does .9 repeating equal one?

[johncmcleod]

This will be hard to show, doing math on a post is kind of hard.

I never really thought about what .9 repeating was, I just said it all the time like "99.9999999999% sure I'm right." But I said that once and remembered that when a number is a repeating you write the digits that repeat over x amounts of nines (x being the amount of digits the numerator has). For example 1/9=.1 repeating, because 1 divided by nine equals .1 repeating. Same thing goes for .16 repeating. It is 16/99 because 16 divided by 99 equals .16 repeating. So that means that .9 repeating would by nine over nine, which is one.

Then I was playing around with some numbers the other day, and here's what I got:

let n=0.9 repeating
10n=9.9999...
-n=-.09999...
---------------------
9n=9
n=1

[Twilight zone music]

Is what I've proved true?

[Asher]

Absolutely not.

[KrazyHorse]

Yes.

Decimal representations and the real numbers are not bijective...

[Asher]

0.99999... ~ 1, but not = 1
It's assumed to be 1 by many people, but it's not really 1.

Am I missing something?

[KrazyHorse]

No, 0.9999... = 1 exactly. They're two different representations of the same real number.

[KrazyHorse]

Come on, you compsci geek. This is elementary Analysis.

[alofatti]

It is true, 0.9999.... is, in the real system of numbers, equal to 1. There are a lots of proofs of this, some of them using analytic concepts of series, other working directly with some basic properties of the real number.

Frogger: decimal representeations and the real numbers are actually bijective (i.e., they have the same cardinal) even though one of them is complete(the real numbers) and the other not.

[Asher]

Originally posted by Frogger
No, 0.9999... = 1 exactly. They're two different representations of the same real number.

Come on, you compsci geek. This is elementary Analysis.

We don't do the pure stuff like that.

We're taught 0.9 repeating is 0.9 repeating.

[IoT]

Definitely not. Both 1.0 and .99999..... have seperate locations on the number line.

Interestingly, Cantor chowed that there are infinitely more irrational numbers (Aleph 1) than there are rational numbers (Aleph 0). (Technically this is the Continuum hypothysis IIRC)

Or to put it another way, the entire space taken up on the number line by rational numbers (Integers, and fractions) is zero.

Some interesting Cantorian mathematics

Xo + Xo = Xo
X1 + Xo = X1
Xo - Xo = Xo
Xo * Xo = Xo

(Xo*Xo = Xo+Xo+ ....)

X0 = Aleph null
X1 = Aleph 1 etc

[Asher]

IoT is thinking along the same lines as me then.

[KrazyHorse]

I meant to say, actually, that the representation is not bijective, since it is onto, but not 1-1.

Please explain your use of word "complete" in this situation.

[KrazyHorse]

Originally posted by IoT
Definitely not. Both 1.0 and .99999..... have seperate locations on the number line

No they don't.

[Asher]

The reason why there's disagreement is because math is such a stupid field.

Like it seems the most common approach these days is to use the logic that something like 1/3 is 0.3 repeating, so 3/3 must be 0.9 repeating, but it's also 1, therefore they're equal...

[Asher]

I have philosophical qualms about assuming that infinitely many .9's equals 1.

[Giant_Squid]

I learned it the following way:

In .333333..., no matter how many threes you list, the threes you haven't listed have a value one third of the way equal to the decimal place before...that is, one third of the way to making the next decimal place a four.

In .666666....no matter how many sixes you list, the sixes you haven't listed have a value two thirds of the way equal to the decimal place before...that is, two thirds of the way to making the next decimal place a seven.

Thus, in .999999....no matter how many nines you list, the nines you haven't listed have a value three thirds of the way equal to the decimal place before, the chain of nines is broken, and the number is exactly equal to one.

[SlowwHand]

Originally posted by Asher
Absolutely not.

Absolutely yes, and I think we may starting a break-through on Asher's problem.

[Asher]

Originally posted by SlowwHand
Absolutely yes, and I think we may starting a break-through on Asher's problem.

ESL?

[alofatti]

Frogger:
when I talk about the decimal representation I am talking about an element which has this form:
xxxxxxxx.xxxxxxx...........
where x is a digit between 0 and 9 and which could extend to the infinite, I am not talking about rational numbers (which are not biyective to the real numbers).

This decimal representation number system has the same cardinal than the real numbers, there is a mapping between them that is both injective and suryective.

The inyection is easy, it is clearly seen that the real numbers are included in this set.
The surjection is somewhat trickier. It is possible to show that this set includes both this sets:
- the set of numbers which do not end with a repeating 9 (which are easily shown to be bijective to the real numbers)
- the set of numbers which do end with a repeating 9 (which are numerable).

Since c (the cardinal of R) + Aleph_O (the cardinal of a numberable set) equals c, then they are both coordinable between themselves.

A complete set is a set which, in an informal way, does not have successions which converge to "ghost" (i.e., inexistent) points. More formally, every Cauchy sequence does have a limit.
I have said in a post before this that the set of decimal representable numbers are not complete, which is not right. First is necesary to define a distance between points to talk about completeness.

[KrazyHorse]

Originally posted by Asher
The reason why there's disagreement is because math is such a stupid field.

Like it seems the most common approach these days is to use the logic that something like 1/3 is 0.3 repeating, so 3/3 must be 0.9 repeating, but it's also 1, therefore they're equal...

