0.999...=1 on wikipedia
Moderators: Moderators General, Prelates, Magistrates
0.999...=1 on wikipedia
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
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
 THE fjafjan
 Posts: 4766
 Joined: Fri Oct 06, 2006 12:22 pm UTC
 Location: Down south up north in the west of eastern west.
 Contact:
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
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
//Yepp, THE fjafjan (who's THE fjafjan?)
Liza wrote:Fjafjan, your hair is so lovely that I want to go to Sweden, collect the bit you cut off in your latest haircut and keep it in my room, and smell it. And eventually use it to complete my shrine dedicated to you.
 Gelsamel
 Lame and emo
 Posts: 8237
 Joined: Thu Oct 05, 2006 10:49 am UTC
 Location: Melbourne, Victoria, Australia
fjafjan wrote: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
 THE fjafjan
 Posts: 4766
 Joined: Fri Oct 06, 2006 12:22 pm UTC
 Location: Down south up north in the west of eastern west.
 Contact:
Gelsamel wrote:fjafjan wrote: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?
//Yepp, THE fjafjan (who's THE fjafjan?)
Liza wrote:Fjafjan, your hair is so lovely that I want to go to Sweden, collect the bit you cut off in your latest haircut and keep it in my room, and smell it. And eventually use it to complete my shrine dedicated to you.
Gelsamel wrote:fjafjan wrote: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?
 fjafjan
 THE fjafjan
 Posts: 4766
 Joined: Fri Oct 06, 2006 12:22 pm UTC
 Location: Down south up north in the west of eastern west.
 Contact:
svk1325 wrote:Gelsamel wrote:fjafjan wrote: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
//Yepp, THE fjafjan (who's THE fjafjan?)
Liza wrote:Fjafjan, your hair is so lovely that I want to go to Sweden, collect the bit you cut off in your latest haircut and keep it in my room, and smell it. And eventually use it to complete my shrine dedicated to you.
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.
Narsil wrote: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 straightforward.
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.
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
 Posts: 541
 Joined: Thu Oct 19, 2006 11:27 pm UTC
 Location: cupar, fife, scotland
 Contact:
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.
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.
sandy wrote:Narsil wrote: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 straightforward.
I feel stupid now. Anyway, if you change ".11111..." to ".000.....01", then doesn't that raise a good point?
Narsil wrote:sandy wrote:Narsil wrote: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 straightforward.
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?
Obligatory link: http://qntm.org/pointnine
 phlip
 Restorer of Worlds
 Posts: 7572
 Joined: Sat Sep 23, 2006 3:56 am UTC
 Location: Australia
 Contact:
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 nonnegative 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.
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 nonnegative 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.
Code: Select all
enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};
void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

 Posts: 39
 Joined: Fri Oct 27, 2006 9:38 pm UTC
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.
Since a mathrelated thread has recently been posted in and I don't feel like making this a new thread, I'll post something totally offtopic 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.
...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
 THE fjafjan
 Posts: 4766
 Joined: Fri Oct 06, 2006 12:22 pm UTC
 Location: Down south up north in the west of eastern west.
 Contact:
that was brilliant, I am fjafjan, and i support this wittyness ^_^
//Yepp, THE fjafjan (who's THE fjafjan?)
Liza wrote:Fjafjan, your hair is so lovely that I want to go to Sweden, collect the bit you cut off in your latest haircut and keep it in my room, and smell it. And eventually use it to complete my shrine dedicated to you.
moopanda wrote:Air Gear wrote:...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.
Who is online
Users browsing this forum: No registered users and 19 guests