On Tuesday, 4 May 2021 at 04:31:09 UTC-4, zelos...@gmail.com wrote:

> "...valid real numbers..."

The fact that you have to precede the object with the adjective "valid" is very telling! :-)

Come on Zelos, you are smarter than the rest of these morons and I want to keep beating your arse. Chuckle.

So, explain to me what multiplication of rational numbers means.... I know you love circularity, but try not to use multiplication before you define it and remember you were asked for a GENERAL definition. You up to it or still searching for answers? LMAO.

I refresh my explanation here to help you! :-)

All operations of arithmetic can be stated strictly in terms of subtraction.. Addition is derived immediately from subtraction.

Division:

A quotient ps/qr is the rational number that is measured in terms of two numbers p/q and r/s.

ps/qr = [ p/q +p/q + ...n times ] / [ r/s + r/s+... n times ] where n=q or n=s.

Example: 2/3 -:- 5/6 = 4/5

p/q = 2/3 and r/s = 5/6

2/3 -:- 5/6 = [ 2/3 + 2/3 + 2/3 ] / [ 5/6 + 5/6 + 5/6] = [2] / [ 5/2] = [2 + 2] / [5/2 + 5/2] = 4 / 5

After the second equals sign, cancellation is again possible by Book 5, Proposition 12.

So the ENTIRE operation is done through geometry using similar triangles. No numbers are even required in the geometric definition, only magnitudes.

Multiplication:

The product (or multiplication) of two numbers p/q and r/s, that is, pr/qs is the quotient of either number
with the reciprocal of the other.

pr/qs = [ p/q +p/q + ...n times ] / [ s/r + s/r+... n times ] where n=q or n=r.

Example: 2/3 x 5/6 = 5/9

p/q = 2/3 and s/r = 6/5

2/3 x 5/6 = [2/3+2/3+2/3] / [6/5+6/5+6/5] = [2] / [18/5] = [ 2+2+2+2+2 ] / [18/5+18/5+18/5+18/5+18/5] = 10 / 18 = 5/9

AND

pr/qs = [ r/s + r/s+... n times ] / [ s/p + s/p+... n times ] where n=p or n=s.

This part is left as an exercise.

NOTE: The above definitions work on ANY magnitude. There is no need to have a separate interpretation for integers and fractions.

All these details are explained in my world famous article:

https://drive.google.com/file/d/0B-mOEooW03iLYTg1TGY4RTIwakU

Compare this clarity with the bullshit of Zelos Malum:

------------------------------------------------------------------------
i. a/b=[(a,b)]
ii. [(a,b)][(c,d)]=[(ac,bd)]
iii. 2.1=21/10=[(21,10)]
iv. 2.2=22/10=[(22,10)]
v. 2.1*2.2=[(21,10)][(22,10)]=[(462,100)]=4.62
-------------------------------------------------------------------------

Step ii is obviously circular in that it assumes multiplication by rational numbers is already defined. The subsequent steps are wrong for too many reasons to address.

>The fact that you have to precede the object with the adjective "valid" is very telling! :-)

Yes, you say they are invalid, yet fail to provide anything that shows them as invalid. Which is an internal contradiction.

>Come on Zelos, you are smarter than the rest of these morons and I want to keep beating your arse. Chuckle.

You have yet to beat me in anything, especially mathematics.

>So, explain to me what multiplication of rational numbers means.... I know you love circularity, but try not to use multiplication before you define it and remember you were asked for a GENERAL definition. You up to it or still searching for answers? LMAO.

I already gave you it
https://proofwiki.org/wiki/Localization_of_Ring_Exists
Read up on it again in the generalized sense.

>I refresh my explanation here to help you! :-)
>Step ii is obviously circular in that it assumes multiplication by rational numbers is already defined. The subsequent steps are wrong for too many reasons to address.

Assuming multiplication on integer is not the same as assuming multiplication on rational numbers.

I did not assume multiplication of rational numbers, only integers.

On Tuesday, 18 May 2021 at 00:51:17 UTC-4, zelos...@gmail.com incorrigible crank wrote:

I had some spare time and decided to entertain myself with this link recommended by Zelos Malum:

> https://proofwiki.org/wiki/Localization_of_Ring_Exists

What's hilarious is how many of the "definitions" in these articles (written by morons of the BIG STUPID or mainstream academia) are taken for granted..

The above link begins with:

Let A be a commutative ring with unity.
If S is a proper **multiplicative** closed subset of A with no element 0, then there exists a localisation (A_s, i) of A at S.

--> Again, notice how telling and deceptive the phrase "with no element 0" is stated. If you had to ask your moronic mainstream lecturer of abstract algebra why 0 is omitted, his response would be "That's the definition, just use it". Well, actually the stupid reptile already knows he will be using fractions in his bullshit, so he wants to avoid "division by 0". Chuckle.

Spoof: The relation (a,s) ~ (b,t) implies atu = bsu where u is in S.

The "proof" or rather Spoof is already circular since it assumes that a/b = s/t where both are fractions. Stating a fact of geometry is NOT a proof.. Rather it is a GIVEN. That's all that happens in the above "localisation proof". A bunch of wankers deluxe! Chuckle.

Critique:

A commutative ring is one where the multiplication operation is commutative in abstract algebra, ie the order of the operands can be commuted (changed) and the result will be the same element. The phrase "with unity" means that 1 is included as an element of A, whatever the fuck "element" means. Mind you, whatever the fuck "set" and "subset" means. LMAO.

Allowing such mental masturbation, the Lemma that follows assumes multiplication has already been defined, but nowhere does this happen. Lemma 1 of this hilarious wiki article then states the operations of addition and multiplication with no further explanation, ie it simply ASSUMES the properties carried over from GEOMETRY.

The purpose of the proof was to show that a localisation exists, ie:

"The localization of a commutative ring R by a multiplicatively closed set S is a new ring S^{-1}R whose elements are fractions with numerators in R and denominators in S."

It's too comical that not only is the multiplication operation assumed, but much more such as fractions, their structure and properties as carried over from GEOMETRY.

This comment is posted in the interests of education and to expose mainstream ignorance and to warn students against the drivel of mainstream mythmatics.

I am the great John Gabriel, the genius who discovered the New Calculus.

