Define a Prime
 

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?

[QUOTE=davar55;99304]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?[/QUOTE]

non-negative integers that are only divisible by themselves and the units in the ring of integers

An integer so that any product equal to the integer must involve the integer itself, units, and nothing else.

Alex

[QUOTE=davar55;99304]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?[/QUOTE]

[spoiler]Let (p,q) be non-negative integers such that N=p*q and p>q

If for all (p,q) p=N, then N is prime.
[/spoiler]

Pau

Here's another try:

[spoiler]A positive integer p is prime if and only if the only solutions to the equations p = a*b = c*d occur when a is not equal to b and a = c and b = d or a = d and b = c. (I almost said that the equation p = a*b has exactly two solutions (a,b) where a and b are positive integers, but then realized that I was violating the terms of the problem!)[/spoiler]

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

[QUOTE=Mini-Geek;99329]that has exactly [B]two[/B] [/QUOTE]

Rule violation. Five yards penalty.

Alex

[QUOTE=TravisT;99307]non-negative integers that are only divisible by themselves and the units in the ring of integers[/QUOTE]

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

Same problem with my definition. Let's try

An integer so that any product equal to the integer must involve the integer itself or one of its associates, any number of units, and nothing else.

Alex

[quote=akruppa;99330]Rule violation. Five yards penalty.

Alex[/quote]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.
:grin: Hooray for loopholes! :grin:

One may try to invent something analytical from
[url]http://mathworld.wolfram.com/LandausFormula.html[/url]

:-)

[QUOTE=Mini-Geek;99339]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.
:grin: Hooray for loopholes! :grin:[/QUOTE]

[QUOTE=davar55]Can you define prime numbers over the non-negative
integers without any explicit reference to 0 or 1 or 2
[b]or any other specific integer[/b]?[/QUOTE]
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

[QUOTE=davar55;99304]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?[/QUOTE]
[spoiler]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.[/spoiler]


Paul

[QUOTE=davar55;99304]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?[/QUOTE]I have another marvelous definition and the margin of this post is only just wide enough to contain it.

[spoiler]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)))
[/spoiler]


Paul

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.)

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

A prime number is one where one can't take the equivelent number of objects and place them into group[COLOR="Red"]s[/COLOR] 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.

[QUOTE=Uncwilly;99465]group[COLOR="Red"]s[/COLOR][/QUOTE]
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 ?

[QUOTE=xilman;99334]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[/QUOTE]


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

Mally

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

[quote=Andi47;99494]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[/quote]

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.

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

Imagine a precisely cut plank of wood "p" inches** long. The width and height do not matter but the shape is a perfect rectangular solid. Along the length, each inch is marked with a line.

If that board can not be cut into pieces along the inch markers ONLY such that you can rearrange them laterally (and without turning the pieces) into a rectangular shape that has different dimensions than the original board (breath!) then p is prime.

** inch is arbitrary of course.

The turning limitation might not be necessary. I'd have to think about that some more.

[QUOTE=ewmayer;99515]An integer p is prime iff no integer x in the range sqrt(p) <= x < p divides p.[/QUOTE]

An integer p is prime if and only if for all A,B, such that A^2 = B^2 mod p,
then A=B mod p or A=-B mod p.

[QUOTE=R.D. Silverman;99523]An integer p is prime if and only if for all A,B, such that A^2 = B^2 mod p,
then A=B mod p or A=-B mod p.[/QUOTE]

Ah, but you had to explicitly reference the number 2 - that's a 10-yard penalty and loss of down for you, Bob. Oh wait, I just referenced 1 and 0 in my description of the penalty - that's [i]log(p)/log(sqrt(p))[/i] minutes in the penalty box for me...

Aren't arbitrary puzzle rules great?

[QUOTE=ewmayer;99526]Ah, but you had to explicitly reference the number 2 - that's a 10-yard penalty and loss of down for you, Bob. Oh wait, I just referenced 1 and 0 in my description of the penalty - that's [i]log(p)/log(sqrt(p))[/i] minutes in the penalty box for me...

Aren't arbitrary puzzle rules great?[/QUOTE]

So change it to A*A and B*B. This isn't an issue of using the
number 2, but rather one of notation.

[QUOTE=R.D. Silverman;99530]So change it to A*A and B*B. This isn't an issue of using the number 2, but rather one of notation.[/QUOTE]I'm amazed that no-one has yet picked me up for using ln(e) in a polynomial to be evaluated at integer values of its variables. Had anyone done so I would have, of course, replaced them all with (x/x)

Paul

[QUOTE=S485122;99478]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 ?[/QUOTE]

Can the high command give a ruling on this? I can reword my answer to say "a person"<->"one".

