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Thread: Every Positive Number Has A Unique Positive Square Root

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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 $\displaystyle 0 < x_{1} < x_{2} $ then $\displaystyle x_{1}^2 < x_{2}^2$ . 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 $\displaystyle a \in \Re $, a > 0, then $\displaystyle a $ has at least one positive square root. For this purpose consider the set

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

    This set is nonempty, since $\displaystyle 0 \in S $, and bounded from above, since if $\displaystyle x > max \{ a,1 \} $ we have $\displaystyle x^2 = x \cdot x > x \cdot 1 = x > a$.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 $\displaystyle x^2$ 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 $\displaystyle y^2 =a$. First, $\displaystyle y > 0 $, for $\displaystyle min \{ 1 , a \} \in S$ , since $\displaystyle ( min \{1 ,a \} )^2 \leq min \{ 1 , a \} \cdot 1 = min \{ 1 , a \} \leq a$. Next, for any $\displaystyle \epsilon $ such that $\displaystyle 0 < \epsilon < y $ we have $\displaystyle 0 < y - \epsilon < y < y + \epsilon $, so

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

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

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

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

    Hence

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

    so

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

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

    Thank you.
    Last edited by Famboozle; May 2nd 2008 at 05:32 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 $\displaystyle a \in \Re $, a > 0, then $\displaystyle a $ has at most [that should be "at least"] one positive square root. For this purpose consider the set

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

    This set is nonempty, since $\displaystyle 0 \in S $, and bounded from above, since if $\displaystyle x > max \{ a,1 \} $ we have $\displaystyle x^2 = x \cdot x > x \cdot 1 = x > a$.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 $\displaystyle x^2$ 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 $\displaystyle x^2>x$, whereas if x<1 then $\displaystyle x^2<x$. If $\displaystyle x>a$ and a≥1, then you can conclude that $\displaystyle x^2>a$. That doesn't work when a<1. For example (taking a=1/4), suppose that $\displaystyle x^2>1/4$. You can't conclude that x>1/4. (In fact, the best you can say is that x>1/2.) But if $\displaystyle x^2>1$ 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 $\displaystyle y^2 =a$. First, $\displaystyle y > 0 $, for $\displaystyle min \{ 1 , a \} \in S$ , since $\displaystyle ( min \{1 ,a \} )^2 \leq min \{ 1 , a \} \cdot 1 = min \{ 1 , a \} \leq a$. Next, for any $\displaystyle \epsilon $ such that $\displaystyle 0 < \epsilon < y $ we have $\displaystyle 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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