Number Systems — Question 4

Let x = 0.99999 ........ (1)

Here one digit 9 is repeating so we multiply both side by 10 in equation (1)

10 x = 9.99999.......... (2)

By subtracting equation (2) - equation (1)

10 x -x = 9.99999........ - 0.99999........

9x = 9

x = $\dfrac{9}{9}$

x = 1

Therefore, (0.99999...) is equivalent to (1), which might seem surprising at first glance, but it makes sense mathematically. This result can be demonstrated by the fact that any number infinitesimally close to 1 but less than 1, when infinitely added to itself, equals 1.
For example, if we take (0.9) and add (0.09), we get (0.99), and if we add (0.009) to that, we get (0.999), and so on. As we keep adding these infinitesimal increments, we approach (1). So, in a way, (0.99999...) is just another way to represent the number (1).