3.14
02-13-2005, 07:10 PM
1 - .99999etc = .000000etc1
But there are an infinite number of those 0's, which means that you never really reach the one, which means that it never really happens, which means that the distance between 0 and .99999etc is 1.
OR
x = .9999etc, so 10x = 9.9999etc
10x - x = 9x -> x = 1
However,
I admit the second proof is a bit dubious, so here's another.
.999~ = (1/10) + (1/100) + (1/1000)+ ... etc.
so you have a geometric sequence.
Plug it into the formula a1/(1-r) where a1 is the first term of the sequence and r is the ratio.
(9/10)/(1-[1/10]) = .9/.9 = 1.
He it goes again, more detailed. If you don't get this, then go away. ~ is my way of saying infinity/the bar thing.
x = .9~
10x = 9.9~
10x - x = 9x
No, substitue with what we said in the first two lines.
9.9~ - .9~ =9
So thus 9x must equal 9.
9x = 9
Solve for x
x = 1
However, I already said that x = .9~. How can it be .9~ and 1 unless they are equivalent?
Without even assigning a value to x, 10x - x = 9x.
But 9.9~ - .9 = 9
So in that case, 9x must equal 9, and hence x must equal 1.
I was not actually substituting x for .9~ but illustrating how you can come to the conclusion that 9x must equal 9.
But there are an infinite number of those 0's, which means that you never really reach the one, which means that it never really happens, which means that the distance between 0 and .99999etc is 1.
OR
x = .9999etc, so 10x = 9.9999etc
10x - x = 9x -> x = 1
However,
I admit the second proof is a bit dubious, so here's another.
.999~ = (1/10) + (1/100) + (1/1000)+ ... etc.
so you have a geometric sequence.
Plug it into the formula a1/(1-r) where a1 is the first term of the sequence and r is the ratio.
(9/10)/(1-[1/10]) = .9/.9 = 1.
He it goes again, more detailed. If you don't get this, then go away. ~ is my way of saying infinity/the bar thing.
x = .9~
10x = 9.9~
10x - x = 9x
No, substitue with what we said in the first two lines.
9.9~ - .9~ =9
So thus 9x must equal 9.
9x = 9
Solve for x
x = 1
However, I already said that x = .9~. How can it be .9~ and 1 unless they are equivalent?
Without even assigning a value to x, 10x - x = 9x.
But 9.9~ - .9 = 9
So in that case, 9x must equal 9, and hence x must equal 1.
I was not actually substituting x for .9~ but illustrating how you can come to the conclusion that 9x must equal 9.