Yes, I do know better! :)

On Tuesday, 18 May 2021 at 11:41:44 UTC-4, Eram semper recta wrote:
> On Tuesday, 18 May 2021 at 00:51:17 UTC-4, zelos...@gmail.com incorrigible crank wrote:
>
> I had some spare time and decided to entertain myself with this link recommended by Zelos Malum:
>
> > https://proofwiki.org/wiki/Localization_of_Ring_Exists
>
> What's hilarious is how many of the "definitions" in these articles (written by morons of the BIG STUPID or mainstream academia) are taken for granted.
>
> The above link begins with:
>
> Let A be a commutative ring with unity.
> If S is a proper **multiplicative** closed subset of A with no element 0, then there exists a localisation (A_s, i) of A at S.
>
> --> Again, notice how telling and deceptive the phrase "with no element 0" is stated. If you had to ask your moronic mainstream lecturer of abstract algebra why 0 is omitted, his response would be "That's the definition, just use it". Well, actually the stupid reptile already knows he will be using fractions in his bullshit, so he wants to avoid "division by 0". Chuckle.
>
> Spoof: The relation (a,s) ~ (b,t) implies atu = bsu where u is in S.
>
> The "proof" or rather Spoof is already circular since it assumes that a/b = s/t where both are fractions. Stating a fact of geometry is NOT a proof. Rather it is a GIVEN. That's all that happens in the above "localisation proof". A bunch of wankers deluxe! Chuckle.
>
> Critique:
>
> A commutative ring is one where the multiplication operation is commutative in abstract algebra, ie the order of the operands can be commuted (changed) and the result will be the same element. The phrase "with unity" means that 1 is included as an element of A, whatever the fuck "element" means. Mind you, whatever the fuck "set" and "subset" means. LMAO.
>
> Allowing such mental masturbation, the Lemma that follows assumes multiplication has already been defined, but nowhere does this happen. Lemma 1 of this hilarious wiki article then states the operations of addition and multiplication with no further explanation, ie it simply ASSUMES the properties carried over from GEOMETRY.
>
> The purpose of the proof was to show that a localisation exists, ie:
>
> "The localization of a commutative ring R by a multiplicatively closed set S is a new ring S^{-1}R whose elements are fractions with numerators in R and denominators in S."
>
> It's too comical that not only is the multiplication operation assumed, but much more such as fractions, their structure and properties as carried over from GEOMETRY.
>
> This comment is posted in the interests of education and to expose mainstream ignorance and to warn students against the drivel of mainstream mythmatics.
>
> I am the great John Gabriel, the genius who discovered the New Calculus.
>
> Yes, I do know better! :)

You see folks, mathematics IS the science of MEASURE and NUMBER.

Set theorists don't know or even understand the meaning of *MEASURE* which is how the Greeks realised NUMBER.

Abstract Algebra is NOT algebra at all, but a failed attempt to generalise algebra. Abstract algebra is for the most part NOT even to be called mathematics because it has ZERO to do with measure.

tisdag 18 maj 2021 kl. 17:47:13 UTC+2 skrev Eram semper recta:
> On Tuesday, 18 May 2021 at 11:41:44 UTC-4, Eram semper recta wrote:
> > On Tuesday, 18 May 2021 at 00:51:17 UTC-4, zelos...@gmail.com incorrigible crank wrote:
> >
> > I had some spare time and decided to entertain myself with this link recommended by Zelos Malum:
> >
> > > https://proofwiki.org/wiki/Localization_of_Ring_Exists
> >
> > What's hilarious is how many of the "definitions" in these articles (written by morons of the BIG STUPID or mainstream academia) are taken for granted.
> >
> > The above link begins with:
> >
> > Let A be a commutative ring with unity.
> > If S is a proper **multiplicative** closed subset of A with no element 0, then there exists a localisation (A_s, i) of A at S.
> >
> > --> Again, notice how telling and deceptive the phrase "with no element 0" is stated. If you had to ask your moronic mainstream lecturer of abstract algebra why 0 is omitted, his response would be "That's the definition, just use it". Well, actually the stupid reptile already knows he will be using fractions in his bullshit, so he wants to avoid "division by 0". Chuckle.
> >
> > Spoof: The relation (a,s) ~ (b,t) implies atu = bsu where u is in S.
> >
> > The "proof" or rather Spoof is already circular since it assumes that a/b = s/t where both are fractions. Stating a fact of geometry is NOT a proof. Rather it is a GIVEN. That's all that happens in the above "localisation proof". A bunch of wankers deluxe! Chuckle.
> >
> > Critique:
> >
> > A commutative ring is one where the multiplication operation is commutative in abstract algebra, ie the order of the operands can be commuted (changed) and the result will be the same element. The phrase "with unity" means that 1 is included as an element of A, whatever the fuck "element" means. Mind you, whatever the fuck "set" and "subset" means. LMAO.
> >
> > Allowing such mental masturbation, the Lemma that follows assumes multiplication has already been defined, but nowhere does this happen. Lemma 1 of this hilarious wiki article then states the operations of addition and multiplication with no further explanation, ie it simply ASSUMES the properties carried over from GEOMETRY.
> >
> > The purpose of the proof was to show that a localisation exists, ie:
> >
> > "The localization of a commutative ring R by a multiplicatively closed set S is a new ring S^{-1}R whose elements are fractions with numerators in R and denominators in S."
> >
> > It's too comical that not only is the multiplication operation assumed, but much more such as fractions, their structure and properties as carried over from GEOMETRY.
> >
> > This comment is posted in the interests of education and to expose mainstream ignorance and to warn students against the drivel of mainstream mythmatics.
> >
> > I am the great John Gabriel, the genius who discovered the New Calculus..
> >
> > Yes, I do know better! :)
> You see folks, mathematics IS the science of MEASURE and NUMBER.
>
> Set theorists don't know or even understand the meaning of *MEASURE* which is how the Greeks realised NUMBER.
>
> Abstract Algebra is NOT algebra at all, but a failed attempt to generalise algebra. Abstract algebra is for the most part NOT even to be called mathematics because it has ZERO to do with measure.

This is your personal opinion and nothing but it

>Again, notice how telling and deceptive the phrase "with no element 0" is stated. If you had to ask your moronic mainstream lecturer of abstract algebra why 0 is omitted, his response would be "That's the definition, just use it". Well, actually the stupid reptile already knows he will be using fractions in his bullshit, so he wants to avoid "division by 0". Chuckle.

The answer any of them would give hten is because you only get the trivial ring then which is not interesting.

>The "proof" or rather Spoof is already circular since it assumes that a/b = s/t where both are fractions. Stating a fact of geometry is NOT a proof.. Rather it is a GIVEN. That's all that happens in the above "localisation proof". A bunch of wankers deluxe! Chuckle.

If you read it, you owuld see it says define the equivalence relation, what it says is that the object a/b and s/t are equivalent if there exists a u such that etc.

There is no ratio to speak of there because it is not dealing with rational numbers, it is a generalization that uses the same notation. Like others you cannot see the difference between notation and objects.

