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.

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[–] dat_math@hexbear.net 10 points 4 months ago* (last edited 4 months ago) (3 children)

and wants to know why 0.999... = 1

\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)

[–] reaper_cushions@hexbear.net 9 points 4 months ago (1 children)

Never thought of employing the geometric series for this, that’s clever.

[–] dat_math@hexbear.net 6 points 4 months ago* (last edited 4 months ago)

^_^ thank you!

[–] Xavienth@lemmygrad.ml 3 points 4 months ago (2 children)

Using i as an index 🤮

[–] mathemachristian@hexbear.net 6 points 4 months ago (2 children)

? Thats pretty standard though right?

[–] dat_math@hexbear.net 6 points 4 months ago* (last edited 4 months ago) (1 children)

in computer programs, yes

not so much in analysis

[–] mathemachristian@hexbear.net 3 points 4 months ago* (last edited 4 months ago)

I'm seeing n and i used as variables in Rudin, predominantly n though, but Im accustomed to n being a constant.

[–] Tomorrow_Farewell@hexbear.net 5 points 4 months ago (1 children)

The symbol 'i' is usually reserved for the imaginary unit.

[–] mathemachristian@hexbear.net 2 points 4 months ago

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.

[–] dat_math@hexbear.net 2 points 4 months ago

Your feedback is valid and I apologize for rendering such an ugly proof

[–] TheWurstman@hexbear.net 2 points 4 months ago (2 children)

And doesn’t this mean 0.99 approaches 1

[–] Tomorrow_Farewell@hexbear.net 8 points 4 months ago

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.

[–] dat_math@hexbear.net 5 points 4 months ago (1 children)

No. 0.99 is 0.9+0.09. The proof I gave shows that 0.99999999999999999999999999999999999(...) is equal to 1.

[–] TheWurstman@hexbear.net 3 points 4 months ago