paulhicks
Joined: Jul 19, 2004
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  Posted:
Oct 04, 2011 - 22:04 |
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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? |
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phil78
Joined: Jul 29, 2004
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  Posted:
Oct 04, 2011 - 22:06 |
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I believe the answer is yes, they are the same but i agree it doesn't seem to make sense |
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WhatBall
Joined: Aug 21, 2008
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  Posted:
Oct 04, 2011 - 22:10 |
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"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. |
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James_Probert
Joined: Nov 25, 2007
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  Posted:
Oct 04, 2011 - 22:11 |
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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). |
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sann0638
Joined: Aug 09, 2010
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  Posted:
Oct 04, 2011 - 22:13 |
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As an example, 1/9 is 0.1 recurring.
Therefore 9/9 is 0.9 recurring.
But obviously 9/9 = 1. |
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paulhicks
Joined: Jul 19, 2004
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  Posted:
Oct 04, 2011 - 22:13 |
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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. |
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Sutherlands
Joined: Aug 01, 2009
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  Posted:
Oct 04, 2011 - 22:13 |
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@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 |
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Sutherlands
Joined: Aug 01, 2009
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  Posted:
Oct 04, 2011 - 22:14 |
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James_Probert wrote: | It is(by definition). | FTFY |
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DukeTyrion
Joined: Feb 18, 2004
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  Posted:
Oct 04, 2011 - 22:16 |
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The problem is, the decimal system can never quite be exact.
Fractions were much better back in the old days. |
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ryanfitz
Joined: Mar 24, 2009
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  Posted:
Oct 04, 2011 - 22:16 |
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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 |
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James_Probert
Joined: Nov 25, 2007
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  Posted:
Oct 04, 2011 - 22:19 |
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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 non-zero (and measurable) probabilities, so not a valid counter-argument.
Yes, I'm aware I've just argued both sides of the arguement |
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dode74
Joined: Aug 14, 2009
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  Posted:
Oct 04, 2011 - 22:22 |
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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 infinite-sum 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. |
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WhatBall
Joined: Aug 21, 2008
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  Posted:
Oct 04, 2011 - 22:23 |
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I've identified the mathematicians by highlighting their inability to laugh at a silly answer to a math related question. |
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GAZZATROT
Joined: Apr 26, 2009
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  Posted:
Oct 04, 2011 - 22:25 |
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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. |
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Sutherlands
Joined: Aug 01, 2009
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  Posted:
Oct 04, 2011 - 22:26 |
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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. |
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