Proof of irrationality of √2, √3, √5

Step 1: The main factor in the equation is (5). Step 2: (p^2=5q^2) usually comes from the proof of (\sqrt{5}). Step 3: To identify the proof, look at the factor in the equation.

Step 1: From (p^2=3q^2), we write (p=3k). Step 2: Substitution gives (9k^2=3q^2), then (q^2=3k^2). Step 3: This correct chain leads to (q) being divisible by (3).

Step 1: After assuming rationality, the number is written as a lowest-form fraction. Step 2: The proof finds a common factor in numerator and denominator. Step 3: This is impossible for a lowest-form fraction.

Step 1: (p=2k) means (p) is even. Step 2: (q=2r) means (q) is also even. Step 3: Both have common factor (2), so they cannot be coprime.

Step 1: (5) is a prime number. Step 2: If a prime divides a square, it also divides the original number. Step 3: This rule is applied to (p) in the proof of (\sqrt{5}).

Step 1: From (q^2=2k^2), (q^2) is even. Step 2: Then by rule (q) is even and can be written as (q=2r). Step 3: Directly writing (q=2k) is a careless step.

Step 1: Assuming rationality, (\sqrt{3}=\frac{p}{q}) is written in lowest form. Step 2: The proof shows both (p) and (q) divisible by (3). Step 3: Such a common factor cannot occur in a lowest-form fraction.

Step 1: We square (\sqrt{n}=\frac{p}{q}) to remove the square root. Step 2: This gives an equation like (p^2=nq^2). Step 3: This equation starts the divisibility and contradiction steps.

Step 1: In contradiction method, the opposite assumption is taken. Step 2: If the rational assumption is false, irrationality is proved. Step 3: Therefore (\sqrt{5}) is irrational.

Step 1: Such proofs begin with the rational assumption. Step 2: At the end, a common factor is found in numerator and denominator. Step 3: In exams, clearly writing the coprime contradiction is most important.

Step 1: In the proof, a rational number is written as a fraction in lowest form. Step 2: In lowest form, numerator and denominator are coprime. Step 3: Later, finding a common factor creates the contradiction.

Step 1: Squaring both sides gives (3=\frac{p^2}{q^2}). Step 2: Clearing the denominator gives (p^2=3q^2). Step 3: After squaring, do not forget to multiply by (q^2).

Step 1: In (m^2=5n^2), the right side has factor (5). Step 2: Therefore (m^2) is divisible by (5). Step 3: First write divisibility of the square, then conclude divisibility of (m).

Step 1: From (a^2=2b^2), (a^2) is even. Step 2: If a square is even, the original integer is even. Step 3: Therefore we write (a=2k), where (k) is an integer.

Step 1: Since (p=3k), (p^2=(3k)^2). Step 2: ((3k)^2=9k^2), so we get (9k^2=3q^2). Step 3: Always remember to square the coefficient.

Step 1: If (p=5k), then (p^2=25k^2). Step 2: From (25k^2=5q^2), we get (q^2=5k^2). Step 3: This leads to the conclusion that (q) is divisible by (5).

Step 1: From (a^2=2b^2), (a^2) is even. Step 2: This gives (a) even, but not directly (a=2b). Step 3: The correct step is to write (a=2k).

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

Step 1: Coprime numbers have no common factor except (1). Step 2: If both are divisible by (3), common factor (3) exists. Step 3: Therefore it contradicts the coprime condition.

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: This shows a common factor in (p) and (q).