Asher, have you ever learnt what a limit is?

Take the limit of the sequence 0.9, 0.99, 0.999, ...

By inspection 0.9999999.... is a limit

By going over the definition of the requirements of being a limit, 1 is also a limit.

By a fairly standard proof, if a sequence has a and b as limits, then a=b

[Asher]

Originally posted by Frogger
Asher, have you ever learnt what a limit is?

Take the limit of the sequence 0.9, 0.99, 0.999, ...

By inspection 0.9999999.... is a limit

By going over the definition of the requirements of being a limit, 1 is also a limit.

By a fairly standard proof, if a sequence has a and b as limits, then a=b

Yes yes, I know that you can formally prove it now.
And yes, I've taken Calculus I and II and courses with combinatorial proofs.

I just don't like assuming repeating numbers are anything else other than repeating numbers. It would have helped if I read his post other than just the title, but it's still a silly math rule to get around how math geeks still haven't come up with a better representation of numbers.

[KrazyHorse]

Originally posted by alofatti
Frogger:
when I talk about the decimal representation I am talking about an element which has this form:
xxxxxxxx.xxxxxxx...........
where x is a digit between 0 and 9 and which could extend to the infinite, I am not talking about rational numbers (which are not biyective to the real numbers).

This decimal representation number system has the same cardinal than the real numbers, there is a mapping between them that is both injective and suryective.

The inyection is easy, it is clearly seen that the real numbers are included in this set.
The surjection is somewhat trickier. It is possible to show that this set includes both this sets:
- the set of numbers which do not end with a repeating 9 (which are easily shown to be bijective to the real numbers)
- the set of numbers which do end with a repeating 9 (which are numerable).

Since c (the cardinal of R) + Aleph_O (the cardinal of a numberable set) equals c, then they are both coordinable between themselves.

A complete set is a set which, in an informal way, does not have successions which converge to "ghost" (i.e., inexistent) points. More formally, every Cauchy sequence does have a limit.
I have said in a post before this that the set of decimal representable numbers are not complete, which is not right. First is necesary to define a distance between points to talk about completeness.

You're absolutely and fundamentally wrong here.

Under the standard decimal representation of the real numbers, every real number has a decimal representation (i.e. Decimal->Real is onto). Unfortunately, the representation is not a bijection, as 0.999999999.... and 1 are both mapped to the same real number.

BTW, when you use the word "complete", most people use the word "closed" to represent a set which contains all of its limit points. And the set of reals and the set of decimal reps are both closed.

[alofatti]

You are not obliged to have o.9999..... equal to 1.
You could very well define your own number system where 0.9999... is not 1 (as well it is correctly defined), though perhaps it would share very little resemblance to the real system.

[KrazyHorse]

Originally posted by Asher

Yes yes, I know that you can formally prove it now.
And yes, I've taken Calculus I and II and courses with combinatorial proofs.

I just don't like assuming repeating numbers are anything else other than repeating numbers. It would have helped if I read his post other than just the title, but it's still a silly math rule to get around how math geeks still haven't come up with a better representation of numbers.

We can have a better representation (i.e. one that's a bijection) simply by eliminating all numerals which end in an infinite string of 9's....

[SlowwHand]

If I was as dull as some of you guys, going into deep analysis of the number 1, I believe I would tie myself in a bag and throw me in the river.
I wouldn't wait for my dad.
Ask Boddington's.
Boddington's needs to come home.

[KrazyHorse]

Originally posted by alofatti
You are not obliged to have o.9999..... equal to 1.
You could very well define your own number system where 0.9999... is not 1 (as well it is correctly defined), though perhaps it would share very little resemblance to the real system.

I know...and that's why I clarified (after you pointed this out) that what I meant was that the standard representation is not a bijection, not that the two sets aren't bijective (though constructing a bijection would be vaguely difficult...)

[KrazyHorse]

Originally posted by SlowwHand
If I was as dull as some of you guys, going into deep analysis of the number 1, I believe I would tie myself in a bag and throw me in the river.
I wouldn't wait for my dad.
Ask Boddington's.
Boddington's needs to come home.

Man, the number 1 is the most interesting number out there.

[Asher]

I had to look up the 1-to-1 and onto terms, since I've always been told they were injective and surjective instead.

[alofatti]

Actually, it is interesting to notice that lots of things are studied around the number 1.
For example, the fact that 0 < 1 requieres a formal proof, which is not obvious (but also not that hard, though).

[alofatti]

Quoted from Frogger:
"Under the standard decimal representation of the real numbers, every real number has a decimal representation (i.e. Decimal->Real is onto). Unfortunately, the representation is not a bijection, as 0.999999999.... and 1 are both mapped to the same real number. "

Not necessarily. One could easily define a map between two sets as he wants to. I am not obliged to send 0.9999... to 1. In terms of cardinality, both sets have the same cardinal. There is a bijection between these sets (which does not send 0.999... and 1 to 1 because that would not be inyective).

Quoted from Frogger:
"BTW, when you use the word "complete", most people use the word "closed" to represent a set which contains all of its limit points. And the set of reals and the set of decimal reps are both closed"

Well, in metrics spaces "closed" is weaker than "complete" (i.e., a space which is complete is closed), but you are right: both sets are closed (in fact, they are complete).

[Asher]

alofatti: What do you do for a living?