If $0\lt y \le 1$, prove that there exists a unique positive real number $x$ such that $x^2=y$ I'm stumped. I don't want an entire solution, just a hint.
If $0\lt y \le 1$, prove that there exists a unique positive real number $x$ such that $x^2=y$
The section in the book I'm on is the least upper bound property
 A: Entire answer. Don't read it all......Let $A=\{z>0 :z^2\leq y\}.$ We have $A\ne \phi$ because $y/2\in A$ because $$0<(y/2)^2=y (y/4)\leq y(1/4)<y.$$   Observe  that $$(1)......(0<z\land z^2>y)\implies A\subset (0,z)\implies $$ $$\implies z \text { is an upper bound for } A.$$  $$\text {(2)...... Now let } x=\sup A.$$ (From Latin supremum, in math it's a synonym for $lub.$) We know $x$ exists because $A\ne  \phi$ and because $1$ is an upper bound for $A$. We show that $x^2$ can't be more than $y$ and can't be less than $y$. First, suppose $y<x^2.$ There exists positive integer $n$ for which $$(x^2-y>1/n)  \land (x>1/n).$$ Let $x'=x-1/(2n)$.  Since $0<x\leq 1$ we have  $$x'^2-y=(x^2-y)-(x/n)+1/(4n^2)\geq (x^2-y)-(1/n)+1/(4n^2)>$$ $$>1/n-1/n+1/4n^2>0.$$ But $x'>0$ so by (1),  $x'$ is an upper bound for $A$ with $x'<x=\sup A$, which is absurd. Therefore we cannot have $y<x^2.$....... Second, suppose $y>x^2$. Then there exists a positive integer $n'$ for which $$y-x^2>1/n'.$$ Let $x''=x+1/(4n')$ .Since $x\leq 1$, we compute $$(x'')^2=x^2+x/(2n')+1/(16n'^2)\leq x^2+1/(2n')+1/(16n'^2)<x^2+1/n'<y $$ which implies $x''\in A.$ But $A$ cannot have a member $x''$ which is greater than $\sup A=x.$ So we cannot  have $y>x^2$..... Therefore $x^2=y$..... The positive number $y$ cannot have two positive square roots $x,x_1$ because if $ [0=y-y=x^2-x_1^2=(x+x_1)(x-x_1)]\land [x+x_1\ne 0]$ then $x=x_1.$  
A: $y(x) = x^2$ is a continuous and monotonically increasing function with $y(0) = 0$ and $y(1) = 1$.  Now apply the intermediate value theorem and for uniqueness use the fact that it is monotonic.  I'm not sure how to directly prove this from the least upper bound property, can you give more context about what you are working with in that section?
A: Let $A=\{a\in\mathbb{R};\;a\geq 0\text{ and } a^2<y\}$ and $B=\{b\in\mathbb{R};\;b\geq 0\text{ and } b^2>y\}$.


*

*$A$ has no largest element;

*$B$ has no least element;

*If $a\in A$ and $b\in B$ then $a<b$.


From 1, 2 and 3 we get the desired conclusion.
Remark: A similar argument shows that given any $y>0$ and $n\in\mathbb{N}$, there exists a unique positive real number $x$ such that $x^n=y$.
A: Assume that there are two possible solutions, say $x_1 > 0$ and $x_2 > 0$ such that
$$x_1^2 = y = x_2^2$$
Can you now proceed to show that if both $x_1$ and $x_2$ are positive, then $x_1 = x_2$?
