326 replies to this topic

[octobclrnts]

All the mathematical demonstrations given have simply proven that the limit of .999999... is 1. None of them have proven that .99999... itself is 1. A limit is the number that is approached but never reached!

[Overlord]

talking about the limit of a number is "stupid", unuseful (the limit of a number is the number itself )
.9999.... is the limit of the seria 0.9 0.09 0.009 0.0009 ..., the limit = 1 so .999.... = 1

[CrimsonCasio]

it still doesnt make sense to me...

[Bija]

0.99999999... is a limite not a defined number
so as the limite is 1, 0.9999...=1

it is infinitely close to 1,always more closer, the 'most' closer is 1

[CrimsonCasio]

by that I take it that your saying that .99999~ is not equal to 1, but infinately close to 1. that I could understand and agree with but I dont see how .9999~ = 1

[Orwell]

0.99999... is the limit of the sequence 0, 0.9, 0.99, 0.999, 0.9999,... and this limit is also equal to 1. (Think at the example of the ball that someone drop against a wall: it first does 90% of the distance, then 99%, then 99,9% etc then finally hit the wall, so it finally does 100% of the way...)

[betoe]

On an university exam, if you have the 3.9999 result, put 4 and look what your teacher will think (here in my uny its bad ).

[Overlord]

if you have 3.9999 don't put 4, it's bad for teachers that like precision.
if you have 3.9999..., you can put 4, and if your teacher says it's wrong just tell him to consult a real math teacher

[2072]

The fact that 0.9999...=1 is a mathematical fact not a "logical" fact. Logical != mathematical

[CrimsonCasio]

logic and math should be equal, else one of them is wrong. (and logic by definition cannot be wrong, though it might not really be logic)

therefor it should be possible to explain this in logical terms.

[Orwell]

(and logic by definition cannot be wrong, though it might not really be logic)

This is absolute nonsense
There are many things you can find "logical" but that are completely wrong (like the words you just said )

[CrimsonCasio]

i dont want this to become a debate of the definetion of logic, but: what im saying is that any statement that is logical must in fact also be true, otherwise it is not really logic (though it may be based on logical thinking)

[qwerty841]

i really started something

[CrimsonCasio]

it happens from time to time

[huhn_m]

some said it really well ...

0.9999... approaches 1 but NEVER actually gets it. it gets closer and closer.

It's like a parable that approaches the x axis but never intersects it.

You can't say "of course it touches the axis because sometimes it gets 0
thats nonsens. Write for such a function in a test that it touches the x axis and
you'll fail the question thats how simply it is.

And if math isn't anymore logic so why to believe in math ...

[Bija]

a parable never intersects the x axis but the limit when x tends towards infinite IS 0 !!

as the same considering the following series :
un=(10^n-1)/(10^n)=1-1/(10^n)

it is clear that the limite of that series is 0.9999...
but as 1/(10^n) tends towards 0, the limite is also 1

it is perfectly logic to me

[huhn_m]

sorry that I have not expressed myself correctly but I'm not to fond of
english math terms

anyways I meant that there are graphs that approach the x axis but NEVER
intersect / touch it. Thats the same with this.

You can always say that the lim to x -> inf. is 0 but you can NEVER say
that y will be 0 at any time.

take 1/(sqrt(x)).
My math teacher would kill me if I said it would be 0 at any time.
Do you now understand what I mean.
You can not say that the limes is the same as the equal sign.

[CrimsonCasio]

I agree with huhn, thats exactly what I was thinking.

[Orwell]

You can always say that the lim to x -> inf. is 0 but you can NEVER say
that y will be 0 at any time.

You can, when time == infinite This "infinite" thing is the key of everything, and that's why we can see that 0.999... = 1 because there is an infinite number of 9

To be more "complete" huhn, you are right when you say that 1/(sqrt(x)) will never be equal to 0, because when we study such functions we generally work with real numbers, and inf & -inf don't belong to this group, so there is no real value for x to have 1/(sqrt(x))==0

[CrimsonCasio]

that still doesnt work, .9999 approches 1 infitately but will never reach it. it cant by its very nature (two diffrent numbers cannot equal each other).

[R00KIE]

0.999(9) WILL NEVER BE EQUAL TO 1
it's very simple, it's just like that.
the only reason that makes us say that 0.999(9) is equal to one is because the error of assuming that isn't important for the operation we do and it's a shorter number to represent and easier to work with.
Another thing that might some people to say that 0.999(9)=1 is because of the finite precision of calculators that after a some calculations that give results that put the floating point precision to test, the calculators mess thing up.

