After all, Q is our general field, R is just one of many completions. Shouldn't this be discussed(if at all) in Q?

Isn't the completion of Q unique?

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Why so much hate? Learning what is not true is just as important as learning what is true.

Isn't the completion of Q unique?

After all, Q is our general field, R is just one of many completions. Shouldn't this be discussed(if at all) in Q?

Isn't the completion of Q unique?

aguacate wrote:Isn't the completion of Q unique?

Only with respect to the norm you're working with on Q.

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mudge wrote:aguacate wrote:Isn't the completion of Q unique?

Only with respect to the norm you're working with on Q.

For those who do not understand what mudge is talking about, the usual norm says that the distance between any two rationals x and y is the absolute value of x-y. This has only one completion, and that is the usual real numbers.

But let's take a prime p. x-y has some number k of factors of p in it. Could be 0, possibly 3 (p

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A++++ Will Read From Again.

Thanks Btilly.

Thanks Btilly.

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aguacate wrote:Why so much hate? Learning what is not true is just as important as learning what is true.After all, Q is our general field, R is just one of many completions. Shouldn't this be discussed(if at all) in Q?

Isn't the completion of Q unique?

Completions are generally unique (hence the term "the completion"), but "the completion of X" is usually shorthand for "the completion of X with respect to Y". So Q has many completions, all of which are uniquely the completion of Q relative to whatever.

For example, R is the (metric space) completion of Q relative to the (usual) metric d(x,y)=|x-y|. Q is also a topological group (in fact, a topological ring), which makes it a uniform space under the usual topology on Q (which is the both the order topology for the usual order, and the metric space topology with the usual metric). R is also the completion of Q in this uniform topology.

One can also talk about other topological structures on Q. In any topological structure in which addition and negation are continuous, Q will be a topological group, and hence be a uniform space and have a completion in the same sense as the above. So we can get all sorts of completions of Q this way. Most notably, the p-adic numbers are obtained in this way: if we choose the right topology corresponding to a given prime p, and take the completion, the result is the p-adic numbers. Alternatively, if we choose the right metric and take the metric space completion, we also get the p-adics.

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When I was in calc, the worst indeterminate form was 1^infinity.

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What he means is, if lim_{x->a}f(x)=1, and lim_{x->a}g(x)=∞, then lim_{x->a}f(x)^{g(x)} can be difficult to compute.

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I'd've thought you could just use some trickery to get it into a form like (1+x/n)^{n} and then you'd be done... but I guess difficulty is subjective.

ameretrifle wrote:Magic space feudalism is therefore a viable idea.

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jestingrabbit wrote:I'd've thought you could just use some trickery to get it into a form like (1+x/n)^{n}and then you'd be done... but I guess difficulty is subjective.

Seriously, I've really got to start using that more often...

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Double conjunction critical hit!

I wonder how standard that is, I would think I'd've heard it used more than once, but I think that's the first time I've ever seen it typed out.

Also, I can see how that would be a tricky problem, anyone got any quick tips?

I wonder how standard that is, I would think I'd've heard it used more than once, but I think that's the first time I've ever seen it typed out.

Also, I can see how that would be a tricky problem, anyone got any quick tips?

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Or use the continuity of log... Still, it's trickier than something you can just apply l'Hopital's to and get an answer right away.

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But like you say, it's only marginally trickier. You just take lim f^{g} = lim exp(ln f^{g}) = lim exp(g ln(f)) = lim exp(ln(f) / (1/g)) = lim exp((f'/f) / (-g'/g^{2}). I guess it's a couple more steps, but I'dn't've thought it was any harder than most indeterminate forms (except the easy 0/0).

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Cosmologicon wrote:But like you say, it's only marginally trickier. You just take lim f^{g}= lim exp(ln f^{g}) = lim exp(g ln(f)) = lim exp(ln(f) / (1/g)) = lim exp((f'/f) / (-g'/g^{2}). I guess it's a couple more steps, but I'dn't've thought it was any harder than most indeterminate forms (except the easy 0/0).

Well, considering many indeterminate forms in calc are something like lim

And I'm sorry, but you've crossed a line with "I'dn't've".

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skeptical scientist wrote:And I'm sorry, but you've crossed a line with "I'dn't've".

I thought that when I saw it, but I'm not so sure now... and yes, using log is a much better idea.

ameretrifle wrote:Magic space feudalism is therefore a viable idea.

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I meant worst as in "why would that be an indeterminate form?"

The answer, obviously is because the log is indeterminate. ln(1^inf)=inf*ln(1)=inf*0.

The answer, obviously is because the log is indeterminate. ln(1^inf)=inf*ln(1)=inf*0.

