1
It was already known in ancient Greece that there are an infinite number of prime numbers. The proof of the following feeling is written in " Principles
Let's say you have only three prime numbers. Let this be $ p $, $ q $, $ r $. When $ pqr + 1 $ is divided by $ p $, $ q $, or $ r $, $ 1 $ is left over. This violates the assumption that the prime numbers are only $ p $, $ q $, $ r $. (This reasoning is the same whether there are 4 or 5 prime numbers. That is, assuming a finite number of prime numbers always leads to a contradiction.)
It uses reductio ad absurdum (a reasoning that derives a contradiction from assumption A and proves A's negation). When I met this proof in junior high school, I was a little moved. The reductio ad absurdum was cool, and I thought it was wonderful to draw a contradiction from the simple formula "total product of all prime numbers plus 1".
2
Around the same time, I learned that there was a theme of "searching for large prime numbers." With the progress of electronic computers, huge prime numbers that could never be found by hand calculation were found one after another, and I saw for some reason that a list was created such as who found what prime number in what year. ..
3
An amateur like me remembers the proof of the "elementary theory" and comes up with such a wrong thing.
You can easily get a huge prime number by multiplying all the found prime numbers and adding 1!
Of course not. The assumption of a finite number only led to the situation that the total product plus 1 was not divisible by any. In reality, there are an infinite number of prime numbers, and just because a finite number of prime numbers are taken out, multiplied, and added to 1 does not mean that they become prime numbers.
4
But is it? The sum of $ n $ prime numbers plus one is not divisible by at least those prime numbers. Therefore, in a sense, it can be said that it is a “number that is difficult to divide”. For that matter, aren't there quite a few prime numbers in them?
Considering the sieve
Therefore, it seems better to arrange the prime numbers in ascending order, take the first $ n $ pieces, and add 1 to their total product to try.
5
First, let's take the first prime number $ 2 $. $ 3 $, which is the sum of this and $ 1 $, is a prime number.
Next. $ 2 \ cdot3 + 1 = 7 $ is a prime number. How nice.
$ 2 \ cdot3 \ cdot5 + 1 = 31 $ is also a prime number. Well, it's in good shape.
How about $ 2 \ cdot3 \ cdot5 \ cdot7 + 1 = 211 $? It's easy to see that neither $ 11 $ nor $ 13 $ is divisible. $ 17 ^ 2 $ will exceed $ 211 $ so you don't have to try it. It turned out to be a prime number. Hmmm, isn't the story too good? So far, I have done my best by hand calculation.
How about $ 2 \ cdot3 \ cdot5 \ cdot7 \ cdot11 + 1 = 2311 $? I gave up the manual calculation. It's not that difficult, but I'm not confident that I won't make a calculation error.
Here to the terminal
irb -r prime
Hit.
This will start the Ruby interactive environment, and the prime number library prime
will be loaded.
And
irb(main):001:0> 2311.prime?
=> true
Oh, is this also a prime number? (You can check that an integer is a prime number at Integer # prime?)
I never thought that only prime numbers would come out so well.
6
Let's examine it systematically in the program. Write as follows in Ruby.
require "prime"
1.step do |n|
m = Prime.take(n).inject(:*) + 1
puts "%3d %s %d" % [n, (m.prime? ? "o" : "x"), m]
end
This program is not written to stop. Formally, it keeps searching indefinitely.
Prime.take (n)
gives the first n
array of prime numbers.
ʻInject (: *)` gives the total product.
And
n
if it is a prime number,
x` if not)Just display and side by side.
7
The results were as follows.
1 o 3
2 o 7
3 o 31
4 o 211
5 o 2311
6 x 30031
7 x 510511
8 x 9699691
9 x 223092871
10 x 6469693231
11 o 200560490131
12 x 7420738134811
13 x 304250263527211
14 x 13082761331670031
15 x 614889782588491411
16 x 32589158477190044731
17 x 1922760350154212639071
18 x 117288381359406970983271
It goes up to $ n = 18 $ at a stretch, but it doesn't go on forever.
It was a prime number up to $ n = 5 $, but then a composite number follows. Hmmm sorry.
When I think that the prime number finally appears at $ n = 11 $, the composite number continues again.
Was it an illusion that I thought, "I think prime numbers are included in a fair proportion?"
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