9, in repeating decimal notation) denotes the repeating decimal consisting of an unending sequence of 9s after the decimal point. This repeating decimal represents the smallest number no less than every decimal number in the sequence (0.9, 0.99, 0.999, ...). This number is equal to 1.

How do you prove that 0.9 recurring is 1?

Identifying The Sequence With Its Limit

= 1 — the sequence of terminating decimals 0.9, 0.99, 0.999, 0.9999, and so on, converges to 1, so the repeating decimal 0.9999... representing the limit of that sequence, is said to be equal to 1.

Is there a number between 0.999 and 1?

As a result, 0.999… = 1 because we don't allow there to be a gap between them (so they must be the same). In other number systems (like the hyperreal numbers), 0.999… is less than 1. Here, infinitely small numbers are allowed to exist, and this tiny difference (h) is what separates 0.999… from 1.