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Well what would it be if it weren't indeterminate? 1? I guess that's reasonable, but at the same time you get introduced to a great counterexample early on: (1+1/x)^{x} as x goes to infinity.

And skeptical scientist, I don't see why you think I'dn't've's going too far.

And skeptical scientist, I don't see why you think I'dn't've's going too far.

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Cosmologicon wrote:And skeptical scientist, I don't see why you think I'dn't've's going too far.

I was being a little tongue-in-cheek*, but as far as I can tell it's unpronounceable. Ayudentave? Plus you hear "I'd've" quite frequently in everyday English, and never hear "I'd'nt've". It's not so bad written out, since it's obvious what you must mean, but try using it in spoken conversation and see where it gets you.

*As I typed this, I stuck my tongue into my cheek. It didn't illuminate anything. What on earth is meant by sticking ones tongue in ones cheek?

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skeptical scientist wrote:*As I typed this, I stuck my tongue into my cheek. It didn't illuminate anything. What on earth is meant by sticking ones tongue in ones cheek?

From phrases.org.uk:

This phrase clearly alludes to the facial expression created by putting one's tongue in one's cheek. This induces a wink (go on - try it), which has long been an indication that what is being said is to be taken with a pinch of salt. It may have been used to suppress laughter. 'Tongue in cheek' is the antithesis of the later phrase - 'with a straight face'.

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Uh, I was just kidding with that last comment. I knew you didn't mean it. It was just an excuse for me to use a quadruple contraction. I thought to myself, if I'dn't've sounds silly, imagine how ridiculous I'dn't've's'll sound!

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Cosmologicon wrote:I'dn't've's'll

This time I'll skip a step and go straight to:

From phrases.org.uk:

Firstly, let's get get clear what word we are talking about here. It's pale, and certainly not pail, - the phrase has nothing to do with buckets. The everyday use of the word pale is as the adjective meaning whitish and light in colour (and used to that effect by Procol Harum and countless paint adverts). This pale is the noun meaning 'a stake or pointed piece of wood'. It is virtually obsolete now except in this phrase, but is still in use in the associated words paling (as in paling fence) and impale (as in Dracula movies).

The paling fence is significant as the term pale became to mean the area enclosed by such a fence and later just the figurative meaning of 'the area that is enclosed and safe'. So, to be 'beyond the pale' was to be outside the area accepted as 'home'.

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Cosmologicon wrote:Uh, I was just kidding with that last comment. I knew you didn't mean it. It was just an excuse for me to use a quadruple contraction. I thought to myself, if I'dn't've sounds silly, imagine how ridiculous I'dn't've's'll sound!

I'dn't've is easy enough, it means "I would not have". But what the hell does the second one mean? I would not have has shall? I had not have is will? i can has cheezburger?

I wonder if finding a sensible meaning for n contractions is an NP-hard problem.

Last edited by Torn Apart By Dingos on Mon Jan 07, 2008 6:35 am UTC, edited 2 times in total.

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Torn Apart By Dingos wrote:But what the hell does the second one mean?

I believe it's supposed to be like:

"Imagine how ridiculous <$word>'ll sound!"

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That explains the 'll, but what does the 's stand for?

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Mathmagic skipped a step. I'dn't've's'll's formed by first taking <word2>=<word1>'s, as in "I don't see why you think <word1>'s going too far," and then doing the above construction with <word2> in place of <word>.

Yes, this post exists solely so that I could write "I'dn't've's'll's".

Yes, this post exists solely so that I could write "I'dn't've's'll's".

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The recursion! It hurts!

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Sorry for being, let's say, platonic; but I must really mention this to you:

You CAN NOT divide by zero!!! and that because 0 does not exists!

If you ask why i'll tell you to read this topic: viewtopic.php?f=17&t=17853

You CAN NOT divide by zero!!! and that because 0 does not exists!

If you ask why i'll tell you to read this topic: viewtopic.php?f=17&t=17853

One: Thanks for necroing a thread to not add anything to it.

Two: When you link to a thread as proof, the first thing I'm going to look for are your posts in it. Which in this case deal with demigods, quantum mechanics, and pretty much just a general blathering about nothing in particular...

Three: You aren't being platonic.

Edit:

This seems contrary to your previous post in this thread

Two: When you link to a thread as proof, the first thing I'm going to look for are your posts in it. Which in this case deal with demigods, quantum mechanics, and pretty much just a general blathering about nothing in particular...

Three: You aren't being platonic.

Edit:

Joemaniax wrote:So... my friends, there you have it... 0 exists and it's quite an important little sucker.

This seems contrary to your previous post in this thread

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Platonic? What is the pure, true form of zero? Nothing!

