We know $$i^2=-1 $$then why does this happen? $$ i^2 = \sqrt{-1}\times\sqrt{-1} $$ $$ =\sqrt{-1\times-1} $$ $$ =\sqrt{1} $$ $$ = 1 $$  EDIT: I see this has been dealt with before but at least with  I'm not making the fundamental mistake of assuming an incorrect definition of $i^2$.
4300
I know there must be something unmathematical in the following but I don't know where it is: \begin{align} \sqrt{-1} &amp;= i \\\\\ \frac1{\sqrt{-1}} &amp;= \frac1i \\\\  \frac{\sqrt1}{\sqrt{-1}} &amp;= \frac1i \\\\  \sqrt{\frac1{-1}} &amp;= \frac1i \\\\  \sqrt{\frac{-1}1} &amp;= \frac1i \\\\  \sqrt{-1} &amp;= \frac1i \\\\  i &amp;= \frac1i \\\\  i^2 &amp;= 1 \\\\  -1 &amp;= 1 \quad !!? \end{align}
27112
Square root of number -1 defined as i, then what is the square root of complex number i?, I would say it should be j as logic suggests but it's not defined in quaternion theory in that way, am I wrong?  EDIT: my question is rather related to nomenclature of definition, while square root of -1 defined as i, why not j defined as square root of i and k square root of j and  if those numbers have deeper meanings and usage as in quaternions theory.
7879
I'm trying to solve this equation:  $$2-x=-\sqrt{x}$$ Multiply by $(-1)$: $$\sqrt{x}=x-2$$ power of $2$: $$x=\left(x-2\right)^2$$ then: $$x^2-5x+4=0$$ and that means: $$x=1, x=4$$    But $x=1$ is not a correct solution to the original equation.  Why have I got it? I've never got a wrong solution to an equation before. What is so special here?
36223
Is there any $2 \times 2$ matrix $A \neq I$ such that $A^3=I$?  In my opinion: no. Thank you very much.
95028
What can be said about$\ f^{-1}(f(S))$ and$\ S$, where$\ f(S):=\{f(x):x \in S\}$?   I feel as if$\ f^{-1}(f(S))$ will be smaller than or equal in size (number of elements) to$\ S$.  For example, if$\ f(x)=x$, then$\ f(\{-1, 0, 1, 2\})=\{-1, 0, 1, 2\}$and$\ f^{-1}(f\{-1, 0, 1, 2\})=\{-1, 0, 1, 2\}$ (same size).  But if$\ f(x)=x^2$, then$\ f(\{-1, 0, 1, 2\})=\{0, 1, 4\}$ and$\ f^{-1}(f\{0, 1, 4\})=\{0, 1, 2\}$, which is a little bit smaller than$\ S$.   What can be said about$\ f(f^{-1}(U))$ and$\ U$, where$\ f^{-1}(U):=\{x \in X:f(x) \in U\}$?   The same size as each other? Always??
24628
What is $i^i$ and $(i^i)^i$. I fond some answers  but is there a clearer explanation of this? Maybe something help me to understand.
34341
What's wrong with this : $$e^{i\pi} = -1$$ $$\therefore e^{2i\pi} = 1$$ $$\therefore \log \left( e^{2i\pi} \right ) = \log(1) = 0$$ $$\therefore 2i\pi = 0$$
6893
this is something that came up when working with one of my students today and it has been bothering me since.  It is more of a maths question than a pedagogical question so i figured i would ask here instead of MESE.  Why is $\sqrt{-1} = i$ and not $\sqrt{-1}=\pm i$?  With positive numbers the square root function always returns both a positive and negative number, is it different for negative numbers?
85708
How can we think of the set$$\{x \in \mathbb{H} : x^2 = -1\}$$geometrically? Is this set finite or infinite? Are there some more geometric ways of thinking about than meets the eye? Here, $\mathbb{H}$ denotes the quaternions.
54348
Is it wrong to say $$ \sqrt{x} \times \sqrt{x} =\sqrt{x^2}= \pm x$$ I am quite sure that  $\sqrt{(x)^2} = \pm(x)$  But, does $\sqrt{x } \times \sqrt{x} =- (x)$ doesn't holds in $\mathbb{R}$ but if we assume $\mathbb{C}$ it holds right?
31848
I understand that the complex numbers have a geometric representation as ordered pairs of real numbers. From what I understand complex numbers are very useful for working in two dimensions mathematically. The fact that $i^2 = -1$ follows from the definition of multiplication of complex numbers:  $(a,b)(c,d) = ac-bd+(bc+ad)i$  My question is this: If one was simply working with ordered pairs of real numbers with no knowledge of $i=\sqrt{-1}$, would this definition of multiplication ever follow logically from a practically arising situation?
9602
If $A$ and $B$ are nonempty sets, prove that $A \times B = B \times A$ if and only if $A = B$.   Proving the first direction of this is easy. That is, if $A = B$ then it is obvious that $A^2 = A^2$. I am wondering how to prove the other direction now. Maybe proving it by contradiction will be easier than directly.
71394
when I apply $f^{-1}$ to both sides $f(f(x))=f(x)$ $(1)$ we get $f(x)=x$ $(2)$.  are $1$ and $2$ equivalent?
79941
With $i^2=-1$, what is $i!$ ?  I'm not sure if it is $i$ or if it is an infinite product $i(i-1)(i-2)(i-3)(i-4)(i-5)\dots$  It would make sense either way.  Is it not defined?
3981
As far as I see, $\sqrt{x^2}$ is not $x$, but $|x|$, meaning the "absolute". I totally get this, because $x^2$ is positive, if $x$ is negative, so $\sqrt{y}$, whether $y = 10^2$ or $y = -10^2$: $y$ is positive.  But then I remember that $\sqrt(x)$ is the same as $x^{1/2}$ and thus, $\sqrt{x^2}$ is the same as $x^{1}$.   So, as far as I get it, $\sqrt{x^2} = x^{2/2}$.  But then, I can cancel $\frac{2}{2}$ to $1$. So: $\sqrt{x^2} = x^1 = x$.  Where does the absolute value in this derivation of the calculation come from?  I get it why its there: Because $(-x)^2 = x^2$, so the radicand is always positive. But from $\sqrt{x^n} = x^{n/2}$ I dont understand how it gets there.
90400
I was trying to solve for $i^i$.  I got distracted and did the following: $e^{i \theta} = \cos(\theta) + i\sin(\theta)$ $e^{i 2\pi} = cos(2\pi) + i\sin(2\pi) = 1$ ||| Take ln() of two sides of equation: $\ln(e^{i2\pi}) = \ln(1)$ $i2\pi = 0$ Clearly this is false. What did I do wrong?
71434
How can I show $i^{-1} = -i$, where $i$ is the imaginary unit?  Here's what I've tried: $i^{-1} = (-1)^{-1/2} = \dots ?$