Thread: Every Positive Number Has A Unique Positive Square Root

1. 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.

2. Originally Posted by Famboozle
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. 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.

Originally Posted by Famboozle
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 y–ε>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.