that your definitions aren't working. 🤷‍♂️
> >>
> >> No, it doesn't work for all "smooth functions" (and you have your own definition of what "smooth" means anyway, so no one will be able to understand anything you say because you re-define words as it suits you). If you use the indeterminates approach and want to extend it to a broader class of functions you need limits. Which is the very thing you want to avoid.
> >>
> >> No, you can't have a unique decomposition f'(x)+Q(h) just from "factors". You need to be able to say something more about Q which makes it unique.. There are infinitely many ways to write f'(x)+h*W(h). So that means neither f'(x) nor Q(h)=h*W(h) are uniquely defined. What you want to say, and I'm stressing this, is that Q(h) goes to 0 as h goes to 0. This would be the correct definition and the one we are using in mainstream mathematics.
> >
> > You get a special mention here, you vile little queer cunt!
> >
> > https://www.youtube.com/watch?v=6uI74Fx8Yec
> This idiotic and abusive video will be removed.

It will be removed because it exposes what a fucking idiots you and your little queer buddy Klyver are?
LMAO. Bwaaaa haaa haaaa.

Oh. Still waiting for the FBI.

What a fucking, vile JEW piece of shit MORON you are.

DIE, you bastard!!!!!! You have caused me a lot of harm. Spitting at you whilst saying this.

I am going to publish a lot more videos on you SHITHEAD!!!! BE prepared for it.

I will RUIN you!!!!

måndag 31 juli 2023 kl. 22:16:19 UTC+2 skrev Eram semper recta:
> On Monday, 31 July 2023 at 12:55:09 UTC-4, Python wrote:
> > Le 31/07/2023 à 15:56, Eram semper recta a écrit :
> > > On Monday, 31 July 2023 at 05:06:33 UTC-4, markus. shithead wrote:
> > >> måndag 31 juli 2023 kl. 10:49:55 UTC+2 skrev Eram semper recta:
> > >>> On Monday, 31 July 2023 at 04:46:09 UTC-4, markus...:
> > >>>> måndag 31 juli 2023 kl. 02:45:51 UTC+2 skrev Eram semper recta:
> > >>>>> On Sunday, 30 July 2023 at 19:29:13 UTC-4, crank markus klyver wrote:
> > >>>>>> söndag 30 juli 2023 kl. 23:05:40 UTC+2 skrev Eram semper recta:
> > >>>>>>> On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
> > >>>>>>>> You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
> > >>>>>>>>
> > >>>>>>>> https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
> > >>>>>>>>
> > >>>>>>>> Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
> > >>>>>>> If queer troll studied Euclid's Elements, then he would know about Euclid's Division Lemma.
> > >>>>>>>
> > >>>>>>>
> > >>>>>>> Consider the expression:
> > >>>>>>> f(x+h) - f(x) = d*h + q
> > >>>>>>>
> > >>>>>>> Where:
> > >>>>>>>
> > >>>>>>> h is the potential factor
> > >>>>>>> d and q are some magnitudes
> > >>>>>>> If q = 0, then f(x+h) - f(x) = d*h
> > >>>>>>>
> > >>>>>>> In this case, f(x+h) - f(x) can be written as h multiplied by the magnitude d.
> > >>>>>>> There is no remainder when dividing f(x+h) - f(x) by the magnitude h.
> > >>>>>>> Therefore, by the generalised Euclidean definition, h would be considered a factor of f(x+h) - f(x) when q = 0.
> > >>>>>>> This aligns with Euclid's Division Lemma, where a number is said to be a factor if there is no remainder when dividing another magnitude by it.
> > >>>>>>>
> > >>>>>>> Like I said, you can't fix stupid.
> > >>>>>>>
> > >>>>>>> Both Jean Pierre Messager (Python, YBM. JPM) and shithead markus klyver are MORONS.
> > >>>>>> Well, let's use actual numbers then. If f(x+h)-f(x)=5 and h=2, then
> > >>>>> That is never the case you fucking idiot. Never. h is a factor of f(x+h)-f(x) meaning that if k = f(x+h)-f(x), we can rewrite k = hd in which case hd/h = d.
> > >>>>>
> > >>>>> <bullshit>
> > >>>> The point is that you can always do that.
> > >>> That is not the point, you idiot. It is known that you can always do that.
> > >>>> Whatever f(x+h)-f(x) is, you can always factorize out h as a real number.
> > >>>>
> > >>>> Ii have pointed out or before, and it seems like you want to see x and g as indeterminates.
