's really a debate of infinite calculation vs. finite calculation. And secondly, real world usage of this principle.
Example 1:
Any decimal may be added to another decimal to make a whole number.
Whereas:
1 - x = y
1 - y = x
But:
x = .999 repeating
1 - x = .000 repeating
x = 1
1 - x = .000 repeating
therefore, .999 repeating = 1
Example 2 (Real World Application)
My second example is more logical. In reality, if you ever had an equation that came out to .999 repeating, you would NEVER get it correct. You would always be off by .000 repeating. Um, again proving the point above, but here saying you will always be off. Hence, in real world applications, we must make the number a finite number. And then .999 repeating will either equal 1 or equal .99... 8 depending on how exact the calculation is.
Example 3 (pirate quote)
Quote:
∞ != ∞
suggestion:
let f(x)=x
let g(x)=2x
what is the relationship between lim x->∞ f(x) and lim x->∞ g(x)? they are both ∞
suggestion:
let f(x)=x
let g(x)=2x
what is the relationship between lim x->∞ f(x) and lim x->∞ g(x)? they are both ∞
Example 4 (pirate quote)
Okay, and just to be fair, here is a theory that supports why .999 repeating does NOT equal 1 using geometry.
Quote:
With invinite numbers no rounding is necessary.
proof 1: let .999... be represented by a geometric series 0.9 + 0.09 + 0.009 + 0.0009 + ...
a geometric series converges iff the ratio r satisfies 0
the series converges
the sum of a converging geometric series is given by a/(1-r) where a is the first term and r is the ratio.
the sum above = 1
proof 2 (from emphy)
let sum (i=1,i=m) x(i) mean the sum of the series x(i)+x(i+1)+x(i+2)+...+x(m-2)+x(m-1)+x(m) from i=1 to i=m.
let epsilon (ep) be a real number 0
let M be a natural number > delta
let m>M
let f(m)=sum(i=1,i=m) 0.9^10(-i);
then f(m)> 1- ep
as ep approaches 0, f(m) approaches 1.
proof 1: let .999... be represented by a geometric series 0.9 + 0.09 + 0.009 + 0.0009 + ...
a geometric series converges iff the ratio r satisfies 0
the series converges
the sum of a converging geometric series is given by a/(1-r) where a is the first term and r is the ratio.
the sum above = 1
proof 2 (from emphy)
let sum (i=1,i=m) x(i) mean the sum of the series x(i)+x(i+1)+x(i+2)+...+x(m-2)+x(m-1)+x(m) from i=1 to i=m.
let epsilon (ep) be a real number 0
let M be a natural number > delta
let m>M
let f(m)=sum(i=1,i=m) 0.9^10(-i);
then f(m)> 1- ep
as ep approaches 0, f(m) approaches 1.
That one's ******** greek to me, but i understand the final answers. Graphically displayed it would be infinite. Like a fractal if you were zooming in to it from the side.
Dang, this only proves the philosophical nature of stoners. ha ha...