>A commutative ring is one where the multiplication operation is commutative in abstract algebra, ie the order of the operands can be commuted (changed) and the result will be the same element. The phrase "with unity" means that 1 is included as an element of A, whatever the fuck "element" means. Mind you, whatever the fuck "set" and "subset" means. LMAO.

Everyone knows what those words mean, just because you're too stupid doesn't that mean it is a invalid.

>Allowing such mental masturbation, the Lemma that follows assumes multiplication has already been defined, but nowhere does this happen. Lemma 1 of this hilarious wiki article then states the operations of addition and multiplication with no further explanation, ie it simply ASSUMES the properties carried over from GEOMETRY.

It doesn't, they define the operations and show they are well-defined. Comon Gerbil, stop being this thick.

On Wednesday, 19 May 2021 at 01:39:13 UTC-4, zelos...@gmail.com wrote:
> >Again, notice how telling and deceptive the phrase "with no element 0" is stated. If you had to ask your moronic mainstream lecturer of abstract algebra why 0 is omitted, his response would be "That's the definition, just use it". Well, actually the stupid reptile already knows he will be using fractions in his bullshit, so he wants to avoid "division by 0". Chuckle.
> The answer any of them would give hten is because you only get the trivial ring then which is not interesting.

LMAO. You truly are a brainwashed bot. There is absolutely NO motivation to omit 0 unless they didn't know it was required for their deception.

Students: Be very careful of mainstream gibberish "this is not interesting" or "which is not interesting", etc. Any time a fucking moron mainstream academic doesn't understand shit, then of course it is "not interesting". Chuckle.

> >The "proof" or rather Spoof is already circular since it assumes that a/b = s/t where both are fractions. Stating a fact of geometry is NOT a proof. Rather it is a GIVEN. That's all that happens in the above "localisation proof". A bunch of wankers deluxe! Chuckle.

> If you read it, you owuld see it says define the equivalence relation, what it says is that the object a/b and s/t are equivalent if there exists a u such that etc.

The equivalence relation changes NOTHING, you fucking Swede IMBECILE!!! Just because it's an "equivalence relation" means the definition is sufficient? LMAO.

If it says "the object a/b and s/t are equivalent if there exists a u such that etc." then it says so because these details are KNOWN from GEOMETRY.

>
> There is no ratio to speak of

LMAO. Hmmm. p/q comes from the ratio p : q where both p and q have a common divisor.

> there because it is not dealing with rational numbers, it is a generalization that uses the same notation.

"...a generalisation", your arse!!! Not even close, you moron! Using the same notation is necessary because the notation is what it means.

> >A commutative ring is one where the multiplication operation is commutative in abstract algebra, ie the order of the operands can be commuted (changed) and the result will be the same element. The phrase "with unity" means that 1 is included as an element of A, whatever the fuck "element" means. Mind you, whatever the fuck "set" and "subset" means. LMAO.

> Everyone knows what those words mean, just because you're too stupid doesn't that mean it is a invalid.

I see. No response again. Tsk, tsk.

> >Allowing such mental masturbation, the Lemma that follows assumes multiplication has already been defined, but nowhere does this happen. Lemma 1 of this hilarious wiki article then states the operations of addition and multiplication with no further explanation, ie it simply ASSUMES the properties carried over from GEOMETRY.

> It doesn't, they define the operations and show they are well-defined.

If wishes were horses...

> Comon Gerbil, stop ...

You don't like it eh? Get used to it crank! You need to be called out for your ignorance, stupidity, dishonesty, arrogance and incompetence.

On Wednesday, 19 May 2021 at 01:39:13 UTC-4, zelos...@gmail.com wrote:

> >The "proof" or rather Spoof is already circular since it assumes that a/b = s/t where both are fractions. Stating a fact of geometry is NOT a proof. Rather it is a GIVEN. That's all that happens in the above "localisation proof". A bunch of wankers deluxe! Chuckle.
> If you read it, you owuld see it says define the equivalence relation, what it says is that the object a/b and s/t are equivalent if there exists a u such that etc.

People, Malum is referring to this:

"The relation (a,s) ~ (b,t) implies atu = bsu where u is in S."

In NO WAY does (a,s) ~ (b,t) imply atu = bsu where u is in S.

An equivalence relation is one between elements of a set that is reflexive, symmetric, and transitive.

Example: The relation "is equal to" denoted by "=" is an equivalence relation on objects that are the same in every possible way. If x, y and z are objects, then

1. x = x means the relation is reflexive. This is incredibly funny because an object is always equal to itself, except perhaps in the bullshit of set theory? LMAO.

2. If x = y then y = x means the relation is symmetrical. A pretty obvious statement, but again in the bullshit of set theory you have equivalence relations on subset operators. Too funny.

3. (Transitivity) if x = y and y = z then x = z This too is hilarious because it was known by Euclid over 2500 years ago:

Common notion 1: Things that are equal to the same thing are equal to each other.

In Malum's asserted "proof", there is NO knowledge of what atu means - the multiplication is ASSUMED.

Chuckle. However, mainstream morons imagine that they are somehow intelligent and feel warm and fuzzy memorising all this WORTHLESS bullshit.

On Wednesday, 19 May 2021 at 09:29:05 UTC-4, Eram semper recta wrote:
> On Wednesday, 19 May 2021 at 01:39:13 UTC-4, zelos...@gmail.com wrote:
> > >The "proof" or rather Spoof is already circular since it assumes that a/b = s/t where both are fractions. Stating a fact of geometry is NOT a proof. Rather it is a GIVEN. That's all that happens in the above "localisation proof". A bunch of wankers deluxe! Chuckle.
> > If you read it, you owuld see it says define the equivalence relation, what it says is that the object a/b and s/t are equivalent if there exists a u such that etc.
> People, Malum is referring to this:
> "The relation (a,s) ~ (b,t) implies atu = bsu where u is in S."
> In NO WAY does (a,s) ~ (b,t) imply atu = bsu where u is in S.
>
> An equivalence relation is one between elements of a set that is reflexive, symmetric, and transitive.
>
> Example: The relation "is equal to" denoted by "=" is an equivalence relation on objects that are the same in every possible way. If x, y and z are objects, then
>
> 1. x = x means the relation is reflexive. This is incredibly funny because an object is always equal to itself, except perhaps in the bullshit of set theory? LMAO.
>
> 2. If x = y then y = x means the relation is symmetrical. A pretty obvious statement, but again in the bullshit of set theory you have equivalence relations on subset operators. Too funny.
>
> 3. (Transitivity) if x = y and y = z then x = z This too is hilarious because it was known by Euclid over 2500 years ago:
>
> Common notion 1: Things that are equal to the same thing are equal to each other.
>
> In Malum's asserted "proof", there is NO knowledge of what atu means - the multiplication is ASSUMED.
>
> Chuckle. However, mainstream morons imagine that they are somehow intelligent and feel warm and fuzzy memorising all this WORTHLESS bullshit.

