Context Free Grammar for {a^ib^j | i,j ≥ 0; i ≠ 2j}
Consider the two languages:
$L_1 = \{a^ib^j \mid i, j \ge 0 \text{ and } i &gt; 2j\}$
$L_2 = \{a^ib^j \mid i, j \ge 0 \text{ and } i &lt; 2j\}$

Convince yourself that $L = L_1 \cup L_2$.

In $L_1$, the number of $a$'s are more than double of $b$'s, so there has to be atleast one $a$ (when there are no $b$'s). Also, for every addition of $b$, atleast 2 $a$'s must be added. You can generate $L_1$ as:

$
S_1 \rightarrow aA \\
A \rightarrow aaAb \mid aA \mid \varepsilon
$

$L_2$ is a little more tricky. The number of $a$'s are less than double the number of $b$'s, so there can be $0$ $a$'s, but non-zero $b$'s. Consider the "base case" of $1$ $b$. The string can be either $b$ or $ab$. We will let the first rule generate this base case. After that, notice that for every addition of $b$, we can increase the number of $a$'s by at most $2$. So we can add $0$, $1$, or $2$ $a$'s for every addition of $b$. We'll let the second rule handle this. So, the CFG for $L_2$ becomes:

$
S_2 \rightarrow Bb \mid aBb \\
B \rightarrow Bb \mid aBb \mid aaBb \mid \varepsilon
$

Note that CFG's are closed under union, that is, the union of two CFG's is also a CFG. So, to get the CFG for $L$, let the starting state $S$ of $L$ lead to either the starting state of $L_1$, or of $L_2$:

$S \rightarrow S_1 \mid S_2$

The rest of the rules remain the same as in the two languages. There may be a simpler grammar, but this was the first that came to mind.