And we can show that via the ontological argument that something is not perfect if it doesn't exist, and therefore... zero can't exist. The Platonic form is perfect, and perfect things exist, and if the perfect form of zero is nothingness! <insert cranky following statements>

Thanks guys, I'll be here all week.

And we can show that via the ontological argument that something is not perfect if it doesn't exist, and therefore... zero can't exist. The Platonic form is perfect, and perfect things exist, and if the perfect form of zero is nothingness! <insert cranky following statements>

Thanks guys, I'll be here all week.

Ok. so I've glanced over all of this, tried to understand most of it and how it is undefined/defined in different fields. But I am not mathcore, so please do not kill me if I've overlooked something. You know how sqrt 2 was invented to solve x^2=0, and i was invented to solve x^2+1=0, why not invent a set of numbers to define division over zero? Notice I say a set, not nullinity or some arbitrary constant. Fundimentally the problem is that division is the inverse function of multiplication, so how about:

a/0 = @ (that's an encircled a,) such that @*0=a?

Where a is any number/function on any field? Eg (3+2i)/0 = 3+2i in a circle (can not draw it here. basically a circle/oval whatever encompassing 3+2i. The notation's not important, the concept is.)

I do believe this hasn't been discussed fully, reading all five pages, and would like to no if this has already been thought of/shot down (and if so, why). Am I right in thinking that i initially had no real world importance until people later found its use in fields like electrical engineering? Don't shoot it down on account of no practical value, give me maths plz.

Awesome.

a/0 = @ (that's an encircled a,) such that @*0=a?

Where a is any number/function on any field? Eg (3+2i)/0 = 3+2i in a circle (can not draw it here. basically a circle/oval whatever encompassing 3+2i. The notation's not important, the concept is.)

I do believe this hasn't been discussed fully, reading all five pages, and would like to no if this has already been thought of/shot down (and if so, why). Am I right in thinking that i initially had no real world importance until people later found its use in fields like electrical engineering? Don't shoot it down on account of no practical value, give me maths plz.

Awesome.

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While the complex numbers present us with a meaningful representation of physical values, which is of course why we actually use them (if they had no application, they would be in the same category as octonions and the like, oddities of mathematics used as an esoteric tool and not taught in secondary education.) So what does this new notation bring to the table? Moreover, do you know whether or not it breaks the consistency of existing mathematics?

And, what information does it provide to the user of that method? The interesting thing about division by zero is that it makes whatever is above the zero irrelevant when performing limits and other such things. Only in the case of 0/0 do you get something interesting in calculus, otherwise, you know your limit is going to approach an infinity.

And, what information does it provide to the user of that method? The interesting thing about division by zero is that it makes whatever is above the zero irrelevant when performing limits and other such things. Only in the case of 0/0 do you get something interesting in calculus, otherwise, you know your limit is going to approach an infinity.

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If we still have the basic properties of arithmetic:

@=@*1=@*(1+0)=@*1+@*0=@+a

subtracting @ from both sides we get a=0. So you're going to lose a lot of the basic properties of arithmetic when you do this. The great then about introducing sqrt(2) and i is that you don't break anything. All of the basic laws of arithmetic hold in the complex numbers just as well as they do in the rationals, and you get a lot more nice properties besides. So sure you can define whatever you want, but in order to get a consistent system you lose a lot, and gain just about nothing, since what good is division by zero if by introducing it, addition, subtraction, multiplication, and division no longer behave as they should?

@=@*1=@*(1+0)=@*1+@*0=@+a

subtracting @ from both sides we get a=0. So you're going to lose a lot of the basic properties of arithmetic when you do this. The great then about introducing sqrt(2) and i is that you don't break anything. All of the basic laws of arithmetic hold in the complex numbers just as well as they do in the rationals, and you get a lot more nice properties besides. So sure you can define whatever you want, but in order to get a consistent system you lose a lot, and gain just about nothing, since what good is division by zero if by introducing it, addition, subtraction, multiplication, and division no longer behave as they should?

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Exactly as Skeptical Scientist said. When coming up with some novel idea, first, does your new idea in mathematics break existing mathematics? Can you use it to gimmick an equation or produce some erroneous statement? Note that this is very, very hard to prove for some things, and that's why there's a huge amount of work that goes into the rigor of mathematics. We have to be certain we aren't undoing the foundations of mathematics every time we create a new method. Second, if your idea does break anything, does it still perform some useful function? Imaginary numbers are useful in physics and that encompasses a very large number of fields, even including the mathematics of sound (though the engineers will try to convince you sqrt(-1) = j instead of i) So where is your encircling a useful method of performing mathematics? What good does it do other than 'solve' a problem by declaring a whole new category of number?

