/sci/ - Science & Math » Thread #12245122

165B, 8x8, ezgif-2-be73b024c3b2.gif

View Same Google iqdb SauceNAO

1/0 is not undefined

Anonymous Sun 18 Oct 11:51:11 2020 No.12245122 View Reply Original Report

Quoted By: >>12245136 >>12245279 >>12245298 >>12245589 >>12246112 >>12246163 >>12246814

I am absolutely certain 1/0 is not undefined.

Assuming
1/0 is undefined.
0/0 is indeterminate.
(x*y)/z = (x/z)*y

(0*1)/0 = 0/0. 0/0 is indeterminate
(1/0)*0 is undefined, as multiplying an undefined number by 0 is still undefined.
Therefore, 0/0 is both undefined and indeterminate, which is a contradiction.

How can something both be undefined and indeterminate?

Anonymous

Anonymous Sun 18 Oct 2020 11:54:22 No.12245128 Report

Quoted By: >>12245135

>How can something both be undefined and indeterminate?
They are synonyms you fucking imbecil

Anonymous

Anonymous Sun 18 Oct 2020 11:56:11 No.12245135 Report

Quoted By:

>>12245128
You are the imbecile. Undefined means there is no answer. Nothing times 0 equals one, so 1/0 must be undefined if it weren't for this contradiction. Anything times 0 is 0, so 0/0 is indeterminate.

I don't know if these are the right words, but the solution set for an undefined thing would be an empty set, while a solution set for an indeterminate thing would be a set of all real numbers.

Anonymous

Anonymous Sun 18 Oct 2020 11:56:34 No.12245136 Report

Quoted By: >>12245145 >>12245489

>>12245122
I envy your stupidity OP.
I wish I could live so simply and content.

Anonymous

Anonymous Sun 18 Oct 2020 12:00:03 No.12245145 Report

Quoted By: >>12245155

>>12245136
Gotta love the way how you just call me a dumbass as opposed to showing how I'm a dumbass.

Anonymous

Anonymous Sun 18 Oct 2020 12:05:01 No.12245155 Report

Quoted By:

>>12245145
awwww it's retarded

Anonymous

Anonymous Sun 18 Oct 2020 12:57:54 No.12245279 Report

Quoted By: >>12246189

>>12245122
>(x*y)/z = (x/z)*y
Not correct when z = 0.

Anonymous

Anonymous Sun 18 Oct 2020 13:06:30 No.12245298 Report

Quoted By: >>12245489 >>12246189 >>12246898

>>12245122
>Therefore, 0/0 is both undefined and indeterminate, which is a contradiction.
Incorrect, all indeterminate forms are undefined. Indeterminate forms are undefined forms which can be represented as limits with multiple values.

Anonymous

Anonymous Sun 18 Oct 2020 14:21:30 No.12245489 Report

Quoted By: >>12245497

This guy >>12245298 is an idiot
This guy >>12245136 is a troll
1/0 is perfectly well-defined. 0/0 is also perfectly well-defined, albeit indeterminate.

Anonymous

Anonymous Sun 18 Oct 2020 14:23:56 No.12245497 Report

Quoted By: >>12245530

>>12245489
Wrong, retard.

https://mathworld.wolfram.com/Undefined.html

Anonymous

Anonymous Sun 18 Oct 2020 14:34:32 No.12245530 Report

Quoted By: >>12245532 >>12245637

>>12245497
>b-b-but big daddy wolfram said so!
1/0 is defined. You can use it like a number, just like zero. Now go do math with it.

Anonymous

Anonymous Sun 18 Oct 2020 14:35:40 No.12245532 Report

Quoted By: >>12245536

>>12245530
No, it's undefined.

Anonymous

Anonymous Sun 18 Oct 2020 14:37:15 No.12245536 Report

Quoted By: >>12245539

>>12245532
No wonder you're so bad at math.

Anonymous

Anonymous Sun 18 Oct 2020 14:38:17 No.12245539 Report

Quoted By: >>12245542

>>12245536
It's undefined regardless of your opinions. Cry more.

