My Math Forum Square root of minus one

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 November 28th, 2012, 03:28 AM #1 Newbie   Joined: Dec 2010 Posts: 8 Thanks: 0 Square root of minus one How is this possible? Am I doing something wrong? sqrt(-1) = (-1)^(1/2) = (-1)^(2/4) = 4th-root((-1)^2) = 4th-root(1) = 1
 November 28th, 2012, 03:55 AM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Square root of minus one $(1)^{2}=(-1)^{2}\not\Rightarrow 1=-1$
 November 28th, 2012, 04:23 AM #3 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: Square root of minus one Here is a common error : $1=\sqrt{1}= \sqrt{-1\cdot -1}=\sqrt{-1} \cdot \sqrt{-1}=i\cdot i =-1$ Another error : $-1=(\sqrt{-1})^2=\sqrt{(-1)^2}=\sqrt{1}=1$
November 29th, 2012, 10:16 PM   #4
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Re: Square root of minus one

Quote:
 Originally Posted by kustrle How is this possible? Am I doing something wrong? sqrt(-1) = (-1)^(1/2) = (-1)^(2/4) = 4th-root((-1)^2) = 4th-root(1) = 1
You won't get the wrong answer if you do the denominator first AND choose the appropriate 4th roots of minus 1 to square.

$\sqrt[4]{-1} \= \ \pm \frac{\sqrt{2}}{2} \ \pm\frac{\sqrt {2}}{2} i$

$$$\frac{\sqrt{2}}{2} \ + \ \frac{\sqrt{2}}{2} i$$^2 \= \ i$

$$$-\frac{\sqrt{2}}{2} \ - \ \frac{\sqrt{2}}{2} i$$^2 \= \ i$

Squaring the other 2 4th roots of minus 1 produces -i

Why don't you try it?

 November 30th, 2012, 06:01 AM #5 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Square root of minus one As with all numbers, -1 has two square roots. However, since the complex numbers do not form an "ordered" field, we cannot simply say "one is positive, the other negative". That is, we have to be very careful to distinguish between the two roots, else we run into the problems zaidalyafey notes.

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