>mfw the majority of /sci/ can't even into basic proofs

File: 12 KB, 694x309, Untitled.png [View same] [iqdb] [saucenao] [google]

Anonymous Wed Jul 27 21:13:57 2016 No.8231759 [Reply] [Original]

>mfw the majority of /sci/ can't even into basic proofs

Anonymous Wed Jul 27 21:19:29 2016 No.8231762

>>8231759
>let x be an even integer
>write x as it's prime factorization 2*p1*p2*...pn
>let y be an integer
>write y as it's prime factorization of p1*p2*....pm
>write xy as it's prime factorization (2*p1*p2*...pn)(p1*p2*...pm)
>now we see that 2 is a factor of xy and therefore it is divisible by 2 and therefore it is even

>>	Anonymous Wed Jul 27 21:20:31 2016 No.8231764 >>8231762 If I were less lazy I would've used better notation, but that is the gist of it. Is that what you had in mind or did you prove it a different way?

>>	Anonymous Wed Jul 27 21:22:17 2016 No.8231768 >>8231762 You didn't prove prime factorization though.

>>	Anonymous Wed Jul 27 21:24:45 2016 No.8231772 >>8231768 you don't have to prove prime factorization it's an axiom

>>	Anonymous Wed Jul 27 21:28:47 2016 No.8231782 Without prime factorization: Let x be even. By definition, this means we can write x = 2m for some m \in \mathbb{Z}. Therefore, xy = 2my = 2(my), where of course my \in \mathbb{Z}. By definition, xy is even.

>>	Anonymous Wed Jul 27 21:29:31 2016 No.8231783 We define even as a number can be written as 2k for some integer k. x = 2k xy = 2ky therefore xy = 2k' thus xy is even

>>	Anonymous Wed Jul 27 21:29:42 2016 No.8231784 >>8231782 this fpbp (b for brainlet)

>>	Anonymous Wed Jul 27 21:54:38 2016 No.8231806 >>8231768 >Wildberger.jpg

>>	Anonymous Wed Jul 27 22:04:01 2016 No.8231816 >>8231759 x is even. Therefore there exists some integer k so that x = 2k Then xy = 2ky = 2(ky) ky is an integer because y and k are integers so xy is an integer of form 2c where c=ky, therefore xy is also even.

Anonymous Wed Jul 27 22:08:12 2016 No.8231818

>>8231772
No it’s not. Go back to >>>/lit/.

Let [math]x \,\in\, \mathbf Z[/math] be even, choose [math]q \,\in\, \mathbf Z[/math] so that [math]x \,=\, 2\,q[/math]. Then, for all [math]y\,\in\, \mathbf Z[/math], [math]x\,y \,=\, 2\,qy[/math] is even, Q.E.D.

>>	Anonymous Wed Jul 27 22:17:45 2016 No.8231823 >>8231772 lmao my dude... it's called the fundamental theorem of artihmetic. if you weren't in a UFD, you wouldn't even have a prime factorization. but you don't even need prime factorication anyway.

>>	Anonymous Wed Jul 27 22:20:34 2016 No.8231825 >>8231816 for simplicity, Integers are a field so it's closed under multiplication.

>>	Anonymous Wed Jul 27 22:20:55 2016 No.8231826 >>8231783 This is the correct answer.

>>	Anonymous Wed Jul 27 22:25:58 2016 No.8231831 >>8231825 >Integers are a field Integers are not even a group under multiplication.

>>	Anonymous Wed Jul 27 22:28:02 2016 No.8231833 >>8231772 >prime factorization is an axiom fucking LMAO

>>	Anonymous Wed Jul 27 22:28:09 2016 No.8231834 >>8231825 The integers most certainly do not form a field. The primary difference between a field and ring is that all elements of a field are units, meaning they have a multiplicative inverse, and only 1 has an inverse in the ring of integers.

>>	Anonymous Wed Jul 27 22:29:53 2016 No.8231836 File: 133 KB, 328x395, 1467533623886.png [View same] [iqdb] [saucenao] [google] >>8231772 >it's an axiom

>>	Anonymous Wed Jul 27 22:30:32 2016 No.8231837 >>8231834 >The integers most certainly do not form a field I am not that guy but certainly the integers must form a field with some operation, Obviously not multiplication.

>>	Anonymous Wed Jul 27 22:32:23 2016 No.8231840 >>8231837 A field is a ring with unique inverses and a ring must necessarily have two operations on it you stupid mong. Lmao how fucking retarded are you? >inb4 merely pretending to be retarded

>>	Anonymous Wed Jul 27 22:35:51 2016 No.8231845 >>8231825 >Integers are a field That sure would make arithmetic geometry a lot easier.

>>	Anonymous Wed Jul 27 22:37:01 2016 No.8231846 >>8231837 >integers are a field >8231772 >prime factorization is an axiom holy FUCK

>>	Anonymous Wed Jul 27 22:37:28 2016 No.8231849 >>8231840 You could define arbitrary operations sufficient to make it a field. Not that it would be useful but it is possible.

>>	Anonymous Wed Jul 27 22:38:44 2016 No.8231851 >>8231849 >You could define arbitrary operations sufficient to make it a field No you can't lmao. Do you even understand how arithmetic is formalized? Do you understand how the integers are constructed? Just stop posting lmao this is so embarassing.

