Proof of irrationality of √2, √3, √5

Step 1: In (p^2=2q^2), the right side has factor (2). Step 2: So (p^2) is even and then (p) is also even. Step 3: In proofs, first write divisibility of the square, then of the number.

Step 1: From (p^2=3q^2), (p^2) is divisible by (3). Step 2: Since (3) is prime, (p) is also divisible by (3). Step 3: Apply the prime factor rule to the correct number.

Step 1: From (p^2=5q^2), (p^2) is divisible by (5). Step 2: Since (5) is prime, (p) is divisible by (5). Step 3: After divisibility, write (p=5k), where (k) is an integer.

Step 1: If (p=2k), then (p^2=(2k)^2=4k^2). Step 2: Substituting in (p^2=2q^2) gives (4k^2=2q^2). Step 3: Writing ((2k)^2) as (2k^2) is a common mistake.

Step 1: Divide both sides of (9k^2=3q^2) by (3). Step 2: We get (3k^2=q^2), that is (q^2=3k^2). Step 3: This leads to (q) being divisible by (3).

Step 1: Divide both sides of (25k^2=5q^2) by (5). Step 2: We get (5k^2=q^2), that is (q^2=5k^2). Step 3: This is used to prove (q) is also divisible by (5).

Step 1: From (p^2=2q^2), we only conclude that (p^2) is even. Step 2: Then (p) is even and (p=2k) is written. Step 3: Writing (p=2q) directly is an algebraic mistake.

Step 1: Coprime numbers have no common factor except (1). Step 2: If both are divisible by (3), they have common factor (3). Step 3: Thus the assumption of being coprime breaks.

Step 1: From (p^2=5q^2), (p^2) is divisible by (5). Step 2: Since (5) is prime, (p) is also divisible by (5). Step 3: The prime-number rule is the backbone of the proof.

Step 1: From (q^2=2k^2), (q^2) is even. Step 2: If a square is even, the original integer is also even. Step 3: Then both (p) and (q) are even and contradiction occurs.

Step 1: Assume (\sqrt{3}) rational and write it in lowest form. Step 2: Squaring gives (p^2=3q^2). Step 3: Finally, show common factor (3) in both and write the contradiction.

Step 1: (p=5k) means (p) is divisible by (5). Step 2: (q=5r) means (q) is also divisible by (5). Step 3: Common factor (5) contradicts the coprime condition.

Step 1: (p^2=2q^2) is only a middle step. Step 2: From it, both (p) and (q) must be shown even. Step 3: The proof is not complete without writing the coprime contradiction.

Step 1: From (p^2=3q^2), first (p^2) and then (p) are found divisible by (3). Step 2: Only after substituting (p=3k) do we get (q^2=3k^2). Step 3: Keeping the order correct makes the proof strong.

Step 1: From (p^2=5q^2), (p^2) is divisible by (5). Step 2: This gives (p) divisible by (5), but not directly (p=5q). Step 3: The correct form is (p=5k).

Step 1: A rational number is written as (\frac{p}{q}). Step 2: In the proof, it is taken in lowest form, so (p) and (q) are coprime. Step 3: Later, a common factor breaks this condition.

Step 1: In the proof of (\sqrt{2}), (p^2=2q^2) appears, so (2) is key. Step 2: In the proof of (\sqrt{3}), (p^2=3q^2) appears, so (3) is key. Step 3: The number under the root becomes the proof factor.

Step 1: In lowest form, numerator and denominator are coprime. Step 2: If both are divisible by (5), common factor (5) exists. Step 3: This contradicts lowest form.

Step 1: An even number is divisible by (2). Step 2: If both are even, (2) is a common factor. Step 3: Coprime numbers cannot have such a common factor.

Step 1: Assume (\sqrt{3}) rational and write it in lowest form. Step 2: The proof shows both numerator and denominator divisible by (3). Step 3: This contradicts the condition of a lowest-form fraction.