Dividing by zero, the truth (this is long!)

[The_root_of_all_evil]

Ahh, dude, there aren't theorems outside 1 & 2 (which are more propositions imo, but whatever).

The answer is yes, and theorem 3 and its remarks should clarify everything.

Theoreum 4 is a counter proposition.

Also he is working in abstract algebra, not symbolic logic or some whacky shit, the additive identity, 0 and zero (as an element) are used as synonims and of course they are there (field, duh).

We call it the zero element because this unique element behaves like the familiar zero of the real numbers.

Like isn't equal. Need a logical equal before we define it as a real number and not just an element.

About him missing some proofs of basic properties of the field? Yeah, and anyone that can read and understand that can also do those, the rest will only read the TL;DR, so i don't see the issue.

Because he's not saying if the fields of addition are equally active in the fields of multiplication. Multiplication isn't a 1 to 1 determinacy because of 0 and -x*-y=x*y.

 

[Xeorm]

Unless I'm mistaken, the issue with dividing by zero has never been that it's impossible per say, but that it's useless to do and means nothing if used.

As in, your definition for a binary operator requires that the function return a value in the set, which is true for 0/0, it just returns every value in the set.

 

[Tanakh]

The answer is yes, and theorem 3 and its remarks should clarify everything.

Theoreum 4 is a counter proposition.

I see him saying that, but I didn't saw the actual wording of the theorem nor it's proof, so assumed everything after the conclusion was just chit-chat. Can you enunciate those two explicitedly plz?

Also he is working in abstract algebra, not symbolic logic or some whacky shit, the additive identity, 0 and zero (as an element) are used as synonims and of course they are there (field, duh).

We call it the zero element because this unique element behaves like the familiar zero of the real numbers.

Like isn't equal. Need a logical equal before we define it as a real number and not just an element.

Mhee, if it quacks like 0, walks like 0, has the algebraic properties of 0 and behaves like 0 under morphisms, I'll call it 0; i know he is skipping the identification of both, but cmon, it's like math 101 but to be fair i always ran from logic like the plague, don't like that field of maths at all, so yeah, strictly speaking he should prove that both things are the same.

Because he's not saying if the fields of addition are equally active in the fields of multiplication. Multiplication isn't a 1 to 1 determinacy because of 0 and -x*-y=x*y.

Yeah, I guess he should specify that, but again, if you know enoguh to comprehend the general idea, those small lacks can be easily filled.


Bear in mind that I am a firmly beliver in our dear God and Saviour the Axiom of Choice and hate rigor unless skipping it makes things muddy (which i dont belive does in this case).

 

[The_root_of_all_evil]

The answer is yes, and theorem 3 and its remarks should clarify everything.

Theoreum 4 is a counter proposition.

I see him saying that, but I didn't saw the actual wording of the theorem nor it's proof, so assumed everything after the conclusion was just chit-chat. Can you enunciate those two explicitedly plz?

He brought forward Theoreum 3 as his answer - Which I assume is the zero element.

Theoreum 4 is the counter proposal that he needs to be able to disprove.

Mhee, if it quacks like 0, walks like 0, has the algebraic properties of 0 and behaves like 0 under morphisms, I'll call it 0; i know he is skipping the identification of both, but cmon, it's like math 101 but to be fair i always ran from logic like the plague, don't like that field of maths at all, so yeah, strictly speaking he should prove that both things are the same.

Yeah, he's proved there is a Zero Element that exists in the real number field which can be regarded as zero - but at the moment it's only self-defined - there needs to be a further step where he proves that the Zero Element IS 0, rather than acts as.

Because he's not saying if the fields of addition are equally active in the fields of multiplication. Multiplication isn't a 1 to 1 determinacy because of 0 and -x*-y=x*y.

Yeah, I guess he should specify that, but again, if you know enoguh to comprehend the general idea, those small lacks can be easily filled.

The obvious point being that while it's still a Zero Element, then we can have a negative Zero Element - and, depending on your strict definition, he's just declared zero as a prime number.

 

[isometry]

You're right that it just comes down to the definition of division. But you don't need fields or rings for any of this. The first direction of your proof of theorem 2 boils down to "if 0 divides x then by the definition of division in the reals there exists a real number r such that 0*r = x, so x = 0 since r*0 = 0 for all real numbers r."

Using rings and fields to prove that only 0 divides and is divisible by 0 is like using a shotgun to kill a mosquito. The weaker the premise, the stronger the proof.

Still, I suppose it's a good exercise for a student to practice these definition. Try using the basic properties of vector spaces to prove r^0 = 1 for all non-zero real numbers r.

