Dividing anything by 0? Watch

Oh I Really Don't Care
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#61
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#61
(Original post by birdsong1)
\infty is misinforming notation. If you notice, the poster was talking about infinite numberS which merely extend the natural numbers. Given an infinite number N, the additive inverse would simply be -N, and the multiplicative inverse 1/N (an infinitesimal).

Associativity of addition is one of the properties that are transferred into this field. With your example, it is not true that -N + 6 = -N because (if I am not mistaken) the statement \forall x \in \mathbb{N}, x + 6 \neq x is transferred.
What are you talking about?
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Dadeyemi
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#62
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1/0 = 0 obviously,

x=1/0=(1^2)/(0^2)= (1/0)^2=x^2 so x=0

It's simple maths.
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birdsong1
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#63
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#63
(Original post by DeanK22)
?? Basically everything - lose additive inverses, multiplicative inverses.
Nah, because
\forall x \in \mathbb{R}, \exists y \in \mathbb{R} : x*y = 1
\forall x \in \mathbb{R}, \exists y \in \mathbb{R} : x + y = 0
are transferred. You might want to check his link.
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Oh I Really Don't Care
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#64
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#64
(Original post by birdsong1)
Nah, because
\forall x \in \mathbb{R}, \exists y \in \mathbb{R} : x*y = 1
\forall x \in \mathbb{R}, \exists y \in \mathbb{R} : x + y = 0
are transferred. You might want to check his link.
the first one of those statements is false ...
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Derringer
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#65
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#65
has anyone ever sucessfully divided something into 1? or is that just an oxymoron?
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ziedj
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#66
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#66
(Original post by birdsong1)
Nah, because
\forall x \in \mathbb{R}, \exists y \in \mathbb{R} : x*y = 1
\forall x \in \mathbb{R}, \exists y \in \mathbb{R} : x + y = 0
are transferred. You might want to check his link.
does 0 not belong to R ?
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birdsong1
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#67
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#67
(Original post by DeanK22)
What are you talking about?
So my previous answer might not have been fair. Basically, there are many models which satisfy the axioms of the real numbers, the most common one being the standard model we all know about. But there are also nonstandard models which contain infintesimal and infinite numbers. These models have to satisfy the axioms with respect to their members (that's the definition of a model) but they might not satisfy them with respect to the standard members. (When the standard members, the usual reals, are considered to be embedded in this nonstandard model.)

For example, the archemedian principle, that \forall x \in \mathbb{R} \exists y \in \mathbb{N} : \frac{1}{y} < x must be satisfied for the nonstandard model, but is not satisfied if y only ranges over the standard integers (x can be infintesimal, so that y needs to be infinite, that is, bigger than any standard integer).

In order to make this work with more than first order sentences, more complicated "superstructures" have to be considered with less obvious embeddings. One result is that not all statements get transferred in cleanly, like for example the least upper bound statement. But that's too much explanation.
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birdsong1
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#68
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#68
(Original post by DeanK22)
the first one of those statements is false ...
Hm, yeah, but I think the corresponding sentence is transferred also (worded \forall x \in \mathbb{R}, \exists y \in \mathbb{R} :  x = 0 \vee xy = 1)
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Oh I Really Don't Care
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#69
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#69
(Original post by birdsong1)
So my previous answer might not have been fair. Basically, there are many models which satisfy the axioms of the real numbers, the most common one being the standard model we all know about. But there are also nonstandard models which contain infintesimal and infinite numbers. These models have to satisfy the axioms with respect to their members (that's the definition of a model) but they might not satisfy them with respect to the standard members. (When the standard members, the usual reals, are considered to be embedded in this nonstandard model.)

For example, the archemedian principle, that \forall x \in \mathbb{R} \exists y \in \mathbb{N} : \frac{1}{y} < x must be satisfied for the nonstandard model, but is not satisfied if y only ranges over the standard integers (x can be infintesimal, so that y needs to be infinite, that is, bigger than any standard integer).

In order to make this work with more than first order sentences, more complicated "superstructures" have to be considered with less obvious embeddings. One result is that not all statements get transferred in cleanly, like for example the least upper bound statement. But that's too much explanation.
You aren't making alot of sense. Up to isomorphism the real numbers (that is a complete linearly ordered field) is unique. You can come up with loads of these things - like fields that are non-archimidean, fields that don't obey the completeness axioms, fields that aren't ordered, etc but no one wants to use them for analysis, roughly speaking. Besides the point you just seem to be rattling on about setting up a model for some axioms, but aren't telling me what the axioms are or telling me what any of your definitions or sets are or what properties half your elelemnts have ...
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birdsong1
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#70
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#70
(Original post by DeanK22)
You aren't making alot of sense. Up to isomorphism the real numbers (that is a complete linearly ordered field) is unique.
Not exactly, because you are missing an essential property, the lub, which "the set of all infintesimals" obviously does not have.

