What is $1 + 999999...$ (an infinite string of $9$s)?

Question

I'm doing a programming assignment in Haskell, and it involves adding "infinite" lists. At the bottom of the assignment, our lecturer has written ... "in some sense an infinite string of $9$s is the additive inverse of the number $1$, so we might say that it is equal to $-1$". I find this to be very confusing, is there actually any sense in this statement?

Answer 1

When the lecturer writes "an infinite string of 9s" he doesn't mean 999..., but ...999, which is infinite to the left. The addition goes

  ...1111111
            1
+ ...99999999
  -----------
  ...00000000

with a carry out of each position. Since every digit of the result is 0, it is to some extent reasonable to say that $1+...999=0$.

Of course, things like "...9999" are not actually numbers, but we can define things like infinite sequences of digits and define some operations on them by analogy to what happens in calculations with actual numbers in base ten. The result turns out to be a ring, commonly called "the 10-adic integers".

Answer 2

If you think of your infinite string as starting on the right and you do the normal addition and carry operation then you will continue carrying all the way down the infinite string. Even though this process never terminates each digit in the string eventually becomes zero and then remains zero for all remaining steps, so I think it's fair to say that the "limit" of this process is the infinite string of zeros.

But of course, this is a very ad hoc explanation of a very ad hoc comment so don't put too much stock in this.

Edit: Apparently, as Hennings answer indicates, you can put stock in this.

Answer 3

It actually makes more sense if you think of the string as infinite to the left: $\ldots 999$. Add $1$ to it: the carry propagates indefinitely, and you get $\ldots 000$. In fact, this is the $10$-adic expansion of $-1$ and in the context of $p$-adic numbers is a perfectly legitimate representation.

Answer 4

Indeed! $\ldots 999999=-1$! I will try to elucidate it.

$$\cdots + 1000x + 100x + 10x + 1x + \frac{x}{10} + \frac{x}{100} + \frac{x}{1000} + \cdots = 0$$

Why is the equation above true? Try multiplying the left-hand side by $10$. As you see, multiplying it by $10$ doesn't change it, therefore it can only be $0$.

Let's plug $x = 9$ into the equation:

$$\cdots + 9000 + 900 + 90 + 9 + \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots = 0$$

or

$$\dots 999999,999999 \ldots = 0$$

or

$$\ldots999999 = -0,999999\ldots$$

We know the famous equation:

$$0,999999\ldots = 1$$

Therefore

$$\ldots 999999 = -1$$