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Math Help - Every Positive Number Has A Unique Positive Square Root

  1. #1
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    Every Positive Number Has A Unique Positive Square Root

    Hello,

    I am currently studying Introduction to Analysis by Maxwell Rosenlicht. On page 28 -29 he gives this proof for Every Positive Number Has A Unique Positive Square Root. Can someone please help me understand a couple of parts of it? I have indicated in bold what I don't understand.


    If  0 < x_{1} < x_{2} then  x_{1}^2 < x_{2}^2 . That is, bigger positive numbers have bigger square roots. Thus any given real number can have at most one positive square root.

    It remains to show that if  a \in \Re , a > 0, then  a has at least one positive square root. For this purpose consider the set

     S = \{ x \in \Re : x \geq 0 , x^2 \leq a \}

    This set is nonempty, since  0 \in S , and bounded from above, since if  x > max \{ a,1 \} we have  x^2 = x  \cdot x > x \cdot 1 = x > a.Hence, y = l.u.b. S exists.

    I don't understand this part with the max function. Is the point of this to show that for all the elements of S there is a minimum  x^2 that may or not be in S that is the least upper bound? Also, why is it the max {a,1} and not just x > a?

    We proceed to show that  y^2 =a. First,  y > 0 , for  min \{ 1 , a \} \in S , since  ( min \{1 ,a \} )^2 \leq min \{ 1 , a \} \cdot 1 = min \{ 1 , a \} \leq a. Next, for any  \epsilon such that  0 < \epsilon < y we have  0 < y - \epsilon < y < y + \epsilon , so

    Also, why is there a min{a,1} here? Why not just drop the min {1,a}?

     ( y - \epsilon )^2 < y^2 < ( y + \epsilon ) ^2

    since bigger positive numbers have bigger squares. By the definition of  y there are numbers greater than  y - \epsilon in  S , but  y + \epsilon \notin S. Again, using the fact that bigger positive numbers have bigger squares, we get

     ( y - \epsilon ) ^2 < a < ( y + \epsilon )^2

    Hence

     ( y - \epsilon )^ 2 - ( y + \epsilon ) ^2 < y^2 - a < ( y + \epsilon ) ^2 - ( y - \epsilon )^2

    so

     \mid y^2 - a \mid < ( y+ \epsilon )^ 2 - ( y - \epsilon )^2 = 4 y \epsilon

    The inequality   \mid y^2 - a \mid < 4 y \epsilon holds for any  \epsilon such that  0 < \epsilon < y, and by choosing  \epsilon small enough we can make  4 y \epsilon less than any preassigned positive number. Thus   \mid y^2 - a \mid is less than any positive number. Since  \mid y^2 - a \mid \geq 0 we must have  \mid y^2 - a \mid = 0, proving  y^2 = a.

    Thank you.
    Last edited by Famboozle; May 2nd 2008 at 06:32 PM. Reason: Correct spelling of elements and change most to least
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  2. #2
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    Quote Originally Posted by Famboozle View Post
    It remains to show that if  a \in \Re , a > 0, then  a has at most [that should be "at least"] one positive square root. For this purpose consider the set

     S = \{ x \in \Re : x \geq 0 , x^2 \leq a \}

    This set is nonempty, since  0 \in S , and bounded from above, since if  x > max \{ a,1 \} we have  x^2 = x  \cdot x > x \cdot 1 = x > a.Hence, y = l.u.b. S exists.

    I don't understand this part with the max function. Is the point of this to show that for all the elemnts of S there is a minimum  x^2 that may or not be in S that is the least upper bound? Also, why is it the max {a,1} and not just x > a?

    At this stage, we're looking for an upper bound for S. It doesn't have to be the least upper bound. So we're looking for a number that's bigger than anything in S. The difficulty is that if x>1 then x^2>x, whereas if x<1 then x^2<x. If x>a and a≥1, then you can conclude that x^2>a. That doesn't work when a<1. For example (taking a=1/4), suppose that x^2>1/4. You can't conclude that x>1/4. (In fact, the best you can say is that x>1/2.) But if x^2>1 then certainly x>1, and 1 in turn is greater than a. The device of using max{a,1} is simply a mechanism for covering the two cases a≥1 and a<1 simultaneously.

    Quote Originally Posted by Famboozle View Post
    We proceed to show that  y^2 =a. First,  y > 0 , for  min \{ 1 , a \} \in S , since  ( min \{1 ,a \} )^2 \leq min \{ 1 , a \} \cdot 1 = min \{ 1 , a \} \leq a. Next, for any  \epsilon such that  0 < \epsilon < y we have  0 < y - \epsilon < y < y + \epsilon , so

    Also, why is there a min{a,1} here? Why not just drop the min {1,a}?

    Similar sort of reason. It is necessary to show that y>0 (because the subsequent section of the proof requires that yε>0, for some ε>0). So we need to find a positive number whose square is less than a. If a≥1 then 1 is such a number; if a<1 then a itself is such a number. In either case, min{a,1} will do the job.
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