Question

2) If p is a prime and p divides (a)?, then & divides a where a isa positive integer ​

Answer 1

Answer:

Let the prime factorisation of a be as follows :

a = p1p2 . . . pn, where p1,p2, . . ., pn are primes, not necessarily distinct. Therefore, a2 = ( p1p2 . . . pn)( p1p2 . . . pn) = p21p22 . . . pn2.

Now, we are given that p divides a2. Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a2. However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of a2 are p1, p2, . . ., pn.

So p is one of p1, p2, . . ., pn.

Now, since a = p1 p2 . . . pn, p divides a.

We are now ready to give a proof that √2 is irrational.

The proof is based on a technique called ‘proof by contradiction’.