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Getting i^3 = i using laws of surds 
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#19
Nov913, 01:46 AM

P: 13

I think Tobias is talking about my recent threads where I showed my resistance to the supposed fact that [itex]\sqrt{ab} = \sqrt{a} \sqrt{b} [/itex] doesn't always hold.
If it makes no sense that [itex]i[/itex] is positive in [itex]i^2[/itex]=1, then [itex]i[/itex] is negative. But does it make sense that a negative number [itex]\times[/itex] a negative number is equal to a negative number ? 


#20
Nov913, 01:52 AM

P: 2,990




#21
Nov913, 02:24 AM

Sci Advisor
P: 834




#22
Nov913, 04:04 AM

P: 13

([itex]i[/itex])*([itex]i[/itex]) = (+[itex]i[/itex])*(+[itex]i[/itex]) = [itex]i^2[/itex] =1 Does it make sense now? A negative number [itex]\times[/itex] a negative number = a negative number. 


#23
Nov913, 04:14 AM

P: 15

@AllyScientific: The equation you just wrote didn't need i in; (x)^{2}=(x)^{2}. So there should be no surprise. You're forgetting perhaps that i=√1; you've sort of cheated as your answer 1 is real, wheras your equation i^{2} is imaginary. I don't think it's possible to claim the laws of signs have been broken... Remember the argand diagram; multiplying by i represents a transformation (of (1,0), if you like) 90° ACW. i is 90° CW. Therefore i^{2} OR (i)^{2} both represent a transformation of 180°  the direction is irrelevent. 


#24
Nov913, 04:17 AM

P: 15




#25
Nov913, 04:26 AM

Sci Advisor
P: 834

You are starting with a false premise: "i must be either positive or negative". Why do you keep persisting with it? I demonstrated it's not even true on the reals! 


#26
Nov913, 05:04 AM

P: 13

are breaking the law of signs. Need I add that they are not infallible? They are breaking their own law, should I not have right to say it ? My point is that we should start with a right premise whatever it may be. 


#27
Nov913, 05:17 AM

Sci Advisor
P: 834

But ##i## is defined to be signless, hence not negative, therefore ##i \times i = 1## doesn't break that law. Similarly, ##i## is defined to be signless, hence not negative, therefore ##(i) \times (i) = 1## doesn't break that law either. 


#28
Nov913, 12:29 PM

Mentor
P: 21,286

Since the OP's concerns have been addressed, and the thread has drifted off into nonsense, I am closing this thread.



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