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People tell me that a prime number is an integer which is divisible only by itself and one. But if 1 is not a prime number, then the only definition which suits prime numbers is "a number with only two factors".
http://zdnet.com.com/2100-1104-949170.html
Something interesting there... but the definition in the opening paragraph made me wonder. Just wondering what other people think.
Ok, just a few things:
1) yes, the general accepted definition is like the one you said:
"A positive integer with two distinct positive integer factors, itself and one."
2) Some people see negative numbers as also primes. (i.e. -2, -3, -5 etc.) In this case, the definition changes to:
"A positive integer with four distinct integer factors, itself, negative itself, minus one and one."
Remember, the factors must be distinct.
Oh, and does anyone know the fastest way the primeshave been listed? (in primes per second, or somthing like that)
well, what I said ("a number with only two factors"), if the factors were the same, then there'd only be one factor, not two, so by default they are distinct anyway, aren't they?
Actually, No.
if a number has two factors which are the same, then in definitions these are still considered two factors. Similar to how every quadratic has two factors, even if they are both the same.
Oh I see. Thanks!
Every time I try to use this definition, people pounce on me, but here is my definition of prime numbers:
A prime number is a number which has only one prime factor.
I realize that this is using the word to be defined inside its own definition, but this eliminates the possibility of 1 being prime.
Yeah, it's like describing the word 'do' without using it.
yeah, i see what you mean. The only problem is, how do you know if the factor is "prime" :p
also: "DO" = many meanings, one being 'the act of completing an action.'
BTW: ever though that a dictionary is in fact quite a silly thing, beacuse EVERY word in it is defined using other words that also have to be defined by using other words that also have to be defined by using other words.........(etc)
i always understood the definition to be
You see the simple elegance of that?(NOTE this was originally in Romanian.. it is not an exact translation because the english language lacks the wording to translate it as such, but it is still the same message)Quote:
A number which has 0 prime factors less than itself.
PS: If you consider negative numbers to be prime, the definition can be changed to "A Number with 0 prime factors less than the absolute value of itself". Beauty:D
well depends on what you mean by factors.Quote:
Originally posted by sql_lall
Actually, No.
if a number has two factors which are the same, then in definitions these are still considered two factors. Similar to how every quadratic has two factors, even if they are both the same.
well, what I said ("a number with only two factors"), if the factors were the same, then there'd only be one factor, not two, so by default they are distinct anyway, aren't they?
x2=x*x
you are right, it does infact has two factors, x and x, although the factors are not distinct, it is still two factors.
x2=1*x*x
now it has three factors. now the factor 1 is different from the factor x. i think you know what the difference is.
also, x2=(1/x)*(x^3)=1*1*1*1*...*1*(2)*(1/2)*(3)*(1/3)*...*e*(1/e)*1.999....*(1/1.999...)*i*(1/i)*(4^2)*(1/4^2)*...*x*x
well you get the point. depends on ur definition of factors. now would all the "1" counts as a factor themselves by the quote or do they count as one single factor :D :D :D
it goes with the numerical numbers too, not just variables. so i guess my point is, what sql_lall did makes sense, his definition was informative.
I see what you mean, thanks for pointing that out. Just a few things though.
I had actually said "positive integer factors" earlier, and was just shortening it to factors now.
Also, i don't know how to explain the factor of 1 thing, apart from maybe that the factor of one is a special case and is only considered once (same as -1)
BTW: That thing about how rational numbers and their even/odd denominators, (1/3 probablility), is that because more ones with even denominators will cancel down than ones with odd denominators??
lol i just found this somewhere and thought it was interesting so i added to my signiture
Firstly, the prime factorization of any number has to be unique, since math is the subject in which each question has one definite answer, which doesn't vary unless the question is varied. 1 is never considered in the prime factorization of numbers, because it ruins the uniqueness of the prime factorization, as it may be counted multiple times.:rolleyes:
1 is not prime because it has only one distinct factor, itself. It should not be included in prime factorization.
