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Darkstar
17-04-2006, 20:36:57
Hello!

I say this math fight over at a different forum. Since we've got some fairly big math nerds here, I thought I would ask: What do you guys think? What side of this math question do you come down on?

Does 0.99 ... equal 1 ?

Beta1
17-04-2006, 20:42:00
depends on how many decimal places your measuring to

Japher
17-04-2006, 20:42:46
significant digits

Provost Harrison
17-04-2006, 20:43:31
0.99 does not equal 1...there's a difference of 0.01. Depending on your rounding criteria, they can be equivalent...

Venom
17-04-2006, 20:44:37
Build a deck 1/10 of an inch off, no big deal. Launch a rocket 1/10 of a degree of, probably a big problem.

Provost Harrison
17-04-2006, 20:46:15
off and it's 1/100th :p

Venom
17-04-2006, 20:49:22
I don't care. I'm 99.99% asleep.

Venom
17-04-2006, 20:52:14
I figured out what this has to do with.

http://www.foxnews.com/story/0,2933,191859,00.html

DORKSTAR must have seen it on a NASA board. Something about the mission being 99.99% successful.

Darkstar
17-04-2006, 21:03:46
No, it's from Egosoft forum from a math fight. Not strangely, the guys with actual 8 year degrees in mathematics over there are staying out the thread entirely. And most of the guys with any decent exposure to math, such as the physics, astronomers, etc. It's just the liberal arts guys who admit that they dropped all their maths the first year arguing it is the exact same thing, versus being approximately the same.

I prefer how SpaceX is claiming that their failed Falcon 1 launch is a complete success (that's, 100, I say, 100 percent success /froghorn-leghorn voice), because they ignited the engine, and the launch craft left the pad (it blew up in the air, IIRC the details).

Venom
17-04-2006, 21:13:26
The launch was successful. The flight...not so much.

Immortal Wombat
17-04-2006, 23:49:53
0.99... to infinite decimal places does equal 1. Please tell me the argument isn't based on a typographical misunderstanding.

KrazyHorse@home
18-04-2006, 06:54:26
Originally posted by Darkstar
Hello!

I say this math fight over at a different forum. Since we've got some fairly big math nerds here, I thought I would ask: What do you guys think? What side of this math question do you come down on?

Does 0.99 ... equal 1 ?

0.9999 repeated does, in fact, equal 1

0.9999 repeated is equal to the infinite series 9/10 + 9/100 + 9/1000 + ... by definition.

That equals 9 * (1/10 + 1/100 + ...) = 9 * (1/(10-1)) = 9*(1/9) = 1

This is a fairly standard fact from introductory real analysis. Anybody who is "arguing" with this isn't much of a math nerd at all.

KrazyHorse@home
18-04-2006, 07:16:36
you could link the original thread, you know...

JM^3
18-04-2006, 07:25:55
you just want to get your smackdown on

JM

KrazyHorse@home
18-04-2006, 07:56:13
:beer:

Funko
18-04-2006, 08:07:22
KH is right. 0.99 doesn't equal 1 but 0.99 recurring does equal 1.

Gary
18-04-2006, 08:59:17
It's that sort of argument / proof that tells me not to trust mathematicians.

As a non-math nerd, show me that 1/10 + 1/100 + ... doesn't just approximate to 1/(10-1), for interest only. (One of these limiting condition things is it ?)

Immortal Wombat
18-04-2006, 11:52:57
Yes.

1/9 is 1.111.... for ever. Each term 1/10, 1/100 adds another 1, so the sum of inifinite terms is 1/9.

BigGameHunter
18-04-2006, 11:57:18
God I hate math.

Venom
18-04-2006, 12:14:27
As long as I can sort of add up the amount of money I spend at a strip club, I know all the math I need.

Gary
18-04-2006, 12:27:00
Amazing. These infinite series have a lot to answer for. Should never have screened the first episode of Coronation Street.

Beta1
18-04-2006, 15:15:44
interesting - coming at it from a different angle. If I get two readings out of the equipment I used to use in the lab that were 0.99 and 1.00 we would count them as the same as we knew the equipment wasn't capable of that sort of precision. I guess we were really saying 0.99+/-0.05 and 1.00 +/-0.05 are not significantly different.

Stupid machine that thinks its better than it really is.

King_Ghidra
18-04-2006, 15:39:40
listening to this argument reminds me of hearing my philosophy logic seminar leader tell me that two parallel lines could meet

Gary
18-04-2006, 15:51:12
Depends on the type of space

Funko
18-04-2006, 16:01:45
Originally posted by King_Ghidra
listening to this argument reminds me of hearing my philosophy logic seminar leader tell me that two parallel lines could meet

At university my mother went to a Maths lecture and the lecturer drew two parallel lines on the board and said "these lines meet at infinity"

Then he drew a point representing infinity on the board and said "if you believe that you'll believe anything."

