This little guy craves the light of knowledge and wants to know why 0.999... = 1. He wants rigour, but he does accept proofs starting with any sort of premise.
Enlighten him.
This little guy craves the light of knowledge and wants to know why 0.999... = 1. He wants rigour, but he does accept proofs starting with any sort of premise.
Enlighten him.
\begin{align} 0.999.... &= 9\cdot(0.1+0.01+0.001+... ) \ &= 9\cdot( 0.1^1 + 0.1^2 + 0.1^3 + ... ) \ &= 9\cdot(\sum\limits_{k=1}^\infty ( \frac{1}{10^k} ) ) \ &= 9\cdot(\sum\limits_{k=0}^\infty ( \frac{1}{10^{(k+1)}}))\ &= 9\cdot(\sum\limits_{k=0}^\infty \frac{1}{10}*(\frac{1}{10^k})) \ &= \frac{9}{10}\cdot (\sum\limits_{k=0}^\infty (\frac{1}{10^k})) \ &= \frac{9}{10}\cdot \frac{1}{(1-(\frac{1}{10}))}\ &= \frac{9}{10}\cdot \frac{10}{9} = 1 \end{align}
The crux rests on a handy result on from calculus: the sum of an infinite geometric series looks likes s = 1/(1-r), when s = \sum\limits_k=0^inf r^k, and |r| < 1.
Sorry for the latex. When will hexbear render latex? This is a bit more readable:
(aesthetic edit for our big beautiful complex analysts)
Never thought of employing the geometric series for this, that’s clever.
^_^ thank you!
Using i as an index 🤮
? Thats pretty standard though right?
in computer programs, yes
not so much in analysis
I'm seeing n and i used as variables in Rudin, predominantly n though, but Im accustomed to n being a constant.
The symbol 'i' is usually reserved for the imaginary unit.
I've seen it used very frequently as a name for a variable, the imaginary unit i usually has a different typeface to distinguish the two is what Im accustomed to.
Your feedback is valid and I apologize for rendering such an ugly proof
And doesn’t this mean 0.99 approaches 1
A real number can't be approaching anything. It is not a function or any other sort of object that can be said to be approaching anything.
No. 0.99 is 0.9+0.09. The proof I gave shows that 0.99999999999999999999999999999999999(...) is equal to 1.
K