February 1st, 2017, 12:10 PM  #1 
Member Joined: Oct 2016 From: Slovenia, Europe Posts: 52 Thanks: 5  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? 
February 1st, 2017, 01:08 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,849 Thanks: 957 
line (4) has no meaning. the domain of the square root function is $[0, \infty)$ 
February 1st, 2017, 01:44 PM  #3 
Senior Member Joined: Aug 2012 Posts: 1,850 Thanks: 508 
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 01:47 PM. 
February 1st, 2017, 04:44 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,268 Thanks: 2434 Math Focus: Mainly analysis and algebra  
February 2nd, 2017, 01:07 PM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,094 Thanks: 846 
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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