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February 1st, 2017, 01:10 PM   #1
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basics --> i*i

(1) $\displaystyle x^2+1=0$

(2) $\displaystyle x^2=-1$

(3) $\displaystyle x=i$

(4) $\displaystyle i=\sqrt{-1}$

(5) $\displaystyle i\cdot i=\sqrt{-1}\cdot \sqrt{-1\:}=\sqrt{\left(-1\right)\cdot \left(-1\right)}=\sqrt{\left(1\right)}=1$

which line is false?
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February 1st, 2017, 02:08 PM   #2
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line (4) has no meaning.

the domain of the square root function is $[0, \infty)$
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February 1st, 2017, 02:44 PM   #3
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Also line 3 should be $x = \pm i$.

And line 5 is not valid if the numbers inside the radical sign aren't real.

Last edited by Maschke; February 1st, 2017 at 02:47 PM.
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February 1st, 2017, 05:44 PM   #4
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Originally Posted by Maschke View Post
And line 5 is not valid if the numbers inside the radical sign aren't real.
I think you mean "if the numbers inside the radical sign aren't negative".
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February 2nd, 2017, 02:07 PM   #5
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This is why it is better not to "define" i as sqrt(-1) (in addition every number has two square roots and there no good way to distinguish between them in the complex numbers).

Better is to define the complex numbers as pairs of real numbers, (a, b), with addition defined "coordinate wise", (a, b)+ (c, d)= (a+ c, b+ d) and multiplication defined by [tex](a, b)*(c, d)= (ac- bd, ad+ bc). Then, since (a, 0)+ (c, 0)= (a+ c, 0) and (a, 0)*(c, 0)= (ac, 0) we can identify the real number, a, as the pair of the form (a, 0).

Also (0, 1)*(0, 1)= (0*0- 1*1, 0*1+ 1*0)= (-1, 0). We identify (-1, 0) with -1 so if we define "i" to be (0, 1) we have i*i= -1 and can write (a, b)= a(1, 0)+ b(0, 1)= a+ bi.
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