> > >>> They are not indeterminates, you monkey. In fact, it doesn't matter at all what is h. You are just too dense.
> > >>>> But then you're derivative definition only works for polynomials.
> > >>> That is an argument by erroneous conclusion because I proved that it works for all smooth functions.
> > >>>> In which case the real definition using limits is clearly superior..
> > >>> Your meds are obviously not helping you. Did you know that now you can get assisted suicide in Switzerland also for your mental condition?
> > >>>
> > >>> Look into it!
> > >> And we are back to you wishing me dead because I'm telling you that your definitions aren't working. 🤷‍♂️
> > >>
> > >> No, it doesn't work for all "smooth functions" (and you have your own definition of what "smooth" means anyway, so no one will be able to understand anything you say because you re-define words as it suits you). If you use the indeterminates approach and want to extend it to a broader class of functions you need limits. Which is the very thing you want to avoid.
> > >>
> > >> No, you can't have a unique decomposition f'(x)+Q(h) just from "factors". You need to be able to say something more about Q which makes it unique. There are infinitely many ways to write f'(x)+h*W(h). So that means neither f'(x) nor Q(h)=h*W(h) are uniquely defined. What you want to say, and I'm stressing this, is that Q(h) goes to 0 as h goes to 0. This would be the correct definition and the one we are using in mainstream mathematics.
> > >
> > > You get a special mention here, you vile little queer cunt!
> > >
> > > https://www.youtube.com/watch?v=6uI74Fx8Yec
> > This idiotic and abusive video will be removed.
> Not a chance, fuck face! sci.math has your details and I am not doxxing you.
>
> You miserable little French Jew Fuck. Gosh, I would love to have just one minute with you and your fuckbuddy Giberlt Strang. I would wring both your necks!
>
> Eat my SHIT and die, you miserable, vile JEW FUCK!!!!!!!!!
>
>
> Oh, and I am still waiting for FBI. LMFAO!!!!!!!!!!!!!!
"Vile Jew"?

måndag 31 juli 2023 kl. 22:20:04 UTC+2 skrev Eram semper recta:
> On Monday, 31 July 2023 at 12:55:09 UTC-4, Python wrote:
> > Le 31/07/2023 à 15:56, Eram semper recta a écrit :
> > > On Monday, 31 July 2023 at 05:06:33 UTC-4, markus.. wrote:
> > >> måndag 31 juli 2023 kl. 10:49:55 UTC+2 skrev Eram semper recta:
> > >>> On Monday, 31 July 2023 at 04:46:09 UTC-4, markus...:
> > >>>> måndag 31 juli 2023 kl. 02:45:51 UTC+2 skrev Eram semper recta:
> > >>>>> On Sunday, 30 July 2023 at 19:29:13 UTC-4, crank markus klyver wrote:
> > >>>>>> söndag 30 juli 2023 kl. 23:05:40 UTC+2 skrev Eram semper recta:
> > >>>>>>> On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
> > >>>>>>>> You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
> > >>>>>>>>
> > >>>>>>>> https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
> > >>>>>>>>
> > >>>>>>>> Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
> > >>>>>>> If queer troll studied Euclid's Elements, then he would know about Euclid's Division Lemma.
> > >>>>>>>
> > >>>>>>>
> > >>>>>>> Consider the expression:
> > >>>>>>> f(x+h) - f(x) = d*h + q
> > >>>>>>>
> > >>>>>>> Where:
> > >>>>>>>
> > >>>>>>> h is the potential factor
> > >>>>>>> d and q are some magnitudes
> > >>>>>>> If q = 0, then f(x+h) - f(x) = d*h
> > >>>>>>>
> > >>>>>>> In this case, f(x+h) - f(x) can be written as h multiplied by the magnitude d.
> > >>>>>>> There is no remainder when dividing f(x+h) - f(x) by the magnitude h.
> > >>>>>>> Therefore, by the generalised Euclidean definition, h would be considered a factor of f(x+h) - f(x) when q = 0.
> > >>>>>>> This aligns with Euclid's Division Lemma, where a number is said to be a factor if there is no remainder when dividing another magnitude by it.
> > >>>>>>>
> > >>>>>>> Like I said, you can't fix stupid.
> > >>>>>>>
> > >>>>>>> Both Jean Pierre Messager (Python, YBM. JPM) and shithead markus klyver are MORONS.
