Proof of irrationality of √2, √3, √5

Correct answer: B. (5)

Explanation: Step 1: The square of (\sqrt{5}) is (5). Step 2: So the left side becomes ((\sqrt{5})^2=5). Step 3: A square and square root cancel each other.

Correct answer: A. If a prime divides a square, it also divides the original number

Explanation: Step 1: Prime factors in a square occur in pairs. Step 2: If a prime divides (p^2), it also divides (p). Step 3: This rule is needed in the proofs of (\sqrt{3}) and (\sqrt{5}).

Correct answer: B. (\sqrt{2}) is irrational

Explanation: Step 1: Assuming rationality makes both (p) and (q) even. Step 2: This contradicts their being coprime. Step 3: Therefore the initial assumption is false and (\sqrt{2}) is irrational.

Correct answer: B. Both (p) and (q) are divisible by (3)

Explanation: Step 1: In the proof of (\sqrt{3}), both (p) and (q) are found divisible by (3). Step 2: But they were assumed coprime at the beginning. Step 3: This is the final contradiction.

Correct answer: A. To show that (q) is also divisible by (5)

Explanation: Step 1: First, (p) is found divisible by (5). Step 2: Substituting (p=5k) gives divisibility by (5) for (q) too. Step 3: A common factor in both creates the contradiction.

Correct answer: B. Because (\frac{p}{q}) is taken in lowest form

Explanation: Step 1: A rational number is taken as a fraction in lowest form. Step 2: In lowest form, numerator and denominator are coprime. Step 3: Finding a common factor later contradicts this condition.

Correct answer: A. They will not remain coprime

Explanation: Step 1: If both are divisible by (5), then (5) is a common factor. Step 2: Coprime numbers should not have a common factor other than (1). Step 3: So this situation goes against being coprime.

Correct answer: A. Assuming rational makes numerator and denominator of a lowest-form fraction both even

Explanation: Step 1: Assume (\sqrt{2}) rational and write it in lowest form. Step 2: The proof gives both numerator and denominator even. Step 3: This contradicts lowest form, so (\sqrt{2}) is irrational.

Correct answer: D. From (p^2=3q^2), (p=3q)

Explanation: Step 1: From (p^2=3q^2), (p^2) is divisible by (3). Step 2: This gives (p) divisible by (3), but we cannot directly write (p=3q). Step 3: The correct way is to write (p=3k).

Correct answer: A. Integer

Explanation: Step 1: An even integer is written as (2) times an integer. Step 2: Therefore in (p=2k), (k) is an integer. Step 3: Mentioning the type of (k) makes the proof clear.

Correct answer: A. Assume rational, square, find both even, contradiction

Explanation: Step 1: First assume (\sqrt{2}) is rational. Step 2: After squaring, both (p) and (q) are found even. Step 3: Both being even contradicts the coprime condition.

Correct answer: A. (p=5k)

Explanation: Step 1: From (p^2=5q^2), (p^2) is divisible by (5). Step 2: So (p) is also divisible by (5) and is written as (p=5k). Step 3: Choose the correct factor according to the number.

Correct answer: A. The original statement is true

Explanation: Step 1: In contradiction, we work with the opposite assumption. Step 2: If that assumption becomes impossible, the original statement is true. Step 3: So when rationality fails, irrationality is proved.

Correct answer: A. In the proof of irrationality of (\sqrt{2})

Explanation: Step 1: In the proof of (\sqrt{2}), we get (p^2=2q^2). Step 2: This makes both (p) and (q) divisible by (2), that is even. Step 3: The common factor (2) creates the contradiction.

Correct answer: A. If (p^2) is even, then (p) is even

Explanation: Step 1: The square of an even number is even and the square of an odd number is odd. Step 2: So if (p^2) is even, (p) is also even. Step 3: This rule is used in the proof of (\sqrt{2}).

Correct answer: A. Therefore (\sqrt{3}) is irrational

Explanation: Step 1: The rational assumption makes both (p) and (q) divisible by (3). Step 2: This contradicts the coprime condition. Step 3: Therefore the final conclusion is that (\sqrt{3}) is irrational.

Correct answer: A. They have a common factor other than (1)

Explanation: Step 1: Coprime numbers have only (1) as a common factor. Step 2: If any common factor other than (1) is found, they are not coprime. Step 3: This contradiction is searched for in irrationality proofs.

Correct answer: A. Their being coprime

Explanation: Step 1: At the beginning, (p) and (q) were assumed coprime. Step 2: If both are divisible by (5), then (5) becomes a common factor. Step 3: So this goes against their being coprime.

Correct answer: A. In all three, we first assume rationality and write a lowest-form fraction

Explanation: Step 1: All three proofs are based on contradiction. Step 2: At the start, the number is assumed rational and written as a lowest-form fraction. Step 3: Then a common factor gives a contradiction.

Correct answer: A. The assumption led to a contradiction, so the given number is irrational

Explanation: Step 1: The proof starts with the rational assumption. Step 2: At the end, that assumption contradicts the coprime condition. Step 3: In the last line, clearly write both the contradiction and irrationality.

Correct answer: A. (q) is even

Explanation: Step 1: From (q^2=2k^2), (q^2) is divisible by (2). Step 2: If the square of an integer is even, the integer is also even. Step 3: So (q) is even, which helps form the contradiction.

Correct answer: A. (q) is divisible by (3)

Explanation: Step 1: From (q^2=3k^2), (q^2) is divisible by (3). Step 2: Since (3) is prime, (q) is also divisible by (3). Step 3: This shows a common factor in (p) and (q).

Correct answer: A. (q) is divisible by (5)

Explanation: Step 1: From (q^2=5k^2), (q^2) is divisible by (5). Step 2: Since (5) is prime, (q) is also divisible by (5). Step 3: Having (5) in both (p) and (q) contradicts the coprime condition.

Correct answer: A. Getting (p=2m) and (q=2n)

Explanation: Step 1: (p=2m) and (q=2n) mean both are divisible by (2). Step 2: Then (2) becomes their common factor. Step 3: A lowest-form fraction should not have such a common factor.

Correct answer: A. Assume rational and show the same prime factor in both numerator and denominator

Explanation: Step 1: Factor (3) works in (\sqrt{3}) and factor (5) works in (\sqrt{5}). Step 2: The rational assumption makes that same factor appear in both numerator and denominator. Step 3: This common factor contradicts lowest form.