Proof of irrationality of √2, √3, √5

Step 1: First, from (p^2=5q^2), (p) is found divisible by (5). Step 2: Then substituting (p=5k) gives (q^2=5k^2). Step 3: Then (q) is concluded divisible by (5).

Step 1: If (a=2k), then (a^2=4k^2). Step 2: From (4k^2=2b^2), we get (b^2=2k^2). Step 3: This proves (b^2) and then (b) are even.

Step 1: All three proofs are based on contradiction. Step 2: So the number is first assumed rational. Step 3: Then this assumption leads to an impossible common factor.

Step 1: To get (3) from (\sqrt{3}), both sides must be squared. Step 2: The correct form is (3=\frac{p^2}{q^2}), not (3=\frac{p}{q}). Step 3: Always square both sides to remove a square root.

Step 1: In lowest form, a fraction cannot be reduced further. Step 2: This means the greatest common divisor of (p) and (q) is (1). Step 3: Later finding (5) in both contradicts this.

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.