>>	Anonymous Wed Jul 27 22:38:53 2016 No.8231852 >>8231849 >1 is even >you could define arbitrary names for numbers, not that it would be useful but it is possible please stop

Anonymous Wed Jul 27 22:40:58 2016 No.8231856

>>8231840
I was just assuming that addition was the first operation.

Holy fuck some people get immediately triggered here. Did I type tumblr instread of 4chan on my browser or something.

Holy shit people like you disgust me. Get the fuck out of this board, faggot.

>>	Anonymous Wed Jul 27 22:42:31 2016 No.8231860 >>8231856 >gets ousted as a complete retard who has no idea how anything works >gets embarrassed and start lashing out at others Grow the fuck up idiot.

>>	Anonymous Wed Jul 27 22:42:44 2016 No.8231861 >>8231856 integers are NOT a field a field has TWO inversible operations you came into a basic math thread pretending to know something you clearly don't now you fucking cry and say it's tumblr because you got BTFO? kill yourself, faggot

Anonymous Wed Jul 27 22:43:29 2016 No.8231864

>>8231851
>No you can't lmao.
Yeah you can, faggot. Do you actually know abstract algebra?

The integers and + form a group so that is the first operation.

Then you need that the integers (except for 0 since that is the identity element for +) form a group too with a second operation.

There are infinitely many operations that suffice this second condition.

Then you need distributivity with both operations, which is also possible.

Do you know that you can even make a field with the naturals, right?

>>	Anonymous Wed Jul 27 22:44:33 2016 No.8231866 >>8231864 >There are infinitely many operations that suffice this second condition. >yet doesn't list any Lmao.

>>	Anonymous Wed Jul 27 22:44:51 2016 No.8231868 >>8231861 >a field has TWO inversible operations Yeah and you can define infinitely many such operations because any relation from ZxZ to Z counts as one such operation. I ask again, do you know fucking abstract algebra?

>>	Anonymous Wed Jul 27 22:44:53 2016 No.8231869 >>8231864 >there are infinitely many operations * that make (Z, +, *) a field show one

>>	Anonymous Wed Jul 27 22:45:54 2016 No.8231871 >>8231866 Here is a thread discussing this. http://math.stackexchange.com/questions/399703/is-it-possible-to-make-integers-a-field Do you know fucking abstract algebra, retard?

>>	Anonymous Wed Jul 27 22:45:56 2016 No.8231872 >>8231868 >Hurr 1 < 0 because there are infinitely many orders in R the integers means (Z, +, *) you idiotic fuck

>>	Anonymous Wed Jul 27 22:46:55 2016 No.8231875 >>8231837 >>8231856 They might, but typically when we talk about rings and fields, we're talking about addition and multiplication. I'm sorry these other fuckers jumped on you.

Anonymous Wed Jul 27 22:47:23 2016 No.8231876

>>8231872
>the integers means (Z, +, *) you idiotic fuck

No!

The integers are a set.

A field is formed with 2 operations and a set.

You can form infinitely many fields in pretty much any set.

You obviously need to know fucking abstract algebra first so you wouldn't have a fucking clue.

>>	Anonymous Wed Jul 27 22:47:24 2016 No.8231877 >>8231871 that's not integer addition. that's Q, you retard. you clearly just googled it to come up with it. come up with ONE operation * that makes (Z, +) a field you imbecile

Anonymous Wed Jul 27 22:47:52 2016 No.8231878

>>8231871
>doesn't know what integers mean
>doesn't know addition is defined via the successor function and multiplication defined in terms of it
>what is basic ring theory
>lists one example where he makes [math]\mathbb{Z} \cong \mathbb{Q}[/math]
Holy shit the retardation is real.

Anonymous Wed Jul 27 22:48:25 2016 No.8231879

>>8231875
>but typically when we talk about rings and fields, we're talking about addition and multiplication.

Then you are wrong about everything.

Fields are a generalization of the concept of our numbers with addition and multiplication.

Now I know no one in this threads knows abstract algebra and they were just told at their calc I class that the reals were a field.

Please kill yourselves.

>>	Anonymous Wed Jul 27 22:48:30 2016 No.8231880 >>8231876 again, the integers means (Z, +, *)

Anonymous Wed Jul 27 22:49:40 2016 No.8231881

>>8231876
You're a fucking idiot. There's a reason people distinguish between "any countably infinite set" and "[math]{\it the}[/math] integers".
>bending precise math terminology to suit his arguments
>hurr I'm not irrational, you're the irrational one

Anonymous Wed Jul 27 22:49:46 2016 No.8231882

>>8231880
>again, the integers means (Z, +, *)

No, Z is the integers.

(Z,+,*) is the field with addition, multiplication and the integers.

What is sad is that you actually insulted me so confidently, thinking that you were right.

Man you people have a lot of killing yourselves to do so you better get started.

>>	Anonymous Wed Jul 27 22:50:52 2016 No.8231884 >>8231882 again, the integers means (Z, +, *). otherwise you just have an arbitrary countable set

Anonymous Wed Jul 27 22:51:35 2016 No.8231887

>>8231881
You can still form infinitely many sets in the integers, retard.

I am just saying that you could also do this anywhere.

>>8231884
>again, the integers means (Z, +, *). otherwise you just have an arbitrary countable set

A set is not a field.

Anonymous Wed Jul 27 22:52:29 2016 No.8231890

>>8231887
>You can still form infinitely many sets in the integers, retard.

You said something else.
>The integers and + form a group so that is the first operation.
>Then you need that the integers (except for 0 since that is the identity element for +) form a group too with a second operation.
>There are infinitely many operations that suffice this second condition.