If there's no application, and it breaks existing mathematics, it's likely to be even harder to look up on Wikipedia than the p-adics. If it has an application but breaks existing mathematics, then it may become a niche tool. If it doesn't break existing theories proved in real or complex numbers and has no application, it's pointless, and even the quaternions have an application (so they've got you there.) And lastly, if it has an application and doesn't break anything, whoohoo, you figured out a clever new way of extending real or complex number systems that doesn't undo everything we've already learned.

If there's no application, and it breaks existing mathematics, it's likely to be even harder to look up on Wikipedia than the p-adics. If it has an application but breaks existing mathematics, then it may become a niche tool. If it doesn't break existing theories proved in real or complex numbers and has no application, it's pointless, and even the quaternions have an application (so they've got you there.) And lastly, if it has an application and doesn't break anything, whoohoo, you figured out a clever new way of extending real or complex number systems that doesn't undo everything we've already learned.

Ok thanks that's starting to make sense. Distributive law for the win! But isn't dividing by @ the same as multiplying by zero? Which would equalise both sides etc etc? Or is this just making it more retarded?

Still confused, but less so. H'ok.

Edit: wait, my bad. You're subtracting. Carry on.

Still confused, but less so. H'ok.

Edit: wait, my bad. You're subtracting. Carry on.

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Basically, it means that in order for you to have a number system that is consistant, you're going to have to lose at least one of:

(The additive inverse and associativity are used implicitly, in skep's last step... to go from "@+a=@" to "(-@)+(@+a)=(-@)+@" to "((-@)+@)+a=(-@)+@" to "0+a=0" to "a=0".)

All of these axioms are part of the definition of a field... any system that loses one of them can't be called a field. And while there do exist systems of numbers that are useful and aren't fields, that usefulness is generally limited somewhat.

[edit] In fact, if you drop one of those axioms, it can't even be a ring, let alone a field...

And that's not even to say that dropping one of those axioms is sufficient to stop the new numbers from breaking stuff... just that it's necessary. There may be other axioms that would conflict, and have to be dropped too.

- Multiplicative identity: There exists a 1 such that for all x, 1*x = x
- Additive identity: There exists a 0 such that for all x, 0+x = x
- Distributive law: For all a, b and c, a(b+c) = ab + ac
- Additive inverse: For every x, there is a -x, such that (-x)+x = x+(-x) = 0
- Additive associativity: For any a, b and c, (a+b)+c = a+(b+c)

(The additive inverse and associativity are used implicitly, in skep's last step... to go from "@+a=@" to "(-@)+(@+a)=(-@)+@" to "((-@)+@)+a=(-@)+@" to "0+a=0" to "a=0".)

All of these axioms are part of the definition of a field... any system that loses one of them can't be called a field. And while there do exist systems of numbers that are useful and aren't fields, that usefulness is generally limited somewhat.

[edit] In fact, if you drop one of those axioms, it can't even be a ring, let alone a field...

And that's not even to say that dropping one of those axioms is sufficient to stop the new numbers from breaking stuff... just that it's necessary. There may be other axioms that would conflict, and have to be dropped too.

Code: Select all

`enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};`

void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

phlip wrote:And while there do exist systems of numbers that are useful and aren't fields, that usefulness is generally limited somewhat.

Pah. Fields may be useful, but rings are far more interesting.

All posts are works in progress. If I posted something within the last hour, chances are I'm still editing it.

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Token wrote:Pah. Fields may be useful, but rings are far more interesting.

Perhaps, but even in a ring, ∀x:0x=0

Code: Select all

`enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};`

void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

phlip wrote:Token wrote:Pah. Fields may be useful, but rings are far more interesting.

Perhaps, but even in a ring, ∀x:0x=0

Indeed. Which is why I was wondering why you immediately jumped to fields instead of rings, and was originally planning to make some comment about you sucking your thumb and clutching the hem of your mother's skirt while cuddling your precious multiplicative inverses. But I digress.

All posts are works in progress. If I posted something within the last hour, chances are I'm still editing it.

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Token wrote:Indeed. Which is why I was wondering why you immediately jumped to fields instead of rings, and was originally planning to make some comment about you sucking your thumb and clutching the hem of your mother's skirt while cuddling your precious multiplicative inverses. But I digress.

Because what little maths I had to do at Uni for my software degree only taught groups and fields, before moving on to vector spaces and the like... so while I have read about the others in my own time (because, let's face it, they're really interesting), those two are more familiar to me. I couldn't remember what the properties of a ring were offhand, and I had to look it up... and once I did look it up, I edited my post to match.

Code: Select all

`enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};`

void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

Which zero goes faster to zero?

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