2
$\begingroup$

How to prove that there is a unique positive real number $x$ such that $x^2 = 2$?

My step: suppose there are $2$ positive real number $x, y$ such that $x^2=y^2=2$, then we take sqrt of both and we get $\sqrt{x^2}=x=y=\sqrt{y^2}$. Is it right?

$\endgroup$
4
  • $\begingroup$ You need a better justification as to why $x^2=y^2$ implies $x=y$. Perhaps, you can say since $x^2=y^2$, therefore $(x-y)(x+y)=0$, now if both $x$ and $y$ are positive, therefore only possibility is $x=y$. $\endgroup$
    – Anurag A
    Jan 12, 2018 at 14:38
  • 2
    $\begingroup$ The existence of a well-defined square root relies on this observation, so until you have finished this proof you do not know that $\sqrt{{}\cdot {}}$ even makes sense as a function, and you therefore cannot use it. $\endgroup$
    – Arthur
    Jan 12, 2018 at 14:38
  • $\begingroup$ Using your approach, taking the square roots leads to $|x|=|y|$, which implies $x=y$, if $x$ and $y$ are non-negative. The existence is easy because $\sqrt{2}$ obviously is a solution in positive real numbers. $\endgroup$
    – Peter
    Jan 12, 2018 at 14:44
  • 2
    $\begingroup$ @Peter I am not sure if the square root function has been defined yet - the uniqueness of a positive solution to the equation $x^2=c$ is required to define this. So that would make this kind of argument circular. $\endgroup$
    – John Doe
    Jan 12, 2018 at 14:50

2 Answers 2

5
$\begingroup$

Better would be: $$x^2=y^2\implies x^2-y^2=0\implies (x-y)(x+y)=0\implies x=\pm y$$Then if $x=-y$, and both are positive, then $x=y=0$, contradiction. So $x=y$. Also, your proof only deals with the uniqueness aspect of the proof, while the question asks for both uniqueness, and existence. Though perhaps you've done this already.

$\endgroup$
3
$\begingroup$

The function $f(x)=x^2-2$ has derivate $2x$, hence is strictly increasing for $x\ge 0$. Because of $f(0)=-2$ and $f(2)=2$ there is a unique real root of $f(x)$ in the interval $[0,2]$, hence a unique real solution in the positive real numbers.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .