Define a Prime

[xilman]

Originally Posted by Mini-Geek

Ok, fine, here's the updated version:

A prime number (or a prime) is a natural number that has exactly (5 - 3) (distinct) natural number divisors, which are (4 - 3) and the prime number itself.
Hooray for loopholes!

Originally Posted by davar55

Can you define prime numbers over the non-negative
integers without any explicit reference to 0 or 1 or 2
or any other specific integer?

I added the emphasis to the original specification of the problem. To the best of my knowledge, 3, 4 and 5 are all specific integers.



Paul

[xilman]

Originally Posted by davar55

Can you define prime numbers over the non-negative
integers without any explicit reference to 0 or 1 or 2
or any other specific integer?

Let p and q be non-negative integers such that p*p>p and q*q>q.

Then p is prime if and only if the smallest value of q such that q divides p! is q=p.


Paul

[xilman]

Originally Posted by davar55

Can you define prime numbers over the non-negative
integers without any explicit reference to 0 or 1 or 2
or any other specific integer?

I have another marvelous definition and the margin of this post is only just wide enough to contain it.

The positive values of this polynomial over the integers are all the primes and nothing but.

(k + (ln(e)+ln(e)))*(ln(e) - (w*z + h + j - q)^(ln(e)+ln(e)) - ((g*k + (ln(e)+ln(e))*g + k + ln(e))*(h + j) + h - z)^(ln(e)+ln(e)) - ((ln(e)+ln(e))*n + p + q + z - e)^(ln(e)+ln(e)) - ((((ln(e)+ln(e))^(ln(e)+ln(e)))^(ln(e)+ln(e)))*(k + ln(e))^(ln(e)+ln(e)+ln(e))*(k + (ln(e)+ln(e)))*(n + ln(e))^(ln(e)+ln(e)) + ln(e) - f^(ln(e)+ln(e)))^(ln(e)+ln(e)) - (e^(ln(e)+ln(e)+ln(e))*(e + (ln(e)+ln(e)))*(a + ln(e))^(ln(e)+ln(e)) + ln(e) - o^(ln(e)+ln(e)))^(ln(e)+ln(e)) - ((a^(ln(e)+ln(e)) - ln(e))*y^(ln(e)+ln(e)) + ln(e) - x^(ln(e)+ln(e)))^(ln(e)+ln(e)) - ((((ln(e)+ln(e))^(ln(e)+ln(e)))^(ln(e)+ln(e)))*r^(ln(e)+ln(e))*y^(ln(e)+ln(e)+ln(e)+ln(e))*(a^(ln(e)+ln(e)) - ln(e)) + ln(e) - u^(ln(e)+ln(e)))^(ln(e)+ln(e)) - (((a + u^(ln(e)+ln(e))*(u^(ln(e)+ln(e)) - a))^(ln(e)+ln(e)) - ln(e))*(n + (ln(e)+ln(e)+ln(e)+ln(e))*d*y)^(ln(e)+ln(e)) + ln(e) - (x + c*u)^(ln(e)+ln(e)))^(ln(e)+ln(e)) - (n + l + v - y)^(ln(e)+ln(e)) - ((a^(ln(e)+ln(e)) - ln(e))*l^(ln(e)+ln(e)) + ln(e) - m^(ln(e)+ln(e)))^(ln(e)+ln(e)) - (a*i + k + ln(e) - l - i)^(ln(e)+ln(e)) - (p + l*(a - n - ln(e)) + b*((ln(e)+ln(e))*a*n + (ln(e)+ln(e))*a - n^(ln(e)+ln(e)) - (ln(e)+ln(e))*n - (ln(e)+ln(e))) - m)^(ln(e)+ln(e)) - (q + y*(a - p - ln(e)) + s*((ln(e)+ln(e))*a*p + (ln(e)+ln(e))*a + p^(ln(e)+ln(e)) - (ln(e)+ln(e))*p - (ln(e)+ln(e))) - x)^(ln(e)+ln(e)) - (z + p*l*(a - p) + t*((ln(e)+ln(e))*a*p - p*(ln(e)+ln(e)) - ln(e)) - p*m)^(ln(e)+ln(e)))



Paul

[R. Gerbicz]

Let n be a non-negative integer, then n is prime if and only if n!*n! isn't divisible by n*n*n. (It's easy to prove this.)

[ewmayer]

An integer p is prime iff division modulo p is defined, i.e. the residue ring Z/pZ is a field.

[Uncwilly]

A prime number is one where one can't take the equivelent number of objects and place them into groups of equal number.
If one is unable have the number of objects or is able to put them into groups with equal number of objects, then it is not prime.

[S485122]

Originally Posted by Uncwilly

groups

Isn't insisting on the plural the same as saying "more than one" and thus an explicit reference to a specific integer ? Or would this be a to restrictive interpretation of the rules ?

[mfgoode]

Originally Posted by xilman

Counter-example.

3 is a non-negative integer

3 is divisible by itself --- meets your definition

3 is divisible by 1, a unit in the ring of integers --- meets your definition

3 is divisible by -1, a unit in the ring of integers --- meets your definition

-3 is an element of the ring of integers

3 is divisible by -3


Therefore, according to your definition, 3 is not a prime.



Paul

I am sure Troels Munkner will be very happy with that derivation.

Mally

[Andi47]

What about this one:

p is prime iff there exist integers a and b with:

p is divisible by a and b, and
a <> b and a <> -b

and for each integer c which divides p :
c = a or c = -a or c = b or c = -b

[davieddy]

Originally Posted by Andi47

What about this one:

p is prime iff there exist integers a and b with:

p is divisible by a and b, and
a <> b and a <> -b

and for each integer c which divides p :
c = a or c = -a or c = b or c = -b

I think it is too obvious that you are referring to
two positive integers. I don't know whether simply
listing the two integers excuses this.

[ewmayer]

An integer p is prime iff no integer x in the range sqrt(p) <= x < p divides p.