Proof of irrationality of √2, √3, √5

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.

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.

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.

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.

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.

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.

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).

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.

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.

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.

Step 1: In (p^2=2q^2), the right side has a factor (2). Step 2: So first we say (p^2) is even, then conclude (p) is even. Step 3: Keep the order of conclusions correct in the proof.

Step 1: Coprime numbers have no common factor other than (1). Step 2: If both are divisible by (3), the common factor is (3). Step 3: This contradiction proves (\sqrt{3}) irrational.

Step 1: From (m^2=5n^2), (m^2) is divisible by (5). Step 2: Since (5) is prime, (m) is also divisible by (5). Step 3: Therefore (m=5k) is the correct next step.

Step 1: Since (p=2r), (p^2=(2r)^2=4r^2). Step 2: Substituting in (p^2=2q^2) gives (4r^2=2q^2). Step 3: Do not forget to square the coefficient.

Step 1: If (a=3t), then (a^2=9t^2). Step 2: From (9t^2=3b^2), divide by (3) to get (b^2=3t^2). Step 3: Do not rush the division step.

Step 1: From (p^2=2q^2), we conclude (p^2) is even. Step 2: This gives (p) even, but not directly (p=2q). Step 3: The correct form is (p=2r), where (r) is an integer.

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

Step 1: If (m=5k), then (m^2=25k^2). Step 2: So (m^2=5n^2) becomes (25k^2=5n^2). Step 3: Writing ((5k)^2) as (5k^2) is a mistake.

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

Step 1: From (a^2=3b^2), (a) is divisible by (3). Step 2: Then substituting (a=3k) shows (b) is also divisible by (3). Step 3: Common factor (3) gives the contradiction.