calios
- 9
- 0
guys i don't understand what's 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 don't 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 don't know, but compare to the calculus spivak for example. its harder to follow ..
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 don't 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 don't know, but compare to the calculus spivak for example. its harder to follow ..