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If 0.999 (etc.) = 1, does 1 - 0.999 = 0?
#1
If 0.999 (etc.) = 1, does 1 - 0.999 = 0?
Obviously, logically the answer is yes. But if so, I feel a little disappointed. 1 - 0.999 (the other side of the asymptote) has always been my favourite number (though people look at me funny when I tell them that Tongue). It's like the other side of infinity. It reminds me of the infinite potential and possibility that can come from something infinitely small. "See the world in a grain of sand".

I tried to read a lot of that 0.999 - 1 thread, and it did seem as if mathematicians leaned towards it equaling 1. So does that mean that the asymptote terminates eventually? That it doesn't get "infinitely close", as I was always taught?

Damn, this is such a buzzkill, lol.
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#2
RE: If 0.999 (etc.) = 1, does 1 - 0.999 = 0?
The number 0.999... never terminates. It also never changes value, so you can't say it "gets infinitely close" to anything. Single numbers aren't asymptotic, so I don't see how you can apply asymptotes here either. Perhaps you are confusing the number 0.999... with a function that gradually approaches 0.999...?
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#3
RE: If 0.999 (etc.) = 1, does 1 - 0.999 = 0?
(March 4, 2012 at 2:00 pm)Tiberius Wrote: The number 0.999... never terminates. It also never changes value, so you can't say it "gets infinitely close" to anything. Single numbers aren't asymptotic, so I don't see how you can apply asymptotes here either. Perhaps you are confusing the number 0.999... with a function that gradually approaches 0.999...?

Sorry about the confusion! I'm not actually very educated in math so my terms are going to be a little bit jumbled and imprecise! But I want to learn more. Smile

A lot of my original post was talking in figurative terms about why I liked the number 1 - 9.999... so much (if it even exists). I know asymptotes are functions and not single numbers. I know such a thing does not technically exist, but I was visualizing it as the "space between" the nonexistent "end" to an asymptotic function and 0, for ease of understanding. It's not really important to my question. More basically I was wondering: does 1 - 9.999... = 0?

Basically I'm talking about the infinitely small number (infinitely close to zero), if that makes it easier? Does it exist?
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#4
RE: If 0.999 (etc.) = 1, does 1 - 0.999 = 0?
Yes. 1-0.9999... =0 because 1 and 0.9999 are the same number and x + (-x) = 0, although the decimal representations are different as I outlined in the other thread. This is because the decimal (base 10) system we use, although sufficient to represent a real number, the same number can have two different decimal representations.
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#5
RE: If 0.999 (etc.) = 1, does 1 - 0.999 = 0?
There are a lot of practical (and aesthetic) reasons to not permit the existence of infinitesimals. Which is why you don't see infinitesimals given much attention.

The most popular modern formulation of analysis with infinitesimals (Abraham Robinson's Hyperreal numbers) involves the construction of an ultrafilter on the natural numbers, and the uniqueness of the Hyperreals as a system of numbers is equivalent to the continuum hypothesis. It's all well-defined and whatnot, but it's a bit further... out there. In order to divide by infinitesimals, you need to have infinitely large numbers. Most statements about the hyperreal numbers are true if and only if a 'standard' version of the statement (about the real numbers) is also true, so even if infinitesimal numbers don't exist, they do still offer an improved methodology for proving things about 'regular' numbers (a lot of difficult results in standard analysis have one- or two-line proofs in the hyperreals).

So I mean, you can talk about infinitesimals and still make sense. But they're a rather unpopular concept, and the machinery you'd need to talk about them is sort of advanced.
So these philosophers were all like, "That Kant apply universally!" And then these mathematicians were all like, "Oh yes it Kan!"
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#6
RE: If 0.999 (etc.) = 1, does 1 - 0.999 = 0?
(March 4, 2012 at 2:52 pm)LastPoet Wrote: Yes. 1-0.9999... =0 because 1 and 0.9999 are the same number and x + (-x) = 0, although the decimal representations are different as I outlined in the other thread. This is because the decimal (base 10) system we use, although sufficient to represent a real number, the same number can have two different decimal representations.

