Mfw the majority of Veeky Forums can't even into basic proofs

No it’s not. Go back to .

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.

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.

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

This is the correct answer.

>Integers are a field

Integers are not even a group under multiplication.

>prime factorization is an axiom
fucking LMAO

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.

>it's an axiom

>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.

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

>Integers are a field
That sure would make arithmetic geometry a lot easier.

>integers are a field
>8231772
>prime factorization is an axiom

holy FUCK

You could define arbitrary operations sufficient to make it a field. Not that it would be useful but it is possible.

>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.

>1 is even
>you could define arbitrary names for numbers, not that it would be useful but it is possible
please stop

I was just assuming that addition was the first operation.

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

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

>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.

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

>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?

>There are infinitely many operations that suffice this second condition.
>yet doesn't list any
Lmao.

>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?

>there are infinitely many operations * that make (Z, +, *) a field
show one

Here is a thread discussing this.
math.stackexchange.com/questions/399703/is-it-possible-to-make-integers-a-field

Do you know fucking abstract algebra, retard?

>Hurr 1 < 0 because there are infinitely many orders in R
the integers means (Z, +, *) you idiotic fuck

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.

>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.

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

>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.

>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.

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

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

>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.

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

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

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

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

A set is not a field.

>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.

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 Veeky Forums
>these are the kinds of people giving you homework advice
>doesn't know what "the" means in English
Pass ESL first, dumbass.

>infinitely many sets
I meant infinitely many fields.

no, the integers means the ordered ring of integers. that's why it's called "THE" integers

>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.

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

>multiplication also forms a group with Z-{0}
Holy shit the dumbassery is oozing out of this idiot fuck.

>multiplication also forms a group with Z-{0}
kek
>And there are many more.
You promise? :^)

please stop it's hurting me

>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.

>expects some idiot who doesn't even understand basic English to know category theory
Lmao

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.

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.

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?

kek

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.

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.

>multiplication
>compatible with the field axioms
kek

one more to the list, welcome back user

>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

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.

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!!!

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:
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 Veeky Forums.

>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

>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!"

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 user 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.

>think game theory here
>he's pretending that the StackExchange posts are his original ideas
hahahaha

The funny thing is you think it's all a samefag, and in reality there are 2-3 people arguing on this side.

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

That user 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.

>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.

multiplication is a general term and it was being used as such, and not the layman "repeated addition" definition.

>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

>>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.

In the integers it is you stupid dumbass lmao

yeah, no. fuck off. you don't know shit, every three posts you say more and more terribly wrong things.

>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.

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

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.

THE integers are not just the set. there's no point calling a generic countable set "the integers" you faggot

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

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.

>CTRL+F you retarded baboon

I mean that that was not me.>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?

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...

>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?

Also, if desired, this countable set can inherit the usual ordering with no extra work.

>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

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.

The ring you mean

>the field
>THE FIELD
>HE'S STILL CALLING Z A FIELD
FUCK OFF. YOU DONT KNOW ANY MATH.

>>>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.

>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?

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

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.

>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

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

Cringed so hard. So this is what true Dunning-Kruger is like.

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

>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.

>hurr I'm just pretending to be stupid
Let this be a lesson to you.

>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