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i don't agree that 0.99999etc=1
because if u multiplie by 10 u don't get 9.999999etc
but 9.9999999(lots of 9s)0
so 9.9999999990-0.9999999999=8.9999999991
that means that it isn't 1
though i hope i don't need to say that 0.9999999etc
is atcually 8.999999(lots of 9s)1
It is commonly accepted by many that 0.999... = 1.
By 0.999... we mean that it has no end; that it is Infinite!
Therefore it won't end in 0, it will just keep going on forever.
You should check out the two 0.999
= 1 threads (although you will need to register) and this page:
http://www.mathsisfun.com/numbers/infinity.html
Also:
9.999... - 0.999... = 9
Last edited by Toast (2006-12-16 03:18:18)
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but a very large number (i don't believe infinity exists) x10 gives a 0 in the end;)
As far as I know if you multiply 0.999... by 10, the only thing that changes is the decimal place.
i.e, When multiplying by 10, move the decimal place in 0.999... one place to the right to become 9.999... . The decimal place changes but the infinity at the end remains.
Last edited by Toast (2006-12-16 05:34:16)
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but a very large number (i don't believe infinity exists) x10 gives a 0 in the end;)
i don't agree that 0.99999etc=1
because if u multiplie by 10 u don't get 9.999999etc
but 9.9999999(lots of 9s)0so 9.9999999990-0.9999999999=8.9999999991
that means that it isn't 1
though i hope i don't need to say that 0.9999999etc
is atcually 8.999999(lots of 9s)1
I spy, with my little eye, an error on MathsIsFun.
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Actually, come to think of it, since ∞-∞ is undefined, and 0.999... is a type if infinity, wouldn't 9.999... - 0.999 be undefined?
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I'll dish out my old catchphrase;
Depends how you think about it.
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Actually, come to think of it, since ∞-∞ is undefined, and 0.999... is a type if infinity, wouldn't 9.999... - 0.999 be undefined?
if u mean by this that you can't say 10xinfinite=infinite with the "," somewhere else, i agree:P
0.999... ≠ 0.999...0. It won't end in a zero.
Explain your basis for saying that 0.999... ends in a zero.
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Actually, come to think of it, since ∞-∞ is undefined, and 0.999... is a type if infinity, wouldn't 9.999... - 0.999 be undefined?
Nope. ∞ is more a 'concept' than a number, and so you can't apply standard arithmetic to it.
9.999... is a perfectly good number that just happens to be infinitely long. But its value is still finite, and so you can do maths with it.
Why did the vector cross the road?
It wanted to be normal.
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Desmond, you happen to watch LOST, don't you?
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