Thanks very much for the answer! Smile

Aw, well that's kind of sad. This is universally agreed on, then? Do you know anywhere I can look at a "layman's terms" explanation of why? (If one exists...I'm probably the only person on earth who both failed high school math and actually cares about the question! Tongue)

EDIT:

(March 4, 2012 at 2:56 pm)Categories+Sheaves Wrote: There are a lot of practical (and aesthetic) reasons to not permit the existence of infinitesimals. Which is why you don't see infinitesimals given much attention.

The most popular modern formulation of analysis with infinitesimals (Abraham Robinson's Hyperreal numbers) involves the construction of an ultrafilter on the natural numbers, and the uniqueness of the Hyperreals as a system of numbers is equivalent to the continuum hypothesis. It's all well-defined and whatnot, but it's a bit further... out there.

Sorry...didn't see your response! Unfortunately, I don't know the meanings of most of those terms yet. My interest in math is very lay at this point. Sorry in advance for any confusion or slowness...it has more to do with my lack of knowledge than actual denseness. Wink
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#7
RE: If 0.999 (etc.) = 1, does 1 - 0.999 = 0?
Hmmm, you need to know properties of real numbers, and how they are constructed to fully understand, specially the definition of dense and countable sets. As to why, I've already outlined it in that post, its the numbering system we use that allows for more than one different representation for the same number. For instance, in the rational number, 1 has infinite fraction representations, e.g. 1/1, 2/2, 3/3,...,n/n.

People tend to think about real numbers in their decimal representations, and that's why some confusion arises on this matter.
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#8
RE: If 0.999 (etc.) = 1, does 1 - 0.999 = 0?
(March 4, 2012 at 3:11 pm)LastPoet Wrote: As to why, I've already outlined it in that post, its the numbering system we use that allows for more than one different representation for the same number. For instance, in the rational number, 1 has infinite fraction representations, e.g. 1/1, 2/2, 3/3,...,n/n.

Oh, I understand. But while I can see that 2/2 obviously equals one, and that 4/2 equals two, I don't see (at this point, probably because I don't understand a lot of the principles of mathematics yet) how 0.999* = 1. I understand that you can write the same number in different ways, but it seems to me that 0.999* is standing for a different idea than 1. But (if there really is mathematical consensus here) I'm sure that's just because I don't understand enough of the principles of math to see why 0.999* clearly equals one, in the way that 2/2 so clearly equals one.

So you're saying understanding the concepts of "dense and countable sets" would help me see clearly why 0.999* and one are the same thing? If so, if it's not too much trouble of course, would you be able to link me to an explanation (of dense and countable sets, and any other mathematical concepts that would help me understand) that is pretty accessible to a layperson?



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#9
RE: If 0.999 (etc.) = 1, does 1 - 0.999 = 0?
No silly... if number minus number = 0.... then the doughnuts have won!

[Image: raised-doughnuts-4.jpg]

See? Every one of them is a 0.
Please give me a home where cloud buffalo roam
Where the dear and the strangers can play
Where sometimes is heard a discouraging word
But the skies are not stormy all day
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#10
RE: If 0.999 (etc.) = 1, does 1 - 0.999 = 0?
(March 4, 2012 at 5:36 pm)Vaeolet Lilly Blossom Wrote: No silly... if number minus number = 0.... then the doughnuts have won!

[Image: raised-doughnuts-4.jpg]

See? Every one of them is a 0.

Naw, but you see, there are other circular things. What about Cherrios? Fruit Loops? There are lots of DVD's and mood rings too but I think they are somewhat less numerous, though they may have the last laugh since they're not biodegradable.

So I think there will be a cereal alliance against the doughnuts, which will result in a fight to the death where the Cheerios would swarm the donuts by sticking to them, thereby suffocating them and killing them all. After that it would be an epic three way battle between Apple Jacks, Cheerios, and Fruit Loops (and any other types of O-shaped cereal I've forgotten, but the Cheerios' sheer numbers will ensure they will emerge victorious).
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