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 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 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?
Attached Images Capture2.PNG (12.3 KB, 5 views) 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 Tags root, square Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post TwoTwo Algebra 43 December 1st, 2013 02:08 AM RichardP Algebra 5 June 12th, 2012 02:28 PM HellBunny Algebra 7 April 15th, 2012 07:06 PM jared_4391 Algebra 3 August 8th, 2007 09:06 AM rain Academic Guidance 1 December 31st, 1969 04:00 PM

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