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The Infamous .999... = 1

Date: 01/12/2002 at 08:14:36
From: Dan T
Subject: The Infamous .999... = 1

Dear Dr. Math,

I have seen answers to the question that .9999... = 1, but some
people in my 8th grade class STILL don't agree. My geometry teacher
says that it is all about giving infinity a value. A student in the
class says I am trying to prove that 1 = 1/infinity.

Can you straighten this out? Thanks.

Dan T.

Date: 01/12/2002 at 09:27:23
From: Doctor Tom
Subject: Re: The Infamous .999... = 1

Hi Dan,

I assume you've seen the Dr. Math FAQ:

0.9999... = 1
http://mathforum.org/dr.math/faq/faq.0.9999.html

If not, take a look.

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.

- Doctor Tom, The Math Forum
http://mathforum.org/dr.math/

Date: 01/12/2002 at 16:40:11
From: Dan T
Subject: The Infamous .999...=1

Sorry, the equation that the student meant was:

1 = 1 - 1/infinity

Date: 01/12/2002 at 16:48:09
From: Doctor Tom
Subject: Re: The Infamous .999...=1

Hi Dan,

Then he's basically right. As you add each new 9 to the expansion, the
errors look more and more like:

1/100, 1/1000, 1/10000, ...

Thus, in a sense, the error begins to look like "1/infinity," which
semms as if it should be zero.

The 1/infinity is meaningless, but the concept of limit is not. We can
say that the limit of the sequence above is zero, and can rigorously
prove it.

- Doctor Tom, The Math Forum
http://mathforum.org/dr.math/

Associated Topics:
High School Number Theory

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