Introduction:
In mathematics, a rational number is a number that can be represented as the quotient or fraction p/q of two integers, where p is the numerator and q is the non-zero denominator. An irrational number is a number that cannot be expressed as a ratio of two integers. In this answer, we will prove that the square root of 2 (√2) is an irrational number.

Proof by contradiction:
We will prove by contradiction that √2 is an irrational number. Suppose √2 is a rational number, then it can be expressed in the form p/q, where p and q are co-prime integers (i.e., they have no common factors other than 1).

Squaring both sides:
Squaring both sides of the equation √2 = p/q, we get 2 = p^2/q^2.

Dividing both sides by 2:
Dividing both sides of the equation by 2, we get p^2/q^2 = 1/2.

Deduction:
This means that p^2 is an even number (since it is the product of 2 and 1/2), which implies that p must be even (since the square of an odd number is odd and the square of an even number is even).

Expressing p as 2m:
We can express p as 2m, where m is an integer. Substituting this value of p in the equation p^2/q^2 = 1/2, we get (2m)^2/q^2 = 1/2, which simplifies to 2m^2/q^2 = 1/2.

Multiplying both sides:
Multiplying both sides of the equation by q^2, we get 2m^2 = q^2/2.

Deduction:
This means that q^2 is an even number (since it is the product of 2 and 2m^2), which implies that q must be even (since the square of an odd number is odd and the square of an even number is even).

Contradiction:
But this contradicts our assumption that p and q are co-prime (i.e., they have no common factors other than 1), since both p and q are even in this case. Therefore, our original assumption that √2 is a rational number is false, and thus √2 is an irrational number.

Conclusion:
Hence, we have proved that the square root of 2 (√2) is an irrational number using the method of proof by contradiction.