prime or not prime

The question is very simple: if we type isprime(12646216567629137) matlab answers '0' which means that the number is not prime. But indeed is a prime (tested with Mathematica). So, what's wrong? Could be something related with the digits of the number?

Answer 1

[Dyuman Joshi]

Not all numbers can be exactly represented in double precision (which is the default numeric data type in MATLAB). The largest consecutive integer possible in floating point format is

s = flintmax;
fprintf('%d',s)

9007199254740992

which corresponds to 2^53.

What's the distance between p and the next largest representable number?

e = eps(s)

e = 2

Thus, the next value that can be represented in double precision is

S = s + eps(s);
fprintf('%d',S)

9007199254740994

If we add something smaller than eps(s) to s, the result does not changes -

fprintf('%d', s+e-0.2)

9007199254740994

As you can see, this is a limitation of double precision numbers (IEEE 754 standard).

Read more here - double, eps

https://in.mathworks.com/help/matlab/matlab_prog/floating-point-numbers.html

https://en.wikipedia.org/wiki/Double-precision_floating-point_format

Alternatively, you can use integers according to the value you have - Data Type - Integers. Note that each integer data types have a limited precision as well.

z1=uint64(12646216567629137)

z1 = uint64 12646216567629137

isprime(z1)

ans = logical
   1

In case the value is so large, that none of the numeric data types can store it exactly, you can use symbolic numbers (Note requires Symbolic Math Toolbox)

digits(20)
z2 = sym('12646216567629137')

z2 = 

12646216567629137

isprime(z2)

ans = logical
   1

Choosing a 20 digit prime number -

z3 = uint64(975319753197531975319)

z3 = uint64 18446744073709551615

As you can see the stored value is not the same as the value provided.

z4 = sym('975319753197531975319')

z4 = 

975319753197531975319

isprime(z4)

ans = logical
   1

When defining symbolic number for a large numerical value, provide the number inside quotation marks.

[B Quintanilla]

Thanks!