# Are these proofs logically equivalent?

Here are two proofs, firstly:

x = 0.999...
10x = 9.999...
= 9 + 0.999...
= 9 + x
9x = 9
x = 1


And secondly:

x = 1 - 1 + 1 - 1 + 1 - 1 ...
= 1 - (1 - 1 + 1 - 1 + 1 ...
= 1 - x
2x = 1
x = 1/2


The fourth line in the first and the third line in the second use the same trick. Does the legitimacy of the first proof guarantee that of the second?

• No. The series x = 1 - 1 + 1 - 1 + 1 - 1 ... is not convergent. The same trick can not be used, if the series is not absolutely convergent. – georg Sep 6 '14 at 7:50
• Why can't it be used? – user117644 Sep 6 '14 at 7:51
• According to the theory of infinite series. – georg Sep 6 '14 at 8:04
• Possible duplicate of this. – Hirak Sep 6 '14 at 9:03

The first proof is also more subtle than it seems. If $x=0.999\dotsc$, then why $10x=9.999\dotsc$? If you have a terminating decimal number, say $x=0.999$, then you can write $$x=\frac{9}{10}+\frac{9}{100}+\frac{9}{1000},$$ and then by well-known rules of arithmetic, you have $$10x=10\left(\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}\right)=\frac{9}{1}+\frac{9}{10}+\frac{9}{100}=9.99.$$ But what about non-terminating numbers? It would seem that if $x=0.999\dotsc$ then you can just write $$x=\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\dotsb,$$ but how do you calculate this infinite sum? It is not obvious that you can do it because for example the sum $$1-1+1-1+\dotsb$$ does not have an obvious or intuitive value. For example, it could be zero: $$1-1+1-1+\dotsb = (1-1)+(1-1)+\dotsb=0+0+\dotsb = 0,$$ or it could be one: $$1-1+1-1+\dots = 1+(-1+1)+(-1+1)+\dotsb = 1+0+0+\dotsb = 1$$ or something else.

Now, once you have a definition for calculating (some) infinite sums, you must prove that with that definition $$10\left(\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}\dotsb\right)=\frac{9}{1}+\frac{9}{10}+\frac{9}{100}+\dotsb=9.999\dotsc$$ So, there is quite a bit of stuff hidden (or forgotten) in the first proof. The above examples about $1-1+1-1+\dotsb$ being zero or one shows that with infinite sums you must prove even basic "intuitive facts" because they might not actually hold!

If with your definition of infinite sums you can prove $$1-1+1-1+\dotsb = 1-(1-1+1-1+\dotsb),$$ then your second proof is also valid. It also shows that whatever definition for $1-1+1-1+\dotsb$ you come up with, if it has the above property, then it must give this sum the value 1/2. However, while the standard definition of infinite sums does give a value for $9/10+9/100+\dotsb$ and has the properties we want, it does not give any value to $1-1+1-1+\dotsb$, so you cannot use that, as others have mentioned.

• Alright now good luck proving $1−1+1−1+\cdots=1−(1−1+1−1+\cdots)$ with whatever new definition you decide to take. – Matt A Pelto Sep 6 '14 at 8:56
• It is your assertion not mine. – Matt A Pelto Sep 6 '14 at 8:59
• No, the LHS has 1 and the RHS has 2, which is no problem as the equation 1/2 = 1 - 1/2 shows :) – nonpop Sep 6 '14 at 9:01
• nonpop clearly has more patience than I. – Matt A Pelto Sep 6 '14 at 9:03
• Nah, while I'm fine with someone using non-standard definitions, if they keep coming up I will demand a reference for them before continuing. – nonpop Sep 6 '14 at 9:07

No. The first series is absolutely convergent, so its terms can be rearranged without changing the sum, while the second series is divergent.

The number $0.\overline{9}$ can be expressed in the form

\begin{align*} 0.\overline{9} & = \sum_{k = 1}^{\infty} \frac{9}{10^k}\\ & = \frac{9}{10} \sum_{k = 0}^{\infty} \frac{1}{10^k} \end{align*} which is a convergent geometric series, so \begin{align*} 0.\overline{9} & = \frac{9}{10} \cdot \frac{1}{1 - \frac{1}{10}}\\ & = \frac{9}{10} \cdot \frac{1}{\dfrac{9}{10}}\\ & = \frac{9}{10} \cdot \frac{10}{9}\\ & = 1 \end{align*}

In fact, the series $$0.\overline{9} = \sum_{k = 1}^{\infty} \frac{9}{10^k}$$ is absolutely convergent since the series $$\sum_{k = 1}^{\infty} \bigg| \frac{9}{10^K} \bigg| = \sum_{k = 1}^{\infty} \frac{9}{10^k}$$ converges. Any rearrangement of an absolutely convergent series has the same sum.

The series

$$1 - 1 + 1 - 1 + 1 - 1 + \cdots = \sum_{k = 0}^{\infty} (-1)^k$$

diverges since the sequence of partial sums $\{1, 0, 1, 0, 1, 0, \ldots\}$ alternates between $1$ and $0$. Since the series never converges to a limit, it is meaningless to say that

$$x = 1 - 1 + 1 - 1 + 1 - 1 + \cdots$$

A series must be absolutely convergent to guarantee that rearranging the terms of the series does not change the sum.

• Can you rearrange the terms to change the sum? That is, to get something other than 1/2. – user117644 Sep 6 '14 at 8:19
• Of course, @mistermarko : $$1-1+1-1+\ldots\stackrel ?=(1-1)+(1-1)+\ldots=0+0+\ldots=0$$ – Timbuc Sep 6 '14 at 10:15
• You're assuming an infinite series has an even number of terms. – user117644 Sep 6 '14 at 11:23

I would say no.

0.999.. is a symbol for a rational number. So for example, $\frac{1}{3}$ is by definition the number with the property $$\frac{1}{3} \cdot 3 = 1,$$ and $\frac{1}{3}$ is just a symbol that we use for this number with this property. Another symbol of the same number is also $0.333...$ All operations on the first proof are justified.

For the second proof $1-1+1-1 ....$ is no real number. It is not clear what you mean with this symbol, and, therefore, you can not add something to some strange symbol that has no meaning.

• You're saying 0.999... is rational? – user117644 Sep 6 '14 at 7:59
• @mistermarko Yes. It is another symbol for the number 1, which is indeed rational. – Adam Sep 6 '14 at 8:01