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.99999 repeating = 1

Name: Anonymous 2007-02-18 15:23

x= .999 repeating
10x = 9.999 repeating
10x-x = 9.999 repeating - .999 repeating
9x= 9
x=1

Name: Anonymous 2007-02-18 18:04

Rather than saying "giving infinity a value," it's perhaps a bit
clearer to say, "giving the concept of a limit of an infinite sequence
of numbers a value."

.9 is not 1; neither is .999, nor .9999999999. In fact if you stop the
expansion of 9s at any finite point, the fraction you have (like .9999
= 9999/10000) is never equal to 1. But each time you add a 9, the
error is less. In fact, with each 9, the error is ten times smaller.

You can show (using calculus or other methods) that with a large
enough number of 9s in the expansion, you can get arbitrarily close to
1, and here's the key:

THERE IS NO OTHER NUMBER THAT THE SEQUENCE GETS ARBITRARILY CLOSE TO.

Thus, if you are going to assign a value to .9999... (going on
forever), the only sensible value is 1.

There is nothing special about .999...  The idea that 1/3 = .3333...
is the same. None of .3, .33, .333333, etc. is exactly equal to 1/3,
but with each 3 added, the fraction is closer than the previous
approximation. In addition, 1/3 is the ONLY number that the series
gets arbitrarily close to.

And it doesn't limit itself to single repeated decimals. When we say:

1/7 = .142857142857142857...

none of the finite parts of the decimal is equal to 1/7; it's just
that the more you add, the closer you get to 1/7, and in addition, 1/7
is the UNIQUE number that they all get closer to.

Finally, you can show for all such examples that doing the arithmetic
on the series produces "reasonable" results:

Since:

1/3 = .333333...
2/3 = .666666...

1/3 + 2/3 = .999999... = 1.

By the way, there is nothing special about 1 as being a non-unique
decimal expansion. Here are a couple of others:

2 = 1.9999...
3.71 = 3.709999999...
2.778 = 2.77799999999999...

...and the student who says you're trying to show that 1 = 1/infinity
is wrong.

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