Please point out what is wrong with this calculation:
$i^2= (i) (i) $
$=\sqrt{-1} \sqrt{-1} $
$= (-1)^{1/2} (-1)^{1/2} $
$=((-1)(-1))^{1/2}$
$=1^{1/2}$
$=1$
Asked
Active
Viewed 78 times
-1
-
It was a typo which has been corrected. Please look at the rest of the calculation – Alpha Centaur Aug 14 '20 at 08:18
-
$(-1)^{1/2} (-1)^{1/2} \neq ((-1)(-1))^{1/2}$. This only works for positive real numbers. – Eminem Aug 14 '20 at 08:20
-
Beware that $i$ is not iota. $\iota$ is iota. – Aug 14 '20 at 08:21
-
This was asked a billion times here. In the complex $a^b a^c=a^{b+c}$ is not always true. – Aug 14 '20 at 08:22
-
Never forget that most complex numbers have two square roots. $i$ is a square root of $-1$ and so $i^2$ is a square root of $1$, but maybe not the one you want it to be. – Angina Seng Aug 14 '20 at 08:25
-
Thanks everyone – Alpha Centaur Aug 14 '20 at 08:26
-
The imaginary number $i$ is "defined" by the relation $i^2=-1,$ where $i$ is a conventional symbol and $-1$ a real number. You cannot assume that $i$ and the complex numbers automatically inherit the properties of the reals. You just found a counterexample. – Aug 14 '20 at 08:27
-
@AnginaSeng: positive real numbers also have two square roots, but we define the square root function which selects the positive root. A square root function can also be defined on the complex. – Aug 14 '20 at 08:30
1 Answers
0
It is due to this ambiguity the definition of separation of root is given as:
$$ \sqrt{ab} =\sqrt{a} \sqrt{b}$$
Only if, a > 0 & b > 0.
For example:
$ \sqrt{3.5}$
=$ \sqrt{(-3).(-5)}$
=$ \sqrt{(-3)}.\sqrt{(-5)}$
=$ \sqrt{(3)}i.\sqrt{(5)}i$
=$ \sqrt{(3)}.\sqrt{(5)}i.i$
=$ (-1)\sqrt{(3)}.\sqrt{(5)}$
=$(-1)\sqrt{3.5}$
Thus x = -x, which is wrong. Therefore separation of a root power is not defined for negative numbers.
-
when you write $\sqrt{-3}$ as $\sqrt{3} i$, are you not violating the rule that separation of root power is not defined for negative numbers ? – Alpha Centaur Aug 16 '20 at 07:43