In 1986 (almost a decade after I left high school), I attended a course on Abstract Algebra and I have to say that it was mostly time wasted. There were a few useful theorems about vector spaces and linear algebra, but other than this, the course was absolute bullshit.

Students, do not waste your time! Take a calculus course and an advanced calculus course, but do not become too frustrated by other garbage such as real ANALysis, metric spaces (most laughable!) and topology.

On Wednesday, 19 May 2021 at 09:29:05 UTC-4, Eram semper recta wrote:
> On Wednesday, 19 May 2021 at 01:39:13 UTC-4, zelos...@gmail.com wrote:
> > >The "proof" or rather Spoof is already circular since it assumes that a/b = s/t where both are fractions. Stating a fact of geometry is NOT a proof. Rather it is a GIVEN. That's all that happens in the above "localisation proof". A bunch of wankers deluxe! Chuckle.
> > If you read it, you owuld see it says define the equivalence relation, what it says is that the object a/b and s/t are equivalent if there exists a u such that etc.
> People, Malum is referring to this:
> "The relation (a,s) ~ (b,t) implies atu = bsu where u is in S."
> In NO WAY does (a,s) ~ (b,t) imply atu = bsu where u is in S.

For starters, the pairs (a,s) and (b,t) can be related in so many different ways of which none have anything to do with the proportional equality, that is, a/b = s/t. It doesn’t get more crankier than this folks! But the art of hand waving is what you might call an art which the morons of mainstream academia have mastered. Simply clothe all the bullshit under the title "localisation"; add some cloaked information about removing the element zero and if queried, respond by saying "It's not interesting with zero"; assume the properties of ratio and proportion which the scumbags inherited from my brilliant ancestors (and might I add never paid any royalties or respected their intellectual rights); and on and on it goes ad nauseam.

You can't make this stuff up folks! LMAO.

>
> An equivalence relation is one between elements of a set that is reflexive, symmetric, and transitive.
>
> Example: The relation "is equal to" denoted by "=" is an equivalence relation on objects that are the same in every possible way. If x, y and z are objects, then
>
> 1. x = x means the relation is reflexive. This is incredibly funny because an object is always equal to itself, except perhaps in the bullshit of set theory? LMAO.
>
> 2. If x = y then y = x means the relation is symmetrical. A pretty obvious statement, but again in the bullshit of set theory you have equivalence relations on subset operators. Too funny.
>
> 3. (Transitivity) if x = y and y = z then x = z This too is hilarious because it was known by Euclid over 2500 years ago:
>
> Common notion 1: Things that are equal to the same thing are equal to each other.
>
> In Malum's asserted "proof", there is NO knowledge of what atu means - the multiplication is ASSUMED.
>
> Chuckle. However, mainstream morons imagine that they are somehow intelligent and feel warm and fuzzy memorising all this WORTHLESS bullshit.

On Wednesday, May 19, 2021 at 9:34:50 AM UTC-4, Eram semper recta wrote:
> On Wednesday, 19 May 2021 at 09:29:05 UTC-4, Eram semper recta wrote:
> > On Wednesday, 19 May 2021 at 01:39:13 UTC-4, zelos...@gmail.com wrote:
> > > >The "proof" or rather Spoof is already circular since it assumes that a/b = s/t where both are fractions. Stating a fact of geometry is NOT a proof. Rather it is a GIVEN. That's all that happens in the above "localisation proof". A bunch of wankers deluxe! Chuckle.
> > > If you read it, you owuld see it says define the equivalence relation, what it says is that the object a/b and s/t are equivalent if there exists a u such that etc.
> > People, Malum is referring to this:
> > "The relation (a,s) ~ (b,t) implies atu = bsu where u is in S."
> > In NO WAY does (a,s) ~ (b,t) imply atu = bsu where u is in S.
> >
> > An equivalence relation is one between elements of a set that is reflexive, symmetric, and transitive.
> >
> > Example: The relation "is equal to" denoted by "=" is an equivalence relation on objects that are the same in every possible way. If x, y and z are objects, then
> >
> > 1. x = x means the relation is reflexive. This is incredibly funny because an object is always equal to itself, except perhaps in the bullshit of set theory? LMAO.
> >
> > 2. If x = y then y = x means the relation is symmetrical. A pretty obvious statement, but again in the bullshit of set theory you have equivalence relations on subset operators. Too funny.
> >
> > 3. (Transitivity) if x = y and y = z then x = z This too is hilarious because it was known by Euclid over 2500 years ago:
> >
> > Common notion 1: Things that are equal to the same thing are equal to each other.
> >
> > In Malum's asserted "proof", there is NO knowledge of what atu means - the multiplication is ASSUMED.
> >
> > Chuckle. However, mainstream morons imagine that they are somehow intelligent and feel warm and fuzzy memorising all this WORTHLESS bullshit.
> In 1986 (almost a decade after I left high school), I attended a course on Abstract Algebra and I have to say that it was mostly time wasted. There were a few useful theorems about vector spaces and linear algebra, but other than this, the course was absolute bullshit.
>
> Students, do not waste your time! Take a calculus course and an advanced calculus course, but do not become too frustrated by other garbage such as real ANALysis, metric spaces (most laughable!) and topology.

I took topology and walked away from it feeling dubious, but i've never gone back on the attack. It doesn't seem to get much shrift here on usenet either. Instantiation is I would think where topology falls on its own sword. Lots of theoretical verbiage and very little instantiation. Just keep throwing another layer atop another layer until you have reached hazy abstraction. Then comes an instance and the pile collapses. So what was the buildup about? Challenging students not to think simply, but to follow the leader. Perhaps I was merely unadaptable just as many here cannot adapt to polysign. As our mimicry encounters previous mimicry it ought to be cause to start again from scratch. If set theory is so weak then everything built atop it must be as weak. How much is merely an attempt at getting to the root versus achieving it? Afaict we are all struggling in the progression. Digression is not a problem though I used to think it had a negative connotation. Interpretation is roughly what we do here. When mathematics attempts a catalog style dismantling of individual works and recompilation/generalization is the result any good? No. Go back to the masters and their original works and you will see a stark difference as one man speaks his own words and motives.

