the distance from 0.99 to 1 is , and so on. The distance to 1 from the th point (the one with 9s after the decimal point) is . Therefore, if 1 were not the smallest number greater than 0.9, 0.99, 0.999, etc., then there would be a point on the number line that lies between 1 and all these points. This point would be at a positive distance from 1 that is less than for every integer . In the standard number systems (the rational numbers and the real numbers), there is no positive number that is less than for all n. This is (one version of) the Archimedean property, which can be proven to hold in the system of rational numbers. Therefore, 1 is the smallest number that is greater than all 0.9, 0.99, 0.999, etc., and so . Discussion on completeness Part of what this argument shows is that there is a least upper bound of the sequence 0.9, 0.99, 0.999, etc.: a smallest number that is greater than all of the terms of the sequence. One of the axioms of the real number system is the completeness axiom, which states that every bounded sequence has a least upper bound. This least upper bound is one way to define infinite decimal expansions: the real number represented by an infinite decimal is the least upper bound of its finite truncations. The argument here does not need to assume completeness to be valid, because it shows that this particular sequence of rational numbers in fact has a least upper bound, and that this least upper bound is equal to one. Rigorous proof The previous explanation is not a proof, as one cannot define properly the relationship between a number and its representation as a point on the number line. For the accuracy of the proof, the number , with nines after the decimal point, is denoted . Thus , , , and so on. As , with digits after the decimal point, the addition rule for decimal numbers implies and for every positive integer . One has to show that 1 is the smallest number that is no less than all . For this, it suffices to prove that, if a number is not larger than 1 and no less than all , then . So let such that for every positive integer . Therefore, which, using basic arithmetic