A repunit is a number consisting solely of ones (such as 11
or 11111).
Let us call p(n) a 10-base integer represented by a
string of n ones, e.g. p(1)=1, p(5)=11111 etc.
Most of the repunit numbers are composite.
2, 19,23,317 are
the first four indices of prime repunits.
Prove: For a prime repunit p(n) to be prime, n has to
be prime.
(In reply to
re: proof by Ady TZIDON)
Ady,
The statements a→b and not-b→not-a are called contrapositives and are equivalent. (If one is true the other is true.)
a = p(n) is prime
b = n is prime
not-a = p(n) is composite
not-b = n is composite
I proved the contrapositive and this implies the truth of the original statement, so a→b is proved.
Pointing out that R(3) is composite is a counterexample to b→a, which is the converse and may or may not be true for any given statement.
|
Posted by Jer
on 2011-01-11 10:18:20 |