 

[Maze1125]

0 is only real if you accept it as the zero element. It's not proven from base logic.
In the stated set of R, you have not shown whether the rules for addition are also available for the multiplication, which leads to Theorem 4.

He says he's talking about fields in general. Which have a zero element by definition.
And, although it's good practice to, he has absolutely no need to prove that R is a field, with 0 as the zero element, as that is accepted mathematical fact. The definition of a field was invented so it could be applied to R no less.

Theorem 3 - unless I'm missing something - isn't up there?

I noticed that too, I assumed he was talking about theorem 2 with a typo.

Theorem 4: X/0 is undefined. As N*0=0 and N<=>X. That directly contradicts Theorem 2.

I'm not sure what you're trying to say there, could you clarify?

 

[Maze1125]

You're right that it just comes down to the definition of division. But you don't need fields or rings for any of this. The first direction of your proof of theorem 2 boils down to "if 0 divides x then by the definition of division in the reals there exists a real number r such that 0*r = x, so x = 0 since r*0 = 0 for all real numbers r."

Using rings and fields to prove that only 0 divides and is divisible by 0 is like using a shotgun to kill a mosquito. The weaker the premise, the stronger the proof.

I disagree that his is the weaker proof.

You're saying "Take division in the reals, then..." he's saying "Take division in a general field, then...".
He uses a more generalised system and so proves more.

 

[targren]

All of this is really mathematical navel-gazing since they ignore one very important point: Division by zero is undefined by definition. Specifically for reasons of things like this.

Even if this proof were accepted, what would the value of 0/0 be? It would be the set of all numbers (since even complex numbers obey r * 0 = 0). It's a meaningless value that tells you nothing. That's why it's "undefined."

UPDATE (now that the stupid site is working again):

Rather, let me be clearer and use an actual (informal) proof by contradiction using OP's assertions:

Let r be some real number such that r = 0/0
Then, by the definition of division, r * 0 = 0
These are the mainstays of the above proof.

Now, by the nature of multiplication (r + 1) * 0 = 0
Or (r + 1) = 0/0

By substitution:
r = r+1

There is no r in the domain of real numbers such that r = r + 1 (trivial proof left as an exercise for the reader). The assertion is disproven by contradiction in the domain of real numbers.

 

[The_root_of_all_evil]

Theorem 4: X/0 is undefined. As N*0=0 and N<=>X. That directly contradicts Theorem 2.

I'm not sure what you're trying to say there, could you clarify?

The original argument was to prove that the zero element can be divided by 0. To come full circle, you'd have to prove that the zero element divided by zero has an answer of zero.

The main legworks been done, he just has to replace theoretical values with real ones.

@Targen: also makes a good point. While there is a theoretical Zero Element, there isn't a practical Zero Element.

 

[Kyber]

so... this is the thread where the smart kids hang out huh?

 

[lacktheknack]

Considering how our previous track record of math threads got us on a Cracked list, I'm very scared of how this thread will go.

OT: Sorry, the university flunkie over here can't contribute.

 

[Groog]

All of this is really mathematical navel-gazing since they ignore one very important point: Division by zero is undefined by definition. Specifically for reasons of things like this.

Well, as briefly touched upon previously we have some people who just are awesome. There was this kid in Germany who really wanted to divide with zero, so when he grew up he defined a complex plane where division by zero is defined. His name was Bernhard Riemann - http://en.wikipedia.org/wiki/Riemann_sphere

 

[Hemothorax]

Any number can be divided by zero, and here's proof:

Divide a cake between 2 persons. You get half a cake.
Divide a cake 'between' 1 person: You get a whole cake.
Divide a cake by 0 persons: Cake is still there.

Ergo : x / 0 = x

 

[lacktheknack]

Any number can be divided by zero, and here's proof:

Divide a cake between 2 persons. You get half a cake.
Divide a cake 'between' 1 person: You get a whole cake.
Divide a cake by 0 persons: Cake is still there.

Ergo : x / 0 = x

No.

If you divide a cake between two persons, each of them gets half a cake.
If you divide a cake 'between' one person, they have a whole cake.
If you divide a cake between no one, then no one has no cake, and there's a cake that sitting in the equation entirely unaccounted for.

Ergo: x / 0 = ???? (undefined)

The problem with your example is that you don't differentiate between each person having cake and a cake sitting lonely on the counter. However, that matters. Having cake sitting on the counter means that the equation isn't done processing.

Also, as we approach zero, the cake analogy breaks. What if you divide a cake up amongst half a person? Apparently, a second cake materializes out of nowhere.

