ITT retarded math
the induction proof is wrong and uses faulty math in subtracting .9r from 9.9r to simplify it to 9
this is WRONG induction on the field of real numbers
The convergence Therm is WRONG too because it says it's approaching 1 but it is NOT 1, it acts LIKE one but is NOT one
Water bottle proof with no math? c'mon now
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>with no math?
it's not like math is required for this.
>"it acts like 1 but it isnt 1 bro"
You sound like a hippy high on pot. You can easily prove the convergence theorem, and it is evident that the formula you derive that's supposedly equal to the number that the infinite series approaches is really equal to the infinite series
>Consider 1
>Divide it by 3
1/3 = 0.333...
>Multiply by 3
1/3 = 0.333...
(×3) (×3)
1 = 0.999...
>Quod Erat Demonstrandum
>Checkmate desu
Smh, mathematicians try and act intellectually superior when they can't even realise that all their autistic formalisations and definitions count for nothing in real life.
ITT: a frustrated high school brainlet posts incorrect assumptions about 0.999... = 1 and blindly expects people to agree with him.
>the induction proof is wrong and uses faulty math in subtracting .9r from 9.9r to simplify it to 9
How is that wrong? If .9r is a number then we can do operations on it right?
>The convergence Therm is WRONG too because it says it's approaching 1 but it is NOT 1, it acts LIKE one but is NOT one
Literally look for the definition of convergence. Limits of sequences are not the same as limits of functions, retardo. Go back to practicing the power rule for calc 2.
>subtracting .9r from 9.9r to simplify it to 9
>this is WRONG
Show me a case where this kind of subtraction results in an incorrect result.
>mulitplying infinity
But how do you know that [math]3\ctimes 0.333...=0.999...[/math], how do you know your multiplication of infinitely long numbers is valid?
Because it's a recurring process
0.3 ×3 = 0.9
0.33 ×3 = 0.99
0.333 ×3 = 0.999
0.333... ×3 = 0.999...
>convergence isn't real because I said convergence isn't real
>I'm so smart
Allow me to prove this easily.
0.99999... is equal to:
[eqn] \sum_{n=1}^{∞} (\frac{9}{10^n}) [/eqn]
And the partial sums are:
[eqn] \frac{9}{10},\frac{99}{100},\frac{999}{1000},... [/eqn]
So the sum is this limit:
[eqn] \lim_{x \to ∞ } \frac{10^x -1}{10^x} [/eqn]
Using L'hopital we get:
[eqn] \lim_{x \to ∞ } \frac{ln(10)10^x}{ln(10)10^x} = \lim_{x \to ∞ } 1 = 1 [/eqn]
The answer is 1. Undeniable mathematical proof.
If you still can't accept it, watch mathologer's non-calc video about it ( Timecode: 8:27 youtu.be
>he can divide exactly in 3 infinitesimal equal parts
this will take infinite time user
you can silmplify it to 9 only if they have the same number of decimals, does it? is infinity=infinity.
And if yo decide a precision is it the same number?
This can be proved very easily by the definition of real numbers via Dedekind cuts
Whats wrong with the infinite series proof? That's how I showed my dad?
>This can be proved very easily by the definition of real numbers via Dedekind cuts
Please do so then, user? thnx
this is equivalent to showing no rational number lies between 0.99999999999999999... and 1.
this is obvious from the fact that any rational has a unique finite decimal expansion
thnx user
>any rational has a unique finite decimal expansion
But you just contradicted yourself because 1 has two different decimal expansions: 1.00000 and 0.99999...
So how would you prove there is no rational number between 0.9999... and 1 without the "unique expansions" argument?
Unique finite decimal expansion that does not terminate with 0s
How do you figure that?
What about:
0.33333...
0.333329999....
These numbers do not end in 0s and they are non-unique expansions for the same number 1/3.
Wait a second I'm fucking retarded, forget that example kek.
I'm just wondering how exactly you know there are no numbers between 1.0 and .999...
What convinced me was the opposite induction.
1 - .999... = N
Solve for N
0.000...
I realized that there never is a 1 at the end of it. It's zeros all the way - forever. And what is another way to say "zeros all the way forever"?
>zero.
So 0.000... = 0
So 0.999... + 0.000... = .999... + 0 = 1
Therefore 0.999... = 1, because 0.999... + 0 = 1
>Dedekind cuts
STOP
[math] \Big(\forall n \in \mathbb{N}:\ 1 - \frac1n \le 0.999... \le 1\Big) [/math] implies [math] 0.999... = 1 [/math] by squeeze theorem.
i meant that you dont need math to know that 0.9... isnt 1
for
It is for almost every number system though. For reals, 0.999... represents the same number as 1. This is why you absolutely need math, because your intuition can be wrong (frequent occurence for brainlets in probability).