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Given any $r \in \mathbb{R}_{>0}$, the number $\sqrt{r}$ is unique in the sense that, if $x$ is a positive real number such that $x^2 = r$, then $x = \sqrt{r}$

I would appreciate any nudge in the right direction. My initial thought would be to prove that this is impossible for $x<\sqrt{r}$ and $x>\sqrt{r}$.

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    $\begingroup$ You might like to see this page: Explanation of uniqueness of square root. $\endgroup$
    – Minus One-Twelfth
    Commented Apr 19, 2019 at 4:47
  • $\begingroup$ The approach most in line with your initial thought is to consider that $x \mapsto x^2$ is strictly increasing on $\mathbb{R}_{>0}$. $\endgroup$
    – jawheele
    Commented Apr 19, 2019 at 4:56
  • $\begingroup$ Can you show that $x\mapsto x^2$ is increasing on $\Bbb R_{>0}$? $\endgroup$
    – Angina Seng
    Commented Apr 19, 2019 at 4:56
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    $\begingroup$ If $x^2 = y^2 = r$ with $x,y>0$, then $(x+y)(x-y) = x^2-y^2 = 0$. As $x+y > 0$, we must have that $x-y=0$, i.e., $x = y$. $\endgroup$
    – amsmath
    Commented Apr 19, 2019 at 5:12
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    $\begingroup$ @amsmath Thank you, your comment cleared things up for me. $\endgroup$
    – Stanley Hudson
    Commented Apr 19, 2019 at 15:13

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