what is the sign of √-1 ?

Question

what is the sign of √-1 ?

Additional Details

4 minutes ago
tell me if its + or -

2 minutes ago
I know that √-1=i !!!!!!!
I want to know if i is positive or negative !!!!!

Answer 1

Positive ... or it may be negative. Both are correct.
Please read reference provided

Answer 2

The question itself is wrong. The square root of minus one is not a real number, but a complex number. The set of real number is called an ordered (sorry, I'm not a native speaker, I mean a set with an order) set, because two numbers are either equal or one is larger then the other. That allows us to define a ">" and a "<" sign and finally we define positive numbers to be >0 and negative numbers to be <0.

Complex numbers are not an ordered set, we can't define any useful "<" or ">", and so we don't have positive or negative complex numbers. The real and imaginary parts of a complex number are again real numbers and these have a sign, but the complex number itself does not.

to or_try_this: The coefficient of i is the imaginary part 1 of the complex number i = 0 * 1 + 1 * i. The real and imaginary part are real numbers, so the *imaginary part* of i has a positive sign, but *not* the complex number. I understand the confusion, because you can change the *relative* sign of a complex number by changing the sing of it's real an imaginary part (z = a + b * i becomes -z = -a - b * i) but that doesn't help you to define an order, you still don't know if i is bigger or less then 0. You can do wicked things with sings in the complex space, for example 1/i = i/(i*i) = i/(-1) = -i, so inverting i according to the product changes it's sign (i * (-i) = 1, 1 is the neutral element of the multiplication: 1 * a = a), but normally (in R, Q, and Z) the minus sign is the additive inverse (a + (-a) = 0, 0 is the neutral element of the addition: a + 0 = a).

So C simply doesn't fulfill some of the rules R, Q and Z do, and one of them is that you simply can't define an order. There is even an extension to C, called the "quaternions" (in which not just the i exists but also a "j" and a "k") in which a*b=b*a isn't true anymore.

Answer 3

Positive

Answer 4

SHORT ANSWER:

it can be EITHER as

-1 x -1 = 1

&

1 x 1 = 1

LONG ANSWER see: Wikipedia

i and − i
Being a 2nd order polynomial with no multiple real root, the above equation has two distinct solutions that are equally valid and that happen to be additive inverses of each other. More precisely, once a solution i of the equation has been fixed, the value −i ≠ i is also a solution. Since the equation is the only definition of i, it appears that the definition is ambiguous (more precisely, not well-defined). However, no ambiguity results as long as one of the solutions is chosen and fixed as the "positive i". This is because, although −i and i are not quantitatively equivalent (they are negatives of each other), there is no qualitative difference between i and −i (that cannot be said for −1 and +1). Both imaginary numbers have equal claim to square to −1. If all mathematical textbooks and published literature referring to imaginary or complex numbers were rewritten with −i replacing every occurrence of +i (and therefore every occurrence of −i replaced by −(−i) = +i), all facts and theorems would continue to be equivalently valid. The distinction between the two roots x of x2 + 1 = 0 with one of them as "positive" is purely a notational relic; neither root can be said to be greater than the other.

Answer 5

When ever you solve a second order equation, you obtain 2 solutions . One is imaginary and the other is real.
So the square root of of 1 will give a + real answer an imaginary answer which is denoted as "i".
1^2 = 1
minus 1^2 =1 so the square root of a negative number is imaginary.
i^2 = -1
i can be plus or minus. and folows the same rules as any other numbers.

Answer 6

By my reckoning, i is neither positive nor negative, but:

the coefficient of i is positive
the coefficient of +i is positive
the coefficient of -i is negative

sqrt(-1) has two answers, +i and -i, and can therefore have either a positive or negative coefficient

i is conventionally defined as the sqrt(-1) having a positive coefficient

Hmm, do you suppose i was first defined by a right-hander or a left-hander?

Answer 7

Whenever the radical is alone, it is positive. So the thing you have in your question (I can't figure out how to make the square root sign in here) is +i.

Answer 8

The root of minus one actually doesnt exist because squares are always +ve ; so i doesnt have any sign it depends on the eqn.