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
Entire answer. Don't read it all......Let $A=\{z&gt;0 :z^2\leq y\}.$ We have $A\ne \phi$ because $y/2\in A$ because $$0&lt;(y/2)^2=y (y/4)\leq y(1/4)&lt;y.$$   Observe  that $$(1)......(0&lt;z\land z^2&gt;y)\implies A\subset (0,z)\implies $$ $$\implies z \text { is an upper bound for } A.$$  $$\text {(2)...... Now let } x=\sup A.$$ (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 $y&lt;x^2.$ There exists positive integer $n$ for which $$(x^2-y&gt;1/n)  \land (x&gt;1/n).$$ Let $x'=x-1/(2n)$.  Since $0&lt;x\leq 1$ we have  $$x'^2-y=(x^2-y)-(x/n)+1/(4n^2)\geq (x^2-y)-(1/n)+1/(4n^2)&gt;$$ $$&gt;1/n-1/n+1/4n^2&gt;0.$$ But $x'&gt;0$ so by (1),  $x'$ is an upper bound for $A$ with $x'&lt;x=\sup A$, which is absurd. Therefore we cannot have $y&lt;x^2.$....... Second, suppose $y&gt;x^2$. Then there exists a positive integer $n'$ for which $$y-x^2&gt;1/n'.$$ Let $x''=x+1/(4n')$ .Since $x\leq 1$, we compute $$(x'')^2=x^2+x/(2n')+1/(16n'^2)\leq x^2+1/(2n')+1/(16n'^2)&lt;x^2+1/n'&lt;y $$ which implies $x''\in A.$ But $A$ cannot have a member $x''$ which is greater than $\sup A=x.$ So we cannot  have $y&gt;x^2$..... 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 $x=x_1.$
Assume that there are two possible solutions, say $x_1 &gt; 0$ and $x_2 &gt; 0$ 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 $x_1 = x_2$?