Solve question 13 in english

Okay, Ayush! Let's solve question 13 in English.

We will prove that $\sqrt{2}$ is an irrational number. A similar proof applies to $\sqrt{3}$ and $\sqrt{5}$.

Proof that $\sqrt{2}$ is irrational:

Step 1: Assume for contradiction that $\sqrt{2}$ is a rational number. If $\sqrt{2}$ is a rational number, it can be expressed in the form $\frac{a}{b}$, where $a$ and $b$ are coprime integers and $b \neq 0$. $\sqrt{2} = \frac{a}{b}$

Step 2: Square both sides of the equation. Squaring both sides gives us: $(\sqrt{2})^2 = \left(\frac{a}{b}\right)^2$ $2 = \frac{a^2}{b^2}$ $a^2 = 2b^2$ This implies that $a^2$ is an even number. If $a^2$ is an even number, then $a$ must also be an even number.

Step 3: Substitute $a$ with $2k$. Since $a$ is an even number, we can write $a = 2k$ for some integer $k$. Substituting this value into $a^2 = 2b^2$: $(2k)^2 = 2b^2$ $4k^2 = 2b^2$ $2k^2 = b^2$ This implies that $b^2$ is an even number. If $b^2$ is an even number, then $b$ must also be an even number.

Step 4: Establish the contradiction. We have found that both $a$ and $b$ are even numbers. This means that $a$ and $b$ both have a common factor of 2. This contradicts our initial assumption that $a$ and $b$ are coprime (i.e., they have no common factors other than 1).

Step 5: Conclude. Since our initial assumption leads to a contradiction, our assumption must be false. Therefore, $\sqrt{2}$ cannot be a rational number. Hence, $\sqrt{2}$ is an irrational number.

Proof for $\sqrt{3}$ and $\sqrt{5}$: The proof for $\sqrt{3}$ and $\sqrt{5}$ being irrational is analogous to the proof for $\sqrt{2}$. For $\sqrt{3}$, we would get $a^2 = 3b^2$, implying both $a$ and $b$ are multiples of 3, contradicting the coprime assumption. For $\sqrt{5}$, we would get $a^2 = 5b^2$, implying both $a$ and $b$ are multiples of 5, contradicting the coprime assumption.

Therefore, $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{5}$ are all irrational numbers.

$\boxed{\text{It is proven that } \sqrt{2}, \sqrt{3}, \sqrt{5} \text{ are irrational numbers.}}$