Poll: Does 0.99 Recurring = 1

Does 0.99 Recurring = 1

  • Yes

    Votes: 225 42.5%

  • No

    Votes: 304 57.5%
  • Total voters
    529
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Originally posted by Deadly Ferret
Well in the same way that in reality, we can't have an infinite number of zeroes after a decimal point with a one on 'the end' of those, we can't have an infinite number of nines after a decimal point. The thing that lets down both concepts is the inclusion of infinity. It only exists in the abstract.
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You realise that the whole idea of an infinite sequence, is that it never ends - i.e there is no end!

:)
 
Originally posted by AcidHell2
so u agree .9r is an infinitly long number if its infinitly long how can it equal 1 it cant.
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1 is an infinitely long decimal too. 1.00000000000000000000000000r, we just have the convention that if you put down 1, you accept there's an infinite number of zeros after it.
 
Originally posted by gambitt
ok.. stopped laughing now.

right, let x = 0.99r
10x = 9.99r

with me so far?

now as x = 0.9999999.... (continue forever)
and 10x = 9.9999999.... (continue forever)

simply subtract the (.999999...) part from 9.999999...

and you get 9.

what is difficult about that?
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that is correct,

so if you subtract 1 from the and 10x = 9.9999999.... (continue forever)

youll get 8.9999999...... (continue forever)
 
Originally posted by w11tho
Hmmm - looked like we were talking about 0.9r to me:

Not 'infinity' as a number.

:p :rolleyes: :D
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But infinity only exists as an abstract, so how can a number that can only exist utilising infinity as part of it's makeup be a number? If thats the case then infinity can also be a number.
 
Originally posted by memphisto
thought crossed my mind while i was trying to catch them out with putting 0.9r through the equation.
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0.9r = 0.99r = 0.999r = etc
Doesn't matter how many 9s you put, since you're just repeating the same thing after the r anyway :)
Originally posted by Gilly
But infinity only exists as an abstract, so how can a number that can only exist utilising infinity as part of it's makeup be a number? If thats the case then infinity can also be a number.
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No. I use an integral to define arcsin (1), which is a number, but an "integral" isn't a number.
 
But X.0 has the same value as X.000 and so on, where as X.9 doesn't have the same value as X.999

X not being anything to do with algebra, just meaning any number.
 
Originally posted by AlphaNumeric
1 is an infinitely long decimal too. 1.00000000000000000000000000r, we just have the convention that if you put down 1, you accept there's an infinite number of zeros after it.
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yes because the 0's are meaningless they hold no value.

hence why you dont say .9999 is 1
you have to say
.9r
 
Originally posted by gambitt
ok.. stopped laughing now.

right, let x = 0.99r
10x = 9.99r

with me so far?

now as x = 0.9999999.... (continue forever)
and 10x = 9.9999999.... (continue forever)

simply subtract the (.999999...) part from 9.999999...

and you get 9.

what is difficult about that?
Click to expand...

x = 0.9r
10 * x = 10 * 0.9r (multiply by 10 (shift the decimal place to the right))
10x = 9.9r

10x - x = 9.9r - x (subtract x which equals 0.9r)
9x = 9 (divide by 9 to remove the common factor)
x = 1

At what point did I do any rounding?

Jokester
 
Originally posted by AlphaNumeric
Thats because 0 is the additive identity, so X + 0 = X. X + 0.009 isn't X.
Your point being?
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hence 0.9r is not 1 as all the 9's hold a value it doesnt matter how far along the infinite chain of 9'su go they all have a value.

hence .9r < 1
 
Originally posted by AlphaNumeric

No. I use an integral to define arcsin (1), which is a number, but an "integral" isn't a number.
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Sorry, you lost me there.

Is there something I can read that explains the difference between infinity and a number incorporating infinty? Shown here as 0.99r
 
Originally posted by AlphaNumeric
Round it up by how much? I think you'll find its 0 ;)
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you cant round it up by any given number cos its infinitley long. But just because theres no number doesnt mean it = 1
 
Originally posted by AcidHell2
hence 0.9r is not 1 as all the 9's hold a value it doesnt matter how far along the infinite chain of 9'su go they all have a value.

hence .9r < 1
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Exactly, since they have value, they are needed, so its important to have the "r" in there. If you took any of them away, its not be 1, but the little contribution by all the 9s makes it 1, has the proof which people have posted about 30 times has shown.
 
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