Show ONE. Fucking imbecile.

Anonymous Wed Jul 27 22:52:32 2016 No.8231891

>>8231882
You do know that the naturals are constructed along with the successor function right? It's not just a nested sequence of empty sets lmao.
>tfw retards like these exist in /sci/
>these are the kinds of people giving you homework advice
>>8231887
>doesn't know what "the" means in English
Pass ESL first, dumbass.

>>	Anonymous Wed Jul 27 22:52:37 2016 No.8231892 >>8231887 >infinitely many sets I meant infinitely many fields.

>>	Anonymous Wed Jul 27 22:53:32 2016 No.8231894 >>8231892 no, the integers means the ordered ring of integers. that's why it's called "THE" integers

Anonymous Wed Jul 27 22:54:21 2016 No.8231895

>>8231890
>Show ONE. Fucking imbecile.

Well, in that sentence multiplication also forms a group with Z-{0}

And there are many more.

But this multiplication does not help us construct a field, but it does construct a group as I was saying.

You are retarded.

>>	Anonymous Wed Jul 27 22:55:22 2016 No.8231896 >>8231895 integer multiplication is NOT a fucking group you fucking retard, it's just a monoid, there's NO inverse you're googling these terms as you go and shoving your foot deeper into your mouth, you colossal sack of shit

>>	Anonymous Wed Jul 27 22:56:16 2016 No.8231898 >>8231895 >multiplication also forms a group with Z-{0} Holy shit the dumbassery is oozing out of this idiot fuck.

>>	Anonymous Wed Jul 27 22:56:23 2016 No.8231899 >>8231895 >multiplication also forms a group with Z-{0} kek >And there are many more. You promise? :^)

>>	Anonymous Wed Jul 27 22:58:18 2016 No.8231902 >>8231895 please stop it's hurting me

Anonymous Wed Jul 27 23:00:14 2016 No.8231906

>>8231879
>Now I know no one in this threads knows abstract algebra and they were just told at their calc I class that the reals were a field.

No one would ever seriously consider the integers as a field. The ring structure on them is extremely important. They are the initial object in the category of rings.

>>	Anonymous Wed Jul 27 23:01:53 2016 No.8231907 >>8231906 >expects some idiot who doesn't even understand basic English to know category theory Lmao

Anonymous Wed Jul 27 23:13:53 2016 No.8231918

>>8231864
To make something good out of this inmense idiocy, here's a proof that (Z,+) can't form a field.

Observation: Since Z is cyclic, multiplication in Z is completely defined by 1*1.
[Note that m*n = ((m-1)+1)*n = (m-1)*n + 1*n and that m*1 = ((m-1)+1)*1 = 1*1 + (m-1)*1, by induction you can write every product as a sum of 1*1 terms. Negative products follow by noticing (-m * n) = - (m*n)]

Now let's say i is the multiplicative identity. Then i*1 = 1, and so if i is not 1 or -1 then:
1 = i*1 = (i-1)*1 + 1*1 = ... = 1*1 + 1*1 + ... + 1*1 , addition i times (or minus that if i is negative)
but 1 is the only cyclic generator and so this doesn't make any sense. Therefore i is 1 or -1.

If i = 1 then we have the usual ring (Z, +, *) which is not a field.
If i = -1 then we have an inverted Z of sorts, where the positives are exchanged with the negatives. This is isomorphic to Z and so is again not a field.

>>	Anonymous Wed Jul 27 23:19:25 2016 No.8231922 >>8231864 How the fuck do you make a group out of the naturals let alone a field? It's certainly not with addition and/or multiplication.

Anonymous Wed Jul 27 23:47:04 2016 No.8231952

Every positive even number can be expressed as 2n

Every positive odd number can be expressed as 2n - 1

Conjecture; if an interger is even, multiplying it by any interger will result in an answer that can be expressed as 2n

2n * (2n-1) = 4n^2 - 2n

2n * 2 n = 4n^2

Since you can express both the above equations as a multiple of 2n, then they're both even. This shows that as long as one factor two factor multiplication an even number then the result will be even.

I think you still have to do some more stuff to technically "prove" it though, right?

>>	Anonymous Thu Jul 28 00:07:22 2016 No.8231981 >>8231831 kek

Anonymous Thu Jul 28 00:13:10 2016 No.8231995

>>8231878
Holy shit, I'm not even him but reading this string of bullshit pisses me off. I've taken 4 abstract algebra courses, some at the graduate level. In abstract algebra, "multiplication" is a GENERAL term that doesn't NECESSARILY mean multiplication AS WE KNOW IT. It can be literally any function $[math]F \times F \rightarrow F [/math] compatible with the field axioms. The guy quoting the stackexchange thread is correct.

>>	Anonymous Thu Jul 28 00:15:49 2016 No.8232000 >>8231995 What's the point on coming back and samefagging like this now? THE integers means (Z,+,) (Z\{0}, ) is not a group (Z,+) can not form a field with any operation That's all. There's nothing else to discuss. Stop.

>>	Anonymous Thu Jul 28 00:17:27 2016 No.8232003 >>8231995 >multiplication >compatible with the field axioms kek one more to the list, welcome back anon

>>	Anonymous Thu Jul 28 00:18:33 2016 No.8232004 >>8231995 >I've taken 4 abstract algebra courses >"multiplication" [...] can be any function compatible with the field axioms >the field axioms dude this is an anonymous board, you didn't have to come back to continue lying

Anonymous Thu Jul 28 00:18:42 2016 No.8232005

>>8231759
Assume x is even (given). That means that there exists an integer (odd or even), k, such that x = 2*k for every even x. Now do two cases.

