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Prove that if $m^2+n^2=0$ then $m=0$ and $n=0$.

Given $m^2+n^2=0$ then $m^2= -n^2$. Because $m$ and $n$ are real numbers, then $m^2 \geq 0$, $n^2 \geq 0$. Therefore, $m=0$ and $n=0$.

Is that correct?

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  • $\begingroup$ Yes, correct. Or just notice that the equation of a circle, $m^2 + n^2 = r^2$, here is for a circle of radius $0$. Done! $\endgroup$ – David G. Stork May 13 at 18:30
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    $\begingroup$ For clarity, you might want to specify that $-n^2\ge0$ and $n^2\ge0$ together imply $n^2=0$, which in turn implies $n=0$. $\endgroup$ – Thorgott May 13 at 18:33
  • $\begingroup$ Yes, it's correct $\endgroup$ – Shubham Johri May 13 at 18:33
  • $\begingroup$ @DavidG.Stork That argument doesn't prove anything. To prove it is a sphere you would need to prove this first. $\endgroup$ – logarithm May 13 at 18:33
  • $\begingroup$ @logarithm: "To prove it is a sphere you would need to prove this first." Huh? $\endgroup$ – David G. Stork May 13 at 21:08
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Hint: Use that $$m^2+n^2\geq 2|mn|$$ so $$|mn|\le 0$$

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