One divided by Zero (1/0)

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Re: One divided by Zero (1/0)

Postby alessandro95 » Sun Feb 09, 2014 7:24 am UTC

exfret wrote:
alessandro95 wrote:That axiom isn't there specifically to exclude ∞, it is there to state a very useful properties of real numbers without which they simply won't work as we wish them to and which happens to exclude infinitesimals and infinities from the field of reals

Okay, but this only means that infinity is only excluded because it's an oddball and not because and not a number, right?

We decide the axioms first and then we call real numbers the field obtained from the axiom, you're doing things in the opposite order, we're not excluding ∞ because it isn't a number, but we're excluding it because it breaks one of the axiom therefore it isn't a real number.
5+3i breaks the oder relation on R so it's not a real number but it can be a number in another system, as well as "apple" can, if you define operations involving it.

exfret wrote:
alessandro95 wrote:R is defined as a field {R,+,*,<}.
Let's not worry about the order relation but only with addition and multiplication for a moment.
The definition of a field is a commutative ring (in this case {R,+,*}), which has a multiplicative inverse for every element but the additive identity, and R contains indeed a multiplicative inverse for number except 0, and it is not supposed to contain one for 0, which is why 1/0 is undefined

I can see that a field is defined for adding, multiplying, and order, (I overlooked that last one) but why do we have to specifically exclude the multiplicative inverse of the additive identity? Even if we should exclude it just so it doesn't violate the real number's properties, why does it have to be excluded from all fields? Also, why must it be undefined just because of this exclusion? It has a value that can be defined. Here: I define x such that x = 1/0. There. Call it meaningless, but it's not undefined. Edit: Call it meaningless but it's not undefined, right?

1/n is the multiplicative inverse of n and since 0 is the additive identity it is defined not to have an inverse or, looking at things the other way around its multiplicative inverse is undefined.
(this is because there isn't a value of x for which 0x=1)
Of course you can say "let x=1/0" but then we are no longer working in a field and we lose a lot of nice properties of fields, there are interesting structures such as wheels where division by 0 is defined but defining the reals as a structure different than a field would bring much more disadvantages than advantages.

exfret wrote:
alessandro95 wrote:1/0 is undefined in the fields of reals for the fields axioms I quoted above

So it's undefined for the reals? But that doesn't make any sense. Edit: But that doesn't make any sense to me. "The Reals" are just an isolated part of mathematics that we created mostly because they can be ordered. 1/0 would be inexistent for the reals. Of course, I still don't understand why you'd have such a restriction to use explicitly only real numbers anyways.

You don't have to restrict yourself to real numbers, but when you work with reals you have to follow the axioms of reals and their implications (1/0 is undefined, ∞ isn't a real number and so on), it is perfectly fine to work in another system (you may want to read about the hyperreals numbers if you didn't already do that) but you can't give a value to things such as 1/0 inside the reals.

exfret wrote:
alessandro95 wrote:I did tell you, it is because of fields axioms

Oh, sorry. I overlooked that because I was expecting different reasons (e.g. zero times any number is always zero). Anyways, the field of axioms would state that it's undefined for a field because it would be the multiplicative inverse of the additive identity (which is what I said myself), is that correct? But, why would it be 'undefined'? It's a number, a defined number, but just one outside of the specific field one might be focusing on. I find it as absurd as calling i undefined when working with reals. Edit: That's just an analogy, I'm not calling your idea absurd.

Of course it could have a defined value inside another algebraic structure, but you need precise axioms describing it and the operations it posses.
I may have expressed myself poorly yesterday (english isn't my language) but what I meant is that 1/0 and ∞ have no meaning inside the field of reals, because there is no way to assign them a value staying inside the field, but that doesn't mean infinity is a meaningless concept, it means ∞ it's not a number but a symbol (when working with the reals of course) and that we cannot evaluate operations involving it in a meaningful way inside the reals.
The primary reason Bourbaki stopped writing books was the realization that Lang was one single person.

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Re: One divided by Zero (1/0)

Postby korona » Sun Feb 09, 2014 12:32 pm UTC

Note that the archimedian property does not only "exclude" a single "number" ∞.
Look at the hyperreal numbers *R. Those are a totally ordered and complete but non-archimedian field. There is not only a single infinite element in *R but there are many infinite elements and many infinitesimal elements (i.e. elements smaller than every real number).
Still 1/0 is not an infinite element but remains undefined because you want the resulting object to carry a field structure.

