You know, this is really getting tiresome and I’m sure anyone following can see what is really going on by now.
You really believed that the conclusion 1- 0 = 1 was based on an assumption and now you’re trying to back pedal. Fine. If you can get anyone to actually believe you at this point, more power to you. But the next time you want to take a pot shot at something I present, don’t think I’m going to be impressed by your supposedly “erudite” posts. I think we both know better at this point.
Wow, really!? :rolleyes:
Also, assumption and axiom are basically the same thing in modern proofs. As used in modern logic, an axiom is simply a premise or starting point for reasoning[…]. I’d say that the premises such as “how the limit operator works” counts as “a starting point for reasoning”. Nitpicking over terminology adds nothing to the discussion.
Like I said, it doesn’t matter if the properties of limits are ground terms or based off of ground terms. They’re still assumptions within the context of the proof you gave. In the problem
x[sup]2[/sup] + 2x -3 = 0
[-2 +/- sqrt(2[sup]2[/sup]-4*-31)]/(21)
x =1,-3
The quadratic formula is an assumption. It doesn’t matter if it can, or has been proven. Within the context of that problem itself, the formula is taken as a given, a.k.a. an assumption that it’s true.
Allotrope, you, Indistinguishable, and I agree. All math is based off of axioms, limits in the Real number system follow from these axioms. You’re getting mad at absolutely nothing except our insistence that, even if limits are provable in the context of the Reals, in the proof you gave specifically they act as an assumption – regardless of how true that assumption is.
Give it a rest dude. I’m in no mood to argue semantics and frankly you’d be out of your league anyway.
Why are you getting so mad? Because we used slightly different terminology than you?
I’m not mad, just irritated at people who try to pretend like they didn’t make a huge mistake. Like a cat running full speed into a class door and then acting like it didn’t happen. Hey, if that’s how you want to play it, go ahead. I just don’t feel like playing anymore today. Maybe tomorrow. How’s that sound?
Right, okay. You won! Good for you! We’re just covering up our stunning failure by pretending we always agreed with you! I haven’t claimed anything inconsistent this entire time (and as far as I can tell, neither has Indistinguishable), but whatever keeps you happy.
Moderator Instructions
allotrope, I want you to dial way back on the snark. If you’re going to continue to play here, you’re going to play nice. You’ve already received one warning. You’ll receive another if you continue in this vein.
Colibri
General Questions Moderator
Just for fun for Jragon (and anyone else who might be amused), I’ll note some examples of where the quadratic formula doesn’t work as normal:
A) In the presence of zero divisors:
In arithmetic modulo 21, we have that 4 is a root of x^2 + 2x - 3, despite being neither 1 nor -3.
B) In noncommutative contexts:
In the quaternions, the quadratic formula would tell us that the roots of x^2 + 2ix - 1 - 2j are -i ± (1 + j). But plugging these into the polynomial will end up yielding ±2k, rather than 0.
C) In characteristic two:
If 2 = 0, we can’t even apply the quadratic formula to x^2 + x - 3, since it would have us divide -1 by 2, which is an impossible division by zero. And even if we could somehow get around that, the quadratic formula would imply that the roots of this equation must be obtainable by field operations and square roots as applied to its coefficients, yet the only things obtainable from field operations and square roots applied to the coefficients will be 0 and 1, neither of which is a root of the equation.
But the zero matrix, identity matrix, [[0, 1], [1, 1]], and [[1, 1], [1, 0]] together comprise a field in arithmetic mod 2, in which the latter two are the roots of x^2 + x - 3 (even though we can’t divide -1 by 2 and even though neither is obtainable from the coefficients by field operations and square roots).
Um… This isn’t well defined. How can it be “infinite except for” something? What is the “1/infinity” decimal place? Define your terms.
It’s like leaving out the blue decimal place, or that oats - grapes = swiss cheese.
Too many holes.
Er, “polynomial”, not “equation”…
We can indeed.
“1 - 0 = 1” is based on assumptions. No need for anyone to backpedal on that claim–it’s plainly true. (And it’s really strange for you to claim it’s not, since a few posts above you said everything in math is based on axioms. Axioms are assumptions.)
You’re the one who has eliminated yourself from everyone’s “take seriously” list, not Indistinguishable.
He basically said that everything is based on axioms (which he roundaboutly acknowledged are akin to assumptions), but then went on to say that things derived from assumptions are not assumptions – even if they’re assumed within the context of a proof. (He says that limits follow naturally from the axioms, and therefore aren’t assumptions in that proof, even though the proof didn’t prove limits from the axioms but merely used the already-proved notion of limits).
He really does agree with you, he’s just getting bizarrely insistent on terminology.
I have told you exactly what I mean in my argument of how to find it using zeno’s paradox… this is really the very point of what’s being dsicussed?
I don’t want to just kepp posting it over and over.
You are assuming you cannot “find” or even that there is an infinite’th decimal place. But I that is the whole point I am making. Move A to B. When you to B there is your infinite’th decimal place.
The precedeing decimal places to the infinite’th are the prior halves you must pass to get to B in Zeno’s Paradox…
What part is not clear about this afte reading the many times I have decribed it?