On Wednesday, 19 May 2021 at 11:43:20 UTC-4, timba...@gmail.com wrote:
> On Wednesday, May 19, 2021 at 9:34:50 AM UTC-4, Eram semper recta wrote:
> > On Wednesday, 19 May 2021 at 09:29:05 UTC-4, Eram semper recta wrote:
> > > On Wednesday, 19 May 2021 at 01:39:13 UTC-4, zelos...@gmail.com wrote:
> > > > >The "proof" or rather Spoof is already circular since it assumes that a/b = s/t where both are fractions. Stating a fact of geometry is NOT a proof. Rather it is a GIVEN. That's all that happens in the above "localisation proof". A bunch of wankers deluxe! Chuckle.
> > > > If you read it, you owuld see it says define the equivalence relation, what it says is that the object a/b and s/t are equivalent if there exists a u such that etc.
> > > People, Malum is referring to this:
> > > "The relation (a,s) ~ (b,t) implies atu = bsu where u is in S."
> > > In NO WAY does (a,s) ~ (b,t) imply atu = bsu where u is in S.
> > >
> > > An equivalence relation is one between elements of a set that is reflexive, symmetric, and transitive.
> > >
> > > Example: The relation "is equal to" denoted by "=" is an equivalence relation on objects that are the same in every possible way. If x, y and z are objects, then
> > >
> > > 1. x = x means the relation is reflexive. This is incredibly funny because an object is always equal to itself, except perhaps in the bullshit of set theory? LMAO.
> > >
> > > 2. If x = y then y = x means the relation is symmetrical. A pretty obvious statement, but again in the bullshit of set theory you have equivalence relations on subset operators. Too funny.
> > >
> > > 3. (Transitivity) if x = y and y = z then x = z This too is hilarious because it was known by Euclid over 2500 years ago:
> > >
> > > Common notion 1: Things that are equal to the same thing are equal to each other.
> > >
> > > In Malum's asserted "proof", there is NO knowledge of what atu means - the multiplication is ASSUMED.
> > >
> > > Chuckle. However, mainstream morons imagine that they are somehow intelligent and feel warm and fuzzy memorising all this WORTHLESS bullshit.
> > In 1986 (almost a decade after I left high school), I attended a course on Abstract Algebra and I have to say that it was mostly time wasted. There were a few useful theorems about vector spaces and linear algebra, but other than this, the course was absolute bullshit.
> >
> > Students, do not waste your time! Take a calculus course and an advanced calculus course, but do not become too frustrated by other garbage such as real ANALysis, metric spaces (most laughable!) and topology.
> I took topology and walked away from it feeling dubious, but i've never gone back on the attack. It doesn't seem to get much shrift here on usenet either. Instantiation is I would think where topology falls on its own sword.. Lots of theoretical verbiage and very little instantiation. Just keep throwing another layer atop another layer until you have reached hazy abstraction. Then comes an instance and the pile collapses. So what was the buildup about? Challenging students not to think simply, but to follow the leader. Perhaps I was merely unadaptable just as many here cannot adapt to polysign. As our mimicry encounters previous mimicry it ought to be cause to start again from scratch. If set theory is so weak then everything built atop it must be as weak. How much is merely an attempt at getting to the root versus achieving it? Afaict we are all struggling in the progression. Digression is not a problem though I used to think it had a negative connotation. Interpretation is roughly what we do here. When mathematics attempts a catalog style dismantling of individual works and recompilation/generalization is the result any good? No. Go back to the masters and their original works and you will see a stark difference as one man speaks his own words and motives.

We cannot throw our hands in the air and simply give up. The future of mathematics is at stake. The worthless courses usually offered as part of a math degree have nothing to do with mathematics. Of all fields of study, more students drop out of math courses because they quickly lose interest and enthusiasm. Such a shame because we need STEM professionals today more than ever. No student wants to spend years learning worthless anti-mathematical bullshit. They are only interested in well-formed concepts that can be applied usefully in situations beneficial to humanity.

No one gives a shit what set theorist or topologist thinks. They are largely irrelevant. We need to remove bullshit from mathematics, to correct the wrong concepts and to teach the New Calculus.

On Wednesday, May 19, 2021 at 7:12:22 PM UTC+1, Eram semper recta wrote:

"No one gives a shit what set theorist or topologist thinks. They are largely irrelevant. We need to remove bullshit from mathematics, to correct the wrong concepts and to teach the New Calculus."

Oh dear, we can only hope that you one day grow out of this Bottom Kitten 115 mindset Sir, it is very disappointing to keep seeing you graffiti these kinds of amateurish blunders on here. Topics like Real Analysis and Topology are useful tools in chaos theory, the study of differential equations and elsewhere. Your 'new calculus' has been read by qualified commentators, such as myself and those far above me in the mathematical hierarchy on here and it has been found not to accomplish what you wish it to. Nice try (well not really) but no cigar.

Thank goodness the mathematical community went down the path laid down for us by geniuses such as Cantor, Weierstrass and Riemann, rather than getting bogged down forever in Euclidean dogma and the pre-limit methods of earlier centuries.

Students would be well advised to seek the guidance of someone far more qualified than yourself on matters like this. You would do well to be working on understanding undergraduate mathematics better.

Still there is hope. See my recent post "Can those with wildly wrong anti-mainstream opinions be cured and go on to better things" for a vision of how things could be, if only enough 'free spirits' wanted it badly enough.

Kind Regards

>In NO WAY does (a,s) ~ (b,t) imply atu = bsu where u is in S.

What they meant is that those statements are DEFINED to be equivalent!

>An equivalence relation is one between elements of a set that is reflexive, symmetric, and transitive.

Which they prove HERE
https://proofwiki.org/wiki/Localization_of_Ring_Exists/Lemma_1

>1. x = x means the relation is reflexive. This is incredibly funny because an object is always equal to itself, except perhaps in the bullshit of set theory? LMAO

But not all relations have that, x < x does not hold, x~x does here.

>2. If x = y then y = x means the relation is symmetrical. A pretty obvious statement, but again in the bullshit of set theory you have equivalence relations on subset operators. Too funny

but does not hold in all relations x<y does not mean y<x for example.

>. (Transitivity) if x = y and y = z then x = z This too is hilarious because it was known by Euclid over 2500 years ago

Again, dipshit, does not hold in all relations so it must be stated.

>In Malum's asserted "proof", there is NO knowledge of what atu means - the multiplication is ASSUMED.

Are you really this stupid? This is Localization OF A RING! It means we have a ring already where multiplication is defined! It is not assumed, it is already provided you moron!

