/sci/ - Science & Math » Thread #11442884

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Anonymous Thu 05 Mar 20:01:00 2020 No.11442884 View Reply Original Report

Quoted By: >>11442887 >>11442912 >>11443085 >>11443116 >>11443302 >>11443340 >>11443348 >>11443349 >>11443511 >>11443515 >>11443556 >>11443558 >>11443580 >>11443768 >>11444026

>cant divide by 0

let me explain it useless cock craving faggots

you got 5 cakes
you distribute them among 0 friends

how many cakes you have left?

checkmate /sci

Anonymous

Anonymous Thu 05 Mar 2020 20:01:55 No.11442887 Report

Quoted By: >>11443340

>>11442884
According to SCIENCE you end up with infinite cakes lmao

Anonymous

Anonymous Thu 05 Mar 2020 20:13:53 No.11442912 Report

Quoted By:

>>11442884
https://en.wikipedia.org/wiki/Wheel_theory

Anonymous

Anonymous Thu 05 Mar 2020 20:17:53 No.11442924 Report

Quoted By:

go ahead and define it to whatever you want and try and build a solid theory of math that reaches as far as the current standard theory. i'll be waiting for your paper

Anonymous

Anonymous Thu 05 Mar 2020 21:23:38 No.11443085 Report

Quoted By: >>11443357

>>11442884
Take five cakes, distribute them among 1/2 a friend, how many cakes do you have?

Anonymous

Anonymous Thu 05 Mar 2020 21:40:21 No.11443116 Report

Quoted By:

>>11442884
how many cakes you have left is not dividing what you got by number of people you distribute cakes
if you want how many cake you have, then its 5/1.
if distributed among 0 people, then its not distribution. low iq fag thread

Anonymous

Anonymous Thu 05 Mar 2020 22:58:12 No.11443302 Report

Quoted By: >>11443310 >>11443326

>>11442884
The inverse of the neutral element of the sum doesn't exist for any field.

Proof:
Let $(F,+, \times)$ be an arbitrary field, let $0 \in F$ denote the neutral Element of the sum operation, that is, if $x \in F \implies x + 0 = x$ .

Lemma: if $x \in F \implies 0 \times x = 0$
Proof:
$0 \times x = (0 + 0) \times x = 0 \times x + 0 \times x$ Definition of neutral element of sum and distributive law.
$0 \times x + 0 = 0 \times x + 0 \times x \implies x = 0$
This step holds true because if $x,y,z \in F$ then it follows that: $x + y = x + z \implies y = z \square$

Now, by definition, the multiplicative inverse of $x$ (written $x^{-1}$ ) is defined such as if $1$ denotes the element neutral to the sum, then $x \times x^{-1} = 1$ .

Choosing $x = 0$ obviously yields a contradiction, as we have proven that for any $x \in F \implies 0 \times x = 0$ so if $x = 0 \implies x \times x^{-1} = 0$ but by definition $x \times x^{-1} = 1$ , and we know by the field axioms that $1 \neq 0$ so we have reached a contradiction, thus we conclude that $0^{-1} \notin F$

$Q.E.D$

Anonymous

Anonymous Thu 05 Mar 2020 23:01:51 No.11443310 Report

Quoted By:

>>11443302
Shit, minor mistake, I meant "1 denotes the element neutral to the multiplication operation" no to the sum.

Anonymous

Anonymous Thu 05 Mar 2020 23:09:20 No.11443326 Report

Quoted By: >>11443447

>>11443302
that just means you can't divide by 0 in an ordered field. What if OP invents a new set theory that discards the old notion of fields?

Anonymous

Anonymous Thu 05 Mar 2020 23:15:46 No.11443340 Report

Quoted By: >>11443345

>>11442884
>>11442887
Division is how many slices of cake you divide it into so that each person gets an equal slice

6 cakes divided by 3 people is 2 cakes each

1 cake div 3 people is 1/3 of a cake each

Anonymous

Anonymous Thu 05 Mar 2020 23:16:54 No.11443345 Report

Quoted By:

>>11443340
5 cakes over 0 people can be argued as infinite slices and 0 slices because you didn't cut the cake so division of 0 is not possible

Anonymous

Anonymous Thu 05 Mar 2020 23:17:24 No.11443348 Report

Quoted By:

>>11442884
>you got 5 cakes
>you distribute it among 1 friend
how many cakes you have left?
checkmate op

Anonymous

Anonymous Thu 05 Mar 2020 23:18:01 No.11443349 Report

Quoted By:

>>11442884
If you divide 5 cakes among 0 friends then you're dividing it by ONE PERSON, yourself, so you get 5/1 = 5

Anonymous

Anonymous Thu 05 Mar 2020 23:19:39 No.11443357 Report

Quoted By: >>11443383 >>11443498

>>11443085
1/2 a friend is not quantifiable

Anonymous

Anonymous Thu 05 Mar 2020 23:27:50 No.11443383 Report

Quoted By:

>>11443357
we're just not that close, ok more like acquaintances.

Anonymous

Anonymous Thu 05 Mar 2020 23:57:24 No.11443447 Report

Quoted By:

>>11443326
The field doesnt need to be ordered in order for my proof to hold.
It holds for any arbitrary field, for example on the complex field, where you can't even define an order.

And inventing a special set where you could divide by 0 could be interesting but probably meaningless. As a field is where you do operations on.

Anonymous

Anonymous Fri 06 Mar 2020 00:11:28 No.11443498 Report

Quoted By:

>>11443357
It is a half
There it is, I just quantified it

Anonymous

Anonymous Fri 06 Mar 2020 00:22:06 No.11443511 Report

Quoted By:

>>11442884
No, it's more like:
>you have 5 cakes
>you split it into 0 groups
impossible

Anonymous

Anonymous Fri 06 Mar 2020 00:25:27 No.11443515 Report

Quoted By:

>>11442884
Sir I praise your trollworthyness
So many have pointed out you are dividing in fact by one
Be most delighted in your efforts

Anonymous

Anonymous Fri 06 Mar 2020 00:38:45 No.11443543 Report

Quoted By:

>division
>"How many do you have left?"
I dont think you understand how division works.

Anonymous

Anonymous Fri 06 Mar 2020 00:43:57 No.11443556 Report

Quoted By:

>>11442884
If YOU have 5 cakes and distribute them among 0 of your FRIENDS, you have still distributed them among 1 person, namely, yourself.

You are pointing out that 5/1 is 5.

Anonymous

Anonymous Fri 06 Mar 2020 00:45:26 No.11443558 Report

Quoted By:

>>11442884
If you have 0 friends to distribute to then you are not dividing it at all therefore you cannot perform this function

Anonymous

Anonymous Fri 06 Mar 2020 00:56:49 No.11443580 Report

Quoted By:

>>11442884
>how many cakes you have left?
None. You ate as much as you could of one of them, but had to throw the leftovers and the others away because they got stale.

Anonymous

Anonymous Fri 06 Mar 2020 02:12:12 No.11443768 Report

Quoted By: >>11443772

>>11442884
The problem isn't that you can't divide by zero. The issue here is zero itself. Yes, I said it /sci/, the additive identity does not exist. Go disprove me by pointing a truly empty sector of space, a "true zero". You can't because nothingness does not exist in this reality as the act of observation introduces reality into that sector of nothingness. You will therefor never be able to observe true nothingness so creating representations of it is idiotic.

>b-but if I ate the apple then there's zero apples
dilate

Anonymous

Anonymous Fri 06 Mar 2020 02:14:23 No.11443772 Report

Quoted By:

>>11443768
>Go disprove me by pointing a truly empty sector of space, a "true zero"
The inside of your skull

Anonymous

Anonymous Fri 06 Mar 2020 02:40:53 No.11443827 Report

Quoted By:

Division by zero produces whatever result I want it to be
Lmao, that is just maths and numbers. They are just a construcy to make reality make sense. Imagine being scared of, ero division in any application.

Anonymous

Anonymous Fri 06 Mar 2020 04:03:37 No.11444026 Report

Quoted By:

>>11442884
>Dividing object among friends
is just a simplified way of understanding division so little kids can grasp
Even then your not counting how much you have left but how much every “person” in the divisor receives
So if I can theoretically slice a cake into an infinite amount of pieces and I distribute said pieces equally among zero people then there is infinite cake slices for those zero people to receive

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