Case 1: let y be even. Thus, y takes the form of y = 2*l for some integer, l. x*y = 4*k*l = 2(2*k*l), which is the definition of an even number.

Case 2: let y be odd, so y = 2*l + 1, so x*y = 4*k*l + 2*k = 2(2*k*l + k), which is the definition of an even number.

Therefore, if x is an even integer, then x*y is even for any integer, y.

Anonymous Thu Jul 28 00:19:05 2016 No.8232008

>>8231922
Here is a hint, plebs.

You know the exclusive or operation? Well, try it on naturals and see what happens.

What is the identity element?

Does every natural have have a 'negative' in the set of the naturals?

I am not even going to fucking tell you the answer. Just go on, take a couple of natural numbers and write them down in binary and then start testing out the XOR operation.

Also, can you prove that XOR is associative? If you can't then you don't belong in this board.

Then if you are not a faggot show that indeed this forms a group.

Now, I WONDER. Is there any related operation (think game theory here) that could serve for our multiplication to make a ring?

Could we make it a field?

Oh, I don't know! I wish I knew abstract algebra!!!

Anonymous Thu Jul 28 00:23:57 2016 No.8232015

>>8232000
Why is it so hard to imagine defining some different functions on the set { ..., -2, -1, 0, 1, 2, ... } that are compatible with the field axioms? Is everyone in this thread honestly asserting that this is impossible?

Honestly, you guys are staring at truth right here:
http://math.stackexchange.com/questions/399703/is-it-possible-to-make-integers-a-field
One of the people who answer in the affirmative has a fucking Ph.D., and another is a grad student at UChicago. I think I'll take what they have to say over anons of questionable credibility on /sci/.

Anonymous Thu Jul 28 00:24:32 2016 No.8232017

>>8232008
>came back after reading the StackExchange thread closely only to spam all the things he read there like buzzwords
>12 sentences to angrily argue that (N, XOR) is a group
kek someone's butthurt

>Could we make it a field?
oh, I've no idea! are you so smart to know? well of course, you just spent 30 minutes reading on nimbers closely from wikipedia in order to come and prove to us you actually know something! WOW!

also, here's a hint. nobody who studies algebra calls it abstract algebra. that's what laymen call it to distinguish it from high school algebra

Anonymous Thu Jul 28 00:26:43 2016 No.8232022

>>8232015
>he posted it again
stop whining, what the fuck do you have to gain by fighting with us here? you already showed clearly that you're an idiotic fuck with your

>(Z/{0},*) is a group

and others. All you're trying to do is trying to cover up for the idiotic

>the integers are a field

by going "hurr I obviously mean with another ring structure :^) I definitely know what fields are!"

Anonymous Thu Jul 28 00:26:54 2016 No.8232023

>>8232017
I never left. After some faggot decided to impersonate me and say that the integers form a group with multiplication I just decided to not answer.

Then anon asked a question and I decided to answer.

You can indeed form a field with the naturals. You can form a field with any set, specially ones as big as countably infinite sets.

You can form a field in any set and you would know this too if you had any idea of what you are talking about.

>>	Anonymous Thu Jul 28 00:28:13 2016 No.8232025 >>8232008 >think game theory here >he's pretending that the StackExchange posts are his original ideas hahahaha

>>	Anonymous Thu Jul 28 00:29:33 2016 No.8232027 >>8232022 The funny thing is you think it's all a samefag, and in reality there are 2-3 people arguing on this side.

Anonymous Thu Jul 28 00:30:29 2016 No.8232028

>>8232025
Hey, this is obviously not my idea. Clearly the guy who came up with this much smarter than any of us.

That anon was doubting the claim that you can form a field with the naturals and I answered and that is just the 'best answer'.

If you want something different then you will probably have to think harder and deal with bijections to nicer sets.

Anonymous Thu Jul 28 00:30:54 2016 No.8232029

>>8232023
>After some faggot decided to impersonate me and say that the integers form a group with multiplication I just decided to not answer

HAHHAHAAHA HOLY SHIT YOU"RE UNREAL
here's more of your best hits:

>The integers and + form a group so that is the first operation.
>Then you need that the integers (except for 0 since that is the identity element for +) form a group too with a second operation.
>There are infinitely many operations that suffice this second condition.

>for simplicity, Integers are a field so it's closed under multiplication.

also

>You can form a field in any set and you would know this too if you had any idea of what you are talking about.

once more, you try and try to show you know what you're talking about, but you don't. you can't form any field from a set of 6 elements. or 12 elements. or 10 elements. YOU would know if you had any idea of what you are talking about.

>>	Anonymous Thu Jul 28 00:32:52 2016 No.8232034 >>8232029 multiplication is a general term and it was being used as such, and not the layman "repeated addition" definition.

>>	Anonymous Thu Jul 28 00:33:04 2016 No.8232035 >>8232028 >the guy who came up with this is much smarter than any of us dude it's the product of copies of F_2 it's not N

Anonymous Thu Jul 28 00:33:16 2016 No.8232037

>>8232029
>>The integers and + form a group so that is the first operation.

They do. This is literally the first thing they show you how to prove in algebra.