Anonymous

Anonymous Sun 18 Oct 2020 14:40:02 No.12245542 Report

Quoted By: >>12245567

>>12245539
still waiting to hear an argument, fag. No, deferring to authority doesn't count.

Anonymous

View Same Google iqdb SauceNAO k0IGUXx.jpg, 101KiB, 785x731

Anonymous Sun 18 Oct 2020 14:50:13 No.12245567 Report

Quoted By:

>>12245542
>makes false claim with no argument
>demands argument in response
What's good for the goose is good for the gander. If you want an argument you can click on division by zero in the link I gave you. It's that too hard for you?

Anonymous

Anonymous Sun 18 Oct 2020 14:56:38 No.12245589 Report

Quoted By: >>12245692

>>12245122
what does "undefined" mean?
what does "indeterminate" mean?

Anonymousz

Anonymousz Sun 18 Oct 2020 15:09:18 No.12245637 Report

Quoted By: >>12246189

>>12245530
> 1/0
It's defined for wheels. It's not defined for rings, such as the integers or fields, such as the rationals, or reals..

Anonymous

Anonymous Sun 18 Oct 2020 15:11:22 No.12245643 Report

Quoted By:

wow.

Anonymous

Anonymous Sun 18 Oct 2020 15:24:32 No.12245692 Report

Quoted By: >>12245914

>>12245589
OP is glossing the distinction between the fact of an expression having an undefined value, which is a statement about the set of definitions currently in use, and the reasons why an expression might be left undefined.
An appeal to authority is entirely appropriate in response to the first question. A general survey of practicing mathematicians would also be appropriate, but impractical. The second question might be responded to by showing that a field or ring that assigns a value to 1/0 can have only one element, if that could be shown.

Anonymous

Anonymous Sun 18 Oct 2020 16:58:38 No.12245914 Report

Quoted By: >>12245961

>>12245692
>The second question might be responded to by showing that a field or ring that assigns a value to 1/0 can have only one element
Could you elaborate on this sentence? it doesn't make sense.
How does assigning value to 1/0 necessitate a single element field/ring, and why is that relevant to whether or not 0/0 is defined?
Did you perhaps mean, to prove that 0 * (1/0) could only possibly be one element?

Anonymous

Anonymous Sun 18 Oct 2020 17:23:23 No.12245961 Report

Quoted By: >>12246064 >>12246120

>>12245914
in any ring, a/b is by definition a*(1/b), where 1/b is by definition the unique element such that it gives 1 when multiplied with b. if you assume that 1/0 exists AND the ring axioms hold, you can easily prove that all elements of this ring are equal.

Anonymous

Anonymous Sun 18 Oct 2020 17:34:26 No.12245999 Report

Quoted By: >>12246054 >>12246120

just define x/0=c lmao problem solved
if they can do it for sqrt(-x) they can do it for x/0

Anonymous

Anonymous Sun 18 Oct 2020 17:49:32 No.12246054 Report

Quoted By: >>12246358

>>12245999
thing is, now you have to define arithmetic operations for c and determine what (if any) field axioms still hold. imaginary unit is useful, because the resulting algebraic structure is still a field, the only thing you lose is ordering.

Anonymous

Anonymous Sun 18 Oct 2020 17:53:25 No.12246064 Report

Quoted By: >>12246100

>>12245961
>if you assume that 1/0 exists AND the ring axioms hold, you can easily prove that all elements of this ring are equal.
equal, yes, that is what I mean here by indeterminate. Loosely speaking, you can say each element has equal "probability" of being 0/0.
At the same time, 0/0 can only equal one element of the set - hence, it is "defined" as a single element in the reals. You should be able to easily prove that too.

Anonymous

Anonymous Sun 18 Oct 2020 18:04:48 No.12246100 Report

Quoted By:

>>12246064
You have no clue what you're talking about.

