The question in the uploaded image asks in Hindi:

"सिद्ध कीजिए कि √5 एक अपरिमेय संख्या है।"
(Translation: Prove that √5 is an irrational number.)

Proof:

To prove that √5 is an irrational number, let us proceed by contradiction.

1. Assume the contrary: Suppose that √5 is a rational number.
   Then, it can be expressed as a ratio of two integers $p$ and $q$, where $p$ and $q$ are coprime integers (having no common factor other than 1), and $q \neq 0$:
   $\sqrt{5} = \frac{p}{q}$

2. Square both sides: Squaring both sides gives: $5 = \frac{p^2}{q^2}$

3. Rewrite the equation: Multiply through by $q^2$ to eliminate the denominator: $p^2 = 5q^2$ This equation shows that $p^2$ is divisible by 5.

4. Conclude $p$ is divisible by 5: If $p^2$ is divisible by 5, then $p$ must also be divisible by 5 (since 5 is a prime number).
   Let $p = 5k$, where $k$ is an integer.

5. Substitute $p = 5k$ into the equation: Substitute $p = 5k$ into $p^2 = 5q^2$: $(5k)^2 = 5q^2$ $25k^2 = 5q^2$ $q^2 = 5k^2$ This equation shows that $q^2$ is divisible by 5.

6. Conclude $q$ is divisible by 5: If $q^2$ is divisible by 5, then $q$ must also be divisible by 5.

7. Contradiction: If both $p$ and $q$ are divisible by 5, this contradicts the assumption that $p$ and $q$ are coprime (since they now have a common factor of 5).

8. Conclusion: Therefore, our assumption that $\sqrt{5}$ is a rational number is false.
   Hence, $\sqrt{5}$ is an irrational number.

Would you like me to clarify or expand on any part of the proof?

5 Related Questions:

1. What is the definition of an irrational number?
2. How do you prove that $\sqrt{2}$ is irrational?
3. Can any square root of a non-perfect square be rational?
4. What is the importance of prime numbers in proofs like this?
5. How can the concept of irrational numbers be applied in real-world scenarios?

Tip:

In proofs by contradiction, always ensure the assumptions you make are logically consistent with the definitions provided.