Chebyshev Said It and Ill Say It Again

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Number of Primes Betwixt n and 2n

  • Thread starter 1MileCrash
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As nosotros travel further upwards the number line, primes become more deficient.

Simply, as n grows larger and larger, the range of numbers between n and 2n grows larger ad larger.

Do these two counteract each other? Does this cause the number of primes between north and 2n to stay relatively consistant, increment, or just shrink regardless?

Answers and Replies

Using the prime number theorem, this can be estimated.
π(n) ~ n/ln(n), so π(2n) - π(northward) ~ 2n/ln(2n) - n/ln(n) which is approximately north/ln(northward).
Chebyshev (and Erdos) said it and I'll say it over again: there'due south always a prime between n and 2n.
Using the prime number theorem, this can exist estimated.
?(n) ~ north/ln(n), so ?(2n) - ?(n) ~ 2n/ln(2n) - north/ln(n) which is approximately n/ln(n).

Then every bit north goes to infinity, the number of primes between n and 2n goes to infinity?
Yes, in the sense that n/logn -> infinity as n -> infinity. (Just use Fifty'Infirmary's to come across this.)
Yes, in the sense that northward/logn -> infinity equally n -> infinity. (Merely utilise L'Hospital'southward to see this.)

Yep, that's what i did.

I don't really similar the quote by chebyshev and erdos because it is bit misleading.:smile:

Well, the quote is by Erdos. See, Chebyshev proved that there was e'er a prime between n an 2n using a rather technical argument. Erdos proved it (I retrieve when he was 19 or so) using a combinatorial argument, so he said "Chebyshev said it I'll say it again, there's always a prime betwixt n and 2n."

But what is misleading virtually it?

Yep, that's what i did.

I don't really similar the quote by chebyshev and erdos considering it is bit misleading.:smile:


I can understand that the quote is entirely inappropriate to the thread, but calling information technology misleading seems far-fetched to me.
Well, the quote is by Erdos. Come across, Chebyshev proved that there was ever a prime between n an 2n using a rather technical argument. Erdos proved it (I think when he was 19 or so) using a combinatorial statement, and so he said "Chebyshev said information technology I'll say it again, there'due south always a prime between northward and 2n."

But what is misleading about information technology?


Well, we agreed in a higher place that the number of primes between n and 2n goes to infinity as n goes to infinlty. Thats a lot of primes. The Erdos quote says in that location is always a prime number, that does not sound like a lot, so IMHO the quote focuses on the triviality of small due north and disregards the greater truth for large northward.:smile:
Well sure, what we said was a sharper result than Bertrand'southward Postulate. I still don't run across how this is misleading (though I will concord it actually doesn't have much to do with the the thread.) The fact that in that location is ALWAYS a prime between north and 2n for all north is interesting on its ain, espicialy when you lot consider that I tin give you an arbirarily long listing of sequent composite integers.

I but posted the quote every bit a joke in the start place.

Well sure, what we said was a sharper effect than Bertrand's Postulate. I all the same don't see how this is misleading (though I will hold information technology really doesn't accept much to practise with the the thread.) The fact that there is ALWAYS a prime betwixt n and 2n for all n is interesting on its own, espicialy when you consider that I can give y'all an arbirarily long listing of consecutive composite integers.

I just posted the quote as a joke in the start place.


Aye merely that list of r sequent composite integers must start at some integer north and my gut tells me that n is much larger than r for big r. For example if r is 100 so north is going to be much larger than 100 and of course 2n-northward is the number of integers in this interval. what would exist 100/n? my gut tells me information technology'south virtually 0

so in a simillar fashion r/n goes to zero as r goes to infinity. Recollect that r is the number of consecutive composite integers and n is the integer where this sequence starts.

Is my gut mistaken?:smile:

It's all skillful, y'all don't take to come across it my fashion, my opinions are not 'etched in rock'

Concluding edited:
You lot're correct; its huge. The sequence begins at (n+1)!.

Only that'south non really the signal. The fact that PNT, in a way, implies Chebyshev'southward theorem does non, in whatever way, mean that Chebyshev's Theorem is "misleading" or unimportant. And it certainly does not have away from the fact that Erdos came upwards with a magnificent proof of information technology.

PNT also implies that there are infinite primes, but this does non have annihilation abroad from Eucilid's proof of the aforementioned.

The sequence begins at (n+1)!.

Not necessarily.

OK....

I was speaking of constructing a sequence like this:

(northward+1)! + 2, (northward+1)! + iii, ..., (due north+1)! + n + 1

This is a listing of n consecutive composites. When you say "not necessarily" do you hateful that you know of another way to construct an arbitrarily long list of blended numbers? If and so, I'd exist interested. Or, do you hateful that not every list of g consecutive composites has to be constructed the way I depict? If this is the case, then I don't really care as I never claimed that the Simply style to construct a list of due north composites was the construction I mentioned higher up. Maybe what I wrote in my last postal service was confusing, but that's simply because I thought we understood that I was referring to creating a list of size northward for a general n, not a specific n. Certainly for "most" due north, there are much lower sequences of n sequent composites; but this doesn't tell us anything almost a general n.

OK....

I was speaking of constructing a sequence like this:

(n+one)! + 2, (n+1)! + 3, ..., (n+1)! + n + 1

This is a list of due north consecutive composites. When you lot say "not necessarily" do you hateful that you know of another style to construct an arbitrarily long list of blended numbers? If and then, I'd exist interested. Or, practice you mean that non every list of 1000 consecutive composites has to be synthetic the way I describe? If this is the case, then I don't really intendance as I never claimed that the ONLY style to construct a list of n composites was the construction I mentioned above. Perhaps what I wrote in my last post was confusing, but that's only because I thought we understood that I was referring to creating a list of size n for a full general n, not a specific n. Certainly for "most" n, there are much lower sequences of northward sequent composites; merely this doesn't tell united states anything about a general n.

In that location are occasions that (north+1)! -i or (n+ane)! +one or both are composite also, so the sequence begins much sooner at (northward+i)! - (n+1), but I don't know if that was what agentredlum had in mind. Too, You must consider agentredlum's prior mail service where he said the sequence of r composite integers begins at some n. I call back he had in mind then the everyman possible n for r consecutive blended integers. Unfortunately your mail was for n consecutive composites rather than for r consecutive composites, just n consecutive composites is more germane to the original posts.
Terminal edited:
From my 'numerical department', the number of primes betwixt n and 2 n is:

100 < #21 > 200
...
1000 < #135 > 2000
...
10000 < #1033 > 20000
...
100000 < #8392 > 200000

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