Not quite there….
0.99999.....(recurring) = 1
EXACTLY. No approximations. No rounding off.
Here's the proof:
(1) Let x = 0.999...
(2) 10x = 9.999...
(3) 10x - x = 9.999... - 0.999...
(4) 9x = 9
(5) x = 1
Steps (2), (3) and (4) might seem dodgy, but they are mathematically impeccable. The confusion (if any) seems to arise because some people tend to think of 0.999... as a r-e-a-l-l-y long series of 9s, rather than an endless series of 9s. The 9s never end. The number doesn't come really close to 1, it IS 1.
To assuage any qualms you might have, let's see another way of approaching the proof:
(1) 1/3 = 0.333... (not an approximation, this is EXACT; try actually dividing)
(2) 3 * 1/3 = 3 * 0.333...
(3) 1 = 0.999...
If you still don't agree, here's something that might set the ball rolling:
0.9, 0.99, 0.999, 0.9999, 0.99999, ...... is an infinite series which approaches 1
0.9999.... is a number (a specific number); it is not almost there, or reaching 1; it is a point, a single number. The dots are just notation.
Very interesting reads on this topic:
http://en.wikipedia.org/wiki/0.999...
http://polymathematics.typepad.com/polymath/2006/06/no_im_sorry_it_.html
http://www.purplemath.com/modules/howcan1.htm
http://qntm.org/?pointnine
All these articles repeat the same points, but the key is in understanding the recurring decimal :o)
February 12th, 2008 - 12:55
Actually it is just another way of writing 1.
it is a different set of nomenclature of maths .
it has no approximations of limits or other sorts.
Yadnesh.