Proof of irrationality of √2, √3, √5

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.

Step 1: From (p^2=3q^2), we get (3\mid p^2). Step 2: Since (3) is prime, (3\mid p). Step 3: Here divisibility by (3), not evenness, is the main point.

Step 1: A rational number is a ratio of two integers. Step 2: The denominator cannot be zero, and the fraction is taken in lowest form. Step 3: Write this complete form at the start of the proof.

Step 1: (2,3,5) are prime. Step 2: Assuming (\sqrt{r}=\frac{p}{q}) finally gives (r\mid p) and (r\mid q). Step 3: This common structure connects all three proofs.

Step 1: In lowest form, numerator and denominator have no common factor except (1). Step 2: If both are even, (2) becomes a common factor. Step 3: So the lowest-form condition fails.

Step 1: Putting (p=5t) gives (25t^2=5q^2). Step 2: Hence (q^2=5t^2), so (5\mid q). Step 3: This makes (5) common to both (p) and (q).

Step 1: A short decimal approximation does not prove irrationality. Step 2: A solid proof assumes rationality and derives a contradiction. Step 3: In exams, write logical proof instead of approximation.

Step 1: (3\mid p) follows from (3\mid p^2) because (3) is prime. Step 2: This cannot be stated the same way for every composite number. Step 3: Mention the word prime in the proof.

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: This impossible situation appears in the proof for (\sqrt{3}).

Step 1: The correct order begins with the rational form. Step 2: Squaring gives (p^2=5q^2), then (5) divides first (p) and then (q). Step 3: Remembering the order makes the proof clear and complete.

Step 1: From (p^2) even, we only get that (p) is even. Step 2: For the full contradiction, (q) must also be shown even. Step 3: Do not stop the proof midway; write until the final conflict.

Step 1: From (p^2=3q^2), we get (3\mid p^2). Step 2: By the prime rule, (3\mid p), so (p=3r) can be written. Step 3: Give the reason before writing such a form.

Step 1: (\frac{5}{1}=5). Step 2: (\sqrt{5}) is the number whose square is (5), so it is not (5). Step 3: Distinguish a number from its square root.

Step 1: The rational assumption is always taken as a ratio. Step 2: It is necessary to write (p,q) coprime and (q\neq0). Step 3: This standard form works in all three proofs.

Step 1: In (p^2=3q^2), the right side is a multiple of (3). Step 2: Since both sides are equal, (p^2) is also a multiple of (3). Step 3: Understand divisibility of the square first, then of the original number.

Step 1: Coprime means there should be no common factor. Step 2: Both being divisible by (5) shows a common factor. Step 3: This contradiction proves (\sqrt{5}) irrational.