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Re: One divided by Zero (1/0)

Postby Derek » Sun Feb 09, 2014 8:32 pm UTC

Exfret, if you haven't already then take a look at the axioms of a field. Here are some of the axioms:
1: If x and y are elements of a field, then x + y is an element of the field (closure of addition).
2: If x, y, and z are elements of a field, then (x + y) + z = x + (y + z) (associativity of addition).
3: Every field has an element 0 such that if x is an element of the field, then x + 0 = x (existence of an additive identity).
4: Every field has an element 1 such that if x is an element of the field, then x * 1 = x (existence of a multiplicative identity).
5: 0 != 1 (additive identity and multiplicative identity are different elements of the field).
6: If x is an element of a field, then -x is an element of the field and x + (-x) = 0 (existence of additive inverse).

Using these axioms, here is a proof that ∞ cannot be part of a field, given the property (P) that 1 + ∞ = ∞ (which you have previously stated in this thread).

From (6) we have ∞ + -∞ = 0. Then 1 + (∞ + -∞) = 1 + 0 = 1, but by (2) we can say 1 + (∞ + -∞) = (1 + ∞) + -∞, and from (P) we have (1 + ∞) + -∞ = ∞ + -∞ = 0, so 1 = 0 which is a contradiction of (5).

There are other ways you could try to define ∞ as well, but in order to satisfy (1) you must have some definition for 1 + ∞. But any definition is either going to violate one of the axioms, or is not going to have the properties of ∞ that you want to have.

So to add ∞ to the real numbers, you have to throw out one or more of the field axioms. But without the field axioms, you lose many of the properties of a field, which makes the resulting number system generally less useful.

But if you're willing to accept not having a field, then here is a concrete example of a number system that includes ∞ that is used every day:

Elements: R + {∞, -∞, NaN}
(Let x equal any element except 0, ∞, -∞, NaN. Let y equal any element)

x/0 = ∞ if x > 0, x/0 = -∞ if x < 0
∞ + x = ∞
∞ * x = ∞ if x > 0, ∞ * x = -∞ if x < 0
∞ + ∞ = ∞
∞ - ∞ = NaN
0/0 = NaN
∞/∞ = NaN
∞*0 = NaN
NaN + y = NaN
NaN * y = NaN
∞ > x
-∞ < x
NaN != y (this includes NaN != NaN)

Someone can correct me if I've left out any important properties.

This is the number system used by computers to compute floating point numbers. Note what properties this system has, and what it doesn't have: Addition and multiplication are closed, and ∞ + 1 = ∞. The operations are commutative, but they are not associative or distributive, so it is not a field. It has additive and multiplicative identities, but it lacks additive and multiplicative inverses. Most surprisingly of all, it lacks the reflexive property of equality, because NaN != NaN. These are fundamental properties that we typically take for granted, but if we want ∞ to work then this is what we have to give up.

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Re: One divided by Zero (1/0)

Postby lalop » Mon Feb 10, 2014 3:52 am UTC

tldr: you can define your own reals with 1/0 = ∞ if you really want; it just breaks many axioms that make proofs simpler, while gaining you nothing most of the time.

notzeb wrote:
Schrollini wrote:A set S is well ordered by the operator < iff ∀xyS, x<y or x>y. The projective line does not satisfy that definition for any operator; therefore, it is not well-ordered. It can't be partially well-ordered or kinda well-ordered, because there ain't no such thing.
What are they teaching in schools these days?

A set S is totally ordered by < if any two unequal elements of S are comparable. Well ordered means something completely different! It means, basically, that you can do induction with it (precise definition is that every subset of S has a least element). The reals are not well ordered by the standard ordering (for instance, there is no least positive real number).

Carry on.

Additionally, every set can be well-ordered.

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Re: One divided by Zero (1/0)

Postby Schrollini » Mon Feb 10, 2014 6:10 am UTC

lalop wrote:Additionally, every set can be well-ordered.

Only if you accept the axiom of choice. And of course,
Jerry Bona wrote:The Axiom of Choice is obviously true, the well–ordering principle is obviously false, and who can tell about Zorn’s Lemma?

That said, I think that my original assertion that the projective real line cannot be totally ordered is just plain wrong. I see no reason that I could define ∞ > a for all other elements a (or for that matter ∞ > a for all a < 27 and ∞ < a for all a ≥ 27). I think what I meant to say was that there is no total ordering of the real projective line that plays nicely with our definitions of addition and multiplication.

I'll let someone else say that in math, or point out that I'm wrong once again.
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