> > >>>>>> Well, let's use actual numbers then. If f(x+h)-f(x)=5 and h=2, then
> > >>>>> That is never the case you fucking idiot. Never. h is a factor of f(x+h)-f(x) meaning that if k = f(x+h)-f(x), we can rewrite k = hd in which case hd/h = d.
> > >>>>>
> > >>>>> <bullshit>
> > >>>> The point is that you can always do that.
> > >>> That is not the point, you idiot. It is known that you can always do that.
> > >>>> Whatever f(x+h)-f(x) is, you can always factorize out h as a real number.
> > >>>>
> > >>>> Ii have pointed out or before, and it seems like you want to see x and g as indeterminates.
> > >>> They are not indeterminates, you monkey. In fact, it doesn't matter at all what is h. You are just too dense.
> > >>>> But then you're derivative definition only works for polynomials.
> > >>> That is an argument by erroneous conclusion because I proved that it works for all smooth functions.
> > >>>> In which case the real definition using limits is clearly superior..
> > >>> Your meds are obviously not helping you. Did you know that now you can get assisted suicide in Switzerland also for your mental condition?
> > >>>
> > >>> Look into it!
> > >> And we are back to you wishing me dead because I'm telling you that your definitions aren't working. 🤷‍♂️
> > >>
> > >> No, it doesn't work for all "smooth functions" (and you have your own definition of what "smooth" means anyway, so no one will be able to understand anything you say because you re-define words as it suits you). If you use the indeterminates approach and want to extend it to a broader class of functions you need limits. Which is the very thing you want to avoid.
> > >>
> > >> No, you can't have a unique decomposition f'(x)+Q(h) just from "factors". You need to be able to say something more about Q which makes it unique. There are infinitely many ways to write f'(x)+h*W(h). So that means neither f'(x) nor Q(h)=h*W(h) are uniquely defined. What you want to say, and I'm stressing this, is that Q(h) goes to 0 as h goes to 0. This would be the correct definition and the one we are using in mainstream mathematics.
> > >
> > > You get a special mention here, you vile little queer cunt!
> > >
> > > https://www.youtube.com/watch?v=6uI74Fx8Yec
> > This idiotic and abusive video will be removed.
> It will be removed because it exposes what a fucking idiots you and your little queer buddy Klyver are?
> LMAO. Bwaaaa haaa haaaa.
>
> Oh. Still waiting for the FBI.
>
> What a fucking, vile JEW piece of shit MORON you are.
>
> DIE, you bastard!!!!!! You have caused me a lot of harm. Spitting at you whilst saying this.
>
> I am going to publish a lot more videos on you SHITHEAD!!!! BE prepared for it.
>
> I will RUIN you!!!!
There's nothing to ruin. Except for your fake math.

On Monday, 31 July 2023 at 16:57:11 UTC-4, markus.shithead klyver wrote:
> måndag 31 juli 2023 kl. 22:20:04 UTC+2 skrev Eram semper recta:
> > On Monday, 31 July 2023 at 12:55:09 UTC-4, Python wrote:
> > > Le 31/07/2023 à 15:56, Eram semper recta a écrit :
> > > > On Monday, 31 July 2023 at 05:06:33 UTC-4, markus.. wrote:
> > > >> måndag 31 juli 2023 kl. 10:49:55 UTC+2 skrev Eram semper recta:
> > > >>> On Monday, 31 July 2023 at 04:46:09 UTC-4, markus...:
> > > >>>> måndag 31 juli 2023 kl. 02:45:51 UTC+2 skrev Eram semper recta:
> > > >>>>> On Sunday, 30 July 2023 at 19:29:13 UTC-4, crank markus klyver wrote:
> > > >>>>>> söndag 30 juli 2023 kl. 23:05:40 UTC+2 skrev Eram semper recta:
> > > >>>>>>> On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
> > > >>>>>>>> You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
> > > >>>>>>>>
> > > >>>>>>>> https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
> > > >>>>>>>>
> > > >>>>>>>> Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
> > > >>>>>>> If queer troll studied Euclid's Elements, then he would know about Euclid's Division Lemma.
> > > >>>>>>>
> > > >>>>>>>
> > > >>>>>>> Consider the expression:
> > > >>>>>>> f(x+h) - f(x) = d*h + q
> > > >>>>>>>
> > > >>>>>>> Where:
> > > >>>>>>>
> > > >>>>>>> h is the potential factor
> > > >>>>>>> d and q are some magnitudes
> > > >>>>>>> If q = 0, then f(x+h) - f(x) = d*h
> > > >>>>>>>
> > > >>>>>>> In this case, f(x+h) - f(x) can be written as h multiplied by the magnitude d.
