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

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?

• 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! – David G. Stork May 13 at 18:30
• For clarity, you might want to specify that $-n^2\ge0$ and $n^2\ge0$ together imply $n^2=0$, which in turn implies $n=0$. – Thorgott May 13 at 18:33
• Yes, it's correct – Shubham Johri May 13 at 18:33
• @DavidG.Stork That argument doesn't prove anything. To prove it is a sphere you would need to prove this first. – logarithm May 13 at 18:33
• @logarithm: "To prove it is a sphere you would need to prove this first." Huh? – David G. Stork May 13 at 21:08

Hint: Use that $$m^2+n^2\geq 2|mn|$$ so $$|mn|\le 0$$