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.