 

[targren]

All of this is really mathematical navel-gazing since they ignore one very important point: Division by zero is undefined by definition. Specifically for reasons of things like this.

Well, as briefly touched upon previously we have some people who just are awesome. There was this kid in Germany who really wanted to divide with zero, so when he grew up he defined a complex plane where division by zero is defined. His name was Bernhard Riemann - http://en.wikipedia.org/wiki/Riemann_sphere

And yet he didn't claim it on real numbers in a trivially disproven method like this. And anything is mathematically possible if you change the definition of the operation in question. So your counterexample is a different system using a different definition, and making my point for me? Reimann defines X/0 as infinity. Again, it's a matter of definition. He also had a justification for creating this new system, rather than yet just another naive attempt to "break math."

 

[Tanakh]

Let r be some real number such that r = 0/0
Then, by the definition of division, r * 0 = 0
These are the mainstays of the above proof.

Now, by the nature of multiplication (r + 1) * 0 = 0
Or (r + 1) = 0/0

By substitution:
r = r+1

You are doing a misstep there. Let me give you a hint, you are canceling a mapping that is not a morphism.

Edit: It might be even clearer where you did that if you write the division as an operator from RxR GL, HF

 

[TheApatheticDespot]

I don't have time right now to go through this carefully, but it looks to me like you've subtly redefined divisibility. I don't recall ever having seen a notion of divisibility in rings which didn't stipulate a nonzero divisor, and the only abstract algebra text I have on hand (Contemporary Abstract Algebra 7th ed. by Gallian) does have that requirement. This is basically like "proving" that 1 is a prime number by showing that its only integer divisors are 1 and itself, but ignoring that primes are greater than one by definition. Under your definition you're correct as far as I can tell, but your proof rests on a difference between your definition and what I understand to be the standard definition. Frankly that puts this argument right on the border of the fallacy of equivocation.

 

[Tanakh]

I don't recall ever having seen a notion of divisibility in rings which didn't stipulate a nonzero divisor, and the only abstract algebra text I have on hand (Contemporary Abstract Algebra 7th ed. by Gallian) does have that requirement.

Clark doesn't seem to ask for it either, but Clark is for pussies! Rotman (master of Rhyming the Rhyme Well) says

If x e G and n is a nonzero integer, then x is divisible by n in G if there is g e G with ng = x.

(bold without italics mine, bold and italics in the original)

So, yeah, strictly speaking all this is nonsense, but it's a fun thread and MUCH MUCH better than the average math one

 

[Maze1125]

The original argument was to prove that the zero element can be divided by 0. To come full circle, you'd have to prove that the zero element divided by zero has an answer of zero.

He doesn't have to do that at all. 0/0 = 0 has no part in what he was trying to prove.

@Targen: also makes a good point. While there is a theoretical Zero Element, there isn't a practical Zero Element.

I don't think targren said anything of the sort.
Regardless, I have to ask, what's the difference between a "theoretical Zero Element" and a "practical Zero Element", and what's the relevance?

I don't have time right now to go through this carefully, but it looks to me like you've subtly redefined divisibility. I don't recall ever having seen a notion of divisibility in rings which didn't stipulate a nonzero divisor, and the only abstract algebra text I have on hand (Contemporary Abstract Algebra 7th ed. by Gallian) does have that requirement. This is basically like "proving" that 1 is a prime number by showing that its only integer divisors are 1 and itself, but ignoring that primes are greater than one by definition. Under your definition you're correct as far as I can tell, but your proof rests on a difference between your definition and what I understand to be the standard definition. Frankly that puts this argument right on the border of the fallacy of equivocation.

It seems that might come down to a difference of "standard definition".

For example, you have the standard definition of a prime as being larger than 1 and having divisors of 1 and itself, while I've always used the definition that a prime is a natural number that has exactly two divisors, 1 and itself, with no stipulation on if it is greater than 1 or not.

Both could certainly be considered "standard", and they are equivalent in the Natural Numbers, but they are nevertheless different.

 

[The_root_of_all_evil]

He doesn't have to do that at all. 0/0 = 0 has no part in what he was trying to prove.

"Can any real number be divided by zero and if so what is the result of that operation?"

That's "real number"/0=x; which he states is 0/0=0.

Regardless, I have to ask, what's the difference between a "theoretical Zero Element" and a "practical Zero Element", and what's the relevance?

A theoretical Zero Element is one that performs the operations. A practical one can be placed on the number line. Relevancy is that he's proven the former exists, but the missing Theoreum 3 probably justifies the latter.


Curiosity though: Why are you defending him when he hasn't come back on this yet? Do you know him?