[Overlord]

and this error is exactly equal to zero when you have an infinite number of '9' in your number, so 0.999... = 1, it's very simple lol

i think each part will have many difficulties to convince the other part... see, we've already written 62 posts

[Daruosh]

I think like OVERLORD.

[huhn_m]

and I think like ROOKIE and Crimson

[Orwell]

0.999(9) WILL NEVER BE EQUAL TO 1
Another thing that might some people to say that 0.999(9)=1 is because of the finite precision of calculators that after a some calculations that give results that put the floating point precision to test, the calculators mess thing up.

<{POST_SNAPBACK}>

We never spoke about calculators in this topic... there is no question about floating point precision here

I have another argument (and for me, you can't refuse it, unless you don't know what the real numbers are )

We have a word in french to characterize the real numbers: we say their group is "dense". That means that between 2 differents real numbers (let's say A and B ), there is allways a third one C between A and B, strictly different from them. (For all A,B reals, there is C real in such a way that A < C < B is verified.)
If you choose 0.001 and 0.002, there is 0.0015 betwen them; between 10^-20 and 2*10^-20, there is 1.5*10^-20 etc...

If you postulate that 0.999... and 1 are two different numbers, then you should be able to find another different number between them, and different from them. But you cannot find a number x in such a way that 0.999999... < x < 1, because of the infinite number of the chiffre '9' This real x should be "closer to 1" than 0.9999... , however we all assumed here that 0.999... was infinitely close.

Thus, if you cannot find this number x (it can't be 0.999... with "more" '9' than 0.999... itself! ), that's because 0.999... and 1 are not two different real numbers.

that still doesnt work, .9999 approches 1 infitately but will never reach it. it cant by its very nature (two diffrent numbers cannot equal each other).

ok for "two diffrent numbers cannot equal each other" .
but who said they were unequals?

[Orwell]

I have to correct something in my last post above: the correct definition for the "density" of the real numbers is that between 2 different real numbers, there is allways a rationnal number different from them."
But the conclusion is the same

[huhn_m]

and so 0,0...1 is equal to zero?
and 1,0....1 is equal to one and so on ... I don't think this is a valid argument
since it fits for an endless amount of infinite numbers

[Orwell]

No, because 0,000...01 and 1,000...01 are finite number (because they have 'a "last chiffre", here the '1', however 0,999... hasn't such a "last chiffre")
You can't have numbers with a first & a last chiffre, with infinite chiffres between them

[CrimsonCasio]

correct me if im wrong, but I dont think that anything that has to do with infinity can be "real", therefor your rule wouldnt apply.

[Orwell]

1/3 = 0.33333... has an infinite number of '3', and is naturally real like 0.9999...

[Overlord]

pi has an infinite number of decimals !

[huhn_m]

so what about 1.9999999... is this equal 2 too and all the other numbers ending with
this.

I'm still on the side that this is only true if you accept a loss of precission.
(see above). I don't know about THIS real number definition you mentioned so
I won't take it into account. I'll ask my physics teacher on thursday since she
really knows a lot ... maybe she can help ...

[CrimsonCasio]

ok, how about this:

to use your rule-
1> .5*(.999oo)+(.999oo) > .999oo

[Orwell]

.5*(.999oo)+(.999oo) < 1? are you sure about that?

it is not a physics teacher we need, it's an algebra or analysys one

[Bija]

I've seen (with a maths teacher :-) in maths that 0.999999..=1

isn't that (and all the demonstrations we gave) enough ?

[CrimsonCasio]

.5*(.999oo)+(.999oo)  < 1? are you sure about that? 

it is not a physics teacher we need, it's an algebra or analysys one 

<{POST_SNAPBACK}>

again, im not a math person but i think you get what im trying to say.

and no, its not enough untill I here an explenation that makes sense.

[Overlord]

5*(.999oo)+(.999oo)  < 1

.5*(.999oo) must be something like (.499oo) no ?

and (.499oo)+(.999oo) surely > 1

[huhn_m]

1)
I think what you ment was 0.5*((.99999...)+(.99999..))<1 because else it would be
approaching 1.5 ... but this way it approaches 1 again You
cracked their "logic" explanation

2nd)
she is a maths/algebra teacher too ... I only don't have her in my maths course ...

[Overlord]

.5*(.99999... + .99999...) = .5*2*(.99999...) = .99999... but is not > .99999... :s as it's the same number

[CrimsonCasio]

.5*(.99999... + .99999...) = .5*2*(.99999...) = .99999... but is not > .99999... :s as it's the same number

<{POST_SNAPBACK}>

that makes absolutely no sense to me...

@huhn: sometimes ignorance is the catalyst of innovation