StarDestroyer.Net BBS

Actually
lim[t->infinity](.9999999~naughty+1/t)= naughty+0

Well, if you think of .9~ as the sum of a series and graph it out as the function f(x), you'd get an asymptote and y=1. Considering that we are taking this to infinity, that means that the y will come infinitely close to y=1. So close that not matter how much we magnify the graph, the graph of f will overlap with the line y=1. Therefore, .9~=1. Or you could go with Nitram's explanation: .3~=1/3 ==> .3~*3=/9~ ==> 1/3*3=1 ==> .9~=1

Of course it does. Consider the following series of subtractions.

1.0 - 0.9 = 0.1
1.00 - 0.99 = 0.01
1.000 - 0.999 = 0.001

Since the digit 1 is appended at the end of each successive result, and an infinite series, by definition, has no end, the 1 never gets appended if we take each term to an infinite number of significant digits. Therefore, we are left with

1.000 ... - 0.999 ... = 0.000 ...

If the difference of two terms is 0, the two terms must be the same. Thus, 1.000 ... = 0.999 ...

Face value: No.
Complex mathematics: Yes.

~ver

Durandal wrote:Of course it does. Consider the following series of subtractions.

1.0 - 0.9 = 0.1
1.00 - 0.99 = 0.01
1.000 - 0.999 = 0.001

Since the digit 1 is appended at the end of each successive result, and an infinite series, by definition, has no end, the 1 never gets appended if we take each term to an infinite number of significant digits. Therefore, we are left with

1.000 ... - 0.999 ... = 0.000 ...

If the difference of two terms is 0, the two terms must be the same. Thus, 1.000 ... = 0.999 ...

The difference of the two terms is not technically zero; it approaches zero as the series approaches infinity. It's a subtle difference, but there.

Darth Wong wrote:The difference of the two terms is not technically zero; it approaches zero as the series approaches infinity. It's a subtle difference, but there.

Not so. Since 0.999... is the limit of the seqence a(n) = 1-(1/10)^n, the difference of 1 and 0.999... doesn't approach anything: it is the number 0. It is only the sequences themselves that can be thought to "approach" things, but the limits of sequences do not--they either exist as numbers or not at all.

Ah the wonders of Cal III. I'd be in it right now if I hadn't gotten behind on my math.

Is this really going anywhere? I can't tell, I'm really confused.. I'll just mod my head and smile.

AnimeJet wrote:Is this really going anywhere? I can't tell, I'm really confused.. I'll just mod my head and smile.

Just remember, when in doubt the answer is C.

muse wrote:Just remember, when in doubt the answer is C.

Unless after A or before B, then the answer's D. Like I said, I haven't taken any math courses in 2 years so as far as I'm concerned the answer is 7.

I get the explanations with fractions and stuff, thats the arguement i'm most familiar with, the others with calculus and stuff though.. riiiight

Kuroneko wrote:

Darth Wong wrote:The difference of the two terms is not technically zero; it approaches zero as the series approaches infinity. It's a subtle difference, but there.

Not so. Since 0.999... is the limit of the seqence a(n) = 1-(1/10)^n, the difference of 1 and 0.999... doesn't approach anything: it is the number 0. It is only the sequences themselves that can be thought to "approach" things, but the limits of sequences do not--they either exist as numbers or not at all.

Yes, the limit of the series equals 1. However, any actual term in the series will never be 1. If we assume the original question to be asking for the limit of the series, then the answer is obviously one, otherwise it is a series which approaches one. Unfortunately, the original question is worded imprecisely.

David wrote:Ah the wonders of Cal III. I'd be in it right now if I hadn't gotten behind on my math.

Actually, this is high-school calculus; basic limits. The only problem is an uncertainty in the way the original question was defined.

This has just reminded me of a whole load of infinities defined by Georg Cantor and that makes my brain die a little bit.

Darth Wong wrote:

Kuroneko wrote:Not so. Since 0.999... is the limit of the seqence a(n) = 1-(1/10)^n, the difference of 1 and 0.999... doesn't approach anything: it is the number 0. It is only the sequences themselves that can be thought to "approach" things, but the limits of sequences do not--they either exist as numbers or not at all.

Yes, the limit of the series equals 1. However, any actual term in the series will never be 1.

Quite right.

Darth Wong wrote:If we assume the original question to be asking for the limit of the series, then the answer is obviously one, otherwise it is a series which approaches one. Unfortunately, the original question is worded imprecisely.

