Getting i^3 = -i using laws of surds


by E_Q
Tags: algebra, imaginary numbers, laws, problem, surds
AllyScientific
AllyScientific is offline
#19
Nov9-13, 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 ?
jedishrfu
jedishrfu is offline
#20
Nov9-13, 01:52 AM
P: 2,477
Quote Quote by AllyScientific View Post
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 ?
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.
pwsnafu
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#21
Nov9-13, 02:24 AM
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Quote Quote by AllyScientific View Post
If it makes no sense that [itex]i[/itex] is positive in [itex]i^2[/itex]=-1, then [itex]i[/itex] is negative.
No. 0 is not positive. Does that mean 0 is now negative?

But does it make sense that a negative number [itex]\times[/itex] a negative number is equal
to a negative number ?
i is not positive nor is it negative. It's signless, just like zero.
AllyScientific
AllyScientific is offline
#22
Nov9-13, 04:04 AM
P: 13
Quote Quote by pwsnafu View Post
i is not positive nor is it negative. It's signless, just like zero.
-[itex]i[/itex] is negative and +[itex]i[/itex] is positive.

(-[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.
E_Q
E_Q is offline
#23
Nov9-13, 04:14 AM
P: 15
Quote Quote by Mark44 View Post
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.
E_Q
E_Q is offline
#24
Nov9-13, 04:17 AM
P: 15
Quote Quote by AllyScientific View Post

(-i)*(-i) = (+i)*(+i) = i2 =-1

A negative number a negative number = a negative
number.
Quote Quote by pwsnafu View Post



i is not positive nor is it negative. It's signless, just like zero.
et voila
pwsnafu
pwsnafu is offline
#25
Nov9-13, 04:26 AM
Sci Advisor
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Quote Quote by AllyScientific View Post
-[itex]i[/itex] is negative and +[itex]i[/itex] 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!
AllyScientific
AllyScientific is offline
#26
Nov9-13, 05:04 AM
P: 13
Quote Quote by pwsnafu View Post
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.
pwsnafu
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#27
Nov9-13, 05:17 AM
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Quote Quote by AllyScientific View Post
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.
Mark44
Mark44 is offline
#28
Nov9-13, 12:29 PM
Mentor
P: 21,008
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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