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 March 1st, 2017, 11:50 AM #1 Newbie     Joined: Feb 2017 From: Netherlands Posts: 8 Thanks: 2 Math Focus: Trigonometry and complex numbers Square rooting to gain negative integers Are there any (complex, real etc.) numbers that satisfy the following equation? $\displaystyle \sqrt{x} < 0$
 March 1st, 2017, 01:38 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2203 No. Thanks from Xxmarijnw
March 1st, 2017, 04:21 PM   #3
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Quote:
 Originally Posted by Xxmarijnw Are there any (complex, real etc.) numbers that satisfy the following equation? $\displaystyle \sqrt{x} < 0$
The square root operation always has two solutions. However by custom the square root of a positive real number is the positive root. However the negative square root is there, but by custom is ignored.

March 1st, 2017, 06:21 PM   #4
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Quote:
 Originally Posted by mathman The square root operation always has two solutions. However by custom the square root of a positive real number is the positive root. However the negative square root is there, but by custom is ignored.
I express this differently.

A positive real number always equals the square of two distinct numbers, one positive and one negative but with equal absolute values.

By definition, $\sqrt{a} > 0\ if\ a > 0.$

So what is the negative root? $-\ \sqrt{a}.$

 March 11th, 2017, 05:30 PM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 "The" square root, of a positive real number, a, is defined to be the positive number whose square is a. But this question is asking for a complex number either of whose square roots is a negative real number. Writing the complex number in polar form, $a= r e^{i\theta}$ where r is a positive real number and $\theta$ is between 0 and $2\pi$, the square roots are $r^{1/2} e^{i\theta/2}$ and $r^{1/2}e^{i(\theta/2+ \pi)}$. Either of those is a negative real number only if $\theta/2= \pi$ or $\theta/2+ \pi= \pi$. Those give $\theta= 2\pi$ and $\theta= 0$ which are really the same positive real number. But that puts us back to the "positive real number" case where $\sqrt{a}$ is, by definition, the positive root. Thanks from Xxmarijnw

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