Division by zero is possible

in the zero ring lol ZOMG!!

division by zero should technically give infinity, but that makes it posibble to make any 2 arbitrary numbers equal so it is defined as "undefined" instead.

>>2

what part of "in the zero ring" do you not understand. gb2/algebra/ lol

oh anonymous, come back when you know galois theory.

1/0=infinity

a/0 is undefined, for a<=R
not the same as infinity...srsly guys.

infinity isn't a member of the reals

>>6
actually a < £$^WRY$£&W%YQ£YA$%&Q%UW$W%&OS*$U+-=-=-=w46q"$^

Also, if you can't understand the above equation you are an idiot.

Besides your argument is completely irrelevant. A simpler counter-argument would have been that no number of 0s added together will make anything other than 0, therefore when you divide by 0 it doesn't make infinity.

In fact a better argument would be that 0 isn't a real number.

wow, you guys are nerds... actually, not even. just a bunch of loser virgins.

>>10
fail college?

>>4

What has Galois Theory to do with anything? In the zero ring, there is only one element 0, which is the multiplcative and additive identity. We have 0/0 = 0 in the ring. This is consistent with multiplication and addition in the 0 ring.

division isn't defined in rings... only fields. fail.

>>11
fail life?

>>13
well i just defined it bitch. what you going to do about it.

you can't define division by the additive identity. ever. even if it makes sense.

>>16
these people are noobs. zero ring is just some faggot idea made up to explain that OMG a/0 IS POSSIBLE RLY! They keep going off using the wrong math and trying to use that to explain something WAAY over their heads.

>>6 is correct, it's just that idiot >>8 read it wrong. it's not a "less than/equal to" sign. It should read as set notation like that small looking e. LOL moron.

>>9 0 belongs to the reals and nobody except failed nerds will argue that.

>>17
so y r u arguing it lol

Also if 0 is a real number you should be able to divide other numbers by it.

Divide something by 0 to prove you are not a fucking idiot and probably gay. Failure to do so will be seen as an admittance of this. You have 24 hours.

Seeing as you are sure you are right this shouldn't be hard. If you fail it probably means you are gay.

K go!

>>16

zomg, 0 in the 0 ring is also a multiplicative identity. pwned faggit.

>>18 is now a failed nerd.

>>20
NO U

>>13

Oh wait, just because it's a ring it can't be a field? GTFO
>>17
Read more, division by 0 is possible in the 0 ring because 0 is the only element in the ring and 0 is it's own multiplicative as well as additive identity.
so we have 0 + 0 = 0
0 x 0 = 0
0/0 = 0

>>22

one of the field axioms is that 0 isn't the same element as 1.

the additive identity never has an inverse...

thus your whole hypothesis falls into the murky sewer waters....

come back when you know more algebra :)

>>23
Finally someone who knows their math!

Anyway, right now it looks like the focus has been shifted to showing that 0 is in the real number system. But smart people already know this and DON'T require proof. *rolls eyes* We'll leave this snack to you noobs. :)

>>23
wait, but I just showed you 0 has an multiplicative inverse in the 0 ring. Prove me otherwise. Just because a ring isn't a field doesn't mean there doesn't exist elements in the ring which are invertible.

>>25
ZOMG you useless cunt. Either shut up or stop breathing. The shit pouring out of your mouth has absolutely no point, even in the already abstract world of maths.

In fact I can say that I create an arbitrary system where division by 0 is a valid operation which results in infinity. Who gives a fuck? It's arbitrary you fucktard.

>>26
We're talking about a ring with only one element you retard. Who gives a shit about infinity.

>>27
Forget that idiot >>26 . He keeps posting wrong shit about infinity because he thinks he elarned math from the wiki.

>>28
WRONG shitface, that was my first post in this thread.

Seriously guys, just stop pretending you can think and shut the fuck up already.
God, why do terrorists target businessmen who actually DO something with their lives, when they should be putting you dumbfucks out of your misery.

well it´s not really hard is it:
lim(1/x)for x->0 results as a positive or negative infinity.
on the other hand: lim(1/x²) for x->0 results only in a positive infinity.

Meanwhile, lim(ax/x) as x->0 is a.

>>30
>>31

Congratulations! You win for using calculus in an algebraic structure with only one element!

>>32
yeah so why make it complicated when it also works simple.

>>32
The Sarcastic Strawman; a bold & interesting move, Anonymous! Let's see how this will affect the discussion..

>>33
>>34
(same person)

so guys I just found a site that explains everything.

http://mathforum.org/dr.math/faq/faq.divideby0.html

just look there and you will find your answer.

>>35

he's talking about an algebraic structure with more than 1 element you noob.

>>36
then tell me why it can be that 0^0=1

>>37

you just have to look at the graph of it:
we all know x^0 is allways 1 and because of this you can say that
x^0=1 for x element of R without 0
and 0^0=1 for x=0

>>37
we're talking about the zero ring here. The 0 ring consists of only 0. if by 1 u mean the multiplicative identity then yes 0/0 = 1 = 0 because in the zero ring 0 is both the additive and multiplicative identity.

>>37
>>38 is a Fag. I have no idea where R came from and it suggested nothing to imply x^0 = 1 when x isn't 0. x^0=? is a mathematical model and ? means what do you have to multiply by itself infinite times in order to get x.

If you want to find what you have to multply by itself 3 time to get x you do x^(1/3), if you want to find what you have to multiply by itself 800 times you do x^(1/800). As you might guess anything divided by infinity is 0, therefore...

x^0=x^(1/infinity)

of course there are certain rules which determine what happens when you multply something by infinity and it revolves around -1, 0 and 1.

infinity*x=f(x)

-1<x<-1, x<0, f(x) = infinity and -infinity
x=-1, f(x) = -1 and 1
x=0, f(x) = 0
1<x<1, 0<x, f(x) = infinity
x=1, f(x) = 1

Seeing as 0 is not a real number, there can be no real answer to x^0, unless x is also not a real number. Yes, everyone in the entire world is wrong except me. However everyone in the world is right about 0^0 = 0

0^0 = 0 because you can multiply 0 infinite times and you still end up with 0

infinity^0 > 1

1 < (infinity and -infinity)^0 < -1

-1^0 doesn't exist like other rooted negative numbers (1/infinity, remember?)

1^0 = 1, 1^anything = 1 of course