guys i dont understand whats logical approach use by tom apostol to proof this theorem :(adsbygoogle = window.adsbygoogle || []).push({});

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 x^{2}≤ 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, a^{2}≤ a(I + a)^{2}and hence a^{2}/(1 + a)2 ≤ a. (this mean to show that a^{2}/(1 + a)2 ≤ a is equal with x^{2}≤ 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 a^{2}/(1 + a)2 ≤ a, we can conclude that three possibilities? )

the following i dont understand :

Suppose b^{2}> a and let c = b - (b^{2}- a)/(2b) = t(b + a/b). Then 0 < c < band

c^{2}= b^{2}- (b^{2}- a) + (b^{2}- a)^{2}/(4b^{2}) = a + (b^{2}- a)^{2}/(4b^{2}) > a. Therefore c^{2}> x^{2}

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 b^{2}> a is impossible.

Suppose b^{2}< a. Since b > 0, we may choose a positive number c such that c < band

such that c < (a - b^{2})/(3b). Then we have

(b + C)^{2}= b^{2}+ c(2b + c) < b^{2}+ 3bc < b^{2}+ (a - b^{2}) ~ 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 b^{2}< a is impossible, and the only remaining

alternative is b^{2}= a.

i dont know, but compare to the calculus spivak for example. its harder to follow ..

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# Logical step to proof

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