An arguement for the equivalence of 1 and 0.9 recurring is as someone said:

Here goes:
let x = 0.9999(recuring)
then 10x = 9.9999(recuring)
then 10x-x = 9.9999(recuring) - x
then 9x = 9
then x = 1

A more shifty little argument though:

Given a positive real number x, find the positive square root of it (calling the positive square root Root(x)). Examining the behavior of Root(x) with different groups of values for x, it seems that:
If x > 1, Root(x) > x > 1
If x = 1, Root(x) = x = 1
If x < 1, Root(x) < x < 1

Now let's say that Root(x) = 0.9 recurring. It's pretty safe to say that x < 0.9 recurring, but we just defined x as being between Root(x) and 1, therefore:
0.9 recurring < x < 1

This pretty much proves that 0.9 recurring = 1.