Oops!:eek:
Sorry, I just went from thinking about factoring to typing about prime factoring in which 1 wouldn't be considered because it isn't prime anyway. I should proofread my posts from now on.;)
don't worry about it. I failed my first English lessons because I misspelled every instance of the word "and" as "canned". Trust me, that was embarrasing, not this.:p
NOTE: it was my teacher's accent that was to blame, I never make mistakes:cool:
seems that even though u post before changing signatures when u view the message it still gives you the new sig.
anywayz i want to prove this:
I've came across this reference:
The probability that a random rational number has an even denominator is 1/3 (Salamin and Gosper 1972)?
i could not find the original proof, I attempted to develop one of my own, but it lacks something in the end:
ok, first lets establish that by adding an integer c to every element of the set of all rational numbers between 0 and 1, we can establish another set of rational numbers S such that each and one of them is between two consecutive integers, namely c and c+1, with no other rational number that is not in the set S but is also bewtween c and c+1. so proving the probability over the whole set of rational numbers is the same as proving the probability over the rational numbers between 0 and 1.
Lets consider a/b, where a and b are integers. if b is odd that accounts for 1/2 of the set.
lets look at a is even and b is even and they reduce down to any/odd. any can be either even or odd.
so lets start from the any even number b. in order to give you odd/odd, numerator has to have at least as many as factor of 2 as denominator.
let f(x) denote (x/2^d)/(x/2), where d is the largest integer such that 2^d divides into x. x/2^d accounts for the number of positive integers that is less than or equal x but is divisible by 2^d. x/2 denote the number of even numbers less than or equal to x. so f(x) is the probablity of all the rational numbers between 0 and 1 that has an even denominator of x such that after reduced will have an odd denominator
btw f(x) reduces to 2/g(x)
then the probability of even/even reducing down to odd/odd is:
(f(2)+f(4)+f(6)+...+f(x))/x
as x gets really large
or
(1+1/2+1+1/4+1+...)/x
I checked this using a computer program: the above the equation should converges to 2/3 which makes the original statement true 2/3 of even/even can be reduced to even/odd -- even/even accounts for 1/4 of the set making (2/3 * 1/4) = 1/6. adding on the odd denominator 1/2 gives you 1/2+1/6 = 2/3, so a random rational number after reduced that gives u an odd denomator gives you 2/3, making a random rational number after reduced that gives u an even denominator 1/3.
can anyone provide me with a mathematical equation?
also, i am looking for another proof:
For a, b, and c any different rational numbers, then
1/(a-b)^2+1/(b-c)^2+1/(c-a)^2
is the square of a rational number (Honsberger 1991).
i just verified using c++:
here is the code. anyone want to give me a mathematical proof?
Quote:
#include<iostream.h>
unsigned long x[32];
void main(){
int i,j,limit;
double prob=0;
cout<<"Enter the limit you want to extend to:"<<endl;
cin>>limit;
x[0]=1;
for (i=1;i<32;i++)
x[i]=x[i-1]*2;
for (i=1;i<=limit;i++){
for (j=1;i%x[j]==0;j++);
prob+=(double) 1/x[j-1];
}
prob/=limit;
cout<<prob<<endl;
}
proof for what?
Did you know that...
The probability that a random rational number has an even denominator is 1/3 (Salamin and Gosper 1972)?
brb lol
well i dont understand anything so im going to live u alone ..
bye :D
As far as I'm aware, the defnintion of a prime number is defined:
A prime number is a positive integer, p > 1 that has no positive integer divisors other than 1 and p itself.
Even if it's not the only definition, you can't argue that it's wrong.
I think it was someone on this forum who posted code for finding if a number is prime or not without loops, using the definition "A prime number is a number which can only be divided by itself and one to return a positive integer". It was something like:
PHP Code:bool IsPrime ( int number )
{
if ( int(number / 1) ) return true;
if ( int(number / number) ) return true;
}
I really hope that's a joke dreamlax. That would return true for any number.:p
you understand what the word only means right Dreamlax.also the definition of a composite (non-prime) number is:Quote:
prime number is a number which can only be divided by itself and one to return a positive integer
Quote:
composite number is a number which can not only be divided by itself and one to return a positive integer, but also can be divided by another positive integer other than 1 and itself and also return a positive integer.
Trust me it's a joke.Quote:
Originally posted by snakeeyes1000
I really hope that's a joke dreamlax. That would return true for any number.:p
Trust me, i'm laughing....internally.
Anyway, have any of you heard about that new function-like thing that finds if a number is prime, faster than any other function. I have had a read of the paper ( a shortened version) and if someone understands what most of it is about and explaination would be very useful.
this is probably what you want, sql_lall
http://www.vbforums.com/showthread.php?threadid=191365
You might like to check out Fermat's Little Theorem . . .