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March 1st, 2017, 11:50 AM   #1
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Math Focus: Trigonometry and complex numbers
Square rooting to gain negative integers

Are there any (complex, real etc.) numbers that satisfy the following equation?

$\displaystyle \sqrt{x} < 0$
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March 1st, 2017, 01:38 PM   #2
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No.
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March 1st, 2017, 04:21 PM   #3
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Quote:
Originally Posted by Xxmarijnw View Post
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.
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March 1st, 2017, 06:21 PM   #4
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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? $-\ \sqrt{a}.$
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March 11th, 2017, 05:30 PM   #5
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"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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