Yes. We are to claim the irrationality of [pmath size=14] sqrt 2 [/pmath].
We will use proof by contradiction.
Assume that [pmath size=14] sqrt 2 [/pmath] is a rational number. Then by definition of rational numbers, we may write that
[pmath size=12] sqrt 2 = n / m [/pmath]
where n, m are integers.
Squaring on both sides: [pmath size=12] (sqrt 2)^2 = (n / m)^2[/pmath], or
[pmath size=12] 2 = n^2 / m^2 [/pmath]
which leads to [pmath size=12] 2 m^2 = n^2 [/pmath] (*)
Let us show that (*) cannot hold true.
Method 1.
Let [pmath size=12] m = 2^p K, n = 2^q L [/pmath], where p, q are integers, K and L are odd integers.
{If either K or L are even numbers, then factor 2 can be extracted iteratively until an odd number (co-factor) is revealed. }
Bringing m, n in the above forms into [pmath size=12] 2 m^2 = n^2 [/pmath], then:
[pmath size=12] 2 ((2^p) K)^2 = (2^q L)^2 [/pmath]
[pmath size=12] 2^{2p+1} K^2 = 2 ^{2q} L^2 [/pmath] (++)
Let us count the number of 2’s on each side. The l.h.s has (2p+1) of factor 2’s (an odd count of 2), and the r.h.s. has (2q) of 2’s (an even count of 2’s). However, it is well established that for any integer, the prime factorization is unique. Therefore, equation (++) cannot holds. Tracing back, we have to revoke the initial assumption
[pmath size=12] sqrt 2 = n / m [/pmath]
Therefore, number [pmath size=12] sqrt 2 [/pmath] is not a rational number.
Method 2.
Note in [pmath size=12] n^2 = 2 m^2[/pmath], the right hand side is even, so the left side is also even. This implies that n must contain factor 2. Let n = 2 N, then
[pmath size=12] 2 m^2 = (2 N)^2[/pmath] — > [pmath size=12] m^2 = 2 N^2 [/pmath]
where the new equation [pmath size=12] m^2 = 2 N^2 [/pmath] implies that number m must contain factor 2. So each of n, m and N contains factor 2.
This process may be applied in an iterative manner, forever. Consequently, both m and n contains an infinite many factors of 2’s. However, this is not possible as either n or m is a finite number.
You got it?
If you got it, you shall be able to prove that [pmath size=12] sqrt 3 [/pmath] is also irrational (i.e. [pmath size=12] sqrt 3 [/pmath] cannot be written as the ratio of two integers.) How? Just use the same idea, but this time, you need to count the number of 3’s.