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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.

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  • $\begingroup$ Counting the number of entries in a different case (rectangular table $64\times10^{16}$) is Problem 466 of Project Euler. But the solutions found there via inclusion-exclusion principle don't easily lead to an estimate of the density. $\endgroup$ – Professor Vector Sep 20 '17 at 6:44
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    $\begingroup$ Peter Humphries found mathoverflow.net/questions/108912/… @mathworker21 You should copy the calculation showing how $\log 2$ appears and do the same for square-free numbers, so we can try to find a way to improve the estimate. And how do you use $\omega(n) \approx \log \log n$ here ? $\endgroup$ – reuns Sep 21 '17 at 3:40
  • $\begingroup$ @reuns there's no point to improving it $\endgroup$ – mathworker21 Mar 28 at 21:39

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