> > > >>>>>>> There is no remainder when dividing f(x+h) - f(x) by the magnitude h.
> > > >>>>>>> Therefore, by the generalised Euclidean definition, h would be considered a factor of f(x+h) - f(x) when q = 0.
> > > >>>>>>> This aligns with Euclid's Division Lemma, where a number is said to be a factor if there is no remainder when dividing another magnitude by it.
> > > >>>>>>>
> > > >>>>>>> Like I said, you can't fix stupid.
> > > >>>>>>>
> > > >>>>>>> Both Jean Pierre Messager (Python, YBM. JPM) and shithead markus klyver are MORONS.
> > > >>>>>> Well, let's use actual numbers then. If f(x+h)-f(x)=5 and h=2, then
> > > >>>>> That is never the case you fucking idiot. Never. h is a factor of f(x+h)-f(x) meaning that if k = f(x+h)-f(x), we can rewrite k = hd in which case hd/h = d.
> > > >>>>>
> > > >>>>> <bullshit>
> > > >>>> The point is that you can always do that.
> > > >>> That is not the point, you idiot. It is known that you can always do that.
> > > >>>> Whatever f(x+h)-f(x) is, you can always factorize out h as a real number.
> > > >>>>
> > > >>>> Ii have pointed out or before, and it seems like you want to see x and g as indeterminates.
> > > >>> They are not indeterminates, you monkey. In fact, it doesn't matter at all what is h. You are just too dense.
> > > >>>> But then you're derivative definition only works for polynomials.
> > > >>> That is an argument by erroneous conclusion because I proved that it works for all smooth functions.
> > > >>>> In which case the real definition using limits is clearly superior.
> > > >>> Your meds are obviously not helping you. Did you know that now you can get assisted suicide in Switzerland also for your mental condition?
> > > >>>
> > > >>> Look into it!
> > > >> And we are back to you wishing me dead because I'm telling you that your definitions aren't working. 🤷‍♂️
> > > >>
> > > >> No, it doesn't work for all "smooth functions" (and you have your own definition of what "smooth" means anyway, so no one will be able to understand anything you say because you re-define words as it suits you). If you use the indeterminates approach and want to extend it to a broader class of functions you need limits. Which is the very thing you want to avoid.
> > > >>
> > > >> No, you can't have a unique decomposition f'(x)+Q(h) just from "factors". You need to be able to say something more about Q which makes it unique. There are infinitely many ways to write f'(x)+h*W(h). So that means neither f'(x) nor Q(h)=h*W(h) are uniquely defined. What you want to say, and I'm stressing this, is that Q(h) goes to 0 as h goes to 0. This would be the correct definition and the one we are using in mainstream mathematics.
> > > >
> > > > You get a special mention here, you vile little queer cunt!
> > > >
> > > > https://www.youtube.com/watch?v=6uI74Fx8Yec
> > > This idiotic and abusive video will be removed.
> > It will be removed because it exposes what a fucking idiots you and your little queer buddy Klyver are?
> > LMAO. Bwaaaa haaa haaaa.
> >
> > Oh. Still waiting for the FBI.
> >
> > What a fucking, vile JEW piece of shit MORON you are.
> >
> > DIE, you bastard!!!!!! You have caused me a lot of harm. Spitting at you whilst saying this.
> >
> > I am going to publish a lot more videos on you SHITHEAD!!!! BE prepared for it.
> >
> > I will RUIN you!!!!
> There's nothing to ruin. Except for your fake math.
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måndag 31 juli 2023 kl. 23:39:12 UTC+2 skrev Ιωάννης Γαβριήλ:
> On Monday, 31 July 2023 at 16:57:11 UTC-4, markus.shithead klyver wrote:
> > måndag 31 juli 2023 kl. 22:20:04 UTC+2 skrev Eram semper recta:
> > > On Monday, 31 July 2023 at 12:55:09 UTC-4, Python wrote:
> > > > Le 31/07/2023 à 15:56, Eram semper recta a écrit :
> > > > > On Monday, 31 July 2023 at 05:06:33 UTC-4, markus.. wrote:
> > > > >> måndag 31 juli 2023 kl. 10:49:55 UTC+2 skrev Eram semper recta:
> > > > >>> On Monday, 31 July 2023 at 04:46:09 UTC-4, markus...:
> > > > >>>> måndag 31 juli 2023 kl. 02:45:51 UTC+2 skrev Eram semper recta:
> > > > >>>>> On Sunday, 30 July 2023 at 19:29:13 UTC-4, crank markus klyver wrote:
> > > > >>>>>> söndag 30 juli 2023 kl. 23:05:40 UTC+2 skrev Eram semper recta:
> > > > >>>>>>> On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
> > > > >>>>>>>> You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
> > > > >>>>>>>>
> > > > >>>>>>>> https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
> > > > >>>>>>>>
> > > > >>>>>>>> Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
> > > > >>>>>>> If queer troll studied Euclid's Elements, then he would know about Euclid's Division Lemma.