>>for simplicity, Integers are a field so it's closed under multiplication.

I never said this.

> you can't form any field from a set of 6 elements. or 12 elements. or 10 elements.

Okay, I see. Galois Theory does not exist.

Yup. I can believe that. Sorry Galois, turns out you died for nothing.

>>	Anonymous Thu Jul 28 00:33:51 2016 No.8232040 >>8232034 In the integers it is you stupid dumbass lmao

>>	Anonymous Thu Jul 28 00:34:06 2016 No.8232041 >>8232034 yeah, no. fuck off. you don't know shit, every three posts you say more and more terribly wrong things.

>>	Anonymous Thu Jul 28 00:34:51 2016 No.8232042 >>8232040 >In the integers it is you stupid dumbass lmao But this is not the point. You can construct so many algebraic structures from the set of integers.

Anonymous Thu Jul 28 00:36:16 2016 No.8232045

>>8232037
the first ridiculous statement is the full block
there is NO operation that makes (Z, +) into a field, you moron. see >>8231918

>>for simplicity, Integers are a field so it's closed under multiplication.
>I never said this.
CTRL+F you retarded baboon

>Okay, I see. Galois Theory does not exist.
>Yup. I can believe that. Sorry Galois, turns out you died for nothing.
once again, you keep lying through your teeth. it's clear you never studied Galois theory, or you would know why there's no field of 6, 10, 12, etc elements.
it's amazing how hypocritical you are, borderline retarded.

Anonymous Thu Jul 28 00:37:23 2016 No.8232046

>>8232042
Except that it is. When you say "the integers" you're referring to a specific object that's a rings constructed from the naturals. What you've been arguing now amounts to "you can make an arbitrary countably infinite set into a field".
Stop posting.

>>	Anonymous Thu Jul 28 00:37:23 2016 No.8232047 >>8232042 THE integers are not just the set. there's no point calling a generic countable set "the integers" you faggot

Anonymous Thu Jul 28 00:39:27 2016 No.8232049

>>8232045
>>8232045
>there is NO operation that makes (Z, +) into a field, you moron. see >>8231918

Yeah, that guy is correct.

In my post I said that there were infinitely many other operations to make a group.

I was actually wondering (pretty much asking) if there could be such a second operation to make it a field.

Then you people told me that making a field of integers is impossible and I said that your claim was retarded, yes you can.

>>8232045
>CTRL+F you retarded baboon

I mean that that was not me.>>8232045
>once again, you keep lying through your teeth. it's clear you never studied Galois theory, or you would know why there's no field of 6, 10, 12, etc elements.

>it's amazing how hypocritical you are, borderline retarded.

I interpreted your words as 'there cannot exist fields with finite elements'. After checking your claim about 6, 10, etc. is true but I did not remember and unfortunately did not process that thought the way I should have.

I suppose... you got me?

>>	Anonymous Thu Jul 28 00:39:45 2016 No.8232050 >>8232047 There is: the entire set is already enumerated for you before you even start, so you can start referring to specific elements right off the bat and even defining functions on that set using arithmetic we're already familiar with...

>>	Anonymous Thu Jul 28 00:40:46 2016 No.8232053 >>8232047 >THE integers are not just the set I am going to be honest. I have never seen someone make this distinction of 'the' integers as the field. Seriously fucking hell. Is this an american thing?

>>	Anonymous Thu Jul 28 00:41:18 2016 No.8232055 >>8232050 Also, if desired, this countable set can inherit the usual ordering with no extra work.

Anonymous Thu Jul 28 00:41:36 2016 No.8232056

>>8232049
>I said something else

THIS IS WHAT YOU SAID YOU INBRED MONKEY
>>The integers and + form a group so that is the first operation.
>>Then you need that the integers (except for 0 since that is the identity element for +) form a group too with a second operation.
>>There are infinitely many operations that suffice this second condition.
IT'S DEAD WRONG

>I did not remember
>that was not me
YEAH YOU TOTALLY KNEW HUH? YOU TOTALLY DIDN'T FUCKING REALIZE THAT 6 WASNT A PRIME HUH? YOU'RE A LYING FUCK THAT'S WHAT YOU ARE. KILL YOUR FUCKING SELF

>>	Anonymous Thu Jul 28 00:42:29 2016 No.8232058 >>8232050 How is that any different than referring to the objects as [math]a_1, a_2,\dots[/math] and so on? Are you seriously this retarded? The integers is defined as the ring, there's no room for debate here.

>>	Anonymous Thu Jul 28 00:42:31 2016 No.8232059 >>8232053 The ring you mean

>>	Anonymous Thu Jul 28 00:42:38 2016 No.8232060 >>8232053 >the field >THE FIELD >HE'S STILL CALLING Z A FIELD FUCK OFF. YOU DONT KNOW ANY MATH.

Anonymous Thu Jul 28 00:43:24 2016 No.8232061

>>8232056
>>>Then you need that the integers (except for 0 since that is the identity element for +) form a group too with a second operation.
>>>There are infinitely many operations that suffice this second condition.

There are infinitely many operations that suffice this second condition.

To make a field you need it to suffice many other conditions and as I can see that would not be possible.

But you can define many more groups with Z and infinitely many operations.

>>8232056
>YEAH YOU TOTALLY KNEW HUH? YOU TOTALLY DIDN'T FUCKING REALIZE THAT 6 WASNT A PRIME HUH? YOU'RE A LYING FUCK THAT'S WHAT YOU ARE. KILL YOUR FUCKING SELF

Nigga I interpreted your words wrong. Why are you so assblasted from this?