Anonymous

Anonymous Sun 18 Oct 2020 18:07:38 No.12246112 Report

Quoted By: >>12246192

>>12245122
1/0 = 0^-1

DUMBBREEDER

Anonymous

Anonymous Sun 18 Oct 2020 18:09:16 No.12246120 Report

Quoted By: >>12246189 >>12246457

>>12245961
That's true for a field. In a ring you can sometimes do a/b even if 1/b isn't a ring element: for example 4/2 is an integer, although 1/2 isn't an integer.
>>12245999
But even in a ring, if 1/x is defined at all, x*(1/x) = (1/x)*x = 1.
Now you can show 0*x = x*0 = 0 for all elements x of a ring: 0*x = 0*x + 0 = 0*x + x + (-x) = 0*x + 1*x + (-x) = (0+1)*x + (-x) = 1*x + (-x) = x + (-x) = 0.
So if 0 * (1/0) = 1, then 0 = 1, and for any y at all, y = 1*y = 0*y = 0.
Thus the only ring in which 1/0 can be meaningful is the ring with only one element.

In wheels they have different axioms, so that some of these steps don't work, and 1/0 can be a valid element of the wheel.

Now all of this is Algebra. I think "indeterminant form" is more of the language of Analysis.

Anonymous

Anonymous Sun 18 Oct 2020 18:23:31 No.12246163 Report

Quoted By: >>12246372

>>12245122
Op here. I forgot to add that i have yet to see a singular contradiction if you say 0/0 and 1/0 are both 0

Anonymous

Anonymous Sun 18 Oct 2020 18:32:18 No.12246189 Report

Quoted By: >>12246205 >>12246247 >>12246346

>>12245298
Indeterminate and undefined are different. While indeterminate is indeed a subset of undefined, the way how 0/0 is undefined and how 1/0 is undefined are polar opposites.
>>12245279
Why does the distributive property just cease working when the denominator is 0?
>>12245637
What actually are wheels and rings? I understand fields i think, but I have never had wheels or rings explained to me.
>>12246120
I'm pretty sure i have seen a proof almost exactly like this, and i'm 99% certain this just proves that 0/0 is not 1, which is completely reasonable.

Anonymous

Anonymous Sun 18 Oct 2020 18:33:01 No.12246192 Report

Quoted By: >>12246199

>>12246112
1/0 is not equal to anything, it's undefined.

Anonymous

Anonymous Sun 18 Oct 2020 18:35:07 No.12246199 Report

Quoted By: >>12246265

>>12246192
It is equivalent to 0^-1. Do you not know how negative powers work? Just because you say something is undefined doesn't mean it is not equal to forms that are equivalent.

Anonymous

Anonymous Sun 18 Oct 2020 18:37:08 No.12246205 Report

Quoted By: >>12246215

>>12246189
>Indeterminate and undefined are different.
I didn't say they weren't.

>the way how 0/0 is undefined and how 1/0 is undefined are polar opposites.
I don't even know what "polar opposite" means in this context but both are undefined because any division by zero is undefined, because any number multiplied by 0 is 0.

Anonymous

Anonymous Sun 18 Oct 2020 18:40:01 No.12246215 Report

Quoted By: >>12246288 >>12246305

>>12246205
0/0 is indeterminate because any number multiplied by 0 is 0. Literally what you said. They are polar opposite as in 1/0 has no answer, while 0/0 could potentially be any answer.

Anonymous

Anonymous Sun 18 Oct 2020 18:48:45 No.12246247 Report

Quoted By: >>12246283

>>12246189
>Why does the distributive property just cease working when the denominator is 0?
Because division by zero is not a valid operation. The distributive property applies to binary operations and operands.

Anonymous

Anonymous Sun 18 Oct 2020 18:51:54 No.12246265 Report

Quoted By: >>12246283

>>12246199
>Do you not know how negative powers work?
I do, negative powers on 0 are undefined. Saying something undefined is
equivalent to anything is gibberish.

