Proof of irrationality of √2, √3, √5

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: Now both (p) and (q) have common factor (5).

Step 1: If both are even, (2) is a common factor. Step 2: This contradicts the assumption that (p) and (q) are coprime. Step 3: Therefore the rational assumption is false and (\sqrt{2}) is irrational.

Step 1: From (p^2=3q^2), (p^2) is divisible by (3). Step 2: This means (p) is divisible by (3), but (p=3q) does not follow directly. Step 3: The correct way is to write (p=3k).

Step 1: A rational number is written as a lowest-form fraction, so (p) and (q) are coprime. Step 2: The proof shows both divisible by (5). Step 3: This gives a clear contradiction to the coprime condition.

Step 1: From (p^2=2q^2), (p^2) and then (p) are proved even. Step 2: But to complete the proof, (q) must also be shown even. Step 3: Only then a contradiction arises through common factor (2).

Step 1: If both are divisible by (3), numerator and denominator have common factor (3). Step 2: Such a fraction can be reduced further by (3). Step 3: Hence it cannot be in lowest form.

Step 1: From (p^2=5q^2), (p) is divisible by (5), so (p=5k). Step 2: Substitution gives (q) also divisible by (5), so (q=5r). Step 3: Common factor (5) contradicts the coprime condition.

Step 1: In (\sqrt{2}), (p^2=2q^2) gives the evenness argument. Step 2: In (\sqrt{3}), the primality of (3) gives the divisibility argument. Step 3: Choose the reasoning according to the number under the root.

Step 1: From (q^2=5k^2), (q^2) is divisible by (5). Step 2: To conclude divisibility of the original number from the square, (5) must be prime. Step 3: Therefore (q) is said to be divisible by (5).

Step 1: Assuming rationality, (\frac{p}{q}) was taken in lowest form. Step 2: If the proof shows it is not in lowest form, the initial assumption is impossible. Step 3: Therefore (\sqrt{2}) is irrational.

Step 1: In a perfect square, exponents of prime factors are even. Step 2: (p^2=3q^2) or (p^2=5q^2) forces the same prime factor into both (p) and (q). Step 3: This common factor contradicts the coprime condition.

Step 1: (p) being even does not automatically make (q) even. Step 2: To prove (q) even, (p=2k) must be substituted in the original equation. Step 3: Writing conclusions without support weakens the proof.

Step 1: From (p^2=5q^2), (p^2) is immediately divisible by (5). Step 2: Then (p) is divisible by (5) and (p=5k) can be written. Step 3: Divisibility of (q) comes after substituting (p=5k), not immediately.

Step 1: Taking (\frac{p}{q}) in lowest form means (\gcd(p,q)=1). Step 2: If both are divisible by (3), their greatest common divisor is at least (3). Step 3: Therefore the condition (\gcd(p,q)=1) breaks.

Step 1: If (p=2k) and (q=2r), both numerator and denominator have common factor (2). Step 2: So (\frac{2k}{2r}) can be reduced to (\frac{k}{r}). Step 3: This shows (\frac{p}{q}) was not in lowest form.

Step 1: (\gcd(p,q)=1) means (p) and (q) are coprime. Step 2: If (p=5m) and (q=5n), (5) is their common factor. Step 3: Therefore this goes against (\gcd(p,q)=1).

Step 1: In lowest form, (p) and (q) should not have any common factor other than (1). Step 2: Finding both divisible by (3) breaks this condition. Step 3: Therefore the rational assumption is impossible and (\sqrt{3}) is irrational.

Step 1: (p) being even may be true, but the reason is not the positivity of (\sqrt{2}). Step 2: The correct reason is that (p^2=2q^2) makes (p^2) even. Step 3: In proof writing, a true statement must have the correct reason.

Step 1: Assuming (\sqrt{5}=\frac{p}{q}) and squaring gives (p^2=5q^2). Step 2: This (5) becomes a common factor in both (p) and (q). Step 3: This identifies the proof of (\sqrt{5}).

Step 1: (p) is an integer and is divisible by (3). Step 2: Therefore (p=3k), where (k) is also an integer. Step 3: Mentioning the type of the new variable makes the proof clear.