0.99999~ =1???
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Corvallis5: So, you mean that if we have an infinite amount of nines, and we take one nine from there, that said number will become smaller? That's is just plain stupid. There is an infinite amount of nines, how can you just take one away? Forum
Off Topic 0.99999~ =1???
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Do you think 0.99999~ =1?
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Corvallis5: So, you mean that if we have an infinite amount of nines, and we take one nine from there, that said number will become smaller? That's is just plain stupid. There is an infinite amount of nines, how can you just take one away? Since, you cant set the exact value for infinite, it just grows and grows...
Devil-Thanh hat geschriebenI think You wrong
0.(9) isn't 1
try this: with 5 characters "9"
a = 0.99999
10a = 10x0.99999 = 9.9999
10a-a = 9a = 9.9999-0.99999 = 8.99991 ( it's different from 9)
a= 9a/9 = 0.99999 (only 0.99999)
so we can write
a=0.99999...
10=9.9999...
10a-a = 9a = 9.9999...-0.99999 = 8.99...991
a = 9a/9 = 0.99999...
so 0.(9) isn't 1
0.(9) isn't 1
try this: with 5 characters "9"
a = 0.99999
10a = 10x0.99999 = 9.9999
10a-a = 9a = 9.9999-0.99999 = 8.99991 ( it's different from 9)
a= 9a/9 = 0.99999 (only 0.99999)
so we can write
a=0.99999...
10=9.9999...
10a-a = 9a = 9.9999...-0.99999 = 8.99...991
a = 9a/9 = 0.99999...
so 0.(9) isn't 1
how old are you?did you study this lesson in your class?
it is 0,(9)(infinite 9 after ,)not a specific number of 9
so then 1 = g(k) + 10^(-k+1). However, k is not merely bounded by the set of integers as euclid proved, the interval between 0.9 and 1 has cardinality larger than that of the set of all integers, meaning that there are uncountably infinite ways of representing one as the sum of g(k) and 10^(-k+1).
So here's our problem:
Let k be the cardinality of the set of Reals, does a^(-k+1) = 0?
Under certain contexts, yes. Under other contexts, no. So we can't say that 10^(-k+1) > 0 or not, hence g(k) can either be 1 or < 1 depending on the context of the problem.
In most of mathematics, we can assume that if k > aleph null, then a^(-k), for all a in the set of integers, will converge to 0 as an axiom, so in most situations where you will need to test whether this is true or not, 0.99999~ = 1 unless you work under a mathematical system under which k > aleph null and a^(-k) =\= 0.
teoretically here is the formule:
0.(x)=x/9
that is the teoretic
so if you have 0.(9) it will be = with 9/9 wich =1
JustARandomPlayer hat geschriebenteoretically here is the formule:
0.(x)=x/9
that is the teoretic
so if you have 0.(9) it will be = with 9/9 wich =1
0.(x)=x/9
that is the teoretic
so if you have 0.(9) it will be = with 9/9 wich =1
I don't think you should be formulating things just yet.
JustARandomPlayer hat geschriebenso another post of mine
teoretically here is the formule:
0.(x)=x/9
that is the teoretic
so if you have 0.(9) it will be = with 9/9 wich =1
teoretically here is the formule:
0.(x)=x/9
that is the teoretic
so if you have 0.(9) it will be = with 9/9 wich =1
this works only for first 9 integers but doesn't work with others for eg:
0.(3)=3/9=0.333333333333
but this won't work on
0.(14) =/= 14/9 =/= 0.1414141414
Easy way to explain anything is u need to use prove on first page because it is applicable for any infinite recurring number.
Or if u are lazy to use it u can do this:
0.(x)=x/(9s times amount of digits)
eg:
0.(27)=27/99=0.2727272727
0.(347)=347/999=0.347347347
If u have non recurring numbers in front recurring ones u need to add "0" for each such number at the end of denominator and subtract that non recurring number from numerator.
eg:
0.4(7)=[47-4]/90=43/90=0.47777777
0.27(12)=[2712-27]/9900=2685/9900=0.2712121212
Navy Seal hat geschriebenintegers
Way out of your league here, bro.
9 / 100 = 0.09
9 / 1000 = 0.009
so
0.9999 ~ = 9 * (1/10 + 1/100 + 1/1000 + ... )
= 9 * ( sum 10^-n for n = 1,infinity)
= 9 * (1/9) = 1
So there you have it kids..
huh i don't get ur joke/statement
@vyn we already proved this a lot of times in this topic lol
This is pointless thread in my opinion.
Navy Seal hat geschriebenThere wasn't a proof what 0.99999 ~ actually was.
Take the first proof.
a = 0.99 .....9 ~
10 a = 9.9 .....9~
10a - a = 8.9.....91
9*a = 8.9.....91
a =8.9....91 / 9 = 0.99.....9 != 1
FAIL
so this one doesnt work when the number of nines doesn't go to infinity. and even if it does, are you still sure 10a - a = 9 ?