>>	Anonymous Thu Jul 28 00:45:07 2016 No.8232063 >>8232061 IM FUCKING ASSBLASTED BECAUSE YOU'RE A FUCKING RETARDED MONKEY FLAILING AROUND WHO CLEARLY DOESN'T KNOW SHIT YOU'RE FUCKING TRIGGERING ME HARD HERE BRO FUCK OFF

>>	Anonymous Thu Jul 28 00:45:29 2016 No.8232064 >>8232061 Just going to answer to myself. It is sad that half of the things I am getting shit over have been things this guy has posted on my behalf. Some people really just love to hate. You do you, you sad sad man.

>>	Anonymous Thu Jul 28 00:46:23 2016 No.8232066 >>8232061 >There is no field with 6 elements >"Hurrr Galois died for nothing hurrr" >You're wrong >"I interpreted it wrong I obviously knew!" kill yourself

>>	Anonymous Thu Jul 28 00:47:25 2016 No.8232067 >>8232064 THERE ARE NO USERNAMES HERE YOU RETARDED BABOON THIS IS AN ANONYMOUS BOARD WE DON'T KNOW AND DON'T GIVE A FUCK ABOUT WHO YOU ARE FUCK OFF

>>	Anonymous Thu Jul 28 00:47:39 2016 No.8232068 >>8232064 >>8232061 Cringed so hard. So this is what true Dunning-Kruger is like.

>>	Anonymous Thu Jul 28 00:49:33 2016 No.8232071 >>8232061 YOU BETTER NOT FUCKING THINK OF MAKING AN ACADEMIC CAREER BECAUSE I'M GOING TO FUCKING TRACK YOU DOWN AND MAKE YOUR LIFE HELL YOU FUCKING INBRED BABOON

Anonymous Thu Jul 28 00:49:35 2016 No.8232072

>>8232067
>See me making a controvorsial point, which was wrong after all
>Posts actually outrageous claims pretending he is me
>Then answers to these posts
>I leave the thread when he starts impersonating me and start playing Rocket League
>Come back, faggot still literally replying to himself I suppose to make him feel good about something
>Decide to intervene on the natural field shit
>Faggot comes back at it with shitting on my posts

Man this board. Unlike you, I think I enjoyed this experience. People can really take trolling far and if anything this shows that no matter how shitty my life gets, it will never be as shitty as yours.

>>	Anonymous Thu Jul 28 00:51:21 2016 No.8232075 File: 148 KB, 941x351, 1454095730261.png [View same] [iqdb] [saucenao] [google] >>8232072 >hurr I'm just pretending to be stupid Let this be a lesson to you.

Anonymous Thu Jul 28 00:51:36 2016 No.8232078

>>8232072
>>8232072
>wrong after all
>WRONG AFTER ALL
you were completely and always wrong about everything
from your perspective it's "wrong after all" because you talked about shit you didn't have a clue about and then you googled it and found out
from everyone else's perspective you're a pretender, a lying hypocritical fuck, and an annoying one at that

>hurr my life won't be as shitty as yours
your brain is full of shit and you're a retarded fucking monkey

Anonymous Thu Jul 28 00:53:14 2016 No.8232083

>>8232075
>Let this be a lesson to you.

I was not pretending to be stupid.

He was pretending to be stupid on my behalf, and I thank him for it. I guess.

>>8232078
I was just wrong about one thing: the fact that you cannot make a field with Z and addition because no second operation that will suffice exists.

To the anon who wrote the proof, thank you. You are the only person in this thread with any fucking decency when it comes to arguments.

>>	Anonymous Thu Jul 28 00:53:26 2016 No.8232084 File: 343 KB, 671x1047, 1466916958797.png [View same] [iqdb] [saucenao] [google] >>8232078 >>>/pol/ Blacks are people too >>8232075 take pedo cartoons to <<<\a/

>>	Anonymous Thu Jul 28 00:54:22 2016 No.8232085 >>8232083 I WROTE THE FUCKING PROOF YOU STUPID RETARDED BABOON ASS MONKEY RETARDED HYPOCRITICAL LYING FUCK

>>	Anonymous Thu Jul 28 00:54:35 2016 No.8232087 >>8232083 Do whatever you need to convince yourself and sleep at night, my dear undergrad.

>>	Anonymous Thu Jul 28 00:56:17 2016 No.8232090 >>8232087 GOD FUCKING SHUT THE FUCK UP FUCKING ASS MONKEY RETARDED HYPOCRITICAL LYING FUCK

>>	Anonymous Thu Jul 28 00:56:49 2016 No.8232091 >>8232090 YOU RETARDED BABOON YOU BETTER NOT FUCKING THINK OF MAKING AN ACADEMIC CAREER BECAUSE I'M GOING TO FUCKING TRACK YOU DOWN AND MAKE YOUR LIFE HELL YOU FUCKING INBRED BABOON

>>	Anonymous Thu Jul 28 00:56:54 2016 No.8232092 >>8232085 Then it is really dumb for you to literally impersonate me. Can we ask japmoot to add those id things /b/ people have? Fucking hell. You are absolutely autistic.