It would be best to treat decimal expansions as the limits of their corresponding partial sums, so that there would be no difference between numbers with finite decimal expansions and those without in as much as they refer to numbers. For example, 0.5 interpreted as simply shorthand for 0.50000... [*], and hence the limit of the trivial sequence (5/10, 5/10+0/100, 5/10+0/100+0/1000, ...). Otherwise, we would be forced to have two separate methods of evaluating decimal expansions depending on whether or not they are terminating, and that would be unnecessarily complex.

I guess the ambiguity would be whether the original "0.9 infinitely" actually refers to that decimal expansion or something else.

[*] Yes, I'm well aware of the concept of the number of significant digits, but that is not a mathematical concept, and thus should not enter a discussion on the mathematical meaning of decimal expansions. There, all numbers are idealized abstractions, and hence exact.

I meant 0.999.. as in the 9's never stop, if that helps XD, not that it really matters, i mean.. boy, i should start focusing on homework -_-

I'm just going to assume that this proof wasn't put up already...

First, 1/3 = 0.333 ad nauseum
(One third equals 0.333 repeating)

So: 1/3 * 3 = 3/3 = 1
(one third times three equals three thirds, otherwise known as one)

0.333 ad nauseum * 3 = .999 ad nauseum
(0.333 repeating times three equals .999 repeating)

0.999 ad nauseum = 3/3 = 1
(Since we already showed that 0.333 repeating equals one third, and we multiplied both by three, their answers must be equal. Therefore, 0.999 repeating equals one.)

Text placed in incase you don't like just reading through math. I know I don't.

AnimeJet: Yes, yes you should.

Technically the best way to explain this is to first determine what makes a number unique. In other words two numbers cannot be unique and equal to each other. One of the clearest ways to determine if two numbers are unique or equivalent is to see if there is a number that exists between the two.

In other words for any two given numbers out there which are NOT equal you can always find a number that is greater than one and less than the other (in other words if we have two numbers x and y they if they are unequal then a number z exists such that x<z<y) However with that same scenario if two numebrs are equal then there exists no such number z. No such number z exists which can be placed between .9 repeating endlessly and 1.

Thus .999... and 1 are not unique compared to each other and are thus equivalent. It doesn't matter the number of terms (as "repeating endlessly" REQUIRES an infinite number of terms in the definition of .999...) they are equal period.

AnimeJet wrote:I meant 0.999.. as in the 9's never stop

Yes, that's equal to one. It doesn't approach one; it is one.

You can also prove it by calculating a geometric series of 9*10^(-n), from n=1 to infinity.

IIRC, there's a simple algebraic proof thingie of this...pity I forget what it is...

Badme wrote:IIRC, there's a simple algebraic proof thingie of this...pity I forget what it is...

I think we've already posted it

Newtonian Fury wrote:

AnimeJet wrote:I meant 0.999.. as in the 9's never stop

Yes, that's equal to one. It doesn't approach one; it is one.

You can also prove it by calculating a geometric series of 9*10^(-n), from n=1 to infinity.

You're forgetting that it is the limit of 1/x which is zero as x approaches infinity, even though we normally express that as the shorthand 1/infinity = zero. As I said, it's a subtle distinction which is normally not made explicit for reasons of practicality.

Darth Wong wrote:

Newtonian Fury wrote:Yes, that's equal to one. It doesn't approach one; it is one.

You can also prove it by calculating a geometric series of 9*10^(-n), from n=1 to infinity.

You're forgetting that it is the limit of 1/x which is zero as x approaches infinity, even though we normally express that as the shorthand 1/infinity = zero. As I said, it's a subtle distinction which is normally not made explicit for reasons of practicality.

No, the distinction is not necessary here. The numerical value of a series is by definition the limit of its sequence of partial sums--any time one references the value of an infinite series, one implicitly uses a limit. There is thus no problem in statements like \sum_{n=1}^\infty 9*10^{-n} = 1 being literally true without any additional qualification, because that qualification is already present in the statement. Making it explicit would be redundant, like qualifying the noun `bachelor' with the adjective `unmarried'.

Edit: typo.

Jeremy wrote:
<picture snipped>

That's about all I have to say on this matter.

What, you're scared of simple sequence problems?

I've been taught that right at the beginning of my maths course

Wait 'till you see some of the more whacked-out integral calculus problems!

Badme wrote:IIRC, there's a simple algebraic proof thingie of this...pity I forget what it is...

I posted that and every one ignored me so i'll post it again.

X= 0.9 Re-accuring which we'll write as 9.999 for this purpose.
10x=9.999999
9x= 9 (9.99999 - 0.99999)
x = 9/9 therefore 1

Simple Math. I Did in Yr10 (first year of GCSE and i don't know what the american equivelent