paulhicks
Joined: Jul 19, 2004

Posted:
Oct 04, 2011  22:04 

A mate of mine was round this evening and claimed something that seemed odd to me. His claim was:
"0.999 (going on for infinity) is exactly the same as 1 in all circumstances".
No i have no idea how we got onto the subject either but frankly i lack a basic understanding of maths beyond my 2 times table. Even so this seemed fundamentaly wrong to me.
I know there are penty of people on here with better maths brains than me so my question is:
Is this correct? 
_________________ Spelling, grammer and sense are for noobs! 

phil78
Joined: Jul 29, 2004

Posted:
Oct 04, 2011  22:06 

I believe the answer is yes, they are the same but i agree it doesn't seem to make sense 


WhatBall
Joined: Aug 21, 2008

"0.999 (going on for infinity) is exactly the same as 1 in all circumstances."
Tell that to the person who won the lottery, or died by being hit by a piece of falling space debris. 
_________________


James_Probert
Joined: Nov 25, 2007

Posted:
Oct 04, 2011  22:11 

It's not (by definition).
But most purposes, it is (as you're going to round the value at some point, due to the fact that you'd run out of universe to count in/with). 
_________________


sann0638
Joined: Aug 09, 2010

Posted:
Oct 04, 2011  22:13 

As an example, 1/9 is 0.1 recurring.
Therefore 9/9 is 0.9 recurring.
But obviously 9/9 = 1. 
_________________
NAF ExPresident
Founder of Cakebowl, Wiltshire's BB League @ cakebowl.co.uk 

paulhicks
Joined: Jul 19, 2004

Posted:
Oct 04, 2011  22:13 

James_Probert wrote:  It's not (by definition).
But most purposes, it is (as you're going to round the value at some point, due to the fact that you'd run out of universe to count in/with). 
See that was what i assumed he was getting at but apparently not. 
_________________ Spelling, grammer and sense are for noobs! 

Sutherlands
Joined: Aug 01, 2009

Posted:
Oct 04, 2011  22:13 

@WhatBall, that doesn't even make sense...
@paulhicks  Yes, those are two ways of representing the exact same number. There are a few ways to prove this.
First, by the real number theorum, between any two distinct real numbers, there is another number which is the average of the two. What is the average of 1 and .99999...? There is no number between them.
Second:
1/9 = .11111....
2/9 = .22222....
3/9 = .33333....
9/9 = .99999....
1 = .99999....
Third:
x = .9999....
10x = 9.9999....
10x  x = 9.9999....  .9999....
9x = 9
x = 1 


Sutherlands
Joined: Aug 01, 2009

Posted:
Oct 04, 2011  22:14 

James_Probert wrote:  It is(by definition).  FTFY 


DukeTyrion
Joined: Feb 18, 2004

Posted:
Oct 04, 2011  22:16 

The problem is, the decimal system can never quite be exact.
Fractions were much better back in the old days. 


ryanfitz
Joined: Mar 24, 2009

Posted:
Oct 04, 2011  22:16 

It is accepted in theoretical math, but while correct for performing mathematical calculations, are not 'exactly the same' for obvious visible reasons.
the idea is that if the .9999 bar goes on indefinitely, you will never reach the 1 chance as there will be infinitely more 9s 


James_Probert
Joined: Nov 25, 2007

Posted:
Oct 04, 2011  22:19 

sann0638 wrote:  As an example, 1/9 is 0.1 recurring. 
It's not.
0.1 recurring is the nearest decimal to 1/9, although the exact value is impossible to give.
sann0638 wrote:  Therefore 9/9 is 0.9 recurring.
But obviously 9/9 = 1. 
the statement I've given above explains why this is a fallacy, although it is accurate for the (very) large majority (if not all currently found) cases.
WhatBall wrote:  "0.999 (going on for infinity) is exactly the same as 1 in all circumstances."
Tell that to the person who won the lottery, or died by being hit by a piece of falling space debris. 
ermmm, those are nonzero (and measurable) probabilities, so not a valid counterargument.
Yes, I'm aware I've just argued both sides of the arguement 
_________________


dode74
Joined: Aug 14, 2009

Posted:
Oct 04, 2011  22:22 

Good explanation from Yahoo:
Quote:  The number "0.9999..." can be "expanded" as:
0.9999... = 0.9 + 0.09 + 0.009 + 0.0009 + ...
In other words, each term in this endless summation will have a "9" preceded by some number of zeroes. This may also be written as:
0.999... = 9/10 + (9/10)(1/10)^1 + (9/10)(1/10)^2 + (9/10)(1/10)^3 + ...
That is, this is an infinite geometric series with first term a = 9/10 and common ratio r = 1/10. Since the size of the common ratio r is less than 1, we can use the infinitesum formula to find the value:
0.999... = (9/10)[1/(1  1/10)] = (9/10)(10/9) = 1
So the formula proves that 0.9999... = 1. 



WhatBall
Joined: Aug 21, 2008

I've identified the mathematicians by highlighting their inability to laugh at a silly answer to a math related question. 
_________________


GAZZATROT
Joined: Apr 26, 2009

Posted:
Oct 04, 2011  22:25 

Erm, I'm CERTAINLY not an expert but if you minus one from this number you would have a negative figure.... therefore it is less than 1. 
_________________ Forever fearless, sometimes stupid. 

Sutherlands
Joined: Aug 01, 2009

Posted:
Oct 04, 2011  22:26 

DukeTyrion wrote:  The problem is, the decimal system can never quite be exact.  Give me a fraction that can't be represented as a decimal, please.
ryanfitz wrote:  It is accepted in theoretical math, but while correct for performing mathematical calculations, are not 'exactly the same' for obvious visible reasons.  I wasn't aware that math used sight to prove something. No, those numbers are exactly the same.
James_Probert wrote:  sann0638 wrote:  As an example, 1/9 is 0.1 recurring. 
It's not.
0.1 recurring is the nearest decimal to 1/9, although the exact value is impossible to give.  It is exactly the same. The exact value is not impossible to give. GAZZATROT wrote:  Erm, I'm CERTAINLY not an expert  Agreed GAZZATROT wrote:  but if you minus one from this number you would have a negative figure.... therefore it is less than 1.  No, you'll have 0. 