Anonymous

Anonymous Sun 18 Oct 2020 18:54:55 No.12246283 Report

Quoted By: >>12246333

>>12246247
How about this. Instead of saying 0, I will say its just x. Distributive property still works when its just 1/x right? So I just do what I did before, but with x instead of 0, and then at the end just replace the x with 0.
>>12246265
Same as above for you.

Anonymous

Anonymous Sun 18 Oct 2020 18:55:49 No.12246288 Report

Quoted By:

>>12246215
That has nothing to do with what indeterminancy means. Something can be indeterminate and undefined, or indeterminate and well-defined. "could be any answer" has nothing whatsoever to do with it.

Anonymous

Anonymous Sun 18 Oct 2020 18:58:56 No.12246305 Report

Quoted By: >>12246317

>>12246215
>0/0 is indeterminate because any number multiplied by 0 is 0.
No, that has nothing to do with being indeterminate. 0/0 is indeterminate because it can be represented by several different forms whose limits are different.

For example as x -> 0

x^2/x -> 0
x/x -> 1
x/x^3 -> inf

>Literally what you said.
Literally not what I said.

Anonymous

Anonymous Sun 18 Oct 2020 19:01:11 No.12246317 Report

Quoted By: >>12246825

>>12246305
Limits are only approximations of what the function does at that point.

I truly do not understand how people can say that indeterminate and undefined are the same thing.

Anonymous

Anonymous Sun 18 Oct 2020 19:04:36 No.12246333 Report

Quoted By: >>12246406

>>12246283
>How about this. Instead of saying 0, I will say its just x.
No, it's undefined.

>Distributive property still works when its just 1/x right?
No, distributive property only applies to operations on valud operands. An undefined number is not a valid operand.

>So I just do what I did before, but with x instead of 0, and then at the end just replace the x with 0.
This is just logically faulty. If it wouldn't work when you use 0 it won't work by substituting 0 for a variable you assume everything is valid for.

Anonymous

Anonymous Sun 18 Oct 2020 19:09:05 No.12246346 Report

Quoted By:

>>12246189
I think Wikipedia has reasonable pages on the usual Algebraic objects like monoids, groups, rings, and fields, and also wheels.

Assuming you are thinking about the algebraic sort of field, and not the physics sort of field...
A ring is like a field, except that nonzero elements don't necessarily have multiplicative inverses. So a typical field might be the rational numbers, while a typical ring might be the integers.
A wheel is a something that I think was invented to make division by zero meaningful. It doesn't satisfy all of the axioms of a ring or field, but I forget exactly how.

Anonymous

Anonymous Sun 18 Oct 2020 19:16:53 No.12246358 Report

Quoted By: >>12246386 >>12246389

>>12246054
>you lose ordering
just count all the imaginary units, then the real ones. There, perfect order.

Anonymous

Anonymous Sun 18 Oct 2020 19:21:58 No.12246372 Report

Quoted By: >>12246406

>>12246163
Indeed, there are no contradictions if you say everything is 0, at least from the point of view of algebraic properties. It's just a very small algebra with only one element.

Anonymous

Anonymous Sun 18 Oct 2020 19:24:37 No.12246383 Report

Quoted By:

"indeterminate" makes sense only in the context of limits, "undefined" is a more general term. all indeterminate expressions are undefined, but for example 1/0 is undefined but not indeterminate.

Anonymous

Anonymous Sun 18 Oct 2020 19:25:27 No.12246386 Report

Quoted By:

>>12246358
Are you proposing a+ib > c+id if and only if either b>d or b=d and a>c?

Anonymous

Anonymous Sun 18 Oct 2020 19:25:39 No.12246389 Report

Quoted By:

>>12246358
what about complex numbers which are not real or imaginary?

Anonymous

Anonymous Sun 18 Oct 2020 19:30:39 No.12246406 Report

Quoted By: >>12246831

>>12246372
Nope, I am saying in a ring or field or whatever with more then one element.

>>12246333
You still haven't shown why I can't manipulate things when I am dividing by 0.

