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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: 17,908 Thanks: 1382 
No.

March 1st, 2017, 04:21 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,305 Thanks: 525  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  
Senior Member Joined: May 2016 From: USA Posts: 785 Thanks: 311  Quote:
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: 2,639 Thanks: 678 
"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, where r is a positive real number and is between 0 and , the square roots are and . Either of those is a negative real number only if or . Those give and which are really the same positive real number. But that puts us back to the "positive real number" case where is, by definition, the positive root. 

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gain, integers, intergers, negative, rooting, square 
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