>>	Anonymous Thu Jul 28 00:57:56 2016 No.8232096 >>8232091 YOU'RE FUCKING TRIGGERING ME HARD HERE BRO FUCK OFF

Anonymous Thu Jul 28 01:00:39 2016 No.8232103

>>8232092
IMPERSONATE WHO YOU FUCKING IDIOT?

>>8231833
>>8231846
>>8231852
>>8231861
>>8231869
>>8231872
>>8231880
>>8231884
>>8231890
>>8231894
>>8231896
>>8231918
>>8232000
>>8232017
>>8232022
>>8232025
>>8232029
>>8232035
>>8232041
>>8232045
>>8232047
>>8232056
>>8232060
>>8232063
>>8232066
>>8232067
>>8232071
>>8232085

witness how you FUCKING DESTROYED MY SANITY SLOWLY

>>	Anonymous Thu Jul 28 01:02:16 2016 No.8232105 >>8232103 I'M GOING TO FUCKING TRACK YOU DOWN AND MAKE YOUR LIFE HELL YOU FUCKING INBRED BABOON

>>	Anonymous Thu Jul 28 01:02:57 2016 No.8232106 >>8232105 YOU RETARDED BABOON YOU'RE A FUCKING RETARDED MONKEY FLAILING AROUND WHO CLEARLY DOESN'T KNOW SHIT FUCK OFF

>>	Anonymous Thu Jul 28 01:06:04 2016 No.8232109 >>8232106 im a chemist i know things

>>	Anonymous Thu Jul 28 01:06:50 2016 No.8232111 >>8232109 loli?

>>	Anonymous Thu Jul 28 01:07:35 2016 No.8232112 >>8232111 I WROTE THE FUCKING PROOF YOU STUPID RETARDED BABOON ASS MONKEY RETARDED HYPOCRITICAL LYING FUCK

Anonymous Thu Jul 28 01:08:53 2016 No.8232116

>>8232112
You are out of the game, that's what you wanted.
I'll spend some of my energy to tell you a few things.
First, you are one adorable monkey to my eyes, a really stupid one, since what you just said is wrong, since you can talk about math using words, one plus one is two, and you can understand nature without understanding math, you drop an apple, it falls and so on :)
a small tiny lecture to you, understand it.

And since you are not in, you don't want to know how to.. for example know how to calculate how to get your dream job? Like I have.. or.. How to be the perfect father?
Those are all states of everything in existence, plausible ones.

And yes, I am aware this is waste of time, like trying to teach algebra to a dog, but to others here who don't get this, be humble, don't attack me, but the theories, we will never be finished, but we can use these for literally anything..

be smart, for once in your life at least.

>>	Anonymous Thu Jul 28 01:09:51 2016 No.8232118 >>8232116 woof woof :3 FUCKING BABOON

>>	Anonymous Thu Jul 28 01:16:04 2016 No.8232125 >>8231759 >hey look i just finished discrete

>>	Anonymous Thu Jul 28 01:20:24 2016 No.8232129 >>8231834 -1 also has an inverse

Anonymous Thu Jul 28 01:21:54 2016 No.8232131

>>8231806
it's a valid fuss because the proof is simple and doesn't require prime factorization. it's like trying to use an algebraic topology result in an introductory complex analysis course. yeah, you can use lifting of covering maps to prove the winding theorem is integral but it's not how you're supposed to do it.

>>	Anonymous Thu Jul 28 02:42:51 2016 No.8232214 >>8231783 underrated post no idea why everyone's using these overpowered FToA proofs

Anonymous Thu Jul 28 02:47:49 2016 No.8232221

>>8231823
>lmao my dude... it's called the fundamental theorem of artihmetic. if you weren't in a UFD, you wouldn't even have a prime factorization.
hey retard, you know that people stick to a UFD only because they like to have the '''''''''''theorem'''''''''''.
fucking undergrads....

>>	Anonymous Thu Jul 28 11:28:58 2016 No.8232901 >>8231768 If /sci/ was good at proofs they would be able to prove prime factorization.

>>	Anonymous Thu Jul 28 11:46:13 2016 No.8232920 This was literally in my exam and I got it right

>>	Anonymous Thu Jul 28 11:46:28 2016 No.8232921 >>8231772 >it's an axiom It most certainly is not an axiom my man.

>>	Anonymous Thu Jul 28 11:47:37 2016 No.8232926 >>8231762 There's no reason to invoke the FTA here. Kind of overkill when all you need is the definition of divisibility.

>>	Anonymous Thu Jul 28 11:51:26 2016 No.8232934 >>8231875 No, they definitely do not. Look at the map from the normal integers into this supposed field.

>>	Anonymous Thu Jul 28 11:54:49 2016 No.8232938 File: 2.64 MB, 400x225, hahahahaha.gif [View same] [iqdb] [saucenao] [google] >>8232037 >Okay, I see. Galois Theory does not exist.

>>	Anonymous Thu Jul 28 12:19:41 2016 No.8232960 >>8231759 if x is even then x = 2p where P is an integer. so xy = 2py where p and y are both integers so 2py is even so xy is even

>>	Anonymous Thu Jul 28 13:01:27 2016 No.8233009 >>8232960 Didn't prove that you can multiple both sides of the equation by the same thing and it will be equal :^)

>>	Anonymous Thu Jul 28 13:10:04 2016 No.8233020 >>8232960 >if x is even then x = 2p proofs?

>>	Anonymous Thu Jul 28 14:09:37 2016 No.8233135 >>8233020 That's the definition of even. You can't prove a definition.

