Prime product pp(n) >/ 5*n for every natural n >/ 5

Terminology:

•  N := {1 2 3 ...} -- the set of all natural numbers
•  P := {2  3  5  7  11  13  17 ... } -- the set of all primes
•  pp(x) := prod{ p in P: p \< x}. -- the product of all primes that are  \<  x

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The product of the empty set, i.e. of zero elements, is equal to 1 (by a definition). This starts the list below:

• pp(1) = 1
• pp(2) = 2
• pp(3) = 2*3 = 6
• pp(4) = pp(3) = 6
• pp(5) = 2*3*5 = 30
• pp(6) = pp(5) = 30
• pp(7) = 2*3*5*7 = 210
• pp(10) = pp(9) = pp(8) = pp(7) = 210
• pp(11) = 2*3*5*7*11 = 2310

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Theorem   For every  n  in  N

•  pp(n) >/ n
•  n >/ 3   ==>   pp(n)  >/  n + 2
  
•  n >/ 5   ==>   pp(n)  >/  5*n

Proof  A simple direct computation shows that the theorem holds for  n = 1 2 3 4 5.  For instance

                      pp(5) = 2*3*5 > 5*5 > 5+2 > 5

Great!

In general

                      n >/5    ==>    5*n  >  n+2  >  n

-- the last of the three theorem's statements implies the previous two for every natural  n >/5.  Thus, in the view of the theorem holding already for  n = 1 2 3 4,  it is enough to prove that last statement only.

That last statement, as noted above, holds for  n=5.  Let's assume now that natural n > 5,  and that that statement holds for every natural  m (in place of  n)  such that. 5 \< m < n  (a proof by induction).

Thus (the induction assumption)  pp(n-1) >/ 5*(n-1).  Consider special cases:

Case A:      pp(n-1) = 5*(n-1)

we see that  n-1  is not divisible by. 5, and that  p | n-1  for every prime  p =/= 5  such that  p \< n-1. Thus  n  is not divisible by any prime  p=/=5 such that  p \< n-1.

It follows that there exists uniquely non-negative intger  k and natural q  such that

                           n =  5^k * q

and  q is not divisible by any prime  \<  n-1, i.e. by any prime  p < n.  Then either  q  is divisible by a prime  t  >/  n  hence  t >/ n,  or  t=1. The first case implies  n = q = t  is a prime hence:

         pp(n)  =  pp(n-1) + n  =  5*(n-1) + n  =  6*n - 5  >/  5*n

and the theorem holds;

or else  q=1, and

                                  n  =  5^k

Then  4 | n-1  while  4 = 2^2  does not divide  pp(n-1) = 5*(n-1) -- a contradiction.

Summary of Case A:  then n  is a prime, and  pp(n) >/ 5*n -- the theorem holds in Case A.

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Case B:      pp(n-1) = 5*(n-1) + r                  where  1 \< r < 5

This means that  pp(n-1)  is not divisible by  5. hence  n-1 < 5, and  n < 6, i.e.  n \< 5. Then the theorem holds.

Summary of Case B:  then  n \< 5 -- the theorem holds in Case B.

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Case C:      pp(n-1)  >/  5*n

This is the remining case. Then

             pp(n)  >/  pp(n-1)  >/  5*n

Summary of Case C:  the theorem holds in Case C.

The theorem holds in all there cases A  C.  End of Proof