Example 1.7. We first show that R does contain at least one number which is missing from Q (so we have filled in at least one hole). Consider the set A := {a R : a 2 < 2}. It is easy to see that this is nonempty (for instance, 1 A) and bounded above (for instance, 2 is an upper bound) and so, by the completeness axiom, s := sup A exists. We claim that s 2 = 2: in other words, s = 2. We prove this using a Goldilocks approach: we show s 2 > 2 (not too hot) and s 2 < 2 (not too cold), so we must have s 2 = 2 (just right). Suppose s 2 > 2, so that s 2 2 > 0 and we can therefore find some n N with n > 4/(s 2 2). It then follows that (s 1/n) 2 = s 2 2s/n + 1/n2 s 2 4/n > 2, (1.1) where the second step is due to the fact that 1/n2 0 for n 1 and s 2 since 2 is an upper bound for A. From (1.1) and the definition of A, it follows that s > s 1/n > a for all a A. However, this contradicts the fact s is the least upper bound for A, and so s 2 > 2. 3 Suppose s 2 < 2, so that 2 s 2 > 0 and we can therefore find some n N with n > 5/(2 s 2 ). It then follows that (s + 1/n) 2 = s 2 + 2s/n + 1/n2 s 2 + 5/n < 2, where the second step is similar to that in (1.1), this time using 1/n2 1/n for n 1. From (1.7) and the definition of A, it follows that s + 1/n A and clearly s + 1/n > s. However, this contradicts the fact s is an upper bound for A, and so s 2 < 2