>>	Sage Thu Jul 28 14:11:57 2016 No.8233145 >>8231836 >>8232075 >>8232084 Take your pedophile cartoons back to >>>/a/. Fucking degenerates.

>>	Anonymous Thu Jul 28 14:16:24 2016 No.8233154 >>8233020 Definition of even.

>>	Anonymous Thu Jul 28 15:32:46 2016 No.8233322 >>8233145 Hello Mr. Sage, you seem to not know that every board is an anime board here on 4chan. Deal with it, newfriend. :^)

>>	Anonymous Thu Jul 28 16:16:07 2016 No.8233407 If x is even then x can be rewritten. x = 2a xy = 2ay 2ay \| 2 QED

>>	Anonymous Thu Jul 28 16:24:09 2016 No.8233418 >>8233407 >2ay \| 2 ayyyy lmao

>>	Anonymous Thu Jul 28 18:56:42 2016 No.8233776 >>8233418 2ayyyyyyy \| lmao

Anonymous Thu Jul 28 19:14:31 2016 No.8233809

So I see there's this Stackexchange link where it's shown that for

X={0, S0, S00, S000, ...}

i.e. an arbitrary countable infinite set (like the set underlying N and Z are), you can put a field structure (X,+',*') on this, by this trick of mapping X to Q, defining +' and *' for X via the field (Q,+,*) and mapping back.

But have you guys clarified the question if we take (N,+,m) or (Z,+,m) with + being the standard addition on those, if there is any binary function m so that these are fields?

>>	Anonymous Thu Jul 28 19:19:01 2016 No.8233815 Prove that the sum of the natural numbers equals -1/12. Keep in mind the natural numbers are all positive. Keep in mind the only way to get a negative number as the result of a sum of two numbers is if one of them is negative.

>>	Anonymous Thu Jul 28 19:19:25 2016 No.8233816 >>8233809 >(N,+,m) (N,+) does not even form a group so that is out of the question. > (Z,+,m) This is the interesting question and it is sad that the guy making it was buried under a bunch of bullshit.

>>	Anonymous Thu Jul 28 19:21:43 2016 No.8233818 >>8233816 It is a good question, and it's been answered by another anon further up the thread. Given (Z, +, m) there is no such binary operation m which gives a field structure here. See >>8231918

>>	Anonymous Thu Jul 28 19:43:56 2016 No.8233853 Let x be an even integer. Then x = 2k for some integer k. Then xy = 2ky. (ky) is an integer by closure of integers under multiplication, thus 2ky is even

Anonymous Thu Jul 28 22:02:51 2016 No.8233995

>>8231759
>all even integers can be divided in half (i.e by 2) to produce another integer (definition)
>ergo, all even integers are multiples of 2, for 2 must be a factor
>any even integer can be expressed as 2a, where a is any integer
>every integer x that is even can be expressed as 2a, when multiplied by an integer y, can be expressed as 2ay
>a can be any integer, so 2ay can be expressed as 2a, making it an even integer

I'll admit it's sloppy, but I'm an engineering pleb.

>>	Anonymous Fri Jul 29 00:34:31 2016 No.8234186 >>8233995 >>all even integers can be divided in half (i.e by 2) to produce another integer (definition) 0 doesn't produce another integer, it comes back with 0.

>>	Anonymous Fri Jul 29 00:47:03 2016 No.8234195 >>8231759 This is easy if you know the definition of an even number. Without that definition, you might not realize that a number is even iff it can be written as 2*k for some integer k. That makes the problem harder.

Anonymous Fri Jul 29 00:50:36 2016 No.8234200

>>8231759
All integers are either even or odd. That is, they can be expressed in either the form 2k or 2k + 1, where k is some integer.
Let m and n be integers.
Suppose y is even. That is, y = 2m, so xy = 2n2m = 2(2nm) = 2k.
Suppose y is odd. That is, y = 2m + 1, so xy = 2n(2m+1) = 2n2m + 2n = 2(2nm + n) = 2k.

>>	Anonymous Fri Jul 29 01:01:57 2016 No.8234214 >>8234186 Ah, messed up there, replace "another" with "an." 0 divided by 2 is 0, so with the correction it stands.

>>	Anonymous Fri Jul 29 02:04:54 2016 No.8234286 >>8231759 Nobody said Y wasn't positive or + and/or -.

>>	Anonymous Fri Jul 29 02:17:28 2016 No.8234302 >>8232037 >Galois Theory does not exist. Finite fields are always of order p^q where p is prime my nigga.

>>	Anonymous Fri Jul 29 11:05:32 2016 No.8234798 >>8234286 Why would it be relevant?

Advanced search
Text to find
Subject [?]Search by post subject. Leave empty for any.
Username [?]Search for user name. Leave empty for any user name.
Tripcode [?]Search for tripcode. Leave empty for any.
Email [?]Search by email. Leave empty for any.
Filename [?]Search by image filename. Leave empty for any.
From Date [?]Enter what date to start searching from. Format is YYYY-MM-DD
To Date [?]Enter what date to start searching until. Format is YYYY-MM-DD
Image hash
Search in	All Posts OPs Only
Deleted posts	Show all posts Show only deleted posts Only show non-deleted posts
Internal posts	Show all posts Show only internal posts Show only archived posts
Order	New posts first Old posts first
Capcode	All Posts Only by Users Only by Mods Only by Admins Only by Developers
Results	Posts Threads
Action	[ Simple ]

/sci/ - Science & Math