+10 0.999... is an integer. amirite?

by Fantastic_Baby4117 4 months ago

It's integer because 0.999... is equal to 1. It might get nuked because this is not an opinion but a fact.

by Fantastic_Baby4117 4 months ago

1/3=0,333…. (1/3)*3=3/3 3/3=1 Thus (0,333…)*3=1 Correct me if I am wrong.

by Rich_Syllabub1435 4 months ago

That is of course correct and the easiest way to understand why it must be true. But it's not a rigorous proof.

by Fantastic_Baby4117 4 months ago

Technically speaking, it 0.999 does not = 1 so this is just factually incorrect

by Anonymous 4 months ago

He means 0.999999999999... to infinity =1

by Mateocummings 4 months ago

Pi is exactly 3!

by Nadersonia 4 months ago

So Pi is 27?

by Fantastic_Baby4117 4 months ago

3! is 6, how did you get 27?

by Low-Medicine 4 months ago

oops, i accidentally cubed it. You may quarter me, my bad.

by Fantastic_Baby4117 4 months ago

Pi is equal to the square root of g

by hvonrueden 4 months ago

Is that a g as in 'gif' or in 'garage'?

by Anonymous 4 months ago

Is this actually unpopular? Theres proofs for it everywhere

by Anonymous 4 months ago

Well the proofs he dismantled are actually not rigorous lol. But he doesn't correctly explain why they arent. Thats why if someone asks me I always go with the rigorous but easy to understand explanation that two real numbers are equal if you can't find a real number between them.

by Anonymous 4 months ago

Not an opinion, it's a fact

by Low-Medicine 4 months ago

shh.

by Fantastic_Baby4117 4 months ago

This is tilting at windmills.

by Fickle-Syrup6135 4 months ago

1/3 = .333… 2/3 = .666… 3/3 = .999… = 1

by Relative_Blueberry 4 months ago

Which explains why it must be so. But mathematically speaking, that isn't a rigorous proof.

by Fantastic_Baby4117 4 months ago

So where's the proof?

by Mrippin 4 months ago

Great read! Thanks!

by Mrippin 4 months ago

Sure. The missing decimal 1 exists somewhere. But it is impossible to find it. Since it is infinitely far away. At what point do we just call it one? We hardly ever need a precision past millionth. And that's not that many places over. Hell, if pi = 3 we can still do pretty well with circles. Past 3.1415927. there's no practical use for more precision until we get on the galactic scale. But even then. We absolutely don't need past 50 digits. And when we know the next digit. That 51st digit of .9 repeating is a 9. And we only need 50. We round up. 9 goes to ten and all the way down until we run out of 9s to turn into 10s. Because ultimately. The difference between .9 repeating and 1 is so pedantically tiny that it doesn't matter. Just like my pp.

by Javier45 4 months ago

That is no proof at all, though.

by Fantastic_Baby4117 4 months ago

You're right. It's just me saying that it doesn't really matter.

by Javier45 4 months ago

Fair point. But too many people are vehemently arguing that every proof if wrong. Always with bad math but often convincing enough that others believe them. Kinda like flat-earthers.

by Fantastic_Baby4117 4 months ago

There's many ways to think about it. But in the end it comes down to the fact that decimal notation has no "neat" way to display 1/3. Sure 0.333... is the representation, but it runs into the problem that there are "infinite" 3s, even though it is a simple fraction. This leads to all kind of complications but in the end, the number system for Real numbers as well as Hyperreal numbers are consistent in showing that 0.999... and 1 are actually the same number, just expressed in a different way.

by Fantastic_Baby4117 4 months ago

And this kids is why we stay in school...

by Anonymous 4 months ago

to learn what exactly?

by Fantastic_Baby4117 4 months ago

nah he's right

by Anonymous 4 months ago

Is the proof using infinite geometric series valid or nah

by Anonymous 4 months ago

Sokka-Haiku by Arbalest15: Is the proof using Infinite geometric Series valid or nah Remember that one time Sokka accidentally used an extra syllable in that Haiku Battle in Ba Sing Se? That was a Sokka Haiku and you just made one.

by Logical-Tree-7540 4 months ago

Yes, it is very valid. It is probably the most rigorous proof we know of. And it extends into the hyperreals as well.

by Fantastic_Baby4117 4 months ago