> > > > >>>>>>>
> > > > >>>>>>>
> > > > >>>>>>> Consider the expression:
> > > > >>>>>>> f(x+h) - f(x) = d*h + q
> > > > >>>>>>>
> > > > >>>>>>> Where:
> > > > >>>>>>>
> > > > >>>>>>> h is the potential factor
> > > > >>>>>>> d and q are some magnitudes
> > > > >>>>>>> If q = 0, then f(x+h) - f(x) = d*h
> > > > >>>>>>>
> > > > >>>>>>> In this case, f(x+h) - f(x) can be written as h multiplied by the magnitude d.
> > > > >>>>>>> There is no remainder when dividing f(x+h) - f(x) by the magnitude h.
> > > > >>>>>>> Therefore, by the generalised Euclidean definition, h would be considered a factor of f(x+h) - f(x) when q = 0.
> > > > >>>>>>> This aligns with Euclid's Division Lemma, where a number is said to be a factor if there is no remainder when dividing another magnitude by it.
> > > > >>>>>>>
> > > > >>>>>>> Like I said, you can't fix stupid.
> > > > >>>>>>>
> > > > >>>>>>> Both Jean Pierre Messager (Python, YBM. JPM) and shithead markus klyver are MORONS.
> > > > >>>>>> Well, let's use actual numbers then. If f(x+h)-f(x)=5 and h=2, then
> > > > >>>>> That is never the case you fucking idiot. Never. h is a factor of f(x+h)-f(x) meaning that if k = f(x+h)-f(x), we can rewrite k = hd in which case hd/h = d.
> > > > >>>>>
> > > > >>>>> <bullshit>
> > > > >>>> The point is that you can always do that.
> > > > >>> That is not the point, you idiot. It is known that you can always do that.
> > > > >>>> Whatever f(x+h)-f(x) is, you can always factorize out h as a real number.
> > > > >>>>
> > > > >>>> Ii have pointed out or before, and it seems like you want to see x and g as indeterminates.
> > > > >>> They are not indeterminates, you monkey. In fact, it doesn't matter at all what is h. You are just too dense.
> > > > >>>> But then you're derivative definition only works for polynomials.
> > > > >>> That is an argument by erroneous conclusion because I proved that it works for all smooth functions.
> > > > >>>> In which case the real definition using limits is clearly superior.
> > > > >>> Your meds are obviously not helping you. Did you know that now you can get assisted suicide in Switzerland also for your mental condition?
> > > > >>>
> > > > >>> Look into it!
> > > > >> And we are back to you wishing me dead because I'm telling you that your definitions aren't working. 🤷‍♂️
> > > > >>
> > > > >> No, it doesn't work for all "smooth functions" (and you have your own definition of what "smooth" means anyway, so no one will be able to understand anything you say because you re-define words as it suits you). If you use the indeterminates approach and want to extend it to a broader class of functions you need limits. Which is the very thing you want to avoid.
> > > > >>
> > > > >> No, you can't have a unique decomposition f'(x)+Q(h) just from "factors". You need to be able to say something more about Q which makes it unique. There are infinitely many ways to write f'(x)+h*W(h). So that means neither f'(x) nor Q(h)=h*W(h) are uniquely defined. What you want to say, and I'm stressing this, is that Q(h) goes to 0 as h goes to 0. This would be the correct definition and the one we are using in mainstream mathematics.
> > > > >
> > > > > You get a special mention here, you vile little queer cunt!
> > > > >
> > > > > https://www.youtube.com/watch?v=6uI74Fx8Yec
> > > > This idiotic and abusive video will be removed.
> > > It will be removed because it exposes what a fucking idiots you and your little queer buddy Klyver are?
> > > LMAO. Bwaaaa haaa haaaa.
