Please help me i dont understand how to find out if numbers are prime how do i have to do to find out

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- Jan 5th 2007, 05:14 PMMoneyMakerPrime Numbers
Please help me i dont understand how to find out if numbers are prime how do i have to do to find out

- Jan 5th 2007, 05:58 PMtopsquark
Say for example you have the number 139. If it is not prime it's divisible by some number less than $\displaystyle \sqrt{139} \approx 12$. So start making a list:

$\displaystyle \frac{139}{2} \neq \text{Integer}$

$\displaystyle \frac{139}{3} \neq \text{Integer}$

.

.

.

$\displaystyle \frac{139}{12} \neq \text{Integer}$

139 is not divisible by any of the numbers 2, 3, ..., 12. Thus 139 is prime. (Actually you don't need to divide by all of 2, 3, ..., 12, you may simply divide by all primes less than 12: 2, 3, 5, 7, and 11.)

Try 527. We need to check divisibility by all numbers less than $\displaystyle \sqrt{527} \approx 23$.

$\displaystyle \frac{527}{2} \neq \text{Integer}$

$\displaystyle \frac{527}{3} \neq \text{Integer}$

.

.

.

$\displaystyle \frac{527}{17} = 31$

Thus $\displaystyle 527 = 17 \cdot 31$ and thus it is not a prime number.

-Dan - Jan 6th 2007, 02:54 PMr_maths
^

gave a thank because i found that helpful :o - Jan 30th 2007, 07:23 AMmahaliabasically...
bear with me, i'm on the other side of the world so i just hope that our math terminologies are the same.

topsquark's post is very helpful, i didn't kow that first part!**basically**a prime number is something that has only two factors, being 1 and itself. our lovely mathematicians are always quarreling over whether or not 1 is prime, but when i last checked, one was**not**prime. just check with your teacher though. - Jan 30th 2007, 11:47 AMCaptainBlack
The definition of a prime natural number is: A natural number is prime if it has exctly

two distinct factors.

Suppose N is not prime then it does not have exactly two distinct factors.

Then it is either 1, which has only one distinct factor:-1, or it has more than

two distinct factors (as it must have at least two, 1 and itself).

Let a be the smallest proper factor of N, then:

N=a.b

for some integer b.

Suppose a>sqrt(N), then as b>=a a.b>sqrt(N).sqrt(N)=N a contradiction.

Therefore the smallest proper factor of a composite number N (composite

number: a non-prime with more than 2 distinct factors) is less than of

equal to sqrt(N).

RonL - Mar 12th 2007, 03:36 PMspanner
there is some formular but it is far too long to write and is basically useless unless ur using numbers greater then 1trillion.

just take the square rute of a number and then its just guess and cheack.

(All factor pairs have one of them under the square rute of the number.)