Proof of irrationality of √2, √3, √5

Step 1: In (p^2=2q^2), the right side has factor (2), so (p^2) is even. Step 2: If the square of an integer is even, the integer is also even, so (p) is even. Step 3: Do not directly write (p=2q); first use divisibility.

Step 1: From (p^2=3q^2), (p^2) is divisible by (3). Step 2: Since (3) is prime, if (p^2) is divisible by (3), then (p) is also divisible by (3). Step 3: Then writing (p=3k) is valid.

Step 1: (p=5k) shows that (p) is divisible by (5). Step 2: Substituting it in (p^2=5q^2) gives (q^2=5k^2), so (q) is also divisible by (5). Step 3: A common factor (5) in both creates the contradiction.

Step 1: If (p=2r), then (p^2=4r^2). Step 2: From (4r^2=2q^2), dividing both sides by (2) gives (q^2=2r^2). Step 3: This proves (q^2), and then (q), is even.

Step 1: (p=3r) and (q=3s) mean both (p) and (q) have common factor (3). Step 2: The numerator and denominator of a lowest-form fraction should be coprime. Step 3: Thus this contradicts the lowest-form assumption.

Step 1: From (p^2=5q^2), first (p^2) and then (p) are proved divisible by (5). Step 2: Only after substituting (p=5k) do we get (q^2=5k^2). Step 3: So directly concluding about (q) is an order mistake.

Step 1: (p=2m) and (q=2n) mean both are divisible by (2). Step 2: This means (2) is their common factor. Step 3: Coprime numbers should not have a common factor other than (1).

Step 1: From (p^2=2q^2), first only (p^2) and then (p) are proved even. Step 2: After substituting (p=2k), (q^2=2k^2) is obtained and then (q) is proved even. Step 3: Skipping order is considered an error in proof writing.

Step 1: In the rational assumption, (\frac{p}{q}) was taken in lowest form. Step 2: (p=3k) and (q=3r) show common factor (3) in both. Step 3: This contradicts lowest form, so the rational assumption is false.

Step 1: Squaring (p=5k) gives ((5k)^2). Step 2: The correct value is (25k^2), not (5k^2). Step 3: Forgetting to square the coefficient can be a major proof error.

Step 1: All three proofs start with the rational assumption. Step 2: At the end, the same prime factor is found common in numerator and denominator. Step 3: This is impossible in a lowest-form fraction, so the proof is completed by contradiction.

Step 1: From (p^2=3q^2), (p) is divisible by (3), so (p=3k). Step 2: Substitution gives (9k^2=3q^2), then (q^2=3k^2). Step 3: This proves (q) is also divisible by (3).

Step 1: From (p^2=5q^2), (p^2) is divisible by (5). Step 2: If a square is divisible by a prime, the original number is also divisible by that prime. Step 3: So (5) being prime is the main basis here.

Step 1: At the start, (\frac{p}{q}) is taken in lowest form, so (p) and (q) are assumed coprime. Step 2: The proof shows both are divisible by (2). Step 3: This is the clear and correct contradiction.

Step 1: To get (3) from (\sqrt{3}), both sides must be squared. Step 2: The square of a fraction is (\frac{p^2}{q^2}). Step 3: Therefore the correct equation is (3=\frac{p^2}{q^2}).

Step 1: If both are divisible by (5), numerator and denominator have common factor (5). Step 2: Such a situation cannot occur in lowest form. Step 3: Therefore the rational assumption is impossible and (\sqrt{5}) is irrational.

Step 1: First (p) is proved even from (p^2=2q^2). Step 2: Substituting (p=2k) gives (q^2=2k^2). Step 3: This proves (q^2), and then (q), is even.

Step 1: If a prime divides a square, it also divides the original number. Step 2: In (\sqrt{3}), this rule is used for (3), and in (\sqrt{5}), for (5). Step 3: This helps get a common factor in numerator and denominator.

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: Therefore writing (q=3r) is correct.

Step 1: Writing (\sqrt{2}=2), (\sqrt{3}=3), or (\sqrt{5}=5) is wrong. Step 2: In the correct proof, the square root is assumed as a fraction and then squared. Step 3: Do not confuse the square root with the number inside it.