Recall the usual version has Achilles chasing a
turtle, where the turtle is given a ten cubit head start and Achilles
runs ten times faster than the turtle. When Achilles runs 10 cubits,
the turtle goes 1 cubit. When Achilles runs 1 cubit, the turtle goes
1/10 of a cubit. It follows that Achilles catches the turtle in
10 + 1 + 1/10 + ... cubits. In modern notation, one can write
these numbers as decimals yielding

10 + 1 + .1 + .01 + ... = 11.1111... = 11 1/9.

It follows that Achilles catches the turtle in 11 1/9 cubits.

The non Archimedean case has the turtle trying to catch Achilles
who is now given a ten cubit head start.

When the turtle goes 10 cubits, Achilles will have run 100 cubits.
When the turtle goes 100 cubits then Achilles will have run 1000
cubits, and so on.

It follows that the turtle goes
10 + 100 + 1000 + .... Ordinarily, this would be considered
as meaningless or ``infinity''. However, one can consider this
purely * formally,* in other words, without caring too much
about the actual meaning of the numbers and just using algebraic
manipulations. So let x = 10 + 100 + 1000+ ..., then

9(x + 1) = 9 + 90 + 900 + 9000 + ... = ... 9999.

Adding 1 to this gives 9( x + 1) + 1

... 9999999

+ 1

= ... 0000000 = 0

where there is never a leading one, as the carries go on forever.
It follows that 9(x+1) + 1 = 0, so x = -1 1/9.

Note that this actually gives a correct answer!

See I.M., Gelfand and A. Shen, * Algebra * Birkhauser, Boston,
1993. Reviewed by R. Askey, Amer. Math. Monthly #102 (1995), 78--81.