Is root of a prime number rational? No, the square root of a prime number is not a rational number. Actually, the square root of a prime number is irrational. But before we answer this question, we know about irrational numbers and prime numbers.

Irrational number: An irrational number is a real number that cannot be expressed as p/q where both p and q≠0 are integers. For example, $\pi$ is an irrational number.

Prime number: A prime number is a positive integer greater than 1 which has exactly two factors; 1 and the number itself. For example, 5 is a prime number

Here, we will answer the following question in this article: If p is a prime number, then √p is irrational.

**Proof of Square Root of Prime is Irrational: **

Let $p$ be a prime number. We will show that root $p$ is an irrational number by the contradiction method. For a contradiction, we assume that $\sqrt{p}$ is rational. By the definition of a rational number, we can write

$\sqrt{p}=\dfrac{a}{b}$ $\quad \cdots (1)$

for some integers $a, b$ with gcd$(a, b)=1$ (greatest common divisor of $a$ and $b$ is $1$). We take squares on both sides of (1). Doing so we obtain

$p=\dfrac{a^2}{b^2}$

$\Rightarrow a^2=pb^2$ $\quad \cdots (2)$

Theorem: Let $p$ be a prime number. Then $p$ divides $A^2$ if and only if $p$ divides $A.$ |

From equation (2), It follows that $p$ divide $a^2.$ Now using the above theorem, we will get that $p$ divides $a.$ So we can write

$a=pk$ for some integer $k$

Now putting this value of $a$ into (2), we have

$(pk)^2=pb^2$

$\Rightarrow p^2k^2=pb^2$

$\Rightarrow pk^2=b^2$.

⇒ $p$ divides $b^2$

⇒ $p$ divides $b$ follows from the above theorem.

Thus we obtain that $p$ divides both $a$ and $b.$ As a result, gcd of $a$ and $b$ cannot be $1.$ Thus we arrive at a contradiction. Hence our initial assumption was not correct. This means that $\sqrt{p}$ is an irrational number.

Conclusion: The square root of primes are not rational numbers. |

Next, we will prove the square root of a prime number is not a rational number using the irreducibility of polynomials.

**Proof of Square Root of Prime is NOT Rational:**

For a prime number $p$, let us consider the polynomial

$x^2-p$

over the field of rational numbers. That is,

$x^2-p \in \mathbb{Q}[x]$

Now, $x^2-p$

$=x^2-(\sqrt{p})^2$

$=(x-\sqrt{p})(x+\sqrt{p})$

Note that $\sqrt{p}$ is not an integer as $p$ is a prime number. We know the following: Polynomial Irreducible over Integers is also Irreducible over Rationals. This implies that the factorisation $x^2-(\sqrt{p})^2$ $=(x-\sqrt{p})(x+\sqrt{p})$ is not over rational numbers. Thus, it follows that $\sqrt{p}$ is not a rational number, and we obtain the result.

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