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Sep 6, 2015 at 6:28 comment added Lucia You're very welcome!
Sep 5, 2015 at 20:20 history edited Lucia CC BY-SA 3.0
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Sep 5, 2015 at 18:12 comment added Lucia That argument was a little too glib, and I need to be a bit more careful. The problem is that if we count numbers of the form $p^{2}q^3$ say then this will be different from $p^3q^2$, but both map down to $pq$. What I wrote is still true, but requires a bit more work. I'll edit it later today if I get a chance.
Sep 5, 2015 at 17:35 comment added Lucia I edited to make that step a little clearer.
Sep 5, 2015 at 17:34 history edited Lucia CC BY-SA 3.0
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Sep 5, 2015 at 5:08 comment added მამუკა ჯიბლაძე ...on the afterthought though, - sorry, seems I still don't understand even the simplest part. I mean, with this argument then also if $p_1p_2\le x$ then also $p_1\le x$, but this does not mean that there are more primes below $x$ than numbers of the form $p_1p_2$, right?
Sep 4, 2015 at 17:46 vote accept მამუკა ჯიბლაძე
Sep 4, 2015 at 17:46 comment added მამუკა ჯიბლაძე Thank you! Should probably figure out the part about $1$'s myself. This allows for simplification in my calculations, maybe I will add more stuff here. In any case obviously this is the right answer to accept.
Sep 4, 2015 at 13:46 history edited GH from MO CC BY-SA 3.0
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Sep 4, 2015 at 13:16 history answered Lucia CC BY-SA 3.0