Theorem 1.3 : Let p be a prime number. If p divides a2, then p divides a,
where a is a positive integer.
*Proof : Let the prime factorisation of a be as follows :
a = p1 p2 . . . pn, where p1, p2. . . . pn are primes, not necessarily distinct.
Therefore, a2 = ( p1 p2 . . . pn )( p1 p2 . . . pn ) = p12 p22 . . . 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 2 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’.
Theorem 1.4 : √2 is irrational.
Proof : Let us assume, to the contrary, that √2 is rational.
So, we can find integers r and s (≠ 0) such that √2 = $ {r}/{s} $ .
Suppose r and s have a common factor other than 1.
Then, we divide by the common factor to get √2=$ {a}/{b} $ ,
where a and b are coprime.
So, b√2 = a.
Squaring on both sides and rearranging, we get
2b2 = a2.
Therefore, 2 divides a2.
Now, by Theorem 1.3, it follows that 2 divides a.
So, we can write a = 2c for some integer c.
Substituting for a, we get 2b2 = 4c2, that is, b2 = 2c2.
This means that 2 divides b2, and so 2 divides b (again using Theorem 1.3 with p = 2).
Therefore, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that √2 is rational.
So, we conclude that √2 is irrational.