You have to write down things mathematically correct.
What is 0.99999 ~?
it's 1 - a, with lim a = 0, ofcourse this is 1.
and
Aura hat geschrieben0.99999~ + 0.111111~ = 1
This is pointless thread in my opinion.
This is pointless thread in my opinion.
Uhh???
0.9999
0.1111
________+
1.1110
1.1110 != 1
1× editiert, zuletzt 22.02.12 18:37:40
vyn: I think he meant 0.9999+0.0001=1. 9*a = 8.9.....91
a =8.9....91 / 9 = 0.99.....9 != 1"
duh U cannot have 91 at the end because it is infinity.
U can get 91 at the end only when u have finite amount of 9s, while when u have infinite amount of 9s u get exactly 9, because amount of 9s after dot in both 9.99999999... and 0.9999999... is infinite, which cancels out when u subtract them.
here is why it is right:
a = 0.(9) = 0.9999999...
10*a = 9.(9) = 9.99999999...
Now when u subtract this:
10a - a = 9.(9) - 0.(9) = 9
9*a = 9
a = 9/9 = 1
SUCCESS :]
but ur solution looks cleaner
Aura hat geschrieben0.99999~ + 0.111111~ = 1
This is pointless thread in my opinion.
This is pointless thread in my opinion.
-.-' 0.(9) + 0.(1) = 1.(1) = 1.11111111...
Please do not post something on mathematical topics before u pass ur math exams at least in year 9 -.-' ur post is even more useless because u did a mistake in it
u cannot have 0.9999999... + 0.000000.....0001 because u need infinite amount of 0s before u write ur 1, which makes it impossible because if u have a 1 in this set of digits then this set of 0s is ot infinite, which contradicts the amount of infinite 9s
Don't explain either, just fucking google your useless 0.99999 + 0.00001 or is 0.9999~=1 and Let it be.
Navy Seal hat geschrieben JustARandomPlayer hat geschriebenso another post of mine
teoretically here is the formule:
0.(x)=x/9
that is the teoretic
so if you have 0.(9) it will be = with 9/9 wich =1
teoretically here is the formule:
0.(x)=x/9
that is the teoretic
so if you have 0.(9) it will be = with 9/9 wich =1
this works only for first 9 integers but doesn't work with others for eg:
0.(3)=3/9=0.333333333333
but this won't work on
0.(14) =/= 14/9 =/= 0.1414141414
Easy way to explain anything is u need to use prove on first page because it is applicable for any infinite recurring number.
Or if u are lazy to use it u can do this:
0.(x)=x/(9s times amount of digits)
eg:
0.(27)=27/99=0.2727272727
0.(347)=347/999=0.347347347
If u have non recurring numbers in front recurring ones u need to add "0" for each such number at the end of denominator and subtract that non recurring number from numerator.
eg:
0.4(7)=[47-4]/90=43/90=0.47777777
0.27(12)=[2712-27]/9900=2685/9900=0.2712121212
yes you are right but for a formule...
oh wait...
i found the formules
so...
0.(x)=x/999 were you put 9 of how many digits of the repeting numer
eg:0.(533)=533/999-there are 3 digits so three 9
if you have 0.y(x) then it will be equal with (yx(not y*x)-y)/90 were you put 9 of how many digits of the repeating number and 0 of how many digits are before the repeting number
eg: 0.74(86)=7486-86/9900
so that's it
2× editiert, zuletzt 22.02.12 19:46:21
JustARandomPlayer hat geschriebenx-the number who is repeating(ex:6)
so the formule:
x/9*10^n were n-the number of how many aaa(idk how it is named :D) is in nvm
here a example
if 0.(143234) there the formule will be 143234/999999
so n is 6 in this case
if 0.(43) then n is 2
so now i hope you will understand
so the formule:
x/9*10^n were n-the number of how many aaa(idk how it is named :D) is in nvm
here a example
if 0.(143234) there the formule will be 143234/999999
so n is 6 in this case
if 0.(43) then n is 2
so now i hope you will understand
and again u have wrong answer.
at least this part is wrong:
x/9*10^n
let n = 6 as u said then:
x/9*1000000 which is wrong because we need 9999999 and not 9000000
represents the series 0.99999...
however 9*g(k) = (10*g(k) - g(k)) is only true when k is bounded by the integers, since this is an implied condition of the sum operator that its bounds cannot be coprime with each other when normalized into integers. Remember that if we define the cardinality (size) of a all integers as aleph-null, then we know from Euclid that the cardinality of both the reals, and any uncountable subset of the reals, are strictly greater than that of the integers. Since k must be bounded between [0, aleph-null], and since in order to represent an infinitely repeating series of 9's in the decimal requires a k* > aleph-null, then k* is technically not closed within reals and hence we can't claim that the above operation is valid since it requires all of its operands to be reals. However where most of arithmetic is concerned, it's acceptable to add the precondition that the summation need not be countably infinite and hence the rest of your proof still stands.