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What is 0.999... + 0.999...?


Date: 09/27/1999 at 20:19:48
From: Darryl Jarman
Subject: What is 0.999... + 0.999...?

Hello,

While discussing the 1 = 0.999... solution, a person asked what is 
0.999... + 0.999...? I think it is a good question - one that I could 
not answer. It should be 2 but how do we show this?

In the archives at

  Getting 0.99999...
  http://mathforum.org/dr.math/problems/dusty4.15.98.html   

you say "Can you figure out why 0.3bar + 0.3bar = 0.6bar? Because 
these numbers go on forever, you will need to use a little logic to 
add them. (The algorithm that you learned for adding numbers doesn't 
work very well when you can't get to the rightmost number.)"

If we copy the idea of limits, we could say that for any number n we 
could start adding at this rightmost point and get the 0.66...6(nth) 
place. And since we can do this for any n then the equation holds. But 
if we try this for 0.9bar + 0.9bar we seem to get something that looks 
less like 2, namely 1.9....8(nth). But if we replace 8 with 9, we seem 
to get a number between 2 and the one we have, which makes me feel as 
if 0.9bar + 0.9bar < 2.

Thank you for your time,
Darryl Jarman


Date: 09/28/1999 at 09:09:45
From: Doctor Rick
Subject: Re: What is 0.999... + 0.999...?

Hi Darryl, thanks for your question. It's an interesting twist on the 
perennial debate about whether 0.999... could really equal 1.

Since that 8 is the nth digit and you take n to infinity, there is no 
digit 8 in the limit. 

Apply the definition of a limit to what you have. We can express it as 
a game: you pick a number epsilon, as small as you want, and I have to 
pick a number n such that

     |2 - 1.999...98| < epsilon

where the 8 is in the nth decimal place. Since

     |2 - 1.999...98| = 2*10^(-n)

I just pick

     n > -log(epsilon/2)

This is no harder than in the case of 0.999...9. The 8 in the last 
place has no effect on the limit; it only delays somewhat the approach 
to the limit, as indicated by that factor of 1/2.

Here is another way to approach the problem, by manipulating infinite 
sums. If you want to make everything explicit, you can write the sums 
as limits of finite sums, and get into the issue of switching the 
limit and the sum as you go through this process.

Let's write 0.999... as an infinite sum:

     0.999... = Sum[n = 1 to infinity](9*10^-n)

Now we can add infinite sums:

  0.999... + 0.999... = Sum[n = 1 to infinity](9*10^-n)
                        + Sum[n = 1 to infinity](9*10^-n)

                      = Sum[n = 1 to infinity](18*10^-n)

                      = Sum[n = 1 to infinity](10*10^-n + 8*10^-n)

                      = Sum[n = 1 to infinity](10^(-n+1))
                        + Sum[n = 1 to infinity](8*10^-n)

                      = Sum[n = 0 to infinity](10^-n)
                        + Sum[n = 1 to infinity](8*10^-n)

                      = 1 + Sum[n = 1 to infinity](10^-n)
                        + Sum[n = 1 to infinity](8*10^-n)

                      = 1 + Sum[n = 1 to infinity]((8+1)*10^-n)

                      = 1 + Sum[n = 1 to infinity](9*10^-n)

                      = 1.999...

I think this is the result you are looking for. Since we know that 
0.999... = 1, the answer is 1 + 1 = 2; but you wanted to see that 
unbroken string of 9's.

By the way, notice that the final 8 never appears in this method. 
That's because we're dealing with infinite sums from the start, so 
there is always a carry from the next digit to the right. The sum of 
10^-n is the sum of the infinite number of carries.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Analysis
High School Number Theory

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