Anonymous

Anonymous Sun 18 Oct 2020 19:46:02 No.12246457 Report

Quoted By: >>12246483

>>12246120
>0 * (1/0) = 1, then 0 = 1,
wtf are you smoking?
this either simplifies to
> $\frac{0}{0} = 1$
or
> $0*1 = 1*0$
you haven't proven shit

Anonymous

Anonymous Sun 18 Oct 2020 19:51:37 No.12246475 Report

Quoted By:

In rings and fields, the primary definitions are in terms of addition and multiplication, the identities for these operations, and the inverses of elements with respect to these elements and these operations. Subtraction and division are secondary.
In a field, division is multiplication by a multiplicative inverse. I.e, a/b == a * (1/b), for nonzero b.
In a ring, you may sometimes want to divide by elements that don't have a multiplicative inverse. Then you might say, a/b is the unique c such that b*c = a, if such an element exists and is unique. If a ring is ordered, for example by absolute value in the case of integers, what happens if we say a/b is the least value c such that bc = a? Something like this is useful in the world of matrix-vector math, where, if B is not of full rank you might want to find the vector c of minimum norm such that Bc = a.
This would give 0/0 =0. What breaks?

Anonymous

Anonymous Sun 18 Oct 2020 19:53:45 No.12246483 Report

Quoted By: >>12246496

>>12246457
That's using the bloody definition of 1/x, as that thing, which, when multiplied by x, gives you 1.

Anonymous

Anonymous Sun 18 Oct 2020 19:56:36 No.12246496 Report

Quoted By: >>12246510

>>12246483
yes, it gives you 1, not 0. Where did the 0 on the LHS come from?

You didnt seriously say "0 times anything, even (1/0), is always 0" right? because that would be a serious brainlet take.

Anonymous

Anonymous Sun 18 Oct 2020 19:59:38 No.12246510 Report

Quoted By: >>12246542 >>12246583

>>12246496
Did your eyes glaze over? I proved it.

Anonymous

Anonymous Sun 18 Oct 2020 20:09:33 No.12246542 Report

Quoted By: >>12246611

>>12246510
ah you mean this?
> 0*x = 0*x + 0
if anything it just highlights the dangers of taking the additive identity for granted.

You are literally saying x = x+1, and then wondering why 0=1.

Anonymous

Anonymous Sun 18 Oct 2020 20:19:24 No.12246583 Report

Quoted By:

>>12246510
>0*x = 0*x + 0 = 0*x + x + (-x) = 0*x + 1*x + (-x) = (0+1)*x + (-x) = 1*x + (-x) = x + (-x) = 0.
That is, we start with 0*x.
We then perform a series of operations that won't change its value.
i) add 0, the additive identity, giving: 0*x + 0
ii) substitute for 0 the expression "x + (-x))", since any x has an additive inverse -x, giving 0*x + x + (-x)
iii) replace one of the "x"s by "1*x", since 1 is a multiplicative identity, giving 0*x + 1*x + (-x)
iv) apply distributive law, giving (0+1)*x + (-x).
v) substitute 1 for 0+1, since 0 is additive identity, giving 1*x + (-x)
vi) substituting x for 1*x, since 1 is multiplicative identity, giving x + (-x)
v) since (-x) is additive inverse of x, this evaluates to 0.
That is 0*x = 0 for ANY element x of a ring. If you want an element x such that 0*x is not 0, you aren't in a ring any more.
(Maybe you are in a wheel.)

Anonymous

Anonymous Sun 18 Oct 2020 20:26:19 No.12246611 Report

Quoted By: >>12246752

>>12246542
What do you mean, "taking ... for granted"? The assertion was made for elements of a ring, so of course the definition of a ring, including its identity elements , was used.

