But that is wrong. Infinite is undefinable by any sum of size.

When you have a problem like 3*5, that's like saying you have 3 sets of 5, which sums up to 15. So for 0*infinity, that's like saying you have 0 sets of infinity (or an infinite number of zeroes!), which would logically be zero because you still have nothing.

Of course, that logic is defining infinity as a number, so it is also inaccurate.

Division by zero is a really simple math problem really.
If I have 100 jelly beans, and I divide them evenly among 0 piles, how many jelly beans are in each pile?
There aren't any piles... so the answer would be 0... wouldn't it?
f(x) = 100/x
f(0) = 100/0 = 0

or maybe it would make more sense this way

lim f(x) x->0- = -inf
lim f(x) x->0+ = inf
inf - inf = 0

MUAHAHAHAH

Post has been edited 3 time(s), last time on Nov 11 2011, 5:53 pm by K_A.

It's just a way of thinking about things. Of course it isn't correct in any rigorous sense.

I'm pretty sure .9 repeating isn't just really close to 1. It is 1. So no, I don't think it's the same thing.

What about .9 repeating forever with an 8 at the end?

But if 9 repeats forever, will there ever be an 8 at the end? The answer is no. You don't even know the meaning of the word forever.

Actually that's technically totally representable with limits...

Represent it.
If you're thinking of something like ' Sigma k=0 to infinity [ 9(1/10)^k ] ' + lim(n-> infinity) 8(1/10)^n ', Then you're not defining 8 at the end of infinite 9s. Represent it now. If you can't, you should take my word for it that you can't.

Limit as x approaches infinity for

1 - (2/x)
;O

Plug in 100... .98

1000, .998
10000, .9998

MISSION ACCOMPLISHED.

Let K be the real number such that the nth digit of K is an 8 and all digits from 1 to n-1 are 9, as n approaches infinity. Just because you can't write the number doesn't mean you can't represent the idea.

That is not a representation for an 8 at the end of infinite 9s under a decimal. First of all, x is all real numbers. You want to subtract 1/10^n, if you want it to have an 8 somewhere in between the infinite 9s. Though, there is no guarantee that 8 will be at the end of all 9s, since there should be an infinite number of 9s, there is no end. With that representation, you're making a number with infinite 9s with an 8 in between the 9s without a certain location in the digits. We'd still see that as 1.

Though, at a glance I see what you mean Vrael.

First off we see it as 1 because .999...=1
X can't actually ever be infinity, but we use limits to describe its motion and predict where it is going.

The limit as x approaches infinity for 1/x, as one would expect, approaches 0.000...001

Of course, one cannot exactly subtract a number such as this from a normal finite number, nor multiply it by two (well you can... but it doesn't do anything really)
Maybe 1 - (1/x) - (1/x) ... or maybe it does work originally. The point is that the expression 2/x was meant to represent two times infinitely small (aka 0.000...1), and therefore represent the idea that you displayed. Of COURSE .999...998 would, in a limit, be represented as a one.

First off, we don't 'describe' motions in limits. Limit is a limit. It's actually not something that is describable with physical representations. We only learn it that way because it is more easier to understand. Limit is simply a logical representation of what will happen if n approaches 'a' infinitely.
You also cannot expect 1/x to approach 0.000001 or something like that. 'x' is a real number. It 'doesnt' approach 0.0000000....01*(e^roote+2+(insert random numbers)). It approaches 0. It is smaller than any real number you can imagine. And not surprisingly, it isn't just represented by 0, the value is 0.
I'm simply saying, most representations of 0.99999...998 fail miserably. Because they happen to be something like 0.99999....99989999...9999..., unless you define a new number like Vrael did.
But, with infinite 9's, it really isn't a new number, but would be 0.99999999...998 = 0.999999999... It would essentially be the same.

A infinitesimal number ending in 8 (or ending in any number at all) will result in 0. I'll proofs it too =3

0.999 ... 9 - 0.999 ... 8
0.000 ... 1 = lim_n→∞ 1/10ⁿ = 0

Replace eight with any real number and you'll get the same result.

You can use the word "describe" (also, you always use double quotes, never single, when there is no nesting going on... just a common english mistake) in that context. If I wanted to be stupid I could say that y=2x describes the graph of the derivative for x^2. It has nothing to do with "understanding" it. It's a matter of semantics and english, which has no place in this discussion.

I admit probably could have used better words, but I don't claim to be the best writer.

I did not say 1/x approached 0.000001, I said it approached 0.000...001.
"X" IS a real number, that's why you can't put infinity in place of it and you have to describe the motion of the graph with limits to find it.
It does approach zero, which is what 0.000...001 is. It's one way of writing zero.
Technically, it's smaller than any positive* real number.

Holy shit where to start.


I don't even know what you're trying to say exactly. You loosely define a limit in an intuitive sense, but then claim that 1/x cannot approach 10e-6. But it can. The lim_x→10e6 1/x is 10e-7, it happens to be defined at 10e6 as well.

You seem to miss that 0.000 ... 1 is 0. So "0.0000000....01*(e^roote+2+(insert random numbers)" is also zero.

And then you continue to say that the value is 0. You've contradicted yourself. This is also just restating what has already been confirmed a page ago in a far less confusing way.



They are the same number: just different representations of the same number.

BeDazed, it seems as if you're out of your element here.

My favorite post in this thread.

A number with a decimal part that contains an infinite number of nines with an eight at the end does not exist. If you've managed to reach the eight, the number of nines was very large but not infinite.

This is nonsense. This claims by extension that all irrational numbers don't exist, take pi for example, the decimal representation is infinite, so if you reach any digit "at the end", the number of digits in the decimal is some large finite number and that number isn't pi. Yet we know pi to exist, we just can't write the whole thing.