Proof of irrationality of √2, √3, √5

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.

Step 1: In (n^2=5k^2), the right side has factor (5). Step 2: So (n^2) is divisible by (5), and by the prime rule (n) is also divisible by (5). Step 3: Apply the correct rule from square to original number.

Step 1: Assume (\sqrt{2}) rational and write it as (\frac{p}{q}). Step 2: Squaring leads to both (p) and (q) being even. Step 3: Both even contradict the coprime condition.

Step 1: In lowest form, numerator and denominator should not have a common factor. Step 2: If both are even, (2) is a common factor. Step 3: Therefore the lowest-form assumption breaks and gives a contradiction.

Step 1: In the proof of (\sqrt{3}), factor (3) is used. Step 2: So if (p^2) is divisible by (3), (p) is divisible by (3). Step 3: Identify the relevant factor in each proof.

Step 1: If a prime factor divides a square, it also divides the original number. Step 2: Since (5) is prime, (p^2) divisible by (5) implies (p) divisible by (5). Step 3: This is the main logic of the proof.

Step 1: After assuming rationality, (p) and (q) were taken coprime. Step 2: Finding common factor (3) makes this assumption impossible. Step 3: So the rational assumption is false and (\sqrt{3}) is irrational.

Step 1: Coprime numbers have no common factor other than (1). Step 2: If both are divisible by (5), (5) is a common factor. Step 3: Therefore this is impossible for coprime numbers.

Step 1: First (p) is proved even. Step 2: If (q) is also proved even, both are divisible by (2). Step 3: Common factor (2) breaks the coprime condition.

Step 1: The proof starts with the rational assumption. Step 2: Squaring gives (p^2=5q^2). Step 3: Finally, a common factor (5) in both gives the contradiction.

Step 1: Coprime means there is no common factor except (1). Step 2: If both are even, (2) is a common factor. Step 3: Therefore it directly contradicts their being coprime.

Step 1: From (a^2=3b^2), (a^2) is divisible by (3). Step 2: Since (3) is prime, (a) is also divisible by (3). Step 3: Therefore writing (a=3k) is valid.

Step 1: In all three, the number is assumed rational and written in lowest form. Step 2: At the end, numerator and denominator share (2), (3), or (5). Step 3: This is the common contradiction.

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: Directly writing (p=5q) is not correct.

Step 1: (p=3r) means (p) is divisible by (3). Step 2: (q=3s) means (q) is also divisible by (3). Step 3: Both have common factor (3), so the coprime condition breaks.

Step 1: First (p) is proved even in the proof. Step 2: If (q=2s), then (q) is also even. Step 3: Both even gives common factor (2) and creates a contradiction.