[QUOTE=xilman;99541]I'm amazed that no-one has yet picked me up for using ln(e) in a polynomial to be evaluated at integer values of its variables. Had anyone done so I would have, of course, replaced them all with (x/x)

Paul[/QUOTE]

Paul,

Using x/x would necessitate x\neq 0.

Prime Number definitions:

1) A positive integer p is a prime or a prime number if it is a whole number larger than unity and its only positive divisors are the unit and itself.
Every prime number has the property that if it divides a product then it must divide at least one of the factors [Euclid c.300 BC.]

2) A positive integer is prime if only *two* but only *two* distinct factors, are itself and unity

Euclid books 7 and 8 regards a number as a line interval compounded of units and
defines a prime as a number which can only be measured by the unit (not itself a number)
It follows from both the above two definitions that unity is not a prime

3) A prime is an irreducible element of a unique factorization domain and is known as prime.
These are for positive primes.

To allow for negative primes we cannot define it without naming some numbers.


Definition: the term can also be used in some other situations where division is meaningful.
For instance in the context of all the integers an integer other than 0+-1 is a prime integer if its only integer divisions are +- 1 and +- n.
Mally.

[QUOTE=mfgoode;99600]Prime Number definitions:[/QUOTE]Mally,

I like to believe that all participants to the thread so far are well aware of the definition of a prime number. But you overlooked the purpose of this thread. The original question was :
[QUOTE=davar55;99304]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?[/QUOTE]

[QUOTE=Zeta-Flux;99599]Paul,

Using x/x would necessitate x\neq 0.[/QUOTE]Correct, and I should have stated that fact in the form "for values of x where x/x is defined".

Thanks for drawing it to my attention.

The purpose behind the formulation of this admittedly kind of
arbitrary question was to try to come up with an alternate,
nonarbitrary explanation for why 1 isn't prime (alternate to
the fundamental theorem of arithmetic's desirable simplicity).
After all, a typical definition of primality in the natural numbers
uses 1 twice:
( n is prime iff n>1 and a|n implies a=n or a=1 )
which begs the question.

But my version of the definition probably still does too:

Among the non-negative integers,
n is prime iff ((n=ab IMPLIES (n=a OR n=b)) AND NOT (n*n=n)).

This excludes 1 (and 0) from primality by using a trivial property
(n*n=n) shared only by 1 and 0, without explicitly mentioning 1
or "unit". So by this definition, 1 is not prime. But as simple as
it may be, it's still a bit arbitrary to put this into the definition, isn't it?
Yet this is all I was getting at when posing the question.

Alternatively, one can define a prime to be any integer p such that Z/pZ is a field.

(However, fields are (by definition) rings with 1\neq 0. So implicit in the definition is p\neq 1. On the other hand, 0 and 1 are implicit in the construction of the positive integers too. So I like my answer.)

[QUOTE=Zeta-Flux;99654]Alternatively, one can define a prime to be any integer p such that Z/pZ is a field.[/QUOTE]

Is there an echo in here? ;)

Is there an echo in here?

Wow! ewmayer, take it as a complement that I missed you post, because your idea is the best one. :)

[quote=Zeta-Flux;99662]Is there an echo in here?

Wow! ewmayer, take it as a complement that I missed you post, because your idea is the best one. :)[/quote]

You mean compliment

No, compl[i]e[/i]ment.

One missed the post; the other posted the missed.

From Merriam-Webster (online):
[quote=http://www.m-w.com/cgi-bin/dictionary?va=complement] Main Entry: [B]com·ple·ment[/B]
Pronunciation: 'käm-pl&-m&nt
Function: [I]noun[/I]
Etymology: Middle English, from Latin [I]complementum,[/I] from [I]complEre[/I] to fill up, complete, from [I]com-[/I] + [I]plEre[/I] to fill -- more at [SIZE=-1]FULL[/SIZE]
[B]1 a[/B] [B]:[/B] something that fills up, completes , or makes perfect [B]b[/B] [B]:[/B] the quantity, number, or assortment required to make a thing complete <the usual [I]complement[/I] of eyes and ears -- Francis Parkman>; [I]especially[/I] [B]:[/B] the whole force or personnel of a ship [B]c[/B] [B]:[/B] one of two mutually completing parts [B]: [SIZE=-1]COUNTERPART[/SIZE][/B][/quote][quote=http://www.m-w.com/cgi-bin/dictionary?va=compliment] Main Entry: [B]com·pli·ment[/B]
Pronunciation: 'käm-pl&-m&nt
Function: [I]noun[/I]
Etymology: Middle French, from Italian [I]complimento,[/I] from Spanish [I]cumplimiento,[/I] from [I]cumplir[/I] to be courteous -- more at [SIZE=-1]COMPLY[/SIZE]
[B]1 a[/B] [B]:[/B] an expression of esteem, respect, affection, or admiration; [I]especially[/I] [B]:[/B] an admiring remark [B]b[/B] [B]:[/B] formal and respectful recognition [B]: [SIZE=-1]HONOR[/SIZE][/B][/quote]I think that compliment is correct, because it being complement in the way Zeta-Flux said it wouldn't make much sense, but it being compliment would.