> > >
> > > Oh. Still waiting for the FBI.
> > >
> > > What a fucking, vile JEW piece of shit MORON you are.
> > >
> > > DIE, you bastard!!!!!! You have caused me a lot of harm. Spitting at you whilst saying this.
> > >
> > > I am going to publish a lot more videos on you SHITHEAD!!!! BE prepared for it.
> > >
> > > I will RUIN you!!!!
> > There's nothing to ruin. Except for your fake math.
> 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You found out how to post emojis?

On 7/31/2023 3:13 PM, markus...@gmail.com wrote:
> måndag 31 juli 2023 kl. 23:39:12 UTC+2 skrev Ιωάννης Γαβριήλ:
>> On Monday, 31 July 2023 at 16:57:11 UTC-4, markus.shithead klyver wrote:
>>> måndag 31 juli 2023 kl. 22:20:04 UTC+2 skrev Eram semper recta:
>>>> On Monday, 31 July 2023 at 12:55:09 UTC-4, Python wrote:
>>>>> Le 31/07/2023 à 15:56, Eram semper recta a écrit :
>>>>>> On Monday, 31 July 2023 at 05:06:33 UTC-4, markus.. wrote:
>>>>>>> måndag 31 juli 2023 kl. 10:49:55 UTC+2 skrev Eram semper recta:
>>>>>>>> On Monday, 31 July 2023 at 04:46:09 UTC-4, markus...:
>>>>>>>>> måndag 31 juli 2023 kl. 02:45:51 UTC+2 skrev Eram semper recta:
>>>>>>>>>> On Sunday, 30 July 2023 at 19:29:13 UTC-4, crank markus klyver wrote:
>>>>>>>>>>> söndag 30 juli 2023 kl. 23:05:40 UTC+2 skrev Eram semper recta:
>>>>>>>>>>>> On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
>>>>>>>>>>>>> You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
>>>>>>>>>>>>>
>>>>>>>>>>>>> https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
>>>>>>>>>>>>>
>>>>>>>>>>>>> Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
>>>>>>>>>>>> If queer troll studied Euclid's Elements, then he would know about Euclid's Division Lemma.
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Consider the expression:
>>>>>>>>>>>> f(x+h) - f(x) = d*h + q
>>>>>>>>>>>>
>>>>>>>>>>>> Where:
>>>>>>>>>>>>
>>>>>>>>>>>> h is the potential factor
>>>>>>>>>>>> d and q are some magnitudes
>>>>>>>>>>>> If q = 0, then f(x+h) - f(x) = d*h
>>>>>>>>>>>>
>>>>>>>>>>>> In this case, f(x+h) - f(x) can be written as h multiplied by the magnitude d.
>>>>>>>>>>>> There is no remainder when dividing f(x+h) - f(x) by the magnitude h.
>>>>>>>>>>>> Therefore, by the generalised Euclidean definition, h would be considered a factor of f(x+h) - f(x) when q = 0.
>>>>>>>>>>>> This aligns with Euclid's Division Lemma, where a number is said to be a factor if there is no remainder when dividing another magnitude by it.
>>>>>>>>>>>>
>>>>>>>>>>>> Like I said, you can't fix stupid.
>>>>>>>>>>>>
>>>>>>>>>>>> Both Jean Pierre Messager (Python, YBM. JPM) and shithead markus klyver are MORONS.
>>>>>>>>>>> Well, let's use actual numbers then. If f(x+h)-f(x)=5 and h=2, then
>>>>>>>>>> That is never the case you fucking idiot. Never. h is a factor of f(x+h)-f(x) meaning that if k = f(x+h)-f(x), we can rewrite k = hd in which case hd/h = d.
>>>>>>>>>>
>>>>>>>>>> <bullshit>
>>>>>>>>> The point is that you can always do that.
>>>>>>>> That is not the point, you idiot. It is known that you can always do that.
>>>>>>>>> Whatever f(x+h)-f(x) is, you can always factorize out h as a real number.
>>>>>>>>>
>>>>>>>>> Ii have pointed out or before, and it seems like you want to see x and g as indeterminates.
>>>>>>>> They are not indeterminates, you monkey. In fact, it doesn't matter at all what is h. You are just too dense.
>>>>>>>>> But then you're derivative definition only works for polynomials.
>>>>>>>> That is an argument by erroneous conclusion because I proved that it works for all smooth functions.
>>>>>>>>> In which case the real definition using limits is clearly superior.
