ahmet

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Posts posted by ahmet


22 minutes ago, wtf said:
1 x 1 = 1. That's the multiplication table for the nonzero elements. What's the problem?
I need to contact my hodja for help.
Could you wait for a while for the clarification. There should be some unclarified details. (And are we confident/sure that Z_{2} contains just two elements. Because this ring is defined by congruence or equivalence (I could not find the correct word to explain very well)
but there are such details in our hand
....
23≡1
22≡0
....
(infinity)
so, is this group (or claimed so) finite?
as I said,I shall contact my hodja and make suitable clarification.
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49 minutes ago, wtf said:
0 isn't invertible in the real numbers either, but the reals are a field. 0 never has a multiplicative inverse.
This is okay. ✔️
but what about the theorem appearing above. In Regard to the multiplication operation, Z_{2}  {0_{z2}}
should satisfy group criteria.(Should be a group for multiplication operation)
2 hours ago, ahmet said:<H,+,.> is a ring. if <H,.> is a group,
we should have at least x element in H such that for 0_{H}≠ b , a.x=b should be satisfied.
is this also being satisfied?
Here H=Z_{2}
maybe you might be right, not sure. Because I am sure that my hodja had made me note that theorem.
But already galois thorem is not specified for specific prime number (to be exempted),therefore you might be right.
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1 hour ago, studiot said:
no, this is not a group for multiplication operation.To be honest Galois's theorem does not specify the case for 2 ,which is prime number. But this prime number is unique prime number that is even number.
There is one a very basic theorem that will not accept any set which has had just two elements to be a field.
That basic theorem is given as below:
Theorem: a ring having just two elements cannot be a group for/across multiplication operation.(Thus <H,+,.> cannot be a field)
Proof:
<H,+,.> is a ring. if <H,.> is a group,
we should have at least x element in H such that for 0_{H}≠ b , a.x=b should be satisfied.
when we care one other theorem ,which states 0_{H.}x=0H , we have b=0_{H and this is a contradiction. }
thus, as any ring that has just two elements cannot be a group in regard to multiplication operation, it will also not be accepted as a field.
I think Galois theorem is invalid specifically for 2.
meanwhile, I think wikipedia might be not suitable for mathematical sciences ,where the issue is upper level than MSc or equivalent.
(for Z_{2} there 0 is not invertible. )
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4 hours ago, joigus said:Raindrops are falling on my head
And just like the guy whose feet are too big for his bed
Nothing seems to fit
Those raindrops are falling on my head, they keep fallingSo I just did me some talking to the sun
And I said I didn't like the way he got things done
Sleeping on the job
Those raindrops are falling on my head, they keep fallingBut there's one thing I know
The blues they send to meet me
Won't defeat me, it won't be long
Till happiness steps up to greet meRaindrops keep falling on my head
But that doesn't mean my eyes will soon be turning red
Crying's not for me
'Cause I'm never gonna stop the rain by complaining
Because I'm free
Nothing's worrying meIt won't be long till happiness steps up to greet meRaindrops keep falling on my head
But that doesn't mean my eyes will soon be turning red
Crying's not for me
'Cause I'm never gonna stop the rain by complaining
Because I'm free
Nothing's worrying meSource: LyricFindSongwriters: Burt Bacharach / Hal David / Burt F. Bacharachwhat is this ?
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would you like to share your opinions in convenience of title?
let us learn which engineering book you mostly enjoyed,pleased to read and why.
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1 hour ago, studiot said:
I agree that the OP proposition does not constitute a field, since it does not contain the required inverses.
However the smallest field contains just two elements, along with appropriate rules for addition and multiplication.
An example would be {1, 0} with a De Morgan/Cayley table.
I think there should be three element at least. lets see
1+(1)=0 but 1 is not element of {1,0} set. Thus we can say that additive inverse does not exist for "1" element in {1,0}
in fact, the second operation should meet/satisfy group criteria to be field, and to be ring; first operation should also satisfy group criteria.
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simply, I see the use of "de morgan" rules.
1 
On 5/3/2020 at 7:22 PM, taeto said:
({0},+,⋅)
this is an ideal (also it is one of trivial ideal) but not a field. Could you explain how you have defined . and + operations here?
what is more ,I can additioally say that there was one theorem claiming that a matehmatical structure containing just 2 element also would not be field.(i.e. not only just one element but just two element is also insufficient to be field.)
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7 minutes ago, swansont said:
Is there any reason to think these are connected?
in normal or direct reflection,no.
but...I mean or claim and also underline "coincidences" , I think that this was high potential.
because it is always probable to find a new thing when you look for something.
but these days ...I am sure that searches are dense. here,in turkey I see some associations (e.g. TUBITAK) also turkish authority ease some conditions
For instance in previous times ethical board was mandatory prior to some researches but now this conditon has been eased.
not only turkey, I predict many countries have high desire to find a cure or vaccine for covid19.
but what if by coincidence ...?
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the cure of AIDS or vaccine of HIV be found by coincidence simultaneously while the studies across covid19 are being performed ?,
what is your opinion?
0 
4 hours ago, iNow said:
Justice is what love looks like in public
ahahahha really right!
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12 hours ago, wtf said:
Don't know what you mean by objects, unless perhaps you mean sets. Or fields. Completeness field is a bad translation and I can't even make a guess as to what it is intended to mean.
as I remember Z and Q are not in the same category. One of them should be different.
Menwhile, to you; which type of fields do you claim? there are many types of fields (E.g. P.F.F. , Euclid field or completeness field etc.)
Anyway, It seems disccussing such contexts under another thread will be better.
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.........................
