With prime number it is very easy to get caught up with to much algebra and complex numbers. A very basic simplification is to realize that prime numbers that occurred become factors to exclude many numbers as possible candidate as prime numbers. 2 for example reduces all likely prime numbers to be odd only. So, walking along odd number line \(n\) is replaced with \(2n+1\).
odd \(3n\) becomes \(6n+3\) with 2 odd numbers between consecutive odd multiples of 3
odd \(5n\) becomes \(10n+5\) with 4 odd numbers between consecutive odd multiples of 5
odd \(7n\) becomes \(14n+7\) with 6 odd numbers between consecutive odd multiples of 7
odd \(11n\) becomes \(22n+11\) with 10 odd numbers between consecutive odd multiples of 11
So, given a odd number \({N}_{x}\) just greater that 11, the next lower prime. We have,
2/3 = probability that \({N}_{x}\) is not a multiple of 3
4/5 = probability that \({N}_{x}\) is not a multiple of 5
6/7 = probability that \({N}_{x}\) is not a multiple of 7
10/11 = probability that \({N}_{x}\) is not a multiple of 11
and so the probability that \({N}_{x}\) is a prime is
\(\prod _{ i }^{ { P }_{ N }<{ N }_{ x }<{ P }_{ N+1 } }{ \frac { { P }_{ i }-1 }{ { P }_{ i } } } \) when \({ P }_{ N }={ N }_{ x }\) the associated probability is 1.
In general, any number being prime is given by the above expression where \(i=1\), \({P}_{1}=2 \) is the first prime.
What is really interesting is as \({ N }_{ x }\rightarrow \infty \)
\({ \prod _{ 1 }^{ { N }_{ x }\rightarrow \infty }{ (\frac { 1 }{ 2 } )(\frac { 2 }{ 3 } )(\frac { 4 }{ 5 } )(\frac { 6 }{ 7 } )...\quad ({ P }_{ i }<1)... } }=0\).
That is to say, given any number, big enough, it is not likely to be a prime number. For example, any number \(99991<{N}_{x}<100003\) has the probability of 0.048752917851015 of being a prime. And any number \(821603<{N}_{x}<821641\) has the probability of 0.0412229294373678 of being a prime. Not bad 4% chance. This probability deceases as \({N}_{x}\) increases. Infinity is not likely a prime.
