# Proof: Why There Is No Rational Number Whose Square is 2?

This article is focused on discussing the proof that there is no rational number whose square is 2. Before starting the proof, let’s get familiar with the basic terms-

**Rational Numbers****: **

A number that can be expressed in the form of p/q, where p and q are integers and q ≠ 0, is known as a rational number. Examples are 0, 1, -1, 5/2, etc.

**Problem Statement:**

There is no rational number whose square is 2.

**Solution:**

Let’s get started with the proof of the above problem statement. Proof in this article will be done using a mathematical technique called Proof By Contradiction.

**Proof by Contradiction-**

It is a mathematical technique, in which first, the assumption that the proposition that needs to be proved is false and then deduces a result using the false proposition, which comes out to be contradictory either to the assumption, or any other commonly known mathematical result. Thus, proving the validity of the proposition. That’s why this technique is known as Proof by Contradiction.

**Proof-**

In this proof, the Proof by Contradiction technique is being used, where first it is assumed that there exists a rational number, whose square equals 2, and then deduce a result using this assumption, which will come out to be contradictory with our assumption. So. Let’s get started with the proof-

1. Assume that there exists a rational number, X = p/q whose square is equal to 2 such that p and q are in their simplest form, i.e. they don’t have any common factor.

X^{2}= 2 (Assumption) (p/q)^{2}= 2 (Since X is a rational number)

2. This implies,

p^{2}/q^{2}= 2 p^{2}= 2q^{2}---(1)

3. From the above equation, it can be said that p^{2} is an even integer as it can be expressed in the form of 2k, where k = q^{2}. Now, it is known that the square of an odd integer is always odd, which means that p cannot be an odd integer, thus p is also an even integer. Hence, p can be expressed in the form 2k, where k is some integer i.e.

p = 2k, for some integer k ---(2)

3. Now, after substituting the value of p from equation (2) in equation (1), the following equation is obtained-

(2k)^{2}= 2q^{2}4k^{2}= 2q^{2}---(3)

3. Dividing both sides by 2, in equation (3), the following equation is obtained-

2k^{2}= q^{2}q^{2}= 2k^{2}---(4)

4. Again, it can be said that q^{2} is an even integer, and since it is known that the square of an odd integer is always odd, thus q cannot be an odd integer. This implies that q is also an even integer.

5. Now, from the above discussion, it can be concluded that p and q are both even integers i.e. they have a common factor of at least 2, but this statement contradicts the assumption at the beginning of this proof that p and q are in their simplest form i.e. they don’t have any common factor.

This contradiction means that the assumption that **there exists a rational number, X = p/q whose square is equal to 2 such that p and q are in their simplest form, is false**. Thus, it proves that there exists no rational number, whose square is equal to 2.

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the **Essential Maths for CP Course** at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**