## More fun with infinite decadic numbers

This is the sixth in a series of posts on the decadic numbers (previous posts: A curiosity, An invitation to a funny number system, What does “close to” mean?, The decadic metric, Infinite decadic numbers).

Last time I left you with some parting exercises; here are the answers, along with some additional commentary.

1. What number is represented by $\dots 11111$? How do you know?

It turns out that $\dots 11111 = -1/9$, and there are several ways to see this. The simplest is to note that adding it to itself nine times yields $\dots 99999$, which we already know is $-1$.

At this point we may also ask, if $\dots 11111$ is $-1/9$, then what is $1/9$? It must be $\dots 88889$, since adding that to $\dots 11111$ yields zero: $\begin{array}{rr} & \dots 11111 \\ + & \dots 88889 \\ \hline & \dots 00000 \end{array}$

Hmm, so $\dots 88889$ is $1/9$, and $\dots 88888$ is $-8/9$ (since it is 8 times $\dots 11111$). This also makes sense because $(-8/9) + 1 = 1/9$: $\begin{array}{rr} & \dots 88888 \\ + & 1 \\ \hline & \dots 88889 \end{array}$

It seems to make sense but it sure is weird! $\dots 88888$ is negative but $\dots 88889$ is positive. Can you figure out a relatively simple way to tell, just by looking at an infinite decadic number, whether it is positive or negative? (I don’t actually know the answer.)

2. How about $\dots 34343434$?

Mark James noted that

If $\dots 111111$ is $-1/9$ then $\dots 343434$ looks like $-34/99$ since $\dots 343434 / 34 = \dots 010101$ and $\dots 010101 \cdot 99 = \dots 99999$.

Indeed, and here’s another way to see the same thing: if $x = \dots 343434$, then $100x = \dots 343400$, so $x - 100x$ is $\begin{array}{rr} & \dots 343434 \\ - & \dots 343400 \\ \hline & \dots 000034 \end{array}$

Hence $-99x = 34$, so $x = -34/99$. As a further check, $9 \cdot (\dots 343434) = \cdots 09090906$, so $99 \cdot (\dots 343434) = 90 \cdot (\dots 343434) + 9 \cdot (\dots 343434) =$ $\begin{array}{rr} & \dots 09090906 \\ + & \dots 90909060 \\ \hline & \dots 99999966 \end{array}$

which ought to be $-34$: sure enough, adding $34$ to $\dots 999966$ yields zero.

In general, we can see that the infinite repeating decadic number $\dots d_1 d_2 \dots d_n d_1 d_2 \dots d_n = \overline{d_1 \dots d_n}$ will be equal to $-\frac{d_1 \dots d_n}{9 \dots 9}$ where the number of $9$‘s in the denominator is the same as the number of digits in the numerator.

3. Can you find an infinite decadic number which represents $-546$? How about $-1/7$? Or $-1/2$? $-546$ is $\dots 9999454$, and $-1/7$ is $\overline{142857}$; by this point I hope you can see why.

However, $-1/2$ is different: we would have to find some number which yields $\dots 99999$ when multiplied by two. Obviously no such number exists: two times anything cannot possibly end in a $9$. The same is true of any fraction with a denominator which is not relatively prime to (i.e. shares a common divisor with) $10$.

However, we can extend our notion of decadic numbers to accommodate such fractions, by allowing digits after a decimal point. So for example $0.5$ will be $1/2$, as usual, and then $-1/2$ will be $-1 + 1/2 = \dots 99999.5$. It’s important to note that we can only allow finite strings of digits after the decimal point: a number like $0.33333 \dots$ is meaningless because the sequence of numbers $0.3, 0.33, 0.333, 0.3333, \dots$ does not converge to anything; in fact, they are getting further and further apart! But any finite number of digits after the decimal point are OK.

In general, we call decadic numbers with no digits after the decimal point (including infinite ones) decadic integers. Obviously all the normal integers are also decadic integers; but so are fractions such as $-34/99$ whose denominators are relatively prime to $10$. Other fractions such as $-1/2$ are not.

Exercises:

1. Show that any fraction whose denominator has only $2$ and $5$ as factors is represented by a finite decimal number.
2. Show that by allowing a finite number of digits after the decimal point, we can represent any fraction as a decadic number. (Hint: factor the denominator into one part consisting only of twos and fives and another part with everything else.)

You may recall that in a previous post I promised to show you a strange decadic number $u$, which is not zero, but is equal to its own square. We’ve now finally seen enough for me to tell you what it is—which I will do in my next post. In the meantime you may want to try discovering it! Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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### 4 Responses to More fun with infinite decadic numbers

1. Lumos! says:

Wait, for Question 1 from the last post, …11111 + …99999 = …00000, right? So wouldn’t that mean that since …99999 = -1, …11111 = 1?

Your explanation of why …11111 = -1/9 makes sense though…

• Brent says:

No, …11111 + …99999 = …11110. After adding the 1 and 9 in the 1’s place you then have to carry a 1 to the next column, and so on.