Density of multiplication table Is there any easy way to show $$\lim_{N \to \infty} \frac{1}{N^2}\#\{ab : 1 \le a,b \le N\} = 0$$ A quick calculation I did shows that the  number of positive integers $\le N^2$ with a prime divisor $p > N$ is at most the order of $(\log 2) \cdot N^2$, so just getting rid of the numbers with a high prime divisor is not sufficient. 
 A: For $N \ge 1$, let $$E_N := \{n \le N : \omega(n) \in [\frac{9}{10}\log\log N,\frac{11}{10}\log\log N]\},$$ where $\omega(n) := \sum_{p \mid n} 1$. An easy double-counting / [second moment method] argument shows that $|E_N| = N-o(N)$ as $N \to \infty$. So, $$\frac{1}{N^2}\#\{ab : 1\le a,b \le N\} = \frac{1}{N^2}\#\{ab : a,b \in E_N\}+o(1).$$ Now, $$\omega(ab) = \omega(a)+\omega(b)-\sum_{p \mid a, p \mid b} 1,$$ so $a,b \in E_N$ and $ab \in E_{N^2}$ imply $\sum_{p \mid a, p \mid b} 1 \ge \frac{7}{10}\log\log N$. Since $|E_{N^2}| = N^2-o(N^2)$, we have $$\frac{1}{N^2}\#\{ab : 1 \le a,b \le N\} \le \frac{1}{N^2}\#\{ab : a,b \in E_N, \sum_{p \mid a, p \mid b} 1 \ge \frac{7}{10}\log\log N\}+o(1).$$ Fix $a \in E_N$. Suppose $b$ shares at least $\frac{7}{10}\log\log N$ prime factors with $a$. The number of $b \le N$ with prespecified $p_1,\dots,p_{\frac{7}{10}\log\log N}$ dividing $b$ is at most $\frac{N}{\prod_{i=1}^{\frac{7}{10}\log\log N} p_i}$, which, if $p_1,\dots,p_{\frac{7}{10}\log\log N}$ denote the first $\frac{7}{10}\log\log N$ primes, is at least $\exp(\frac{6.99}{10}\log\log N\log\log\log N)$ by the prime number theorem. Therefore, since $a$ has at most $\frac{11}{10}\log\log N$, the number of $b \le N$ with $\sum_{p \mid a, p \mid b} 1 \ge \frac{7}{10}\log\log N$ is at most $${\frac{11}{10}\log\log N \choose \frac{7}{10}\log\log N}\frac{N}{\exp(\frac{6.99}{10}\log\log N\log\log\log N)},$$ which is $o(N)$, since ${\frac{11}{10}\log\log N \choose \frac{4}{10}\log\log N} \le (\frac{11}{10}\log\log N)^{\frac{4}{10}\log\log N} \le \exp(\frac{4.01}{10}\log\log N\log\log\log N)$. We conclude $$\frac{1}{N^2}\#\{ab : a,b \in E_N, \sum_{p \mid a, p \mid b} 1 \ge \frac{7}{10}\log\log N\} = o(1).$$