On Thursday, 20 May 2021 at 01:43:08 UTC-4, zelos...@gmail.com wrote:
> >In NO WAY does (a,s) ~ (b,t) imply atu = bsu where u is in S.
> What they meant is that those statements are DEFINED to be equivalent!

Man, you get dumber by the minute. That is not at all the case because that statement is part of your spoof which I see you modified since yesterday in the Wiki. Alas, it is still delusional as ever and wrong as ever.

You cannot define an equivalence relation as "Equivalent" to another given equivalence, because if you do this, then by implication you ACCEPT all the properties of that relation. YOU UTTER IMBECILE! Chuckle.

LMAO.

> >An equivalence relation is one between elements of a set that is reflexive, symmetric, and transitive.
> Which they prove HERE

YOU do NO such thing. I can tell that Wiki has your signature on it - clearly the work of an idiot.

> https://proofwiki.org/wiki/Localization_of_Ring_Exists/Lemma_1
> >1. x = x means the relation is reflexive. This is incredibly funny because an object is always equal to itself, except perhaps in the bullshit of set theory? LMAO
> But not all relations have that, x < x does not hold, x~x does here.
> >2. If x = y then y = x means the relation is symmetrical. A pretty obvious statement, but again in the bullshit of set theory you have equivalence relations on subset operators. Too funny
> but does not hold in all relations x<y does not mean y<x for example.
> >. (Transitivity) if x = y and y = z then x = z This too is hilarious because it was known by Euclid over 2500 years ago
> Again, dipshit, does not hold in all relations so it must be stated.

It holds in ALL mathematics. I never talk about your BULLSHIT, only mathematics. LMAO again.

> >In Malum's asserted "proof", there is NO knowledge of what atu means - the multiplication is ASSUMED.
> Are you really this stupid? This is Localization OF A RING! It means we have a ring already where multiplication is defined!

You have ringworm in your brain, idiot! You have been shown that "localisation of a ring" does not provide any description of multiplication, that it merely ASSUMES these properties from GEOMETRY. The stupid moron is once again YOU!

> It is not assumed, it is already provided you moron!

If wishes were horses, ...

Chuckle.

On Thursday, May 20, 2021 at 1:43:08 AM UTC-4, zelos...@gmail.com wrote:
> >In NO WAY does (a,s) ~ (b,t) imply atu = bsu where u is in S.
> What they meant is that those statements are DEFINED to be equivalent!
> >An equivalence relation is one between elements of a set that is reflexive, symmetric, and transitive.
> Which they prove HERE
> https://proofwiki.org/wiki/Localization_of_Ring_Exists/Lemma_1
> >1. x = x means the relation is reflexive. This is incredibly funny because an object is always equal to itself, except perhaps in the bullshit of set theory? LMAO
> But not all relations have that, x < x does not hold, x~x does here.
> >2. If x = y then y = x means the relation is symmetrical. A pretty obvious statement, but again in the bullshit of set theory you have equivalence relations on subset operators. Too funny
> but does not hold in all relations x<y does not mean y<x for example.
> >. (Transitivity) if x = y and y = z then x = z This too is hilarious because it was known by Euclid over 2500 years ago
> Again, dipshit, does not hold in all relations so it must be stated.
> >In Malum's asserted "proof", there is NO knowledge of what atu means - the multiplication is ASSUMED.
> Are you really this stupid? This is Localization OF A RING! It means we have a ring already where multiplication is defined! It is not assumed, it is already provided you moron!

I do see localization as more verbiage accumulation.
Already the numerical form of large numbers is established via the modulo form and radix representation.
That this same mechanisms develops smaller numbers implies that this information could have been reused.
That these modulo forms are rationals goes ignored by this third class mathematics which would prefer the option to balloon, thus making more room for more publication by more PhDs. They must leave ordinary numbers behind why? Back two sentences and never halt; just return.

On Thursday, 20 May 2021 at 09:08:37 UTC-4, timba...@gmail.com wrote:
> On Thursday, May 20, 2021 at 1:43:08 AM UTC-4, zelos...@gmail.com wrote:
> > >In NO WAY does (a,s) ~ (b,t) imply atu = bsu where u is in S.
> > What they meant is that those statements are DEFINED to be equivalent!
> > >An equivalence relation is one between elements of a set that is reflexive, symmetric, and transitive.
> > Which they prove HERE
> > https://proofwiki.org/wiki/Localization_of_Ring_Exists/Lemma_1
> > >1. x = x means the relation is reflexive. This is incredibly funny because an object is always equal to itself, except perhaps in the bullshit of set theory? LMAO
> > But not all relations have that, x < x does not hold, x~x does here.
> > >2. If x = y then y = x means the relation is symmetrical. A pretty obvious statement, but again in the bullshit of set theory you have equivalence relations on subset operators. Too funny
> > but does not hold in all relations x<y does not mean y<x for example.
> > >. (Transitivity) if x = y and y = z then x = z This too is hilarious because it was known by Euclid over 2500 years ago
> > Again, dipshit, does not hold in all relations so it must be stated.
> > >In Malum's asserted "proof", there is NO knowledge of what atu means - the multiplication is ASSUMED.
> > Are you really this stupid? This is Localization OF A RING! It means we have a ring already where multiplication is defined! It is not assumed, it is already provided you moron!
> I do see localization as more verbiage accumulation.

No. Not just verbiage accumulation but ASSUMPTION of ideas from GEOMETRY. Only an ignoramus would say something like:

"I define p & q to be (p)(q) or pq" thus assuming the prior definition of pq.

> Already the numerical form of large numbers is established via the modulo form and radix representation.

Irrelevant.

> That this same mechanisms develops smaller numbers implies that this information could have been reused.
> That these modulo forms are rationals goes ignored by this third class mathematics which would prefer the option to balloon, thus making more room for more publication by more PhDs.

Chuckle. It's easy to add shit to a heap that is already full of shit.

> They must leave ordinary numbers behind why? Back two sentences and never halt; just return.

