Proof of irrationality of √2, √3, √5

Step 1: From (p^2=2q^2), factor (2) first appears in (p). Step 2: Later factor (2) also appears in (q). Step 3: Common factor (2) contradicts the coprime condition.

Step 1: We have (p=3k). Step 2: Squaring gives (p^2=(3k)^2=9k^2). Step 3: The coefficient (3) must also be squared to (9).

Step 1: In the proof, both (p) and (q) are found divisible by (5). Step 2: This means their common factor is (5). Step 3: This breaks the condition of being coprime.

Step 1: Coprime numbers are defined as having only (1) as common factor. Step 2: Finding any common factor other than (1) is impossible. Step 3: Irrationality proofs show exactly this impossible situation.

Step 1: (p) and (q) were assumed coprime at the start. Step 2: Both even shows common factor (2). Step 3: This is a contradiction, so (\sqrt{2}) is irrational.

Step 1: If both are divisible by (3), the fraction has common factor (3). Step 2: Such a fraction can be reduced further. Step 3: So it contradicts the lowest-form assumption.

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: This is the basis for writing (p=5k).

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 (p^2=2q^2), saying (p^2) is even is correct. Step 2: But it is not the final conclusion; both (p) and (q) must then be shown even. Step 3: Complete the proof up to contradiction.

Step 1: From (q^2=5k^2), (q^2) is divisible by (5). Step 2: By the prime rule, (q) is also divisible by (5). Step 3: Then (p) and (q) both have common factor (5).

Step 1: Writing (\sqrt{2}=2), (\sqrt{3}=3), or (\sqrt{5}=5) is wrong. Step 2: The correct method assumes rationality, writes a fraction, and squares. Step 3: Do not treat a square root as equal to the number inside.

Step 1: If both are even, numerator and denominator have common factor (2). Step 2: So the fraction can be reduced further by (2). Step 3: This contradicts the lowest-form assumption.

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

Step 1: (5) is rational but not a perfect square. Step 2: Since it is not a perfect square, (\sqrt{5}) is not rational. Step 3: Check a number and its square root separately.

Step 1: The rational assumption makes both (p) and (q) even. Step 2: This contradicts their being coprime. Step 3: So the final sentence should state both contradiction and irrationality.

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.