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Here's how to prove each number is irrational: a) Prove that (1)/(sqrt(2)) is irrational. Step 1: Assume, for the sake of contradiction, that (1)/(sqrt(2)) is a rational number. If (1)/(sqrt(2)) is rational, then it can be written in the form (a)/(b), where a and b are coprime integers and b ≠ 0. (1)/(sqrt(2)) = (a)/(b) Step 2: Rearrange the equation to isolate sqrt(2). sqrt(2) = (b)/(a) Since a and b are integers and a ≠ 0, (b)/(a) is a rational number. Step 3: Identify the contradiction. This implies that sqrt(2) is a rational number. However, it is a known fact that sqrt(2) is irrational. This contradicts our initial assumption. Step 4: Conclude the proof. Therefore, our assumption that (1)/(sqrt(2)) is rational must be false. Hence, (1)/(sqrt(2)) is irrational. b) Prove that 7sqrt(5) is irrational. Step 1: Assume, for the sake of contradiction, that 7sqrt(5) is a rational number. If 7sqrt(5) is rational, then it can be written in the form (a)/(b), where a and b are coprime integers and b ≠ 0. 7sqrt(5) = (a)/(b) Step 2: Rearrange the equation to isolate sqrt(5). sqrt(5) = (a)/(7b) Since a and b are integers and 7b ≠ 0, (a)/(7b) is a rational number. Step 3: Identify the contradiction. This implies that sqrt(5) is a rational number. However, it is a known fact that sqrt(5) is irrational. This contradicts our initial assumption. Step 4: Conclude the proof. Therefore, our assumption that 7sqrt(5) is rational must be false. Hence, 7sqrt(5) is irrational. c) Prove that 6+sqrt(2) is irrational. Step 1: Assume, for the sake of contradiction, that 6+sqrt(2) is a rational number. If 6+sqrt(2) is rational, then it can be written in the form (a)/(b), where a and b are coprime integers and b ≠ 0. 6+sqrt(2) = (a)/(b) Step 2: Rearrange the equation to isolate sqrt(2). sqrt(2) = (a)/(b) - 6 sqrt(2) = (a - 6b)/(b) Since a and b are integers and b ≠ 0, (a - 6b) is an integer and b is an integer. Therefore, (a - 6b)/(b) is a rational number. Step 3: Identify the contradiction. This implies that sqrt(2) is a rational number. However, it is a known fact that sqrt(2) is irrational. This contradicts our initial assumption. Step 4: Conclude the proof. Therefore, our assumption that 6+sqrt(2) is rational must be false. Hence, 6+sqrt(2) is irrational. Send me the next one 📸