On Thursday, May 20, 2021 at 9:52:18 AM UTC-4, Eram semper recta wrote:
> On Thursday, 20 May 2021 at 09:08:37 UTC-4, timba...@gmail.com wrote:
> > On Thursday, May 20, 2021 at 1:43:08 AM UTC-4, zelos...@gmail.com wrote:
> > > >In NO WAY does (a,s) ~ (b,t) imply atu = bsu where u is in S.
> > > What they meant is that those statements are DEFINED to be equivalent!
> > > >An equivalence relation is one between elements of a set that is reflexive, symmetric, and transitive.
> > > Which they prove HERE
> > > https://proofwiki.org/wiki/Localization_of_Ring_Exists/Lemma_1
> > > >1. x = x means the relation is reflexive. This is incredibly funny because an object is always equal to itself, except perhaps in the bullshit of set theory? LMAO
> > > But not all relations have that, x < x does not hold, x~x does here.
> > > >2. If x = y then y = x means the relation is symmetrical. A pretty obvious statement, but again in the bullshit of set theory you have equivalence relations on subset operators. Too funny
> > > but does not hold in all relations x<y does not mean y<x for example.
> > > >. (Transitivity) if x = y and y = z then x = z This too is hilarious because it was known by Euclid over 2500 years ago
> > > Again, dipshit, does not hold in all relations so it must be stated.
> > > >In Malum's asserted "proof", there is NO knowledge of what atu means - the multiplication is ASSUMED.
> > > Are you really this stupid? This is Localization OF A RING! It means we have a ring already where multiplication is defined! It is not assumed, it is already provided you moron!
> > I do see localization as more verbiage accumulation.
> No. Not just verbiage accumulation but ASSUMPTION of ideas from GEOMETRY. Only an ignoramus would say something like:
>
> "I define p & q to be (p)(q) or pq" thus assuming the prior definition of pq.

The product does not bear out through pure Euclidean geometry. Without the specification of unity the product cannot be generated. That geometers will have to assign a unit length to arrive with arithmetic product, whereas no such need arises in performing sums; now we have a powerful chasm between sum and product; something that the ring definition denies.

In effect geometers are forced to plug numbers into their purely numberless system. Really then doesn't this pose another product? It is as I wrote elsewhere a base product, where the base is arbitrarily chosen. I suppose this means that in a base b system
a b = a
b b = b
and then the puzzle becomes how to resolve
ac [base b]
but to assign b as unity is to stuff numbers into geometry, no? Weren't numbers and geometry supposed to operate in parallel without this need? This base system would do that. It is not so inappropriate to write:
a [b] c
in that the b is like a pivot of a scale that would balance when a is small and c is large relative to b.

An area analysis will have to select a unit area as well, so is the solution to specify unity or is it to confess that there is a base lurking?

When physics begets forces they are product relations and are inversely related to distance. This places nearby objects as more than far away objects, so to what degree do we have distance backwards? upside down? The influence of distant objects is naught. The influence of adjacent objects is large. Clearly the product as we study it means very little physically speaking. Whereas the sum retains perfect correspondence vectorially, superpositionally, etc.
> > Already the numerical form of large numbers is established via the modulo form and radix representation.
> Irrelevant.

So here you are in line with the status quo. That already as they work in N they can instantiate a large value such as
1001
we see a mechanism that can get us to the other side of unity; a mechanism already taught to school children who have no need of all this higher 'purity'; a place where PhDs get their publications made thus guaranteeing an accumulated system.
No. The next phase will be a paring down. Eventually there will be a generation of PhDs who ground the pile and witness their own interpretation of mathematics in their own terms.

No matter what base of counting we are in a few more digits go a long way better, and epsilon/delta is as close as anybody ever got to the continuum. Close enough is good enough. Instantiate anybody's workings. Don't they all boil out to the same thing? The rest is sheer verbosity.

> > That this same mechanisms develops smaller numbers implies that this information could have been reused.
> > That these modulo forms are rationals goes ignored by this third class mathematics which would prefer the option to balloon, thus making more room for more publication by more PhDs.
> Chuckle. It's easy to add shit to a heap that is already full of shit.
> > They must leave ordinary numbers behind why? Back two sentences and never halt; just return.

On Friday, 21 May 2021 at 07:06:01 UTC-4, timba...@gmail.com wrote:
> On Thursday, May 20, 2021 at 9:52:18 AM UTC-4, Eram semper recta wrote:
> > On Thursday, 20 May 2021 at 09:08:37 UTC-4, timba...@gmail.com wrote:
> > > On Thursday, May 20, 2021 at 1:43:08 AM UTC-4, zelos...@gmail.com wrote:
> > > > >In NO WAY does (a,s) ~ (b,t) imply atu = bsu where u is in S.
> > > > What they meant is that those statements are DEFINED to be equivalent!
> > > > >An equivalence relation is one between elements of a set that is reflexive, symmetric, and transitive.
> > > > Which they prove HERE
> > > > https://proofwiki.org/wiki/Localization_of_Ring_Exists/Lemma_1
> > > > >1. x = x means the relation is reflexive. This is incredibly funny because an object is always equal to itself, except perhaps in the bullshit of set theory? LMAO
> > > > But not all relations have that, x < x does not hold, x~x does here..
> > > > >2. If x = y then y = x means the relation is symmetrical. A pretty obvious statement, but again in the bullshit of set theory you have equivalence relations on subset operators. Too funny
> > > > but does not hold in all relations x<y does not mean y<x for example.
> > > > >. (Transitivity) if x = y and y = z then x = z This too is hilarious because it was known by Euclid over 2500 years ago
> > > > Again, dipshit, does not hold in all relations so it must be stated..
> > > > >In Malum's asserted "proof", there is NO knowledge of what atu means - the multiplication is ASSUMED.
> > > > Are you really this stupid? This is Localization OF A RING! It means we have a ring already where multiplication is defined! It is not assumed, it is already provided you moron!
> > > I do see localization as more verbiage accumulation.
> > No. Not just verbiage accumulation but ASSUMPTION of ideas from GEOMETRY. Only an ignoramus would say something like:
> >
> > "I define p & q to be (p)(q) or pq" thus assuming the prior definition of pq.

> The product does not bear out through pure Euclidean geometry.

Nonsense. It does! I gave you the example of the angles subtended on the same circle arc from which similar triangles are constructed and both the quotient and product are then beautifully defined through use of Book 5, Prop.. 12.

> Without the specification of unity the product cannot be generated.

Again, you display your ignorance and lack of education. The very idea of unity is the reason Euclid covered the concepts in Books 1 - 6 before introducing the abstract unit in book 7, itself a concept derived from a ratio of equal magnitudes uses as a standard measure.

> That geometers will have to assign a unit length to arrive with arithmetic product, whereas no such need arises in performing sums;

That is what I explained so many times in several threads on this newsgroup! Algebra is a weaker form of geometry where the abstract unit whose size is irrelevant is used with the same geometric properties (proportion, ratio, etc) to perform arithmetic.

> now we have a powerful chasm between sum and product; something that the ring definition denies.

Of course. Nothing Malum writes ever makes any sense. He simply parrots what his idiot lecturers drum into his brain.

>
> In effect geometers are forced to plug numbers into their purely numberless system.

Rubbish. Geometers are never forces to do any such thing. Algebra was invented by the Ancient Greeks so that they could talk about measure without referring to the fundamental concepts of geometry.

