Ahh damn just read exapno’s pos… and I think that’s a great idea. I had no idea this had been posted b4 (although I should have guessed).
Oh and also, please allow guests to edit posts, I’m planning on subscribing anyway and it just ends up making ppl double post like this.
So you’ve accepted that .3… = 1/3, right?
I think you’ll agree that .3… + .3… + .3… = .9…, and that 1/3 + 1/3 + 1/3 = 1. Doesn’t that imply that .9… = 1?
Man, not even Charter Members can edit posts. And the mods think this is a feture.
FWIW, I agree: If people were allowed to edit posts, you’d get jokers posting inflammatory nonsense, garnering huge amounts of follow-ups to what was originally written, and then altering their post to make everyone else look stupid. A five-minute editing window might allay this, but everyone has access to the Preview button.
So, do you now accept that 0.999~ = 1.000~?
Careful: Statements like this have been confusing in the past. It’s better to say that if the sequence has a limit, the sequence as a whole is equal to the limit. That the sequence is infinite isn’t relevant, as summing a sequence isn’t a process in the usual sense.
How does this work? I have an intuitive grasp of limits and other basic calculus, but I don’t get how this sequence can sum to zero.
The sequence doesn’t sum to zero - it approaches zero. If I’m not mistaken, the sequence sums to one.
Derleth, I think you are confusing “sequence” and “series”. (It’s easy to do, and I just checked in a mathematics dictionary to make sure I got it right.) When you are tralking about sums, you are talking about a series. A series is the sum of the terms given in a sequence. For example, the sequence that I gave:
{1/2, 1/4, 1/8, 1/16. 1/32, … }
can be treated as a series, where the partial sums (the sums of the first n terms) are the sequence:
{1/2, 3/4, 7/8, 15/16, 31/32, … }
And, in this case, it is a convergent series, with limit 1.
Right. And the value of .99999 … is the limit of the series {.9, .99, .999, .9999 …}, which is 1. And Skadi, if you want, I’ll show you the formal epsilon-delta proof of that; but if the discussion is going over your head already, you may not want to see it. (Though it’s just college freshman calculus stuff, so if you’re planning on getting to that educational level, if may be worth your while.) (Actually, a college math prof here posted that they’d removed that from the coverage in that course at his school, which leaves me flabbergasted. Teaching calculus without teaching the basic theoretical underpinnings of limits???)
If .999… != 1, then there is some real number y such that .999… < y < 1. What is y?
Indeed, which floored me when I learned I wasn’t supposed to teach it in my Multivariate courses, specially since they hadn’t seen it in one variable.
Of course, all this means is that I circulate it as samizdat.
The sum is 1, and the product is 0. How weird is that?
The sum is 1, and the product is 0. How weird is that?
I was discussing this with my dad and brother last night (my brother and I saying that 0.999…=1 and my dad saying they aren’t equal). My dad had an interesting observation that I couldn’t answer, but I thought someone else here could.
If two number are equal, you should be able to divide one by the other and get 1. So he did some long division for 1 divided by 0.999… and ended up with a remainder. He reasons that even though the answer will be 1.000… the fact that you have a remainder shows they are not equal.
Any takers?
You can’t divide by an infinite number of digits–you need to use a finite number.
What algorithm did he use to divide 1 by 0.999… ?
Let me suggest one, given that 0.999… represents the limit of an infinite sequence {0.9, 0.99, 0.999, … }:
Given an infinite sequence {S1, S2, S3, …} with a limit other than 0, the result of dividing 1 by the limit of that sequence is the limit of the sequence {1/S1, 1/S2, 1/S3, … }.
That gives you the sequence of recurring decimal fractions:
1/0.9 = 1.1111 …
1/0.99 = 1.01010101 …
1/0.999 = 1.001001001001 …
1/0.9999 = 1.0001000100010001 …
It’s pretty clear that the limit of that sequence is 1, so 1 divided by 0.999 … is equal to 1. QED
It’s the same old problem, not understanding the difference between an infinite number of nines and a very large number of nines. The “remainder” is, by definition, smaller than any positive number, and is therefore zero.
Not very. The sequence of partial products of any summable sequence of real numbers is 0. For an infinite products to converge to a nonzero limit the sequence of factors must converge to 1.
Figured as much. My dad’s main argument was that if you can’t write it down, you can’t prove it for certain. I asked him to show me a case anywhere where 3+6!=9 and he accused me of asking him to prove a negative.
Sigh…
Of course, that should read as 3 + 6 != 9 rather than 3 + 6! = 9. :smack:
Some mathematicians would share your father’s belief – they don’t accept anything which depends on infinity. But for them, the expression 0.999… would be meaningless, because it depends on infinity and limits and such stuff – not different from 1, just meaningless. (I suspect that they have trouble with other stuff, like the square root of 2 and pi, as well, because working with these involves infinity as well – not accepting infinity means not accepting a large part of mathematics).
Well, it’s subtler than that. The so-called “constructivists” insist that everything be “constructible”. They’re cool with the sequence of partial sums to the extent that any given element of it can be constructed, but they have misgivings about the whole sequence. Constructivists have their own version of the real numbers, which works sufficiently like everyone else’s that they don’t disagree until you get to some really weird things like the Banach-Tarski paradox.
The flip side of this is that with the rise of structuralism in the philosophy of mathematics and the rise of topos theory in categorical mathematics, there is a very real sense in which constructivists aren’t doing the same thing at all. “Standard” mathematics is performed in one topos (commonly called Set) while constructivist mathematics is performed in another. One of the major differences, for instance, is that the logical structure of the constructivist topos is not a Boolean algebra, but a Heyting algebra. This means that the law of the excluded middle is rejected. P^!P is not trivially true, so !!P is not equivalent to P, although !!!P is equivalent to !P. For further reading, you could try to read Saunders Mac Lane and Ieke Moerdijk’s Sheaves in Geometry and Logic: A First Introduction to Topos Theory, but it might be simpler just to get a Ph.D. in math at a category-friendly school.