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I thought I could be able to find an explanation of this (and perhaps I knew it a long time ago) but I cannot find an answer now.

If I assume that
$b^2 \leq |x|$, then is it true that both
$b \leq \sqrt{|x|}$ and
$b \geq - \sqrt{|x|}$
?

I've informally convinced myself that it is true, but I'd like to know for sure.

Where I really get confused is if I assume that $b^2 \leq x$, so that the argument to sqrt can be negative. Do the inequalities above still hold in this case with the complex numbers?

Thanks.

Monica Sendi Afa
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Dan
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1 Answers1

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You are right, the inequality $b^2\le |x|$ is equivalent to the inequality $-\sqrt{|x|}\le b\le \sqrt{|x|}.$

Note that if you have the inequality in the way $b^2\le x$ then actually you have $x\ge 0$ since $x\ge b^2\ge 0.$ So the argument of the square root is nonnegative.

The same argument doesn't work with complex numbers, because in the set of complex numbers there is no a order compatible with the usual operations.

mfl
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