Anyway, I just googled the question and the first result tells me that in normal Euclidian geometry there's no such thing as infinity and parallel lines never meet but you can define non-Euclidian geometries where there is infinity and all lines also meet at it as well as any intersections in normal space.

So there you go, a definitive answer.

:clueless:

King_Ghidra
18-04-2006, 16:08:40
:lol: thank you

The Mad Monk
18-04-2006, 18:18:10
In terms of significant figures, it depends if you are adding/substracting, or multiplying/dividing.

If the two were added, the 0.99 would be rounded up to 1.0, for a total of 2.0.

If they were multiplied, both values would be taken unrounded, and the answer would be rounded to two sig figs, giving 0.99.

KrazyHorse@home
18-04-2006, 18:39:05
Originally posted by Gary
It's that sort of argument / proof that tells me not to trust mathematicians.

As a non-math nerd, show me that 1/10 + 1/100 + ... doesn't just approximate to 1/(10-1), for interest only. (One of these limiting condition things is it ?)

http://mathworld.wolfram.com/GeometricSeries.html

Gary
18-04-2006, 19:03:47
Arrhhh !!! I prefer IW's explanation. Not spotted the deliberate error yet but ...

KrazyHorse@home
18-04-2006, 19:30:38
Originally posted by Gary
Arrhhh !!! I prefer IW's explanation. Not spotted the deliberate error yet but ...

IW's explanation is incorrect, because it presumes that you already accept what the repeating decimal means. The argument is circular.

KrazyHorse@home
18-04-2006, 19:34:55
Originally posted by Gary
Not spotted the deliberate error yet but ...

There's no error (well, actually there's no error if you assume that the series converges). Proving that it does converge is pretty easy, though.

Immortal Wombat
18-04-2006, 19:58:11
Originally posted by KrazyHorse@home
IW's explanation is incorrect, because it presumes that you already accept what the repeating decimal means. The argument is circular. It wasn't an argument, it was a "here is maths that you can do on your calculator" demonstration of the principle.

Darkstar
18-04-2006, 20:42:55
See, I always just do the 1/9 thing in base 9. That takes care of the problem and keeps the accuracy.

Base 9:
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 10
1 * 10 = 10
10 / 10 = 1

Everything works out just dandy.

You know, I've been very warped by how computers do math. Literally, 0.999... does not equal 1. Standard computer math functionality always end up cutting short any value if it is a repeating value, so the math doesn't work out properly because the computer is using values less then proper.

Darkstar
18-04-2006, 20:44:23
Originally posted by KrazyHorse@home
There's no error (well, actually there's no error if you assume that the series converges). Proving that it does converge is pretty easy, though.

So, that proves there aren't any values between 0.999 ... and 1? or that the values in-between are part of the convergance? IIRC, certain number systems can demonstrate values between 0.999... and 1. Such as Red-Blue HackenBush or even HackenStrings.

Venom
18-04-2006, 20:58:15
Originally posted by KrazyHorse@home
0.9999 repeated does, in fact, equal 1

0.9999 repeated is equal to the infinite series 9/10 + 9/100 + 9/1000 + ... by definition.

That equals 9 * (1/10 + 1/100 + ...) = 9 * (1/(10-1)) = 9*(1/9) = 1

This is a fairly standard fact from introductory real analysis. Anybody who is "arguing" with this isn't much of a math nerd at all.

Wait a minute....that almost makes sense.

Except for this part: 9 * (1/10 + 1/100 + ...) = 9 * (1/(10-1)) = 9*(1/9) = 1

Venom
18-04-2006, 20:59:03
And this part: 0.9999 repeated is equal to the infinite series 9/10 + 9/100 + 9/1000 + ... by definition

KrazyHorse@home
18-04-2006, 22:13:48
Originally posted by Darkstar
See, I always just do the 1/9 thing in base 9. That takes care of the problem and keeps the accuracy.

Base 9:
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 10
1 * 10 = 10
10 / 10 = 1


Huh?


You know, I've been very warped by how computers do math. Literally, 0.999... does not equal 1. Standard computer math functionality always end up cutting short any value if it is a repeating value, so the math doesn't work out properly because the computer is using values less then proper.

Not true. Floating points are rounded properly when they reach the limit of accuracy.

KrazyHorse@home
18-04-2006, 22:15:06
Originally posted by Darkstar
So, that proves there aren't any values between 0.999 ... and 1? or that the values in-between are part of the convergance? IIRC, certain number systems can demonstrate values between 0.999... and 1. Such as Red-Blue HackenBush or even HackenStrings.

What the fuck are you talking about? There are no values between 0.999 repeated and 1.

KrazyHorse@home
18-04-2006, 22:23:08
All right, I just looked up hackenstrings

All it is is a different number system. everything you prove in mathematics is only valid within the confines of your formal system. "Hackenstring arithmetic" is a completely different formal system.

There's no point to bringing it up.

Koshko
19-04-2006, 03:53:39
I want my fucking penny.

The Mad Monk
19-04-2006, 07:06:05
She dosen't want you.