Q:

Prove that if $c>0, x>0$, and $x^2=c$ then $x $ is unique.

Prove that if $c>0, x>0$, and $x^2=c$ then $x$ is unique. 
Here is how I start.  Assume $x$ is not unique.  Let, $x>0, y>0$, and $c>0$ such that $x^2=c$ and $y^2=c$.  
(1)By the transitive property we have $x^2=y^2$.  
(2)Since $x$ and $y$ are positive, taking the square root of both sides we show $x=y$.  
(3)Therefore x is unique.  
I'm sure step (2) is not valid... Any hints?

A:

$x^2=y^2$
$(x-y)(x+y)=0$
$x-y=0$ since $x+y>0$
$x=y$
Your argument at 2) is circular (you use as hypothesis that $u^2=v$ has an unique solution; the same thing that you need to prove)

A:

I agree that step 2 is not valid.  Here's my approach:
Suppose that "$x$ is not unique".  Then there exist $0<x<y$ with $x^2 = y^2 = c$.  However,
$$
c = x^2 = xx < yx < yy = y^2 = c
$$
But $c<c$ is a contradiction.

