# Getting i^3 = -i using laws of surds

 P: 13 I think Tobias is talking about my recent threads where I showed my resistance to the supposed fact that $\sqrt{ab} = \sqrt{a} \sqrt{b}$ doesn't always hold. If it makes no sense that $i$ is positive in $i^2$=-1, then $i$ is negative. But does it make sense that a negative number $\times$ a negative number is equal to a negative number ?
P: 2,803
 Quote by AllyScientific I think Tobias is talking about my recent threads where I showed my resistance to the supposed fact that $\sqrt{ab} = \sqrt{a} \sqrt{b}$ doesn't always hold. If it makes no sense that $i$ is positive in $i^2$=-1, then $i$ is negative. But does it make sense that a negative number $\times$ a negative number is equal to a negative number ?
The problem here is that you are applying Real number notions that you learned to an imaginary number i and by extension to the complex number plane where i can be +i or -i.
P: 820
 Quote by AllyScientific If it makes no sense that $i$ is positive in $i^2$=-1, then $i$ is negative.
No. 0 is not positive. Does that mean 0 is now negative?

 But does it make sense that a negative number $\times$ a negative number is equal to a negative number ?
i is not positive nor is it negative. It's signless, just like zero.
P: 13
 Quote by pwsnafu i is not positive nor is it negative. It's signless, just like zero.
-$i$ is negative and +$i$ is positive.

(-$i$)*(-$i$) = (+$i$)*(+$i$) = $i^2$ =-1

Does it make sense now? A negative number $\times$ a negative number = a negative
number.
P: 15
 Quote by Mark44 I'll go out on a limb here and posit that what E_Q really meant but didn't get across clearly was this, "It's a mistake to think that √4=±2, that often doesn't get picked up on..."
Sorry D H, ultra-bad phrasing. Cheers Mark, you must have got used to my flawed writing ^^

@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 i2 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 i2 OR (-i)2 both represent a transformation of 180° - the direction is irrelevent.
P: 15
 Quote by AllyScientific (-i)*(-i) = (+i)*(+i) = i2 =-1 A negative number × a negative number = a negative number.
 Quote by pwsnafu i is not positive nor is it negative. It's signless, just like zero.
et voila
P: 820
 Quote by AllyScientific -$i$ is negative and +$i$ is positive.
False. i is signless. Similarly -i is also signless.

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!
P: 13
 Quote by pwsnafu You are starting with a false premise: "i must be either positive or negative". Why do you keep persisting with it?
I suspect that the mathematicians themselves are starting with a false premise, because they
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.
P: 820
 Quote by AllyScientific I suspect that the mathematicians themselves are starting with a false premise, because they are breaking the law of signs.
The "law" you are inferring to is "if ##x## is a negative number and ##y## is a negative number then ##xy## is positive". Am I correct?
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.

 Need I add that they are not infallible? They are breaking their own law, should I not have right to say it ?
They aren't.

 My point is that we should start with a right premise whatever it may be.
They are, and you are not.
 Mentor P: 21,215 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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