Prove to me that 0.999... does not equal 1

prove to me that 0.999... does not equal 1

Wouldn't that imply that .000... = 0?

But it does....
This is also true. An infinite amount of zeroes equals zero

Do people that constantly make these threads just have never been through calculus? Not understand convergence?

op here
i do believe that 0.999... = 1
but im curious to see people try to disprove it

>implying proof by contradiction is valid

You literally posted an algebraic proof of it as your image, don't pretend this wasn't a clickbait thread for (you)s.

Prove to me that it does without using erroneous approximations of fractions.

You might as well ask me to prove it using toothpicks

What is 0.999... + 0.111...?

1.1 repeating

What happens when you subtract 1 from that?

same thing as subtracting 0.999...

you get 0.111...

>a number with infinite significant figures has the same value as one with one significant figure
>people actually believe this

I'm going to take the bait.

1. If you accept the properties of real numbers then you are forced to accept that .999...=1. Why? Because there is no number between .999... and 1.

2. Convergence of geometric series.

3. The supremum and infimum of .999... are both 1. But this is impossible unless .999... = 1.

Fuck you. I hate the use of the word brainlet in pretty much any context, but anyone who claims that .999...!= 1 is an actual, true to life fucking brainlet.

You are what's wrong with Veeky Forums. Can't we clean this board up a bit, please?

So 9.999999....=10 ?

9.99...=9+0.99...=9+1=10
Did you even need to ask?

does 999=1000?

No, but 999.999..=1000

Does that mean 3.1415...99999... = pi?

So the square root of .9999999999...
Equals the square root of 1?

could be on to something here

(1-0.99...) = 0.000..1
(1-0.99...) + (1-0.99...) = 0.00...2
0.00...2 * inf = 2

but if 0.999... == 1 then
(1-0.99...) = 0
(1-0.99...) + (1-0.99...) = 0
0.00...2 * inf = 0

>0.00...2 * inf = 0
should be 0 * inf = 0

Can we just add the fraction proof to this and make this an automated answer for these threads so we can shut them down immediately?

suppose [math]0.\overline{9}=1[/math]. then let [math]\epsilon>0[/math]. if [math]0.\overline{9}+\epsilon=1[/math], then [math]\epsilon=0[/math]. but [math]\epsilon>0[/math], which is a contradiction.