guys i dont understand whats logical approach use by tom apostol to proof this theorem : Every nonnegative real number a has a unique nonnegative square root. proof by apostol: 1.Proof If a = 0, then 0 is the only square root. <- okey 2.Assume, then, that a > 0. Let S be the set of all positive x such that x2 ≤ a. ( i think this one mean to show that irrational number is in set S, right?) . 3.Since (1 + a)2 > a, the number 1 + a is an upper bound for S. Also, S is nonempty because the number a/(1 + a) is in S; in fact, a2 ≤ a(I + a)2 and hence a2/(1 + a)2 ≤ a. (this mean to show that a2/(1 + a)2 ≤ a is equal with x2 ≤ a ? 4.By Axiom 10, S has a least upper bound which we shall call b. Note that b ≥ a/(1 + a) so b > O. There are only three possibilities: b2 > a, b2 < a, or b2 = a. (this mean, from b ≥ a/(1 + a) and a2/(1 + a)2 ≤ a, we can conclude that three possibilities? ) the following i dont understand : Suppose b2 > a and let c = b - (b2 - a)/(2b) = t(b + a/b). Then 0 < c < band c2 = b2 - (b2 - a) + (b2 - a)2/(4b2) = a + (b2 - a)2/(4b2) > a. Therefore c2 > x2 for each x in S, and hence c > x for each x in S. This means that c is an upper bound for S. Since c < b, we have a contradiction because b was the least upper bound for S. Therefore the inequality b2 > a is impossible. Suppose b2 < a. Since b > 0, we may choose a positive number c such that c < band such that c < (a - b2)/(3b). Then we have (b + C)2 = b2 + c(2b + c) < b2 + 3bc < b2 + (a - b2) ~ a. Therefore b + c is in S. Since b + c > b, this contradicts the fact that b is an upper bound for S. 1herefore the inequality b2 < a is impossible, and the only remaining alternative is b2 = a. i dont know, but compare to the calculus spivak for example. its harder to follow ..