Does .9-repeating equal 1?

[loinburger]

If somebody is using a number system other than the real number system then ".9-repeating /= 1" follows from the axioms for their number system. However, I'd prefer that they provide their axioms and rationales behind said axioms before they demonstrate their proof that .9-repeating /= 1. It's somewhat more glamorous if somebody says "this is a number system that I've developed to simplify thus-and-such or improve the efficiency of this-and-that, and oh by the way this number system entails that .9-repeating /= 1" than if somebody says "I set out to prove that .9-repeating /= 1, and here is a number system that proves this, and I don't know what else this number system is good for but BY GOD I've proved that within this number system my irrational biases are complete justified by virtue of the fact that I've turned my biases into axioms."

[Adalbertus]

0.999... != 1 for any number system other than the decimal system.

[Ari Rahikkala]

Originally posted by OzzyKP
Wow, 231 posts, this very well may be the nerdiest thread in the history of Apolyton.

A truly nerdy thread can't be as full of dumbness as this one is.

[Saras]

Can someone please close this?

[Kuciwalker]

Originally posted by Adalbertus
0.999... != 1 for any number system other than the decimal system.

Sure it does, as long as the system uses the real numbers.

[Lul Thyme]

Originally posted by Adalbertus
0.999... != 1 for any number system other than the decimal system.

you have to make the dinstinction between numbers and their representation.
0.99.... and 1 are two decimal representations of the same number (assuming we're working with real numbers).

AS KH pointed out, many numbers have 2 different representations in many representation systems.
For example, in binary, 0.111111.... =1

[Cort Haus]

I can't believe this thread is still going. I guess it is somewhere to escape all the election threads. :shrug:

[Heresson]

How much is 0,999(...) + 0,999(...)?

[Adalbertus]

Originally posted by Kuciwalker


Sure it does, as long as the system uses the real numbers.

No, it doesn't in the binary expansion of real numbers, where there is no 9, nor it does in the sedecimal system, where it is 9/F (or 9/15 = 0.6 in decimal representation).



@Heresson:
In what representation?

[Lawrence of Arabia]

Let a = b

a^2 = ab

a^2 +a^2 - 2ab = ab + a^2 - 2ab
2(a^2 - ab) = a^2 - ab

2 = 1

does TWO now equal ONE???

[Kuciwalker]

No, it doesn't in the binary expansion of real numbers, where there is no 9

In binary .111111111111111111111111 = 1

[Adalbertus]

Originally posted by Lawrence of Arabia
Let a = b

....

does TWO now equal ONE???

Nice try



@Kuciwalker: I suppose the thread is about 9 repeating and not (base - 1) repeating.

[Darius871]

Not gonna read this entire thread, but to all the ladeeda high-and-mighty non-libarts majors calling any opponents idiots, I've seen plenty of post-grad mathematicians dispute this unquestioned dogma as well. A few of the same points they bring up can be seen in these two debates, for anyone obsessed enough to read them:

http://polymathematics.typepad.com/polymath/2006/06/no_im_sorry_it_.html

http://polymathematics.typepad.com/polymath/2006/06/the_saga_contin.html

(Not taking a side here myself, just noting that even among actual "experts" there still is considerable debate.)

[Kuciwalker]

Originally posted by Adalbertus
Nice try

@Kuciwalker: I suppose the thread is about 9 repeating and not (base - 1) repeating.

The two are basically equivalent.

[BeBMan]

Polymathematics? I thought that's what we already see at work here. IMO this special kind of math should be labeled just "Polymathics"