1/3 = .333...

but 3*.333...=.999...

1 =/= 0;

1 =/= .999...

.999... + .000...1 = 1

Get it right fags.

(1/3+0.000...1/3)*3=1 then?


...or something. Fail either way.

>>3

0.000...1 is a number such that 0.000...1 * n < 1 for all n in N. Meaning it is non-archimedian number (infinitesimal in that case).

The only such number in the reals is 0.

a=b
Multiply by a
a^2=ab
add a^2
2(a^2)=(a^2)+ab
Subtract 2ab from each side
2(a^2)-2ab=(a^2)-ab
2((a^2)-ab)=(a^2)-ab
divide by ((a^2)-ab)
2=1

Fuck you.

>>1
Obviously something is wrong with your first assumed statement.  So just stop fretting about this bullshit.

>>1
The thing is: 1 = 0.999...

Basic fucking math analysis

It's not hard at all to show that the repeating decimal 0.9999... equals 1.

Geometric sequences.

.999= 9/10 + 9/100 + 9/1000 + ...

The decimal .999... is a geometric series with the first term 9/10 and common ratio 1/0

Using the forumala a/(1-r), you get .999... = (9/10)/(1-(1/10)) = (9/10)/(9/10)=1

QED

Fags.

Prove me wrong.

>>9
Holy fucking Picard.

>>9
listen, just because you can show it, does not mean that it is actually true in any sense other than the bullshit sense

you are wrong. .999... does not equal 1.

.999... is not a real number, so theoretically it could prove that it equals 1, but in the real world, it could never exist, and thus it cannot equal 1 which is a real number.

real =/= unreal

double fag.

>>12

Numbers are abstract concepts. None of them are real, nigger.

>>13

Even if i wanted accept that response (which I suppose is true, but I think you understand the difference between a real number and an unreal number) .999.... still does not equal 1.

.999.../1 =/= 1

.999... = 99.99....% of 1

.999... + (1-.999...) = 1
1-.999... = 1-.999...
.999... = .999...

it is what it is

nothing more.

now shut the fuck up.

1-.999.... + 1 *10^-infinity.

so fuck you

>>14

>> .999... + (1-.999...) = 1
>> 1-.999... = 1-.999...
>> .999... = .999...

All of those are also true if 0.999... = 1, so I'm not sure what your point is.

>>16
They're also true if .999... = 9000, so it sounds like a long exercise in circular logic.

>>14
lrn2math

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>>6
For the retards, because I've actually had stupid people show me this in real life who act like they were geniuses. It's really pissing me off. Whoever started it is a master troll.

a = b
Using both a and b just confuses stupid people, so I'll just substitute a. Our example will set a to 3 for illustration.
a = a

ex. a = 3, 3 = 3

Multiply by a
a^2 = ab
Let's further simplify the exponents, as they confuse stupid people.
aa = aa

ex. 9 = 9

add a^2
2(a^2) = (a^2) + ab
2(aa) = aa + aa

ex. 18 = 18

Subtract 2ab from each side
2(a^2) - 2ab = (a^2) - ab
2(aa) - 2aa = aa - aa

ex. 0 = 0
(A step was actually skipped on the right side, going from "aa + aa - 2aa" to "aa - aa".)

2((a^2)-ab)=(a^2)-ab
2(aa - aa) = aa - aa

ex. 2(0) = 0, 0 = 0

divide by ((a^2) - ab)
Divide by "(aa - aa)", which is 0.

ex. 2(0) / 0 = 0 / 0, undefined.

That's the trick. It's valid, if misleading, until the last step, which is simple division by 0. Which isn't allowed and doesn't work.

>>12
>>14
i think you mean, "rational" and "irrational"... dont make up terms at your wim, please

>>20

Rats, foiled again.