Better use exp(ix)= cos(x)+ i sin(x) and then me that 1=1.

Juan Weisz

Maybe ,but My objective was to use vector calculus if I understand correctly F.ex (1 0 0)x(0,i,0)=(0,0,i)

Im not arguing your point just trying to explain why I chose (-1,0,1) base Otherwise Im fine with every opinion which is mathematically correct Thanks

Juan Weisz

In principle you are right:

'Better use exp(ix)= cos(x)+ i sin(x) and then me that 1=1.'

I just don't get this part 'and then me that 1=1.'Maybe misprint or something could you expand on that and give some comment

Thank you

Well, to conclude i=0 is not correct.

Vector calculus not needed here, that i can see.

Dont know what you intend with cross product.

You know series expansions for exp, sin, cos?

Juan Weisz

exp(ix)= r(cos(x)+ i sin(x)) →(r(sin(x),i,sin(x+π)):|sin(x)∈Z)=exp(ix)

Here is vector calculus right there

btw what was "and then me that 1=1." part in your answer?

Comment :

'Well, to conclude i=0 is not correct.'

My take on it:

e^(i*π) = -1→e^(2i*π)=1→2i*π=0;

since π≠0→i=0

Where is the mistake?

Maybe presenting data in the vector form is not accurate That is another question That is why I opened this discussion and I welcome all thoughts on the subject

A cos or sin function may be zero

In an infinite no of places, or take

a value of 1 or -1 in other places

Infinite times. Look at Euleŕ formula.

Exp(0)=1 also. exp(n 2pi i)=1 , n integer.

r=1, there is just a complex number right there!

Certainly Sir

That's what I was looking for:

eip+1=0→(eip,1,0) For me that's enough

It was nice talking to;)

It only means p=pi, 3pi, 5pi, etc or-pi,-3pi,...

A complex no. only has 2 components.

A vector in 3D space can be written in component form, ( 𝑥 , 𝑦 , 𝑧 ) , or in terms of the fundamental unit vectors, 𝑥 ⃑ 𝑖 + 𝑦 ⃑ 𝑗 + 𝑧 ⃑ 𝑘 So what's your point?

Btw I didn't say that the vector is (e^(2n+1)pi,1,0) or (e^-(2n+1)pi,1,0) at that

Ok I give you a credit for that Hamilton did not find the three dimensional complex numbers Does it mean that they don't exist? I leave it open for the discussion

Take the diagonal matrix entries:

(-i)^2,e^πi,πi; and another:

e^2πi,e^πi,πi;

Compare their determinants (one is real and another one is not) and tell me that imaginary plane does not exist in a 3 D space I'd gladly accept any arguments Thank you

Dont know

You are very dispersed

You can find commutative rings with 3 variables, or more,

But not fields.

One is x+ty +ttz, ttt=1

Very techincal terms.

You wrote

Exp(ip)+1=0

What does that imply?

I answered p=pi, 3pi, ...

J.H Christ !

It only implies one thing ln(e^(2ip))=0 I thought it's clear from a get go

Speaking of dispersion

ip=0

To be perfectly honest Im absolutely not interested in technicalities My point was to show that the existance of ternary sets with imaginary parts is possible But by no means am I going to prove that or get into the details of the question why and why not If people think that tripple sets with imaginary parts are nonsense so be it but a plane is characterized by its area and if its a real number then perhaps a tripple set with an imaginary part exists That's the point I was trying to make

Get along well with your math teacher?

You do need to know ring and field.

Then answer your question.

In a ring yes.

In a field no.

I can not make head or tail from the disjointed answers that make very little sense

First things first What does my math teacher has to do with anything? It sounds rather like Chat GPT automatic response

Further on What makes you think I need to know a ring and field?

Then why should I answer my own question?

Next what question did I ask? and how can I possibly answer my own questions

What does the answer 'In a ring yes.

In a field no.' mean?

All those 'responses ' are just phrases that are incomprehensible for a human being Please give comprehensive comments to your last response which I suspect has very little substance

Thank you very much again

Any book on abstract algebra will emphasize basic structures in algebra, these being

Fields, rings , groups.

In a ring basically you allow addition, multiplication, all comon things like associativity or distributivity.

But sometimes you cannot invert some nonzero element.

You can in a field such as R orC.

Im just wondering how your educational system handles things, and what your level is.

Ok In response to your last question Part of my studies back when I was a university student was AI theory and if my answer to your question is not complete I can tell you this I can recognize machine generated text e.g chatbot ai response etc. Like I said before I received sufficient ammount of information and I want to thank you for your time spent Goodbye

Just a brief note on what was being said I do value your comments and to conclude our conversation I would like to attach my first work on associativity and distribution Perhaps it could shed some light on how I handle the problem Like I said before I thank your for the feedback and apologize for not answering further questions

See the attached file pls

Regards

The work may be better in 3 way logic, not in algebra.

You have a long way to go in algebra.

What makes you think so, Sir?

