0.999...=1 on wikipedia

[SpitValve]

Wikipedia's featured article for today:

http://en.wikipedia.org/wiki/0.999...

It is a hard concept to explain to people... read through the talk pages, lots of confusion there

[fjafjan]

but why not just point out htat 0.99999999... = 3/3

or, 0.3333..... = 1/3

0.33333 * 3 = 3/3

and also one could do say "where should we round this number?" and no matter where you chose you end up with 1...


It is a strange concept at first though, true

[ulnevets]

what is the largest number that is less than 1, but not 1?

[kira]

what is the largest number that is less than 1, but not 1?

That's pretty much the equivalent of saying "what's the largest number that is not infinity?"

[Gelsamel]

but why not just point out htat 0.99999999... = 3/3

or, 0.3333..... = 1/3

0.33333 * 3 = 3/3

and also one could do say "where should we round this number?" and no matter where you chose you end up with 1...


It is a strange concept at first though, true

Because someone with an objection to .999... = 1 may easily have an objection to .333... = 1/3.

[fjafjan]

but why not just point out htat 0.99999999... = 3/3

or, 0.3333..... = 1/3

0.33333 * 3 = 3/3

and also one could do say "where should we round this number?" and no matter where you chose you end up with 1...


It is a strange concept at first though, true

Because someone with an objection to .999... = 1 may easily have an objection to .333... = 1/3.

But that is just the way one writes it, ultimatly, if ... = infinite 3s, then 0.33... is 1/3 .... Or are these people saying you cannot write 1/3 with decimals?

[mezz]

The largest number that is not 1 ?

<1 ?


lim 1 - [1/(n)] ??
n->infinity


lim n
n->1-

How do you say it mathematically?

[Gelsamel]

Or are these people saying you cannot write 1/3 with decimals?

Yes.

[svk1325]

Or are these people saying you cannot write 1/3 with decimals?

Yes.

Not in base 10 anyways

You may be able to explain that 0.999... = 1. You may even be able to mathematically prove it. Some people still won't believe you and just say "but they're written differently" or something to that effect.

Wasn't there some thread like this under "Logic Puzzles", or am I just imagining things?

[Peshmerga]

I'm fairly convinced that that "fact" .999... = 1 is a Jewish conspiracy started to cripple world governments and online forums.

[Gelsamel]

I prefer...

Let x = .999....
x=.999....
10x = 9.999....
10x - x = 9.999.... - .999....
9x = 9
x = 1

Therefore 1 = .999...

[wisnij]

what is the largest number that is less than 1, but not 1?

No such real number exists. The real numbers are dense, meaning that for any number n < 1, there will always be a number m such that n < m < 1.

[SpitValve]

Wasn't there some thread like this under "Logic Puzzles", or am I just imagining things?

true...

[fjafjan]

Or are these people saying you cannot write 1/3 with decimals?

Yes.

Not in base 10 anyways

decimal

so it was sort of implied

[Narsil]

This can be mathematically proven false by the simple question of "what happens if I show up at a dollar store with 99 and .99999999.... pennies?" Obviously I'll be .11111111..... penny short, so I won't be able to buy anything. Of course, a "Take a penny, leave a penny" jar could in theory solve that.

[sandy]

This can be mathematically proven false by the simple question of "what happens if I show up at a dollar store with 99 and .99999999.... pennies?" Obviously I'll be .11111111..... penny short, so I won't be able to buy anything. Of course, a "Take a penny, leave a penny" jar could in theory solve that.

In what universe does .999... + .111... = 1.0? Remember to carry the 1 when you add. I think what you mean to say is .999... + .000...1 = 1.0, but you can't do that -- you can't just say I have infinite zeros and then a 1 at the end. You should check out the logic puzzles thread, the various proofs of .999... = 1 are fairly straight-forward.

[Hamorad]

A few ways of putting it that I like:

0.999... = 0.9 + 0.09 + 0.009 + 0.0009 ...
= 9*0.1 + 9*0.01 + 9*0.001 + 9*0.0001 ...
= 9*10^-1 + 9*10^-2 + 9*10^-3 + 9*10^-4 ...
n=N
= lim ∑ 9*10^-n
n=1
= lim (9*10^-1 + 9*10^-2 + ... + 9*10^-N)
N->infinity
= lim (9*0.1 + 9*0.01 + ... + 9*0.00...01)
N->infinity
= lim (0.9 + 0.09 + ... + 0.00...09)
N->infinity
= lim (.999....99)
N->infinity
= lim (1 - 0.000...001)
N->infinity
= lim (1 - 10^-N)
N->infinity
= lim (1) - lim (10^-N)
N->infinity N->infinity
= 1 - 0
= 1

Another way:

For any two unique real numbers, it is possible to pick another real number between them.
In the case of 0.999... and 1, it is impossible to pick another number in between them. Therefore they must be the same number.

[thomasjmaccoll]

this thread has got my head working about maths again for the first time in a year : )

[Pathway]

We take .999... and 1 to be real numbers.

But what are real numbers, exactly?

Well, most of the axioms defining them are pretty familiar, such as the additive identity axiom (for real x and n, there exists one and only one number n such that x + n = x) or the multiplicative identity axiom (for real x and n, there exists one and only one number n such that n*x = x).

But the weird one is the Least Upper Bound axiom. "Every nonempty subset S of the real numbers, if S is bounded above, has a least upper bound."

In other words, there exists a number M for every set of real numbers such that if there is no number in S greater than some number N, then M is less than or equal to N.

So: if we take the set of numbers from some arbitrary number less than 1 to 1, not including the right endpoint, S = (n<1, 1) , then S has a least upper bound M = 1.

