Poll: Does 0.99 Recurring = 1
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Soldato
No it wouldn't. If you don't accept the infinite decimal expansion of 0.9r, you can't possibly accept 0.0r1 corresponding to anything. After all, if you accept 0.0r1 exists, and 1 exists, then their difference is 0.9r, which therefor must exist. Its all or nothing I'm afriad
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Well in my world 3x4 is 1873268942032498 - 453543/23497
Totals
-54375498589
now some one tell me thats wrong
Originally posted by AlphaNumeric
No it wouldn't. If you don't accept the infinite decimal expansion of 0.9r, you can't possibly accept 0.0r1 corresponding to anything. After all, if you accept 0.0r1 exists, and 1 exists, then their difference is 0.9r, which therefor must exist. Its all or nothing I'm afriad
ive excepted it exists
where havent i??
Soldato
Is root 2 finite? root 2 < 2 (obviously), but its got an infinite expansion, just likt 0.9r.
Besides, how can 0.9r be infinite, you just said it was <1, and 1 isn't infinite.
Yet again, you've contradicted yourself nicely
Soldato
True.
AcidHell, whats sin(pi) ? (feel free to calculator it)
Sin is an infinite expansion polynomial, and pi is a never ending, non repeating number. Yet I can tell you exactly what sin(pi) is, with no room for error at all.
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Originally posted by memphisto
mathematically its wrong
philisophically .............
If i'm honest the onlt time i worry about maths is when i click the check out button on overclockers
0.1/2 = 0.05
0.09r/2 = 0.049r5
Now surely, that shows there is a difference. I really can't understand how numbers can be the same, when they are different values. I'll keep reading though.
0.09r/2 = 0.049r5
Now surely, that shows there is a difference. I really can't understand how numbers can be the same, when they are different values. I'll keep reading though.
Soldato
There's the error. Double you 0.45r and you don't get 0.09r, so your initial division was incorrect.
0.45r = 0.455555.......
Doubled is 0.09111111111
0.9r/2 = 0.49r.
Originally posted by AlphaNumeric
Is root 2 finite? root 2 < 2 (obviously), but its got an infinite expansion, just likt 0.9r.
Besides, how can 0.9r be infinite, you just said it was <1, and 1 isn't infinite.
Yet again, you've contradicted yourself nicely
yes it will always be less than 1
even though its infinite its still less than 1.
if root 2 is not finite it cant be used in maths it has to be rounded.
or expressed vie an expresion like .9r
No. r is not a quantitly that is multiplied by 0.09. It's an indication that the 9s continune indefinitely.
So 0.09r/2 = 0.049r
Understood.
Didn't bother reading past the 3rd page as it was starting to look like too many people had too much spare time mixed with too big an urge to be right this afternoon
But voted no. I was always taught that the only way any number ending with .99 recurring could end up a whole number would be if you rounded it up.
But voted no. I was always taught that the only way any number ending with .99 recurring could end up a whole number would be if you rounded it up.
Soldato
You now are talking nonsense. Even those in this thread with no maths at all realise infinite is a lot bigger than 1!
Sure it can.
x = root 2
I have x^2 + 1 = (root 2)^2 + 1 = 5
No error, no rounding up, no problem.
Soldato
Who taught you?This is basic A level stuff!
Originally posted by VDO
This is basic A level stuff!
Mine was up to GCSE Level
And tbh it's basic common sense surely? Ask anyone on the street how the hell 0.99 could become 1 without rounding up and they're going to say it cant be.Also if it's so definite according to you that it does equal 1 how come there's a poll and a huuuuuuuge debate about it?
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