Carom 1-16 - s253053503.websitehome.co.uk

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Transcript Carom 1-16 - s253053503.websitehome.co.uk

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Activity 1-16:
The Distribution Of Prime Numbers
Suppose we write down the first few prime numbers.
2, 3, 5, 7, 11, 13, 17, 19....
So of the first 20 positive whole numbers, eight are prime.
That is quite a lot!
As we keep writing down prime numbers,
they become much more spread out.
If we write down the prime numbers
between 1001 and 1020, we get three:
1 009, 1 013, 1 019
If we write down the prime numbers
between 1 000 001 and 1 000 020, we get one:
1 000 003
Task: show how to construct a
string of consecutive whole numbers
length m containing no primes
where m is as large as we like.
Consider the string of five numbers
4x5x6x7x8 + 4
4x5x6x7x8 + 5
4x5x6x7x8 + 6
4x5x6x7x8 + 7
4x5x6x7x8 + 8
Clearly none of these is prime.
Can we use this technique in general?
Consider the string of numbers
n(n+1)(n+2)...(n+m) + n,
n(n+1)(n+2)...(n+m) + (n+1),
n(n+1)(n+2)...(n+m) + (n+2),
...
n(n+1)(n+2)...(n+m) + (n+m)
So how are the primes distributed?
This is question that has attracted mathematicians
for centuries, and which is still very much alive.
What we do know is that there are always more primes –
the list of primes never comes to a halt.
The proof of this below is due to Euclid,
created more than two thousand years ago.
Suppose we have a list of primes, say 2, 3, 7, 11.
Consider the number 23711 + 1.
Certainly 2, 3, 7, 11 all fail to divide into this number,
because they all leave a remainder 1.
In fact, 23711 + 1 = 463, which is prime,
so we have found a new prime that is not in our list.
Does this always work?
Suppose instead we have the list of primes 2, 5, 7, 11.
Consider the number 25711 + 1.
This is 771 = 3  257.
So this time we get a product of primes,
none of which are in our original list.
Either way, it seems that given a list of primes,
we can always extend that list if we need to.
Euclid made this a little more general...
Suppose that we have a list of prime numbers, p1, p2, ... pn.
Consider P = p1p2...pn+1.
It is clear that none of p1, p2 ... divide P exactly:
they all leave remainder 1.
So either P is prime, or else some prime not in our list divides P.
Either way, we have found a new prime
that is not on our original list,
so our list can always be extended,
and the number of primes is infinite.
Let us call π(x) the number of primes less than x.
So π(20) = 8.
If we were to draw a graph of π(x) against x,
what would it look like?
We can tell roughly from our work above
that it will look roughly like this:
But for mathematicians,
being approximately right is not enough here:
what is the equation of this line?
Let’s get some more data with which to work.
Prime Number
Spreadsheet
http://www.s253053503.websitehome.co.uk/
carom/carom-files/carom-16.xlsm
We can use Autograph
to plot π(x) against x for x up to 500.
π(x)
What equation might fit this curve?
x
The Prime Number Theorem
(proved by Hadamard and de la Vallée Poussin in 1896)
tells us that π(x) tends towards
as x gets large, where k is a constant.
Task: try adding the equation
y=
to your Autograph plot.
Can you adjust k to get a good fit?
π(x
We have a reasonable fit for small x here,
but we need to remember that infinity
Is a very long way away!
x
With
k = 1.18.
The number of digits in a number n is proportional to
So the Prime Number Theorem tells us
that the probability that
x, a number chosen at random, is prime
is inversely proportional to the number of digits in x.
A beautiful rule, n’est ce pas?
The number
1000000000000066600000000000001
is prime.
That’s a 1, 13 zeros, 666, 13 zeroes and a 1.
This number has 31 digits.
(This is called Blephegor’s prime,
after one of the seven princes of Hell.)
The chance of a random three digit number being prime is
0.158888…
So the chance of a random 31 digit number being prime is
(3/31)(0.158888…) = 0.01538, or 1.5% of the time.
With thanks to:
Wikipedia, for another excellent article.
Carom is written by Jonny Griffiths, [email protected]