>>>>>>>> Your meds are obviously not helping you. Did you know that now you can get assisted suicide in Switzerland also for your mental condition?
>>>>>>>>
>>>>>>>> Look into it!
>>>>>>> And we are back to you wishing me dead because I'm telling you that your definitions aren't working. 🤷‍♂️
>>>>>>>
>>>>>>> No, it doesn't work for all "smooth functions" (and you have your own definition of what "smooth" means anyway, so no one will be able to understand anything you say because you re-define words as it suits you). If you use the indeterminates approach and want to extend it to a broader class of functions you need limits. Which is the very thing you want to avoid.
>>>>>>>
>>>>>>> No, you can't have a unique decomposition f'(x)+Q(h) just from "factors". You need to be able to say something more about Q which makes it unique. There are infinitely many ways to write f'(x)+h*W(h). So that means neither f'(x) nor Q(h)=h*W(h) are uniquely defined. What you want to say, and I'm stressing this, is that Q(h) goes to 0 as h goes to 0. This would be the correct definition and the one we are using in mainstream mathematics.
>>>>>>
>>>>>> You get a special mention here, you vile little queer cunt!
>>>>>>
>>>>>> https://www.youtube.com/watch?v=6uI74Fx8Yec
>>>>> This idiotic and abusive video will be removed.
>>>> It will be removed because it exposes what a fucking idiots you and your little queer buddy Klyver are?
>>>> LMAO. Bwaaaa haaa haaaa.
>>>>
>>>> Oh. Still waiting for the FBI.
>>>>
>>>> What a fucking, vile JEW piece of shit MORON you are.
>>>>
>>>> DIE, you bastard!!!!!! You have caused me a lot of harm. Spitting at you whilst saying this.
>>>>
>>>> I am going to publish a lot more videos on you SHITHEAD!!!! BE prepared for it.
>>>>
>>>> I will RUIN you!!!!
>>> There's nothing to ruin. Except for your fake math.
[cut shit]
> You found out how to post emojis?

Trim the shit? ;^)

On Monday, 31 July 2023 at 18:53:58 UTC-4, Chris M. Thomasson wrote:
> On 7/31/2023 3:13 PM, markus...@gmail.com wrote:
> > måndag 31 juli 2023 kl. 23:39:12 UTC+2 skrev Ιωάννης Γαβριήλ:
> >> On Monday, 31 July 2023 at 16:57:11 UTC-4, markus.shithead klyver wrote:
> >>> måndag 31 juli 2023 kl. 22:20:04 UTC+2 skrev Eram semper recta:
> >>>> On Monday, 31 July 2023 at 12:55:09 UTC-4, Python wrote:
> >>>>> Le 31/07/2023 à 15:56, Eram semper recta a écrit :
> >>>>>> On Monday, 31 July 2023 at 05:06:33 UTC-4, markus.. wrote:
> >>>>>>> måndag 31 juli 2023 kl. 10:49:55 UTC+2 skrev Eram semper recta:
> >>>>>>>> On Monday, 31 July 2023 at 04:46:09 UTC-4, markus...:
> >>>>>>>>> måndag 31 juli 2023 kl. 02:45:51 UTC+2 skrev Eram semper recta:
> >>>>>>>>>> On Sunday, 30 July 2023 at 19:29:13 UTC-4, crank markus klyver wrote:
> >>>>>>>>>>> söndag 30 juli 2023 kl. 23:05:40 UTC+2 skrev Eram semper recta:
> >>>>>>>>>>>> On Saturday, 29 July 2023 at 21:21:58 UTC-4, Eram semper recta wrote:
> >>>>>>>>>>>>> You may think this is a trivial question, however, you might be surprised to know that mathematics graduates in Sweden and France do not know the meaning. In fact, most graduates of mathematics do not understand the concept very clearly. The following link refers:
> >>>>>>>>>>>>>
> >>>>>>>>>>>>> https://www.academia.edu/105057161/What_exactly_is_a_factor_in_mathematics
> >>>>>>>>>>>>>
> >>>>>>>>>>>>> Therein you will learn where the first mention was made and in what context. It's also interesting from a historical point of view.
> >>>>>>>>>>>> If queer troll studied Euclid's Elements, then he would know about Euclid's Division Lemma.
