### Theory:

In grade $$9$$, we have learnt about the concept of irrational numbers and their properties. We have also learnt how to plot irrational numbers in the number line. In this topic, we are going to learn how to prove irrational numbers.
A number $$x$$ is said to be an irrational number if it cannot be expressed in the form of $$\frac{p}{q}$$ where $$q \neq 0$$.
Example:
$$\sqrt{2}$$, $$\sqrt{5}$$, $$\sqrt{7}$$ are some examples of irrational numbers.
Important!
In this topic, we are going to prove that $$\sqrt{p}$$ is irrational where $$p$$ is a prime. Now, we shall prove the theorem using the fundamental theorem of arithmetic.
Theorem:
Let $$p$$ be a prime number. If $$p$$ divides $$a^2$$, then $$p$$ divides $$a$$, where $$a$$ is a positive integer.

Proof: The prime factorization of $$a$$ can be written as:

$$a = p_1p_2 ….p_n$$

Where $$p_1, p_2, …., p_n$$ are unique prime factors of $$a$$.

Then, the prime factorization of $$a^2$$ can be written as:

$$a^2 = a \times a$$

$$a^2 = (p_1p_2 ….p_n)(p_1p_2 ….p_n)$$

$$a^2 = p_1^2 p_2^2 ….p_n^2$$

It is given that $$p$$ divides $$a^2$$. We know that from the fundamental theorem of arithmetic that $$p$$ is one of the prime factors of $$a^2$$. That is, $$p$$ is one of ($$p_1,p_2, ….,p_n$$).

Since $$a = p_1p_2 ….p_n$$ and $$p$$ is one of ($$p_1,p_2, ….,p_n$$), then $$p$$ divides $$a$$.

Hence, we proved.
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