How to make a program to check if a number is a prime?

Hello I'm just starting with matlab and I have to make a program to check if a number is a prime. My pgrogramme looks like that but it ain't working.

    function prime(x)
    i= 3:2:sqrt(x) ;
     if mod(x,2) == 0
        disp ('is not a prime')
     if mod(x,i) > 0
        disp('is not a prime')
     else
        disp ('is a prime')
     end
    end

Can you help me make my code work please.

[Waseem Wasi]

First condition is true but in second you to have write if(mod(x,2)==1) then this will be show a prime number

Answer 1

[Sean de Wolski]

doc isprime

[Marin]

I know there's a program to do that, I have to make my own...

[Jan]

@Marin: It is useful to mention such important details in the question already. Otherwise the forum users waste the time for mentioning the built-in functions.

Answer 2

[Jan]

Do you see the difference between:

if mod(x,2) == 0
  disp ('is not a prime')

and

if mod(x,i) > 0
   disp ('is not a prime')

In the first test you check, if the modulo returns 0. In the second, you want the result to be non-zero to decide, that x is not a prime. Suspicious.

Then the variable i is a vector. In consequence mod(x, i) replies a vector also. But the if-condition must be a scalar, as the English term "if" implies more or less. The functions all() and any() are useful in such cases.

You find a lot of useful links, when you search in the net for prime number algorithms. E.g. ask WikiPedia for Eratosthenes.

Answer 3

[Anmol singh]

Prime number - a number can devide by itself or 1 so now we can iterate loop 2 to n/2 , if n is devide by any number from 2 to n/2 then it is not prime .

loop : i = 2 <= n/2 
        if mod(n/i == 0)
    disp ('is not a prime');
    return ;
      end if
 i++;
end loop 
    disp ('it is  a prime');
return ;

It is very nice trick to check prime number till n/2 .Not need to iterate 2 to n-1

Refrence :

How to Check number is prime or not

[Walter Roberson]

That is not MATLAB code. It is also not C or C++ code. I do not know what programming language it is. Certainly not java as might be hinted by the link to a java blog.

There is no need to iterate to n/2: you can stop iterating at floor(sqrt(n))