> >>>>>>>>>>>>
> >>>>>>>>>>>>
> >>>>>>>>>>>> Consider the expression:
> >>>>>>>>>>>> f(x+h) - f(x) = d*h + q
> >>>>>>>>>>>>
> >>>>>>>>>>>> Where:
> >>>>>>>>>>>>
> >>>>>>>>>>>> h is the potential factor
> >>>>>>>>>>>> d and q are some magnitudes
> >>>>>>>>>>>> If q = 0, then f(x+h) - f(x) = d*h
> >>>>>>>>>>>>
> >>>>>>>>>>>> In this case, f(x+h) - f(x) can be written as h multiplied by the magnitude d.
> >>>>>>>>>>>> There is no remainder when dividing f(x+h) - f(x) by the magnitude h.
> >>>>>>>>>>>> Therefore, by the generalised Euclidean definition, h would be considered a factor of f(x+h) - f(x) when q = 0.
> >>>>>>>>>>>> This aligns with Euclid's Division Lemma, where a number is said to be a factor if there is no remainder when dividing another magnitude by it.
> >>>>>>>>>>>>
> >>>>>>>>>>>> Like I said, you can't fix stupid.
> >>>>>>>>>>>>
> >>>>>>>>>>>> Both Jean Pierre Messager (Python, YBM. JPM) and shithead markus klyver are MORONS.
> >>>>>>>>>>> Well, let's use actual numbers then. If f(x+h)-f(x)=5 and h=2, then
> >>>>>>>>>> That is never the case you fucking idiot. Never. h is a factor of f(x+h)-f(x) meaning that if k = f(x+h)-f(x), we can rewrite k = hd in which case hd/h = d.
> >>>>>>>>>>
> >>>>>>>>>> <bullshit>
> >>>>>>>>> The point is that you can always do that.
> >>>>>>>> That is not the point, you idiot. It is known that you can always do that.
> >>>>>>>>> Whatever f(x+h)-f(x) is, you can always factorize out h as a real number.
> >>>>>>>>>
> >>>>>>>>> Ii have pointed out or before, and it seems like you want to see x and g as indeterminates.
> >>>>>>>> They are not indeterminates, you monkey. In fact, it doesn't matter at all what is h. You are just too dense.
> >>>>>>>>> But then you're derivative definition only works for polynomials.
> >>>>>>>> That is an argument by erroneous conclusion because I proved that it works for all smooth functions.
> >>>>>>>>> In which case the real definition using limits is clearly superior.
> >>>>>>>> Your meds are obviously not helping you. Did you know that now you can get assisted suicide in Switzerland also for your mental condition?
> >>>>>>>>
> >>>>>>>> Look into it!
> >>>>>>> And we are back to you wishing me dead because I'm telling you that your definitions aren't working. 🤷‍♂️
> >>>>>>>
> >>>>>>> No, it doesn't work for all "smooth functions" (and you have your own definition of what "smooth" means anyway, so no one will be able to understand anything you say because you re-define words as it suits you). If you use the indeterminates approach and want to extend it to a broader class of functions you need limits. Which is the very thing you want to avoid.
> >>>>>>>
> >>>>>>> No, you can't have a unique decomposition f'(x)+Q(h) just from "factors". You need to be able to say something more about Q which makes it unique. There are infinitely many ways to write f'(x)+h*W(h). So that means neither f'(x) nor Q(h)=h*W(h) are uniquely defined. What you want to say, and I'm stressing this, is that Q(h) goes to 0 as h goes to 0. This would be the correct definition and the one we are using in mainstream mathematics.
> >>>>>>
> >>>>>> You get a special mention here, you vile little queer cunt!
> >>>>>>
> >>>>>> https://www.youtube.com/watch?v=6uI74Fx8Yec
> >>>>> This idiotic and abusive video will be removed.
> >>>> It will be removed because it exposes what a fucking idiots you and your little queer buddy Klyver are?
> >>>> LMAO. Bwaaaa haaa haaaa.
> >>>>
> >>>> Oh. Still waiting for the FBI.
> >>>>
> >>>> What a fucking, vile JEW piece of shit MORON you are.
> >>>>
> >>>> DIE, you bastard!!!!!! You have caused me a lot of harm. Spitting at you whilst saying this.
> >>>>
> >>>> I am going to publish a lot more videos on you SHITHEAD!!!! BE prepared for it.
> >>>>
> >>>> I will RUIN you!!!!
> >>> There's nothing to ruin. Except for your fake math.
> [cut shit]
> > You found out how to post emojis?
> Trim the shit? ;^)

No. The idea is to make him scroll down and experience a little flavour of his behaviour.