Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Logical step to proof

  1. Jul 6, 2012 #1
    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? :confused: )

    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 ..
  2. jcsd
  3. Jul 6, 2012 #2


    User Avatar
    Science Advisor

    Hey calios and welcome to the forums.

    The idea is that if you have a bound that is greater than a least upper bound, then the least upper bound is not actually the least upper bound and hence a contradiction.

    The way that we think about lower and upper bounds is that we look at the least upper bound and the greatest lower bound.

    If we have an upper bound that is higher than the least upper bound then the least upper bound must be the one that's greater. Similarly for lower bound, if we have a lower bound that is lower than the greatest lower bound then again it must not actually be the greatest lower bound.

    It's like saying the lowest upper bound is 2 and then saying that an upper bound of 3 exists which is a contradiction. Similarly with a lower bound of -5, if we have a lower bound of -3 then it means that there is a contradiction.

    That's the basic idea and the goal is to use these definitions to unambigously show what the actual bounds should be which are used in proofs since a contradiction between these will show a contradiction in the proof itself.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook