View Full Version : Does 0.99 ... equal to 1.0 ?

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...

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.

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.

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

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

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.

vBulletin® v3.8.2, Copyright ©2000-2018, Jelsoft Enterprises Ltd.