0.(9) ≠ 1

0.999...9 + 0.000...1 = 1
0.999...9 = 1 ->
-> 0.000...1 = 0
Infinitesimal = 0

I have already read everything and many times

>1/3 -> 0.(3)
so
1/3 -> 0.(3) |x3
3/3 -> 0.(9)
1 -> 0.(9) hmmm
but 0.(9) < or = 1.
so 0.(3) x3 ≠ 0.(9) or/and 1/3

If 0.(9) != 1
It must be whether 0.(9) > 1 or 0.(9) < 1
But 0.(9) > 1 is wrong because integer part of 0.(9) , which is 0, is smaller than 1

Also, since both are real number there must be 1 another real number between 0.(9) and 1

So let's say 0.(9) < a < 1 and a exists
Since a < 1, a = 0.b1b2b3b4.....bn
If b1 < 9, 0.(9)

...

just 0.(3) -> 1/3 ok

>0.999...9 + 0.000...1 = 1

wow so 1+0=1
ty mathsuperman

1) If 0.(9) = 1, then infinitely small = 0.
But this is absurd.

2) If 0.(0)1 does not belong to the set of real numbers, then and 0.(9) does not belong to the set of real numbers.
They simultaneously cease to belong to the set of real numbers.

>1) If 0.(9) = 1, then infinitely small = 0.
But this is absurd.
Prove it then

>2) If 0.(0)1 does not belong to the set of real numbers, then and 0.(9) does not belong to the set of real numbers.
They simultaneously cease to belong to the set of real numbers.

0.(0)1 is fucking real number you retard

Do you know why is sqrt(2) a real number?

Above all, you didn't falsify my argument; you just sperg out another autistic claiming