Well, you started off with some mad idea in algebra, then to

Vector problem, etc pseudo vector actually.

What is so 'mad' according to your definition about a balanced ternary set?

Call it a pseudo vector It must have three components they are (-1,0,1) There is always a problem with 0 because it always translates a vector in space onto the plane Substituting it ( a 0 part by an imaginary part ) allows to calculate an area which in most cases will be expressed as an imaginary number however if you substitute two components then the answer is not so trivial as the area becomes a real number However as you mentioned correctly pseudo vectors will not work for the fields or whatever you referred to Im working on it now Any other questions or comments? One remark only if you keep insulting me I will simply ignore your questions Do I make myself clear?

And I mean it

Not really, about i=0, thats mad.

Of course a cosine may vary between -1 and 1, thats ok.

A cross product is a pseudo vector

A lot of unrelated stuff...in my mind..

From there to imagine generalizing C.

Thats a big jump. Good question.

Hope i answered well "rings"

I will certainly report this content one way or another

Now I would like you to face the truth

According to you 'Not really, about i=0, thats mad'

Now do a simple math and tell me who is really mad:

e^(iπ) + 1 = 0

You asked me about what I was taught Here is the simpliest math ever:

(-1)^2→e^(2iπ)=1

Now take the ln of the left and the right side of the above equation and you have 2iπ=0→i=0

Now you must know Mark Twain who is probably closer to your culture than mine

He once said:

"Never argue with stupid people, they will only drag you down to their level and then beat you with experience"

Im not sure if you understand but I made it perfectly clear that my conversation with you is over You noticed probably that I blocked your name hoping that I will not get back to it again

Well I hope you've learnt something cause I did, definitely Im only curious what makes people so persitent Btw your contribution into this conversation is equal i

In slightly better language i is (0,1) in (x,y) language.

But that is not (0,0)

You may mean

Real part of i=0

Anyway, your formula with i=0 would mean 2=0,

Since exp(0)=1

What you are doing is taking a log of a complex no.

It reads log(z) = log abs(z) +i arg(z)

Abs(z)= sqrt( xx+yy)

Arg(z)= arc tan (y/x)

If i did not make any mistake.

Its a multi valued function arg is indeterminate within 2 pi n

'Anyway, your formula with i=0 would mean 2=0,

Since exp(0)=1'

This is a mighty conclusion Yes powerful Big applause Thanks

I only wonder if anyone else has any comments on this Cause I don't

'Anyway, your formula with i=0 would mean 2=0,

Since exp(0)=1'

For the record

The so-called 'my formula' is what Euler's formula says:

e^iπ+1=0→e^(2iπ)=1→2iπ=0→i=0

but it doesn't mean: i=0 would mean 2=0 Since exp(0)=1'

The fact that

e^(2iπ)=1↛2=0 That's a wrong conclusion right there

Now the first line of the above statement says:

'Anyway, your formula with i=0 would mean 2=0,'

In other words paraphrazing the above quote:

e^(2iπ)=1→2=0

From the standpoint of logic this is a wrong statement for a simple reason and that is:

e^(2iπ)=1↛2=0

I quote:

'Juan Weisz added a reply

8 hours ago

In slightly better language i is (0,1) in (x,y) language.

But that is not (0,0)

You may mean

Real part of i=0

Anyway, your formula with i=0 would mean 2=0,

Since exp(0)=1

What you are doing is taking a log of a complex no.

It reads log(z) = log abs(z) +i arg(z)

Abs(z)= sqrt( xx+yy)

Arg(z)= arc tan (y/x)

If i did not make any mistake.

Its a multi valued function arg is indeterminate within 2 pi n'

In conclusion Mr Weisz Please remember one basic thing of algebra Mathematics etc.

zero is both real, imaginary, and complex!

2 is not and neither is π giving 2π approx 6.28318530718 which multiplied by 0 gives 0

I should admit though that my conclusion regarding the value of i is wrong It has always been the sqr(-1) (Do not mistake it for an apology though)

In vector algebra: (0,e^πi)(0,e^πi)=e^2πi→lne^2πi=1→2pi=0 but then again it brings us to the start of a discussion

but e^2πi=1^x→2πi=0x→0=(2πi)/x etc. etc. etc For me its easier to stop and admit that i=sqr(-1) which is the definition of imaginary

I found my mistake in reasoning Did you?

I actually wanted to apologize to myself i=sqr(-1) by definition not by any logical deduction

Ring rules

(X1, y1)+(x2,y2)=(x1+x2, y1+y2)

(X1, y1)(x2,y2)= (x1x2-y1y2, x1y2+x2y1)

So

(0,1)(0,1)=(-1,0)

That what it all means.

You are a tough customer.

Try the system x+ty tt=1 ( not -1), " hyperbolic" not " complex".

The def. i=sqrt(-1) can trick you.

apologies misprint (e^ip-1)(e^ip+1)→e^2ip-1=0

What??

;))