Proof of irrationality of √2, √3, √5

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

Step 1: In these proofs, the square root is first assumed rational. Step 2: Then an impossible situation appears because numerator and denominator get a common factor. Step 3: Hence this is called proof by contradiction.

Step 1: Both have (3) as a common factor. Step 2: So the fraction can be reduced by (3), meaning it is not in lowest form. Step 3: This becomes the contradiction in the proof for (\sqrt{3}).

Step 1: (\sqrt{2}) is assumed rational and written as a fraction in lowest form. Step 2: The proof shows that numerator and denominator are both even. Step 3: This is impossible in lowest form, so (\sqrt{2}) is irrational.

Step 1: If a prime number is a factor of a square, it is also a factor of the original number. Step 2: Therefore (5\mid b^2) gives (5\mid b). Step 3: Instead of writing it without reason, mention that (5) is prime.

Step 1: (\frac{p}{q}) was taken in lowest form. Step 2: The proof shows that both (p) and (q) are divisible by (2). Step 3: So the form is no longer lowest, and the assumption fails.

Step 1: In (\sqrt{3}), the common factor obtained is (3). Step 2: In (\sqrt{5}), the common factor obtained is (5). Step 3: The idea is the same; only the prime number changes.

Step 1: If (x) were odd, then (x^2) would be odd. Step 2: Since (x^2) is given even, (x) must be even. Step 3: This rule is used immediately in the proof of (\sqrt{2}).

Step 1: (5\mid a) only tells divisibility of (a). Step 2: Later (5\mid b) is also obtained, creating a common factor. Step 3: So coprimality is not proved; a contradiction is obtained.

Step 1: (2,3,5) are prime numbers and not perfect squares. Step 2: Assuming their square roots rational creates a common factor in the coprime numerator and denominator. Step 3: Therefore (\sqrt{2}), (\sqrt{3}), and (\sqrt{5}) are all irrational.

Step 1: From (p^2=2q^2), we get (2\mid p^2). Step 2: Since (2) is prime, (2\mid p). Step 3: In such proofs, state the prime-factor rule clearly.

Step 1: A rational number is written as a ratio of two integers. Step 2: In lowest form, the numerator and denominator are coprime. Step 3: Later, getting a common factor contradicts this condition.

Step 1: (x^2=5y^2) shows that (x^2) has (5) as a factor. Step 2: Since (5) is prime, (x) is also divisible by (5). Step 3: Conclude about (x) first, then move to (y).

Step 1: Put (p=2r) in (p^2=2q^2). Step 2: Then (4r^2=2q^2), so (q^2=2r^2). Step 3: This step completes the proof that (q) is even.

Step 1: In lowest form, numerator and denominator must be coprime. Step 2: The proof forces both to have (3) as a common factor. Step 3: Coprimality and a common factor cannot occur together.

Step 1: If a prime factor appears in a square, it appears in the original number too. Step 2: Since (5) is prime, (5\mid x^2) implies (5\mid x). Step 3: This rule is the main base of the proof for (\sqrt{5}).

Step 1: Both being even means both have (2) as a common factor. Step 2: But (p) and (q) were taken coprime. Step 3: Therefore the assumption that (\sqrt{2}) is rational is false.

Step 1: From (a^2=3b^2) and (a=3k), we get (9k^2=3b^2). Step 2: Simplifying gives (b^2=3k^2), so (3\mid b). Step 3: This shows (3) in both numerator and denominator.

Step 1: Putting (x=5m) gives (y^2=5m^2). Step 2: Hence (5\mid y^2), and by the prime-factor rule (5\mid y). Step 3: This gives the final common factor.

Step 1: If (p^2) is even, then (p) is even. Step 2: Calling (p) odd violates the parity rule. Step 3: In error-based questions, check small rules carefully.