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April 7th, 2014, 10:53 PM  #1 
Senior Member Joined: Oct 2013 From: Far far away Posts: 422 Thanks: 18  square root of 1
What I would say is that in the first instance we took 1 as the square root of 1 In the last step 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 Is my explanation correct? is there a better way to explain this error in reasoning? 
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.

April 8th, 2014, 08:06 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,634 Thanks: 2620 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 $\pm\sqrt{a^2} = a$. 

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