Proof of irrationality of √2, √3, √5

Step 1: From (p^2=2q^2), we immediately get (p^2) even. Step 2: Only after substituting (p=2k) do we get (q^2=2k^2). Step 3: Skipping the order makes the proof weak.

Step 1: If both are divisible by (5), numerator and denominator share (5). Step 2: Such a fraction can be reduced further. Step 3: This contradicts the assumption of lowest form.

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

Step 1: If both are even, both have common factor (2). Step 2: This cannot happen for coprime numbers. Step 3: This final reason proves (\sqrt{2}) irrational.

Step 1: (5) is rational but not a perfect square. Step 2: The square root of a non-perfect square need not be rational, and (\sqrt{5}) is irrational. Step 3: Check a number and its square root separately.

Step 1: In lowest form, numerator and denominator should not have a common factor. Step 2: Finding (2) in both shows the fraction can be reduced. Step 3: Therefore the initial lowest-form statement becomes false.

Step 1: After substituting (p=5k), we get (q^2=5k^2). Step 2: This makes (q^2) divisible by (5). Step 3: By the prime rule, (q) is said to be divisible by (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: In (p^2=2q^2), the key factor is (2). Step 2: Finding both (p) and (q) divisible by (2) identifies the proof of (\sqrt{2}). Step 3: This gives contradiction to the coprime condition.

Step 1: In the proof, both (p) and (q) are proved divisible by (3). Step 2: But they were assumed coprime at the start. Step 3: This contradiction completes the proof.

Step 1: From (p^2=5q^2), (p) is found divisible by (5). Step 2: Substituting (p=5k) gives (q^2=5k^2). Step 3: This proves (q) is divisible by (5).

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

Step 1: Squaring (p=3k) gives ((3k)^2). Step 2: Its correct value is (9k^2), not (3k^2). Step 3: Do not forget to square the coefficient.

Step 1: Assuming (\sqrt{5}) rational gives (p^2=5q^2). Step 2: This proves both (p) and (q) divisible by (5). Step 3: This contradicts coprime condition, so (\sqrt{5}) is irrational.

Step 1: The proof starts with the rational assumption. Step 2: At the end, contradiction comes from the coprime condition. Step 3: Clearly writing the reason for contradiction helps in exams.

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

Step 1: The proofs of (\sqrt{2}), (\sqrt{3}), and (\sqrt{5}) use contradiction. Step 2: First assume rationality, write a lowest-form fraction, and square. Step 3: Finally, a common factor gives contradiction.

Step 1: In the proof, the rational assumption leads to an impossible common factor. Step 2: This contradicts the assumption. Step 3: So the final line should clearly state contradiction and irrationality.

Step 1: If both are even, (a) and (b) have common factor (2). Step 2: In a lowest-form fraction, numerator and denominator should not have a common factor other than (1). Step 3: So this contradicts the rational assumption and proves (\sqrt{2}) irrational.

Step 1: In (\sqrt{3}), the prime factor is (3), and in (\sqrt{5}), the prime factor is (5). Step 2: When (p^2) is divisible by that prime, (p) is also divisible by the same prime. Step 3: This idea later shows a common factor in (p) and (q), creating contradiction.