(Original post by DeanK22)
You can come up with loads of these things - like fields that are non-archimidean, fields that don't obey the completeness axioms, fields that aren't ordered, etc but no one wants to use them for analysis, roughly speaking.
A transfer principle of statements between the two models helps a lot in this regard.

(Original post by DeanK22)
Besides the point you just seem to be rattling on about setting up a model for some axioms, but aren't telling me what the axioms are or telling me what any of your definitions or sets are or what properties half your elelemnts have ...
Basically, not only would it be way too long, but I don't even know half this stuff (although admittedly I should look it up and try to explain to you as an exercise). For more immediate results, however, look it up: nonstandard analysis.
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Saichu
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#71
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#71
(Original post by birdsong1)
(although admittedly I should look it up and try to explain to you as an exercise)
(Original post by DeanK22)
You aren't making alot of sense. Up to isomorphism the real numbers (that is a complete linearly ordered field) is unique. You can come up with loads of these things - like fields that are non-archimidean, fields that don't obey the completeness axioms, fields that aren't ordered, etc but no one wants to use them for analysis, roughly speaking.
I randomly came across this thread a few weeks ago, and I believe I have beaten you both to the punch.

http://www.math.tamu.edu/~saichu/Nonstandard.html*

(I've been meaning to write such an introduction for a while now.)

To answer your question, Dean, what distinguishes the hyperreals from the reals is that not all subsets have the lub property, only the so-called internal subsets.

The distinguishing factor is not, as common misconception puts it *cough Ian Stewart's revision of "What is Mathematics" cough*, that the hyperreals simply don't satisfy the reals' second-order statements. If they didn't, they'd be worthless. But they satisfy them in a limited way, and this limited way is good enough to make them still fruitful in analysis.



*Still taking suggestions/revisions
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lfcwes605
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#72
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#72
(Original post by Dadeyemi)
1/0 = 0 obviously,

x=1/0=(1^2)/(0^2)= (1/0)^2=x^2 so x=0

It's simple maths.
1/0 doesn't equal 0,
if it did then you would get
1 = 0^2
which would lead to 1 = 0
Your simple maths is wrong, sorry
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JAKstriked
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#73
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#73
(Original post by lfcwes605)
1/0 doesn't equal 0,
if it did then you would get
1 = 0^2
which would lead to 1 = 0
Your simple maths is wrong, sorry
You don't think it's possible he was joking?
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lfcwes605
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#74
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#74
(Original post by JAKstriked)
You don't think it's possible he was joking?
Haha good point....hmmm well if anyone believes 1/0 can equal 0 they have an answer there for why it doesn't
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mustang1972
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#75
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#75
These type of threads have a number of responses

(i) People blindly quoting cliches they have no idea about
(ii) People not reading messages before them, then posting **** despite it clearly not being the case
(iii) Some good mathematical discussion

Shame the majority doesnt fit into the last part, but well done to those who have pushed the thread the right way.

IM 100% SURE SOME FOOL WILL POST AFTER THIS ABOUT \iT BEING INFINITY
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planetearth
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#76
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#76
Dividing anything by 0 = infinite. Not 0. Most mathematicians should know that. It is that simple.

EDIT: I am gladly that fool that stated it was infinite ^^
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scubafreak
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#77
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#77
Dividing is just repeated subtraction. 10/2 is 10 - 2, -2, -2, -2, -2 until you get 0. Then your answer is how many times you've taken away. You can do 10 - 0 -0 -0 til you're blue in the face and still end up with 10. So I guess you could say it's infinity, you can go on forever and not get an answer.
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planetearth
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#78
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#78
(Original post by scubafreak)
Dividing is just repeated subtraction. 10/2 is 10 - 2, -2, -2, -2, -2 until you get 0. Then your answer is how many times you've taken away. You can do 10 - 0 -0 -0 til you're blue in the face and still end up with 10. So I guess you could say it's infinity, you can go on forever and not get an answer.
Exactly.
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birdsong1
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#79
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#79
(Original post by Saichu)
I randomly came across this thread a few weeks ago, and I believe I have beaten you both to the punch.

http://www.math.tamu.edu/~saichu/Nonstandard.html*

(I've been meaning to write such an introduction for a while now.)

To answer your question, Dean, what distinguishes the hyperreals from the reals is that not all subsets have the lub property, only the so-called internal subsets.
Personally, I wouldn't have used typed theory. Superstructures can be analyzed fine within the context of model theory. But an interesting take on Nonstandard Analysis, nonetheless.
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Mobix
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#80
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#80
0/0 = 1 there you go problem solved;p
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