Nearly one
Prove that 0.9999999999..... = 1.
Labels: mathschallenge
posted by Chris at
11:49 AM
Tuesday, December 22, 2009
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11 Comments:
If 0.99999...=x
1000x=999.9999999999...
999x=999
x=1
1/3 = .33333333333......
if 1/3 x 3 = 1
and .333333333333.... x 3 = .999999999999....
then .999999999..... = 1
9/9=1
9*1/9=1
9*.1111.....=1
.9999...=1
Cam
Really it depends on the tolerance specifications.
Generally +/- 0.001 is within tolerance, so 0.9999999... is equal to one, but in specifically in math it is NOT one.
So it depends on the usage.
That was quick.
The first three solutions look OK to me.
Personally I was going to suggest:
x = 0.999999.....
10x = 9.99999999.....
So 10x - x = 9x = 9 => x = 1.
I'm pretty sure that this is mathematicaly exact, the tolerance is precisely 0. The error is not even an infinitesimal.
Anonymous, no it doesn't. 0.99999.... = 1.
For the curious, the deleted post broke the kid sister rules. The poster was also a typical idiot who claimed that we were the idiots and then went on to show he hadn't a clue.
My idea is this there is no number s.t to multiply and divide with other numbers and give us the 0.9999999999999 number.It means that there is no ROOt of no to give us 0.9999999999... and no such number in IR.
Hi walidjan. I'm sorry, but I haven't understood what you said. Also, what is IR?
Q: How many mathematicians does it take to screw in a lightbulb?
A: 0.999999….
I find the result (.9999... =1) rather disturbing. I can't help but feel that the number of 9s is countably infinite, yet we seem to have a continuity which I would associate with the uncountably infinite. It makes me think that Georg Cantor's infinities are more axiomatic than I've previously understood them to be.
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