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/sci/ - Science & Math

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1785921 No.1785921 [Reply] [Original]

If .999... is really less than 1, then taking higher and higher powers of it will make it smaller and smaller, right? Like how if you multiply .5 x .5 x .5 x .5 ... it approaches zero.

Therefore if you calculate .9999... ^ infinity, you'll get zero if .9999... < 1 and one if .9999... = 1.

So... someone should totally do that.

>> No.1785932

If by ".999..." you mean (10^n-1)/(10^n) when n approaches infinity (infinite 9's), that IS equal to 1.

>> No.1785934
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1785934

http://www.wolframalpha.com/input/?i=.999999^infinity

.999999^infinity = 0

http://www.wolframalpha.com/input/?i=.999999999999999999999999999999^infinity

.999999999999999999999999999999^infinity = 0

http://www.wolframalpha.com/input/?i=.99999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999999999999999999999999999999999999999999999999999999^infinity

.999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999^infinity = 0

Continuing in this manner, no matter how many 9's you enter, you'll always get an answer of zero. Therefore .9999....^infinity = 0 and .9999.... < 1.

QED MOTHERFUCKERS

>> No.1785935

.999...=1

there's even a thread on wiki about it for you idiots.

/thread.

>> No.1785936

.9999999999999999999999999999999999999^2 = 1

>> No.1785942
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1785942

>>1785934

>> No.1785953

0.999...^2 = 0.81818181...

>> No.1785963

.999 only equals 1 because of the flaws in our own definitions of mathematics.

>> No.1785979

>>1785934
that's not how induction works

>> No.1785995
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1785995

>>1785979

>> No.1785999
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1785999

>>1785921

>> No.1786001

>>1785934
Consider the sequence S = {.9, .99, .999, .9999, ...}.
Given a positive integer m, we pick out a number s_m in S (so that s_m is the decimal point followed by m '9's, then infinity '0's). Fix an arbitrarily small positive epsilon, we seek N such that 1 - s^N < epsilon. Obviously as m increases to infinity, so does N.

>> No.1786006

>>1785934

>comparing a number less than one with .999...
>knows nothing of mathematics, or logic for that matter

>> No.1786008

>>1786001
I meant s^N < epsilon not 1 - s^N < epsilon
durr

>> No.1786014

>>1786008
I meant s_m instead of s durr

>> No.1786022

>>1786006
as the numbers .999999, .9999999999999, .999999999999999999, etc. approach .999999..., the values of .999999^infinity will approach .9999....^infinity by continuity. since these numbers are all 0, their limit, which is .9999...^infinity, is also zero.

>> No.1786023

Protip: 100% of threads about .999... are troll threads. sage goes in the email field.

>> No.1786035

>>1786022

> implying that the limit as n -> infinity of .999...^n = 0

>> No.1786040
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1786040

Don't you mean 99.99...%?

>> No.1786044

>>1786022
Learn what a limit is before you use it. http://en.wikipedia.org/wiki/Limit_of_a_sequence

>> No.1786051

>>1785921
>If .999... is really less than 1
Flawed premise.

>> No.1786052

>>1786044
lrn2 calculus, n00btard

>> No.1786054

>>1786051
Flawed reading comprehension.

>> No.1786059

>>1786022
>implying that you can speak of a sequence as being "continuous"

>> No.1786073

>>1785934
> no matter how many 9's you enter, so long as it is a finite number of 9s
Fixed that for you. lrn2infinity

>> No.1786082

>>1786052
lrn2 epsilon delta, idiot who thinks he knows calculus because he can spell "continuity".