(It suffices to show that pi(x)=o(x), which is a quite weak statement about the distribution of the primes that can be proved by a completely elementary sieving process. Am just thinking of the cleanest way to write this. Will post shortly.)
(It suffices to show that pi(x)=o(x), which is a quite weak statement about the distribution of the primes that can be proved by a completely elementary sieving process. Am just thinking of the cleanest way to write this. Will post shortly.)
Ahh yes, that was the other way to do it, but that was meant to be the second part to the problem. My intended solution was to use the fact that n! has n-1 obvious factors, and exploit that to construct an arithmetic progression strictly consisting of composite numbers, which is a lot easier to prove.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
(And yes, this is strictly more work than one needs to do to answer your original question, where a construction using a factorial explicitly gives us prime gaps of length blah. I just did it this way because I personally wanted to see how "low-tech" I could make a proof of the fact that the primes have density zero. )
Stronger statements about prime distribution are also very much within the scope of MX2 level techniques.
One such example is Bertrand's postulate, which states that there is at least one prime strictly between n and 2n for any positive integer n greater than 1.
This will be harder for a HS student to do without guidance, but I encourage any interested student to think about it / have a crack at it.
Last edited by seanieg89; 27 Jan 2016 at 12:35 PM.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
This wording means that the question is now valid, before you said that any subinterval of length n contained a subinterval of k primes (pi(x)>k>0), which was clearly not true and hence I was confused. I would normally post my solution, but Seanieg89 has already done so and mine was pretty similar except in the part where you had to prove that there exists a subinterval with 0 primes.
Last edited by Blast1; 30 Jan 2016 at 12:12 AM.
A bit on the easier side, but just to get the ball rolling in this thread again:
A function g(n) defined on the non-negative integers satisfies:
i) g(0)=g(1)=0.
ii) g(p)=1 if p is prime.
iii) g(mn)=m*g(n)+n*g(m) for all non-negative integers m and n.
Find all n such that g(n)=n.
---------------------------------------
A somewhat harder followup question is proving that the sequence obtained by:
u(0)=63
u(n+1)=g(u(n))
tends to infinity.
Last edited by seanieg89; 1 Feb 2016 at 11:48 PM.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
Last edited by Paradoxica; 5 Feb 2016 at 9:36 PM.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
Well I'm not sure if my mathematical rigor is exact, but here goes.
Suppose we had a solution with more than one of the numbers not equal to zero. However, if we multiply all the denominators up onto the other side, we would have an arbitrarily large set of congruences that are all equal to zero on the right hand side, but not equal to zero on the left hand side, as each term on the left hand side is not divisible by one of the primes on the right hand side. this leads to an impossible set of residues, and we have a contradiction. Thus, the only solutions that can exist are for exactly one of the integers equal to it's denominator, and every other term equal to zero. This verifies the class of solutions I mentioned earlier, and we are done.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
Exactly, although worded slightly funny. The point is that the j-th term on the LHS MUST be divisible by p_j as every other term trivially is.
This means that each a_j must be a non-negative integer multiple of p_j. (Because we do not get a prime factor of p_j in this term from multiplying out the denominators.)
I.e we have a finite ordered collection of non-negative integers summing to 1. The only way for this to happen is for one of these integers to be 1 and the others to be zero.
This lets us draw the desired conclusion.
Last edited by Paradoxica; 7 Feb 2016 at 6:38 PM.
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
I'm pretty new to this stuff, but I think I can do 3.
edit: Now that I think about it, Euler's Theorem is total overkill. Just note that x can be congruent to -2, -1, 0, 1 or 2 mod 5. After dealing with the 0 case, the +-1 cases have x^4 = 1, and the +-2 cases have x^4 = 16 which is congruent to 1 mod 5.
Also, since this is supposed to be a an elementary thread, if anyone hasn't done modular arithmetic yet, note that it is sufficient to make sure the statement holds true for x = 0,1,...,14. I don't think it can get any more elementary than that.
Last edited by KingOfActing; 6 Feb 2016 at 10:11 PM.
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Last edited by InteGrand; 11 Feb 2016 at 2:36 PM. Reason: Typo
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
This establishes the claim for all n larger than 6.
Verifying the inequality for the five values of n between 2 and 6 with a calculator or by hand completes the proof.
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