Assume (1)/(sqrt(2)) is a rational number.

Here's how to prove that the given numbers are irrational: Part (I): Prove (1)/(sqrt(2)) is irrational. Step 1: Assume (1)/(sqrt(2)) is a rational number. If (1)/(sqrt(2)) is rational, 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. sqrt(2) = (b)/(a) Squaring both sides: (sqrt(2))^2 = ((b)/(a))^2 2 = (b^2)/(a^2) 2a^2 = b^2 (Equation 1) Step 3: Analyze the implications of Equation 1. From 2a^2 = b^2, it means that b^2 is divisible by 2. If b^2 is divisible by 2, then b must also be divisible by 2. So, we can write b = 2k for some integer k. Step 4: Substitute b=2k into Equation 1. 2a^2 = (2k)^2 2a^2 = 4k^2 a^2 = 2k^2 (Equation 2) Step 5: Analyze the implications of Equation 2. From a^2 = 2k^2, it means that a^2 is divisible by 2. If a^2 is divisible by 2, then a must also be divisible by 2. Step 6: Reach a contradiction. From Step 3, b is divisible by 2. From Step 5, a is divisible by 2. This means that a and b have a common factor of 2. However, in Step 1, we assumed that a and b are coprime integers (meaning they have no common factors other than 1). This contradicts our initial assumption. Step 7: Conclude. Since our assumption leads to a contradiction, our initial assumption that (1)/(sqrt(2)) is rational must be false. Therefore, (1)/(sqrt(2)) is an irrational number. Part (I) is irrational --- Part (II): Prove 7sqrt(5) is irrational. Step 1: Assume 7sqrt(5) is a rational number. If 7sqrt(5) is rational, 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) Step 3: Analyze the implication. Since a and b are integers, a is an integer and 7b is a non-zero integer. Therefore, (a)/(7b) is a rational number. This implies that sqrt(5) is a rational number. Step 4: Reach a contradiction. However, we know that sqrt(5) is an irrational number. This contradicts our conclusion from Step 3. Step 5: Conclude. Since our assumption leads to a contradiction, our initial assumption that 7sqrt(5) is rational must be false. Therefore, 7sqrt(5) is an irrational number. Part (II) is irrational --- Part (III): Prove 6 + sqrt(2) is irrational. Step 1: Assume 6 + sqrt(2) is a rational number. If 6 + sqrt(2) is rational, 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) Step 3: Analyze the implication. 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. This implies that sqrt(2) is a rational number. Step 4: Reach a contradiction. However, we know that sqrt(2) is an irrational number. This contradicts our conclusion from Step 3. Step 5: Conclude. Since our assumption leads to a contradiction, our initial assumption that 6 + sqrt(2) is rational must be false. Therefore, 6 + sqrt(2) is an irrational number. Part (III) is irrational