But we also know that there is no number x in S = (n<1, 1) such that x > .999......, because any real number added to .999... would be at least equal to 1. That means, by the definition of least upper bound, that .999... is also the least upper bound M of S, so since M = M, .999... = 1.

[Narsil]

This can be mathematically proven false by the simple question of "what happens if I show up at a dollar store with 99 and .99999999.... pennies?" Obviously I'll be .11111111..... penny short, so I won't be able to buy anything. Of course, a "Take a penny, leave a penny" jar could in theory solve that.

In what universe does .999... + .111... = 1.0? Remember to carry the 1 when you add. I think what you mean to say is .999... + .000...1 = 1.0, but you can't do that -- you can't just say I have infinite zeros and then a 1 at the end. You should check out the logic puzzles thread, the various proofs of .999... = 1 are fairly straight-forward.

I feel stupid now. Anyway, if you change ".11111..." to ".000.....01", then doesn't that raise a good point?

[SpitValve]

This can be mathematically proven false by the simple question of "what happens if I show up at a dollar store with 99 and .99999999.... pennies?" Obviously I'll be .11111111..... penny short, so I won't be able to buy anything. Of course, a "Take a penny, leave a penny" jar could in theory solve that.

In what universe does .999... + .111... = 1.0? Remember to carry the 1 when you add. I think what you mean to say is .999... + .000...1 = 1.0, but you can't do that -- you can't just say I have infinite zeros and then a 1 at the end. You should check out the logic puzzles thread, the various proofs of .999... = 1 are fairly straight-forward.

I feel stupid now. Anyway, if you change ".11111..." to ".000.....01", then doesn't that raise a good point?

ok, let's talk about this 0.00...001 character. How would you more mathematically describe it?

0.1 = 10^-1
0.01 = 10^-2
0.00...001 = limit as (n->infinity) of 10^-n

agreed?

Let's say that 0.00...001 is not zero. Then we can multiply it by a _finite_ number to get 1.

a * 0.00...001 = 1

so

a = 1/(0.00...001)

a = 1/(limit as (n->infinity) of 10^-n)
a = limit as (n->infinity) of 1/(10^-n)
a = limit as (n->infinity) of 10^n
a = infinity

which is not finite... so the initial assumption is wrong: 0.00...001 must be equal to zero.


In terms of pence - if you have 99.99... pence, you have 100p. If you have less than 100p, then you have less than 99.99.... pence.

Practially speaking, how on earth are you going to remove an infitesmal part of a penny anyway?

[mezz]

1 - .898989... = .101010...

Very interesting, but I can't think of any time I will need to use it.

[moopanda]

1 - .898989... = .101010...

Very interesting, but I can't think of any time I will need to use it.

Whenever you need to calculate 99/99 - 89/99 = 10/99, obviously!

[shobadobs]

Obligatory link: http://qntm.org/pointnine

[phlip]

There is a rule with the real numbers:

If:
(1) x ≥ 0
(2) x < ε for all ε > 0
Then:
x = 0

It's closely related to limits, and all that fun stuff. It basically says there's no infinitesimals in the real numbers. If you can say that a non-negative number is less than every positive number, then it must be zero.

You can try to debate this through philosophy, but you'll be debating the definition of a "real number"... it can be proven that the real numbers have no infinitesimals.

Now, 1 - 0.999... < ε. There is no real number that is less than 1 - 0.999... but greater than 0. Therefore, 1 - 0.999... = 0; 0.999... = 1.


The thing to remember is that the decimal representations of numbers are just that. Representations. The number 16 isn't defined somewhere as "the number represented in decimal by a 1 followed by a 6"... A 1 followed by a 6 is just a simple way of representing the number. 0.999... and 1 are simply two ways of representing the same number in decimal, the same way that 1/1 and 2/2 are two ways of representing the same number as fractions.

[Gemini25RB]

I'll just take the simple(?) route and say that since the Real number system is complete, so between any two distinct numbers there exists a third. Now consider 0.999999... (note, the ellipsis designates that the series of nines repeats forever). Now consider 1. Can you name a number between the two? No, because there are an infinite number of nines, so we cannot tack on another number.

[Pathway]

Wow. That qntm site is absolute gold. His fiction reminds me of Asimov. No joke. He's good.

[henre]

Hahahahahahahaha

"I didn't understand that proof."

Try an earlier one.

hahahahahahaha

[Air Gear]

Since a math-related thread has recently been posted in and I don't feel like making this a new thread, I'll post something totally off-topic here...

...the worst math joke ever.

Yup, somebody today, in my analysis class, made the worst math joke ever and now you will hear it.

Basically, my prof brought pumpkin pie for the class...one remark she made, "I thought, 'There are twenty people in the class? Should I have gotten MORE than two?'"

And the response, almost instantly, by a guy in the back...

"We know by radian measure that two pie is enough to go around."

Though surely people have said this about that guy before, he seriously needs to be stopped.

[fjafjan]

that was brilliant, I am fjafjan, and i support this wittyness ^_^

[Hawknc]

I lol'd.

[Gelsamel]

Joke

Ahahah that was freakin' awesome. I can't believe he said that in response so quickly, I've had a few moments like that, but I actually took a few seconds to make up the joke.

[moopanda]

...the worst math joke ever.

I'm not so sure, check out the Nerdy jokes thread on this forum. And the paper I linked from it. It has many many contenders for that coveted title.

[Air Gear]

...the worst math joke ever.

I'm not so sure, check out the Nerdy jokes thread on this forum. And the paper I linked from it. It has many many contenders for that coveted title.

I dunno. The ones on that paper weren't quite as bad; I REALLY like the puns along the lines of "Zorn's lemon", "the real lime", and "simple pole in the complex plane"...though I will admit that the one with the constant, e^x, and the derivative operator walking around is a lot worse, so ok, it's not the worst, but it's really bad.