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 (a)/(b), where a and b are coprime integers and b ≠ 0. sqrt(2) = (a)/(b) Step 2: Square both sides of the equation. Squaring both sides gives us: (sqrt(2))^2 = ((a)/(b))^2 2 = (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. It is proven that sqrt(2), sqrt(3), sqrt(5) are irrational numbers. That's 2 down. 3 left today — send the next one.