My Math Forum square root of -1

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

April 7th, 2014, 10:53 PM   #1
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square root of -1

What I would say is that in the first instance $-1=(\sqrt{-1})^2=\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)}$ we took -1 as the square root of 1

In the last step $\sqrt{1}=1$ we took +1 as the square root of 1

In a way the equality -1 = 1 is "TRUE" in the sense that (-1)^2 = (1)^2 but of course $-1\neq 1$

Is my explanation correct? is there a better way to explain this error in reasoning?
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 April 7th, 2014, 10:58 PM #2 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra The rule $\sqrt a\sqrt b=\sqrt{ab}$ applies only to nonnegative real numbers. Applying it to $-1$ isn’t valid. Thanks from shunya
 April 8th, 2014, 08:06 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra $i\sqrt{6} = \sqrt{-2}\sqrt{3} = \sqrt{-6} = i\sqrt{6}$ The equality applies to all complex numbers. I think Shunya's answer is more or less spot on. Th equation $a^2 = b^2$ has two solutions $a = b$ and $a = -b$ (or more succinctly $|a| = |b|$. But $a$ cannot both positive and negative in one equation, it it either one or the other. Similarly for $b$. You have to be careful that the roots are valid for the problem that you are solving. Usually we write that $\sqrt{a^2} = a$, the positive root. If we want the negative root we explicitly say so, using $-\sqrt{a^2} = a$ or if both are permissible $\pm\sqrt{a^2} = a$. Thanks from shunya

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