Primality test: Difference between revisions

Note that the largest factor, 50, is half of 100. This holds true for all ”n”: all divisors are less than or equal to ”n”/2.

:$1\; times\; 100,\; ;\; 2\; times\; 50,\; ;\; 4\; times\; 25,\; ;\; 5\; times\; 20,\; ;\; 10\; times\; 10,\; ;\; 20\; times\; 5,\; ;\; 25\; times\; 4,\; ;\; 50\; times\; 2,\; ;\; 100\; times\; 1$.

Notice that products past $10\; times\; 10$ are the reverse of products which appeared earlier. For example, $5\; times\; 20$ and $20\; times\; 5$ are the reverse of each other. Note further that of the two divisors, $5\; leq\; sqrt\{100\}\; =\; 10$ and $20\; geq\; sqrt\{100\}\; =\; 10$. This observation generalizes to all $n$: all divisor pairs of $n$ contain a divisor less than or equal to $sqrt\{n\}$, so the algorithm need only search for divisors less than / equal to $sqrt\{n\}$ to guarantee detection of all divisor pairs.

:{{math|2 × 50, 4 × 25, 5 × 20, 10 × 10, 20 × 5, 25 × 4, 50 × 2}}

Also notice that 2 is a prime dividing 100, which immediately proves that 100 is not prime. Every positive integer except 1 is divisible by at least one prime number by the [[Fundamental Theorem of Arithmetic]]. Therefore the algorithm need only search for ”prime” divisors less than / equal to $sqrt\{n\}$.

Notice that products past 10 × 10 merely repeat numbers which appeared in earlier products, merely [[commutativity|commuted]]. For example, 5 × 20 and 20 × 5 consist of the same numbers in opposite order. This holds true for all ”n”: all unique divisors of ”n” are numbers less than or equal to {{sqrt|”n”}}, so we need not search past that. (In this example, $sqrt\; n$ = $sqrt\; \{100\}$ = 10.)

For another example, consider how this algorithm determines the primality of 17. One has , and the only primes $leq\; sqrt\{17\}$ are 2 and 3. Neither divides 17, proving that 17 is prime. For a last example, consider 221. One has , and the primes $leq\; sqrt\{221\}$ are 2, 3, 5, 7, 11, and 13. Upon checking each, one discovers that $221\; /\; 13\; =\; 17$, proving that 221 is not prime.

All even numbers greater than 2 can also be eliminated: if an even number can divide ”n”, so can 2.

In cases where it is not feasible to compute the list of primes $leq\; sqrt\{n\}$, it is also possible to simply (and slowly) check all numbers between $2$ and $sqrt\{n\}$ for divisors. A rather simple optimization is to test divisibility by 2 and by just the odd numbers between 3 and $sqrt\{n\}$, since divisibility by an even number implies divisibility by 2.

An example is to use trial division to test the primality of 17. We need only test for divisors up to $sqrt\; n$, i.e. integers less than or equal to $scriptstyle\; sqrt\{17\}\; approx\; 4.12$, namely 2, 3, and 4. 4 can be skipped because it is an even number: if 4 could evenly divide 17, 2 would too, and 2 is already in the list. That leaves 2 and 3. Divide 17 with each of these numbers, and we find that neither divides 17 evenly—both divisions leave a remainder. So, 17 is prime.