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Does i^2 equals to -1 or 1?

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Space_0
"i" in math stands for square root of -1,


if we use the common way of solving the question, it would be (√-1)^2, cancelling both root and exponent to get -1


But if we use the square root rule, we'll get √-1 * √-1, which equals to √(-1 * -1), which ultimately equals to √1 and thus 1.


So which is it?
ClevelandsMyBro
ok disclaimer; im still a math nerd noob so i may suck explaining this, but...

Space_0 wrote:

But if we use the square root rule, we'll get √-1 * √-1, which equals to √(-1 * -1), which ultimately equals to √1 and thus 1.
is wrong. to explain it simply, the rule for radicals is that only positive factors can go in or come out of them.

square roots are a factor of a number that when multiplied by itself gives the original number. √-1 doesn't have any solutions because;
(-1) x (-1) = 1
which contradicts the "gives the original number" part.
Karmine

ClevelandsMyBro wrote:

ok disclaimer; im still a math nerd noob so i may suck explaining this, but...

Space_0 wrote:

But if we use the square root rule, we'll get √-1 * √-1, which equals to √(-1 * -1), which ultimately equals to √1 and thus 1.
is wrong. to explain it simply, the rule for radicals is that only positive factors can go in or come out of them.

square roots are a factor of a number that when multiplied by itself gives the original number. √-1 doesn't have any solutions because;
(-1) x (-1) = 1
which contradicts the "gives the original number" part.
Yeah, that's why i is an imaginary number
Corne2Plum3
The function square root is only defined for positive real numbers, you can't use √ for negative or complex numbers
ClevelandsMyBro
also i love and hate myself for learning imaginary numbers just for a post in an online forum. my school haven't reached it yet lmao.
WitherMite
roots can also be written as fractional exponents. In the case of square roots, 1/2, so

i^2 = (-1^1/2)^2 = (-1^1/2)(-1^1/2) = -1^1 = -1

square root rule is a simplified shortcut of doing the above for real numbers, as that math gets more complicated when the numbers in the roots are different. it works because all real squares are positive, so the roots can be pairs of positive or negative numbers that you can do math on. and doesnt work on complex numbers, where the square is negative and has imaginary roots you can't do math on.

examples:

(3^1/2)^2 = (3^1/2)(3^1/2) = 3^1 = 3

can also be solved as

x = (3^1/2)(3^1/2)
x^2 = [(3^1/2)(3^1/2)]^2
x^2 = 3*3
x = (3*3)^1/2
square root rule is these three above steps at once.
x = √9
x = 3

with different, non square bases you have to take the algebra route as you cant easily simplify them:

x = (5^1/2)(2^1/2)
x^2 = [(5^1/2)(2^1/2)]^2
x^2 = 5*2
x = (5*2)^1/2
x = √10

eventually you have to do math with the root if they are different, which you cant do with imaginary numbers, because of this you cant get the -1 out of the root like in step two of the above examples. all you can do with i is simplify to isolate it so you can do math on the real part of complex numbers. i^2 just happens to be a real number, -1.
abraker

Space_0 wrote:

But if we use the square root rule, we'll get √-1 * √-1, which equals to √(-1 * -1), which ultimately equals to √1 and thus 1.
This is wrong

√x * √x is not equal to √(x * x) for negative numbers

If you use the relation √x = x^0.5

then,

√-1 * √-1 = (-1)^0.5 * (-1)^0.5 = (-1)^(0.5 + 0.5) = (-1)^1 = -1
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