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 March 3rd, 2014, 08:09 PM #1 Senior Member     Joined: Oct 2013 From: Far far away Posts: 431 Thanks: 18 i am imaginary $i=\sqrt{-1}$ Is i < or > or = 1? 1) Since i is not on the real number line this question is meaningless ON THE OTHER HAND 2) Since $i^{2}=-1$ and -1 < 1 it follows that i < 1 What is the right answer?
 March 4th, 2014, 11:54 AM #2 Global Moderator   Joined: Dec 2006 Posts: 21,026 Thanks: 2257 If i is not a real number, none of the choices mentioned in the question applies. The second suggestion is incorrect, as what would follow is i² < 1, which doesn't imply i < 1.
March 4th, 2014, 12:06 PM   #3
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Re: i am imaginary

Quote:
 Originally Posted by shunya $i=\sqrt{-1}$ Is i < or > or = 1? 1) Since i is not on the real number line this question is meaningless ON THE OTHER HAND 2) Since $i^{2}=-1$ and -1 < 1 it follows that i < 1 What is the right answer?
i cannot be negative because negative times negative number is positive number.
i cannot be positive because positive number times positive number is positive number.

i is neither negative nor positive.

But what are +i and -i ?
Is +i positive and -i negative?

 March 4th, 2014, 01:03 PM #4 Global Moderator   Joined: Dec 2006 Posts: 21,026 Thanks: 2257 Both +i (which equals i) and -i are imaginary, not real. Hence neither should be called positive or negative. However, the imaginary part of +i is 1 (which is positive) and the imaginary part of -i is -1 (which is negative).
 March 4th, 2014, 01:29 PM #5 Senior Member   Joined: Nov 2013 Posts: 160 Thanks: 7 Re: i am imaginary How is this possible? Does i have two values: $i= \sqrt{-1}$ $i= - \sqrt{-1}$ Because $i^2= (\sqrt{-1})^2 = (- \sqrt{-1})^2 =((-1)^2 \cdot \sqrt{-1})^2 = -1$
 March 5th, 2014, 01:49 AM #6 Global Moderator   Joined: Dec 2006 Posts: 21,026 Thanks: 2257 No, just as 3² = (-3)² = 9 doesn't mean that 3 has two values. Like any non-zero number, -1 has two square roots, i and -i. A positive real has one positive and one negative square root. Any other non-zero number has one square root with a positive imaginary part and one square root with a negative imaginary part.
March 5th, 2014, 02:30 AM   #7
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Re:

Quote:
 Originally Posted by skipjack No, just as $3^2= (-3)^2 = 9$ doesn't mean that 3 has two values. Like any non-zero number, -1 has two square roots, i and -i.
If we write $(+3)^2= (-3)^2 = 9$, why this could not be understood that
3 has two values +3 and -3. Isn't it only a general agreement that 3=+3 ? Why we could not instead use a convention
that 3=-3 ?

If -1 has two square roots, i and -i
$i= \sqrt{-1}$
$-i= \sqrt{-1}$
it is the same as
$i= \sqrt{-1}$
$i= - \sqrt{-1}$
meaning that i has two values: +i and -i, which we can plot on the complex plane on the opposite sides
of unit circle.

Last edited by TwoTwo; March 28th, 2014 at 11:33 PM.

 March 5th, 2014, 03:14 AM #8 Global Moderator   Joined: Dec 2006 Posts: 21,026 Thanks: 2257 With such a convention, adding 3 = 3 to 3 = -3 would give 6 = 0, etc., which doesn't seem useful. In the case of +i and -i, you have acknowledged these correspond to opposite sides of a unit circle, and this wording acknowledges that the opposite sides are not the same side, so it would be confusing to hold that +i and -i are the same or equal.

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