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1. Prove that sqrt(5) is irrational. Step 1: Assume, for the sake of contradiction, that sqrt(5) is a rational number. If sqrt(5) is rational, it can be expressed as a fraction (a)/(b), where a and b are integers, b ≠ 0, and a and b are coprime (their greatest common divisor is 1). sqrt(5) = (a)/(b) Step 2: Square both sides of the equation. (sqrt(5))^2 = ((a)/(b))^2 5 = (a^2)/(b^2) Rearrange the equation: 5b^2 = a^2 This implies that a^2 is a multiple of 5. If a prime number p divides a^2, then p divides a. Thus, a is a multiple of 5. Step 3: Since a is a multiple of 5, we can write a = 5c for some integer c. Substitute a = 5c into the equation 5b^2 = a^2: 5b^2 = (5c)^2 5b^2 = 25c^2 Step 4: Divide both sides by 5. b^2 = 5c^2 This implies that b^2 is a multiple of 5. Therefore, b is also a multiple of 5. Step 5: From Step 2, we found that a is a multiple of 5. From Step 4, we found that b is a multiple of 5. This means that a and b have a common factor of 5. Step 6: This contradicts our initial assumption in Step 1 that a and b are coprime. Therefore, our initial assumption that sqrt(5) is rational must be false. Hence, sqrt(5) is an irrational number. Proven 2. Prove that 3 + 2sqrt(5) is irrational. Step 1: Assume, for the sake of contradiction, that 3 + 2sqrt(5) is a rational number. If 3 + 2sqrt(5) is rational, it can be expressed as a fraction (a)/(b), where a and b are integers, b ≠ 0, and a and b are coprime. 3 + 2sqrt(5) = (a)/(b) Step 2: Isolate the sqrt(5) term. Subtract 3 from both sides: 2sqrt(5) = (a)/(b) - 3 2sqrt(5) = (a - 3b)/(b) Step 3: Divide by 2 to completely isolate sqrt(5). sqrt(5) = (a - 3b)/(2b) Step 4: Analyze the right-hand side of the equation. Since a and b are integers, a - 3b is an integer, and 2b is a non-zero integer. Therefore, (a - 3b)/(2b) is a rational number. Step 5: This implies that sqrt(5) is a rational number. However, we know that sqrt(5) is irrational. Step 6: This is a contradiction. Our initial assumption that 3 + 2sqrt(5) is rational must be false. Therefore, 3 + 2sqrt(5) is an irrational number. Proven 3. Prove that the following are irrationals: (i) (1)/(sqrt(2)) Step 1: Assume, for the sake of contradiction, that (1)/(sqrt(2)) is a rational number. If (1)/(sqrt(2)) is rational, it can be expressed as a fraction (a)/(b), where a and b are integers, b ≠ 0, and a and b are coprime. (1)/(sqrt(2)) = (a)/(b) Step 2: Rearrange the equation to isolate sqrt(2). Take the reciprocal of both sides: sqrt(2) = (b)/(a) Step 3: Analyze the right-hand side of the equation. Since a and b are integers, and a ≠ 0, (b)/(a) is a rational number. Step 4: This implies that sqrt(2) is a rational number. However, it is a known mathematical fact that sqrt(2) is irrational. Step 5: This is a contradiction. Our initial assumption that (1)/(sqrt(2)) is rational must be false. Therefore, (1)/(sqrt(2)) is an irrational number. Proven (ii) 7sqrt(5) Step 1: Assume, for the sake of contradiction, that 7sqrt(5) is a rational number. If 7sqrt(5) is rational, it can be expressed as a fraction (a)/(b), where a and b are integers, b ≠ 0, and a and b are coprime. 7sqrt(5) = (a)/(b) Step 2: Isolate sqrt(5). Divide both sides by 7: sqrt(5) = (a)/(7b) Step 3: Analyze the right-hand side of the equation. Since a and b are integers, a is an integer, and 7b is a non-zero integer. Therefore, (a)/(7b) is a rational number. Step 4: This implies that sqrt(5) is a rational number. However, we know that sqrt(5) is irrational. Step 5: This is a contradiction. Our initial assumption that 7sqrt(5) is rational must be false. Therefore, 7sqrt(5) is an irrational number. Proven (iii) 6 + sqrt(2) Step 1: Assume, for the sake of contradiction, that 6 + sqrt(2) is a rational number. If 6 + sqrt(2) is rational, it can be expressed as a fraction (a)/(b), where a and b are integers, b ≠ 0, and a and b are coprime. 6 + sqrt(2) = (a)/(b) Step 2: Isolate sqrt(2). Subtract 6 from both sides: sqrt(2) = (a)/(b) - 6 sqrt(2) = (a - 6b)/(b) Step 3: Analyze the right-hand side of the equation. Since a and b are integers, a - 6b is an integer, and b is a non-zero integer. Therefore, (a - 6b)/(b) is a rational number. Step 4: This implies that sqrt(2) is a rational number. However, it is a known mathematical fact that sqrt(2) is irrational. Step 5: This is a contradiction. Our initial assumption that 6 + sqrt(2) is rational must be false. Therefore, 6 + sqrt(2) is an irrational number. Proven Got more? Send 'em 📸