does 0.9... = 1?

[TruffleSpark]

0.9 receding and 1, are they the same thing?

 

[ditto]

0.9 recurring does not exist.

 

[Sean Shemmy]

77+33=100.

 

[Tablet County]

No because then it'd be 1.

 

[Randy Facalding]

it's some faggotshit to do with "limits" idk, haven't been in calculus for a while

 

[stares at error messages]

Why not? 2.99 is the way you write 3.00 these days.

 

[Sparkletor 2.0]

Yes. If you have .9 of a candy bar you still have 1 candy bar.

 

[Fat Gollum]

If it is then you shouldn't have a problem if I leave you with 0.9... of your penis.

 

[Toji Suzuhara]

Yes, pi=3 as well

 

[Temmie]

Depends on if your CPU supports floats.

 

[Penis Drager 2.0]

What's 0.3... × 3?

 

[bashe]

it's some faggotshit to do with "limits" idk, haven't been in calculus for a while

What's 0.3... × 3?

That pretty much hits the nail on the head. you can also use the formula for the geometric series for proof that .9 repeating is 1

 

[We Are The Witches]

So because:
0.333... * 3 = 0.999...
0.333... = 1/3
1/3 * 3 = 1
0.333... * 3 = 1

Has both solutions.
Or maybe not... maybe it's something else and the formulas are wrong. Maybe 1/3 is not 0.333... and it's instead 0.333... + (?).

 

[W'rkncacnter]

I don't think you'd ever multiply 0.333.... by anything, you'd convert it to a fraction and then multiply. I'd say 0.333... is an imperfect way of writing 1/3 so therefore 0.999... is an imperfect way of writing 1.

Spoiler

This wouldn't happen if we had a base 12 numerical system like god intended.

 

[Penis Drager 2.0]

Or maybe not... maybe it's something else and the formulas are wrong. Maybe 1/3 is not 0.333... and it's instead 0.333... + (?).

Let's say that is the case.
If 0.9... was not 1, that would imply some number, 0.0...01, exists which makes up the difference.

Such is the case where existing math is unsuitable to do what we're trying to do, we'll do what all good mathematicians do and make shit up. We'll call this infinitesimal value "j." So 0.9... + j = 1.
1/3, then, is 0.3... + j/3. That value gives us some interesting results:
Evaluating j/3, we get "0.0...033... + j/3"
We divided this number and it got larger. And now there's recursion to boot.

We could have even more fun with this, noting that 0.9.../9 = 0.1... while 1/9 would be 0.1...+j/9.
j/9 evaluates to "0.0...01... + j/9"
Congrats, we have just invented a brand new field of infinitesimal mathematics and all it really does is show that infinitely repeating numbers repeat infinitely. Be my guest if you want to do math this way.

 

[We Are The Witches]

Let's say that is the case.
If 0.9... was not 1, that would imply some number, 0.0...01, exists which makes up the difference.

No, I would not treat it like that.

If 0.000...1 was the one that made the difference, it would mean 0.999... has an end to it, which isn't the case, because after adding the first with the latter (let's say 0.000001 to 0.999999) you'd seemingly get to 1, but then remember that the 9s continue, so you'd get 1.000000999...

For that reason I'd leave it similarly to what you've said: 0.999... + j = 1

However this is symbolic, you would not try to divide "j" by anything, it simply means that 0.999... cannot reach 1 ever by itself, and that's that. Just like (0.333... + j) would simply mean that (0.333...) cannot reach (0.333...4) by itself.

This is how I would do things, purely for theoretical purposes, you would not use this for anything in real life.

 

[Geddy Lee's Fee and See]

Yes. Just because one number has multiple representations doesn't mean those representations correspond to a different thing. It's like how you're the same person even though you've gotten older or moved to a different place. I am still me and you are still you even if some extraneous features regarding our appearance has changed.


Anyways a simple way to prove this is to find the sum of the geometric sum of 9*(1/10)^n (where n >=1). That's all that 0.999.... is and you get (0.9)/(1 -0.1) = 0.9/0.9 = 1.