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/sci/ - Science & Math

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10075991 No.10075991[DELETED]  [Reply] [Original]

>0.999... = 1

>> No.10075999

>It's this brainlet again

>> No.10076007

does 0,999... = 1 imply that an infinitesimal is zero?

>> No.10076045

0.999... is equal to 0.999... but if you're an engineer 1 is easier to work with.

A more extreme case would be using 3 for pi.

>> No.10076050

>>10075991
Name one (1) rational between 0.99999.... and 1.

>> No.10076059

Define epsilon such that
1 != 1 - epsilon

>> No.10076084
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10076084

>>10076050
>0.99999... < r < 1
>0.999999... = r
dumbfuck

>> No.10076092

>>10075991
For very large values of 0.999...

>> No.10076108

>>10076084
>the next element of the series
Y'all niggas insist on disappointing me.

>> No.10076111

>>10075991
this same thread has been going on since like 2005

>> No.10076140

>>10076108
I mean...

1 < c < [math]\infty[/math]
why wouldn't [an] acceptable answer for c be the next element, aka 2?

>> No.10076143

>>10076084
How the fuck
>0.999...<r
And
>0.999...=r
?

>> No.10076152

>>10076140
Because 0.999.... is the limit of the sequence, not an element of it. I can't say that the number after 0.99999...... is between it and one.

>> No.10076167

>>10075991
Let s be a series such that

s = \sum _{n=1}^{\infty }\left(\frac{9}{10^n}\right)

by doing the summation formula

=9\cdot \sum \:_{n=1}^{\infty \:}\frac{1}{10^n}
=9\left(\sum _{n=0}^{\infty \:}\frac{1}{10^n}-\frac{1}{10^0}\right)
=9\left(\frac{10}{9}-\frac{1}{10^0}\right)
=1

>> No.10076173

If 1/3 = 0.333… , 2/3 = 0.666…, then 3/3 = 0.999… , but 3/3 = 1, so 0.999 = 1 surely?

>> No.10076178

>>10076007
In standard analysis infinitesimals do not exist.

>> No.10076197

>>10076152
You can't say any number under any circumstance could come after 0.999... if that is your rationale. Its supposed to be infinite, ain't it? Kinda fuckin your own ideas up if you presume something can come after infinity.

mathleticians have a divine misunderstanding of infinity.