Anonymous

Anonymous Sun 18 Oct 2020 21:05:28 No.12246752 Report

Quoted By: >>12246803

>>12246611
>including its identity elements
0 is not an additive identity element here

Anonymous

Anonymous Sun 18 Oct 2020 21:20:26 No.12246803 Report

Quoted By: >>12246854

>>12246752
What is the additive identity element here, then?
Usual terminology for a ring is 0 for the additive identity element, 1 for the multiplicative identity element, unless you are constructing the ring elements from something else that already defines these terms. That's how I was using the terms. Please feel free to read my claim as: If you have a ring where 1/0 is defined, where 1 is the multiplicative identity and 0 the additive identity, then the ring only has one element, and these identity elements are the same.
Do you have an idea for a ring where 1/0 has a meaningful value, and 1 and 0 have other meaningful values? If you say "1" is just a funny way of writing "two", and "0" is just a funny way of writing "one", I will laugh, briefly.

Anonymous

Anonymous Sun 18 Oct 2020 21:23:27 No.12246814 Report

Quoted By:

>>12245122
The expression 1/0 is syntactically correct and semantically meaningless, which is why it appeals to retards.

It means, literally, "That number which, when multiplied by 0, equals 1".

Since no such number exists, it's crap. Nothing; nada; meaningless, mu.

Anonymous

Anonymous Sun 18 Oct 2020 21:25:19 No.12246825 Report

Quoted By: >>12246898

>>12246317
>Limits are only approximations of what the function does at that point.
LOL no. Limits are not approximations and the functions don't have to actually reach their limits. It's like you have a 12 year old's understanding of math.

>I truly do not understand how people can say that indeterminate and undefined are the same thing.
No one said they are. Learn how to read.

Anonymous

Anonymous Sun 18 Oct 2020 21:26:31 No.12246831 Report

Quoted By: >>12246898

>>12246406
>You still haven't shown why I can't manipulate things when I am dividing by 0.
Because dividing by 0 is an undefined operation. You might as well manipulate 1 divided by cow.

Anonymous

Anonymous Sun 18 Oct 2020 21:26:33 No.12246832 Report

Quoted By:

these threads are why /sci/ is so shit. stinks of underage highschool pseuds.

Anonymous

Anonymous Sun 18 Oct 2020 21:32:48 No.12246854 Report

Quoted By: >>12246924

>>12246803
>What is the additive identity element here, then?
I don't know. But it definitely isn't 0.
>1 and 0 have other meaningful values
1 is still the multiplicative identity, and 0 is both signless and an absorbing element. That good enough?

Anonymous

Anonymous Sun 18 Oct 2020 21:45:45 No.12246898 Report

Quoted By: >>12248079

>>12246825
>>12245298
>Incorrect, all indeterminate forms are undefined. Indeterminate forms are undefined forms which can be represented as limits with multiple values.
If that's not you then someone did.

>>12246831
I can do whatever I want to 1/cow. Doesn't mean I will get a useful answer out, but 2*(1/cow) is the same as (2*1)/cow. Commutative property still works, and it makes no sense either way.

Anonymous

Anonymous Sun 18 Oct 2020 21:53:12 No.12246924 Report

Quoted By:

>>12246854
>That good enough?
Nope.

Anonymous

Anonymous Mon 19 Oct 2020 04:17:59 No.12248079 Report

Quoted By:

>>12246898
>If that's not you then someone did.
That specifically says they are different. I'm sorry you don't understand basic English and logic and think that a set and a subset of that set are the same thing.

>I can do whatever I want to 1/cow
Yes, and you can do whatever you want to 1/0. That doesn't mean that whatever you do is mathematically valid.

>but 2*(1/cow) is the same as (2*1)/cow.
Gibberish. Saying the opposite has as much validity, which is none.

Capcode	All Only User Posts Only Moderator Posts Only Admin Posts Only Developer Posts
Show Posts	All Only With Images Only Without Images
Deleted Posts	All Only Deleted Posts Only Non-Deleted Posts
Ghost Posts	All Only Ghost Posts Only Non-Ghost Posts
Post Type	All Only Sticky Threads Only Opening Posts Only Reply Posts
Results	All Grouped By Threads
Order	Latest Posts First Oldest Posts First

Your latest searches