You are just not accpeting the fact that there is one, and that we can find it, so this is why you do not understand.
But does this notation map onto anything meaningful?
For example, which of these statements is true?
.000…1 > .000…5
.000…1 < .000…5
What’s one half of .000…1?
What’s five times .000…1?
I’m sure you could construct some sort of consistent mathematical system where this notation make sense, but if you just toss it out there without any superstructure of axioms to explain how it works, it’s meaningless.
That’s why I asked you upthread how you would solve this equation:
x = 10 * 0.9…
I know what x is in the math I use. What is the answer in the math you use?
There is no infinite’th decimal place.
There can’t be an infinite’th decimal place, by definition. If you insist that one exists, you don’t understand the concept of infinity.
Ok, I’m going to try to get answer again.
Is there any reason to use the notation .999… besides showing that an arithmetic operation gets stuck in a loop?
Also, for those wondering about .999…1, an infinite series followed by something else, it occurs when you try to alphabetize mixed numbers and words. If the numbers come first, followed by letters, 1 2 3 … A B C, then there are infinite numbers that can be ordered before the first letter. This doesn’t seem to confuse people looking up movie listing on the EPG.
Coders: Can’t Live With Them; Can’t Bend, Fold or Spindle Them. A Turing Play in One Act by Exapno Mapcase
Exapno Mapcase: Welcome to Mapcase Computing. I like to talk to all of our new hires their first day on the job.
erik150x: I’ve been looking forward to this talk.
EM: Your personnel records show that you’re highly qualified.
erik: I have an MS in Computer Science, also studied physics for quite some time and have always had an avid interest in all the sciences including math. I know perfectly well intuition is always right.
EM: [stunned, whispers to himself] Intuition is always right? [aloud] Anyway, I’m sure you’ll love working here. We’ve developed the Mapcase Super Infinity Computer. Not only does it use Mapcase hardware but we wrote Mapcase, a language to use specifically on it, just to handle infinities. Mapcase will allow users to solve problems that were intractable on any other system. In fact, Mapcase is built right into the firmware for maximal efficiency. There aren’t even provisions for allowing other compilers, so no other language will work on it.
erik: Great, so you don’t mind if I do all my work in Fortran, right?
EM: Fortran? I’m sorry, but while that may have been a fine language for its time what we’ve developed since far outstrips any capabilities it has. We can get quick answers to problems that would take forever in Fortran. Besides, Fortran just won’t run on this computer. You have to write in Mapcase.
erik: No problem, Fortran it is.
EM: Fortran it isn’t. I told you, it can’t work on this system. Even if you somehow got it running the answers spit out would be gibberish.
erik: Not so. I did a few test runs and here are my answers.
EM: [reads the printouts] But these are gibberish.
erik: Oh, you may think so, but when I read them I think they provide a much better answer.
EM: Not for me, and I run the place and lay down the rules. I say if you want to work here you have to write in Mapcase.
erik: Fortran. Got you.
EM: Not Fortran. If you think you can come up with better answers go find a machine with a Fortran compiler and do all the work you want in Fortran. But seriously, erik, it can’t work here. That’s the way things are set up. All my workers agree, and believe me, they’re experts. Even the guy who takes naps with his head under the desk and his feet in the aisles that you’ll keep tripping over is an expert. Some of them are old enough to know Fortran too and they’ll also tell you it won’t work here on these problems.
erik: I hear you. Just answer one question.
EM: [sighs] All right. What’s your question?
erik: Since the answers I get in Fortran are so superior to your answers, why don’t I just keep working in Fortran until the day I die?
EM: [Bends, folds, and spindles erik.] There’s your contradiction that makes this all false.
The end.
Nitpick: I said this thinking of the quadratic formula the way I like to think of it, as giving the roots of x^2 + bx + c as -p ± sqrt(p^2 - c), where p = b/2. But I suppose the way people usually phrase the formula is as giving the roots of ax^2 + bx + c as (-b ± sqrt(b^2 - 4ac))/2a. In which case, the problem won’t be division of 1 by 0, but division of 0 by 0. [When 2 = 0, the numerator and denominator of the quadratic formula will both always be 0].
The rest of what I said about this case still holds, though. When 2 = 0, the roots may exist but not be obtainable from the coefficients by any combination of addition, subtraction, multiplication, (well-defined) division, and square roots, as seen in the provided example.
No reason I know of; I think that saying .999… = 1 is used mainly as an example of how limits are considered to work.
Just wanted to pop in to interject this:
I’m OK with notation like .999…5 where you have a digit in the infinitieth place. It’s definitely not a real number, but it makes sense: Just index the digits by the ordinal omega+1. Of course, you can extend the digits further with higher ordinals.
One thing that isn’t clear to me, however. How should arithmetic work with this notation? For example, what is:
.000…5 + .000…5 ?
One obvious candidate would be .000…10, but that seems unsatisfying. It doesn’t seem to preserve the order structure in any natural way. Is this number larger or smaller than .000…5?
Anyway, I’m not saying it can’t be done. I am saying it doesn’t seem interesting to me unless this can be rectified in some way.