Is root 2 rational? No, root 2 is not a rational number? Before we answer this, let us first understand what a rational number is. A real number is called a rational number if it can be written as p/q where both p and q≠0 are integers with gcd(p, q)=1. Otherwise, that number is called an irrational number. In this section, we will learn whether √2 is a rational number or not.
The negation of √2 is irrational: √2 is rational.
Table of Contents
First Proof of Root 2 is Irrational:
At first, we will prove that root 2 is an irrational number by the contradiction method. Suppose that √2 is rational. So we can write
$\sqrt{2}=\dfrac{p}{q}$ $\quad \cdots (1)$
for some relatively prime integers $p, q$, that is, gcd of $a$ and $b$ is $1$. Squaring both sides of the equation (1), we get that
$p^2=2q^2$ $\quad \cdots (2)$
This implies that $p^2$ is an even number. Now we will use the following fact:
| $A^2$ is even if and only if $A$ is even. |
As $p^2$ is even, we can conclude that $p$ is even. As $p$ is even, we write $p=2k$ for some integer $k$. From equation (2), we obtain that
$(2k)^2=2q^2$
$\Rightarrow 4k^2=2q^2$
$\Rightarrow 2k^2=q^2$.
This shows that $q^2$ is even. Hence, we deduce that $q$ is also even.
As both $p$ and $q$ are even numbers, we conclude that $2$ divides both $p$ and $q$. Thus, we arrive at a contradiction to the fact that gcd$(p,q)=1$.
So our initial assumption was wrong. In other words, we have proven that $\sqrt{2}$ is an irrational number.
Second Proof of Root 2 is Irrational:
Note that $1<2<4$. Taking square root on both sides, we have
$\sqrt{1} < \sqrt{2} < \sqrt{4}$.
It follows that $1 < \sqrt{2} <2$. As $1$ and $2$ are consecutive integers, we conclude that $\sqrt{2}$ is not an integer.
Therefore, we suppose that $\sqrt{2}$ is a rational number and we can write
$\sqrt{2}=\dfrac{a}{b}$ $\quad \cdots (3)$
where $a, b$ are relatively prime integers (i.e. gcd of $a$ and $b$ is $1$ ) with $b>1$. Squaring both sides of (3), we get that $\dfrac{a^2}{b^2}=2$. This implies that $\dfrac{a^2}{b}=2b$.
As $\gcd(a,b)=1$, we have $\gcd(a^2, b)=1$. In other words, $b$ cannot divide $a^2$. Thus, $\dfrac{a^2}{b}=2b$ is not an integer, a contradiction to the fact that $b$ is an integer. Hence our assumption $\sqrt{2}$ is rational is not correct. It means that $\sqrt{2}$ is an irrational number.
Interesting Facts:
- Root 2 is not a rational number
- √2 = 1.414 up to 3 decimal places
- Multiply root 2 with itself we will get a rational number. See that √2 × √2 = √(2×2) = 2.