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1 hour ago, wtf said:
If x is nonzero then xH is an ideal. If it's not a proper ideal then it's all of the ring, including 1. Hence every nonzero element is invertible. So a commutative ring with no proper ideals is a field. If the ring is not commutative a similar proof would show that it must be a division ring, but you'd have to check both left and right ideals for the proof.
Note that our ring is required to have 1. If not, I'm not sure if this result is still true.
welcome @wtf,
.I think I have done a mistake. (But some terms changing between languages, so I am not sure whether I would express all terms correctly (or equiavlently)
Description: if <H,+,.> has 1H and has no zero divisors, and is commutative, then we call this ring with " completeness field" (I translated from turkish into english that term)
Description: if H is commutative and has 1_{H} and for every x element H{0_{H}}, at least y element H{0_{H}} such that x.y=1H conditions are being satisfied then we call this completeness field specifically with "object" (but not sure about the correctness of this term)( or equivalently this ring should satisfy group criteria for the second operation (multiplication))
All Q,R,C are objects.
also, all finite completeness fields are also objects.
Q,R and C should have no more than their trivial ideals
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15 minutes ago, studiot said:
This is the last thing you should do.
heyy ! I was just joking , but generally my jokes have potentiality to be also real. The thing I know I do not lie.
ok,as they do not prefer to reply in the current status,I think I can postpone some operations.
meanwhile,I was to use their materials about html+css ..(also maybe js)
but I am still not sure whether interferring their specific samples would also provide me another copyright (?) (my own copyright)
Thanks
(Could you also respond: assume please I have contacted to the author of that or any else book. Also assume that the author kindly gives me permission to use the content. Should I also take the permisson from the publisher??)
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25 minutes ago, swansont said:
Permission requires a response. You don't have it until they actually respond in the affirmative.
ok. this is already rational idea.
but contacting over the phone,will not this be non sense? because they will be able to refute later?
I think I need written (material)/ permission (like an e mail)
ok,now I decide to postpone somethings.
(This applies to more than copyright.)
furthermore,as far as I know copyright is not like a patent , and is valid only in the published country. But I do not really dominate all the relevant contents
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maybe,I will try that but this will be a bit time later. Furthermore, do you have information about the questions on this issue ,for instance , what does require me to request a permission and not.
(Specifically: I do not know whether I have to take permission on constructing a website or websites. I intend to use almost all of the codes they provided in some samples, but I also know that I would definitely interfere to that codes, for instance I shall change css part, and will add some new codes.
however,I shall not use any extra (for instance I shall not copy paste any else section of the book and will not use the information provided in example (appearing on that sample of website)
but I shall use just the codes succintly.
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5 minutes ago, iNow said:
No, all you know is that they have not responded. Lack of response should not be interpreted as implicit permission.
and what to do now,should I send a complaint form? inquiring: why don't you reply?!
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I have tried to contact wiley online library customer care and requested permission to use some parts in some of the books published by this publication but they still have not responded although two mails forwarded.(I mean,the same request two times sent)
So, can I decide that they have already allowed me to use as they do not care or are in silent mode?
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hi, once again;
I would ask a new question on this issue:
DO YOU BELIEVE THAT I WOULD LOSE THE SMARTNESS OF MY VOICE IF I GET OLDER? (if we compare the case with young ages)
now I am 31 (of course proudly feel myself as 18 , but I do not know what would happen at my for instance 45 or 60)
I am sure I am trying to calculate rather non mathematical probability ahahaha
all in all it seems normal, because we have emotions ..
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12 minutes ago, studiot said:
All of this categorisation goes a lot further into pure maths than I normally like to.
But I know you work more on the pure side.
hi, I do not know what exactly caused you to think so.
but no problem for now.
meanwhile I do not think that that description was available here.
a reminding: with trivial ideals I meant for instance in <H,+,.> ring, H and {0H} are trivial ideals. (There should be no more ideals for this ring)
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1 hour ago, studiot said:
Yes there is a great deal more to algebra than is taught in school, indeed there is more than one 'algebra'.
As regards number systems,
N and Z do not form algebraic fields
N fails for additive inverses and Z fails for multiplicative inverses.
ahahaha @studiot ,you seem like you criticize the sets pahaha rather than criticizing people ,well ,very well. I really appreciate the tone of your tongue. haha.
(like saying; Z is not a successfull set for multiplicative inverses and N already seems like just a stupid set, he even cannot contain additive inverses )
Meanwhile,
I saw @wtf ' s explanations (provided in the link) but I also thought that there would be many extension fields.
mmm may I ask a question , with what are you expresing a ring satisfying all the conditions given below in english? (in fact bourbaki has provided that definition but I could not immediately /quickly find)
>> with no zero divisor.
>> commutative
>> contains unitary element of multiplication
one more question: what do you mathematically call a specific ring with, that has not had any else ideals except its trivial ideals?
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A field with one element.
in Linear Algebra and Group Theory
Posted · Edited by ahmet
hi,
my lovely hodja replied today. But probably..
as I understand she implies that if except unitary element (of first operation) any ring contains just two elements ,then this will not satisfy group criteria according to second operation.
To be honest her email is a bit mixed. I am not sure whether I should recontact her. But I see "loves" conclusional word instead something more formal (e.g. regards, best regards also sincerely etc. Generally I had seen "sincerely" if someone would act across me rather formal ,especially when they do not know me, but loves is a bit intimate (not like warmly "kind regards" (I mean this last one seems like both positive and negative or at the centre of these approaches )
she also accepts that just {1} set is a group according to multiplication operation and it is obvious.
mmm,yes I reread the taken email and decided that she would express that if except the unitary element of first operation(I mean the ring's zero) ,if we have just two element,then this (according to the second operation) will not meet/satisfy group criteria. Z_{2} satisfies group criteria , because this ring has just one element except ring's zero.