How can I change height of line break in math mode? I need to add more space between each two lines of the equation.  \begin{gather*}     EFF_a = \sum_{j=0}^{2} (\frac{ \sum_{i=1}^{4} (EFF_{{a_j}_i} = {\{ p_i; q_i}\}  \rightarrow p_i \cdot q_i)}{ \sum_{i=1}^{4} q_i } \cdot W_{EFF_{a_j})} =     \\ = \frac{1 \cdot 15 + 2 \cdot 15 + (-1) \cdot 2}{15+15+2} \cdot 0.5 + \frac{1 \cdot 314 + 2 \cdot 101 + (-1) \cdot 33 + (-2) \cdot 4}{314+101+33+4} \cdot 0.4 +     \\ + \frac{1 \cdot 23 + (-1) \cdot 33 }{23+33} \cdot 0.1 = 1.0743718394437423 \end{gather*}   This produces:
Is there a way to consistently increase the vertical spacing between equation lines in a gather (or gather*) environment from amsmath?  Specifically, I want to produce something like  \begin{gather*} a = b \\[2ex] c = d \\[2ex] e = f \end{gather*}   without having to write [2ex] at the end of each equation line.  I can think of some hacks that redefine \\ to automatically add the spacing, but is there some cleaner/better way to achieve this?
What my thought was that the bijection was a piecewise function:  $f(x) = 1/n+1$ if $x = 1/n$ for some $n \in \Bbb N$ and $x=1/n$ if $x \neq1/n$ for some $n \in \Bbb N$.  However, the textbook says this is incorrect. I don't actually see why. Was it a matter of brackets or the open variable $(0,1]$, which is what I tend to think. Inequality about infinite cardinals The following is a question I'm interested in solving:    The idea is to solve it without using AC. As @Asaf Karagila pointed out in the comment section:  "Even if $\kappa$ is an infinite cardinal, you can't assume that $\kappa \geq \omega$ without using AC".  I am still trying to think this through: My reasoning being, that as $\kappa &gt; n \forall n \in \omega$, then $\kappa \geq \sup\{n:n\in \omega\}=\omega$ as $\omega$ is a limit ordinal.  Where does my reasoning fail? And also, a detailed solution would be very appreciated. I'm not sure how to prove these basic theorems in propositional calculus. 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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 $i^2$. Indexing redundantly named vector in R