What follows is indicative of your poor understanding of geometry, number and mathematics. Sorry, I do not have time to address all your gibberish. To converse on my level, you first need to educate yourself. I have given you links to sufficient articles. Study them!

> Really then doesn't this pose another product? It is as I wrote elsewhere a base product, where the base is arbitrarily chosen. I suppose this means that in a base b system
> a b = a
> b b = b
> and then the puzzle becomes how to resolve
> ac [base b]
> but to assign b as unity is to stuff numbers into geometry, no? Weren't numbers and geometry supposed to operate in parallel without this need? This base system would do that. It is not so inappropriate to write:
> a [b] c
> in that the b is like a pivot of a scale that would balance when a is small and c is large relative to b.
>
> An area analysis will have to select a unit area as well, so is the solution to specify unity or is it to confess that there is a base lurking?
>
> When physics begets forces they are product relations and are inversely related to distance. This places nearby objects as more than far away objects, so to what degree do we have distance backwards? upside down? The influence of distant objects is naught. The influence of adjacent objects is large. Clearly the product as we study it means very little physically speaking.. Whereas the sum retains perfect correspondence vectorially, superpositionally, etc.
> > > Already the numerical form of large numbers is established via the modulo form and radix representation.
> > Irrelevant.
> So here you are in line with the status quo. That already as they work in N they can instantiate a large value such as
> 1001
> we see a mechanism that can get us to the other side of unity; a mechanism already taught to school children who have no need of all this higher 'purity'; a place where PhDs get their publications made thus guaranteeing an accumulated system.
> No. The next phase will be a paring down. Eventually there will be a generation of PhDs who ground the pile and witness their own interpretation of mathematics in their own terms.
>
> No matter what base of counting we are in a few more digits go a long way better, and epsilon/delta is as close as anybody ever got to the continuum.. Close enough is good enough. Instantiate anybody's workings. Don't they all boil out to the same thing? The rest is sheer verbosity.
> > > That this same mechanisms develops smaller numbers implies that this information could have been reused.
> > > That these modulo forms are rationals goes ignored by this third class mathematics which would prefer the option to balloon, thus making more room for more publication by more PhDs.
> > Chuckle. It's easy to add shit to a heap that is already full of shit.
> > > They must leave ordinary numbers behind why? Back two sentences and never halt; just return.

On 5/21/2021 6:05 AM, Timothy Golden wrote:
> On Thursday, May 20, 2021 at 9:52:18 AM UTC-4, Eram semper recta wrote:
>> On Thursday, 20 May 2021 at 09:08:37 UTC-4, timba...@gmail.com wrote:
>>> On Thursday, May 20, 2021 at 1:43:08 AM UTC-4, zelos...@gmail.com wrote:
>>>>> In NO WAY does (a,s) ~ (b,t) imply atu = bsu where u is in S.
>>>> What they meant is that those statements are DEFINED to be equivalent!
>>>>> An equivalence relation is one between elements of a set that is reflexive, symmetric, and transitive.
>>>> Which they prove HERE
>>>> https://proofwiki.org/wiki/Localization_of_Ring_Exists/Lemma_1
>>>>> 1. x = x means the relation is reflexive. This is incredibly funny because an object is always equal to itself, except perhaps in the bullshit of set theory? LMAO
>>>> But not all relations have that, x < x does not hold, x~x does here.
>>>>> 2. If x = y then y = x means the relation is symmetrical. A pretty obvious statement, but again in the bullshit of set theory you have equivalence relations on subset operators. Too funny
>>>> but does not hold in all relations x<y does not mean y<x for example.
>>>>> . (Transitivity) if x = y and y = z then x = z This too is hilarious because it was known by Euclid over 2500 years ago
>>>> Again, dipshit, does not hold in all relations so it must be stated.
>>>>> In Malum's asserted "proof", there is NO knowledge of what atu means - the multiplication is ASSUMED.
>>>> Are you really this stupid? This is Localization OF A RING! It means we have a ring already where multiplication is defined! It is not assumed, it is already provided you moron!
>>> I do see localization as more verbiage accumulation.
>> No. Not just verbiage accumulation but ASSUMPTION of ideas from GEOMETRY. Only an ignoramus would say something like:
>>
>> "I define p & q to be (p)(q) or pq" thus assuming the prior definition of pq.
>

> The <snip mumbo jumbo crap>

On Monday, 24 May 2021 at 12:22:26 UTC-4, Meritocracy wrote:
> On 5/21/2021 6:05 AM, Timothy Golden wrote:
> > On Thursday, May 20, 2021 at 9:52:18 AM UTC-4, Eram semper recta wrote:
> >> On Thursday, 20 May 2021 at 09:08:37 UTC-4, timba...@gmail.com wrote:
> >>> On Thursday, May 20, 2021 at 1:43:08 AM UTC-4, zelos...@gmail.com wrote:
> >>>>> In NO WAY does (a,s) ~ (b,t) imply atu = bsu where u is in S.
> >>>> What they meant is that those statements are DEFINED to be equivalent!
> >>>>> An equivalence relation is one between elements of a set that is reflexive, symmetric, and transitive.
> >>>> Which they prove HERE
> >>>> https://proofwiki.org/wiki/Localization_of_Ring_Exists/Lemma_1
> >>>>> 1. x = x means the relation is reflexive. This is incredibly funny because an object is always equal to itself, except perhaps in the bullshit of set theory? LMAO
> >>>> But not all relations have that, x < x does not hold, x~x does here.
> >>>>> 2. If x = y then y = x means the relation is symmetrical. A pretty obvious statement, but again in the bullshit of set theory you have equivalence relations on subset operators. Too funny
> >>>> but does not hold in all relations x<y does not mean y<x for example.
> >>>>> . (Transitivity) if x = y and y = z then x = z This too is hilarious because it was known by Euclid over 2500 years ago
> >>>> Again, dipshit, does not hold in all relations so it must be stated.
> >>>>> In Malum's asserted "proof", there is NO knowledge of what atu means - the multiplication is ASSUMED.
> >>>> Are you really this stupid? This is Localization OF A RING! It means we have a ring already where multiplication is defined! It is not assumed, it is already provided you moron!
> >>> I do see localization as more verbiage accumulation.
> >> No. Not just verbiage accumulation but ASSUMPTION of ideas from GEOMETRY. Only an ignoramus would say something like:
> >>
> >> "I define p & q to be (p)(q) or pq" thus assuming the prior definition of pq.

The above line is precisely what YOU claim Zelos Malum. LMAO.
You will always be a moron.