[quote=cheesehead;99672]No, compl[I]e[/I]ment.

One missed the post; the other posted the missed.[/quote]

And who's pissed the most?

[QUOTE=davieddy;99676]And who's pissed the most?[/QUOTE]

:lol: Alex K. is probably kicking himself for letting you beat him to posting that zinger. (He PMed me the same quip.)

In any event, perhaps the fact that your last name is "Eddy" tells us the answer to your question. "Let it flow through you you must, young Whiz-taker..."

If I ever need a dose of humble pie, I know where to go!

(Now you can talk about mixing metaphors.)

[quote=Mini-Geek;99673]I think that compliment is correct, because it being complement in the way Zeta-Flux said it wouldn't make much sense[/quote]I was trying to show [I]why[/I] Zeta-Flux's "compl[I]e[/I]ment" did make sense ([b]1c[/b]) :geek: , but my alliteration and brevity seem to have obscured my meaning. :surrender

[QUOTE=S485122;99602]Mally,

I like to believe that all participants to the thread so far are well aware of the definition of a prime number. But you overlooked the purpose of this thread. The original question was :[/QUOTE]

[QUOTE=Mally] 2) A positive integer is prime if only *two* but only *two* distinct factors, are itself and unity

Euclid books 7 and 8 regards a number as a line interval compounded of units and
defines a prime as a number which can only be measured by the unit (not itself a number)
It follows from both the above two definitions that unity is not a prime[/QUOTE]

Well S4852122:

I admit I faulted in my second def:that's why I emphasised the word Two

In the next line I clarified grandscorpion's post by giving it straight from Euclid

Do all posters know this?

In the final line I gave a very reasonable explanation of why 1 is not considered a prime and this conclusion cannot be reached with out the use of the word two in the first def:

If you re-read the original problem the restriction is not only on 0 ,1 ,or 2 but *any other specified integer* Well where does that lead us?

Moreover my aim, as always, is not to appear a wisecrack with witty answers but to disseminate knowledge, fundamental at that, so people can pick up the bits and pieces of gems thrown in. Unfortunately the modern generation lacks a firm base and foundation!

Mally.

My p cents worth:

p is prime iff the only two factors of p are p and p/p provided
p <> p - p and
p <> p/p

Regards
Patrick

[quote=Patrick123;99753]My p cents worth:

p is prime iff the only [B]two[/B] factors of p are p and p/p provided
p <> p - p and
p <> p/p

Regards
Patrick[/quote]Violation.

[QUOTE=Mini-Geek;99756]Violation.[/QUOTE]

Whoops let's rephrase that:

p is prime iff its only factors are p and p/p provided
p <> p - p and
p <> p/p and
p is positive.

Regards
Patrick

[QUOTE=R.D. Silverman;99523]An integer p is prime if and only if for all A,B, such that A^2 = B^2 mod p,
then A=B mod p or A=-B mod p.[/QUOTE]

So, 1 and 6 are primes then?

[QUOTE=Jushi;99870]So, 1 and 6 are primes then?[/QUOTE]How do you define an equality modulo 1 ?

[QUOTE=S485122;99874]How do you define an equality modulo 1 ?[/QUOTE]

Usually, "a = b mod c" means "c divides (b-a)". Then any two integers are equal mod 1, because 1 divides any integer.

[quote=S485122;99874]How do you define an equality modulo 1 ?[/quote]
That's not much worse than equality mod [tex]2\pi[/tex] which everyone knows from study of trigonometric functions... "x = y (mod z)" is defined for any number z (including 0, where it gives equality) as he wrotes, or [tex]x-y \in z\,\mathbb Z[/tex] where Z are the integers, i.e. zZ are all numbers of the form zk with k some integer.