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
[Math] If $0\lt y \le 1$, prove that there exists a unique positive real number $x$ such that $x^2=y$
analysisproof-writingreal-analysis
Best Answer
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 $x=x_1.$