Proof of irrationality of √2, √3, √5

Step 1: In (n^2=5k^2), the right side has factor (5). Step 2: So (n^2) is divisible by (5), and by the prime rule (n) is also divisible by (5). Step 3: Apply the correct rule from square to original number.

Step 1: Assume (\sqrt{2}) rational and write it as (\frac{p}{q}). Step 2: Squaring leads to both (p) and (q) being even. Step 3: Both even contradict the coprime condition.

Step 1: In lowest form, numerator and denominator should not have a common factor. Step 2: If both are even, (2) is a common factor. Step 3: Therefore the lowest-form assumption breaks and gives a contradiction.

Step 1: In the proof of (\sqrt{3}), factor (3) is used. Step 2: So if (p^2) is divisible by (3), (p) is divisible by (3). Step 3: Identify the relevant factor in each proof.

Step 1: If a prime factor divides a square, it also divides the original number. Step 2: Since (5) is prime, (p^2) divisible by (5) implies (p) divisible by (5). Step 3: This is the main logic of the proof.

Step 1: After assuming rationality, (p) and (q) were taken coprime. Step 2: Finding common factor (3) makes this assumption impossible. Step 3: So the rational assumption is false and (\sqrt{3}) is irrational.

Step 1: Coprime numbers have no common factor other than (1). Step 2: If both are divisible by (5), (5) is a common factor. Step 3: Therefore this is impossible for coprime numbers.

Step 1: First (p) is proved even. Step 2: If (q) is also proved even, both are divisible by (2). Step 3: Common factor (2) breaks the coprime condition.

Step 1: The proof starts with the rational assumption. Step 2: Squaring gives (p^2=5q^2). Step 3: Finally, a common factor (5) in both gives the contradiction.

Step 1: Coprime means there is no common factor except (1). Step 2: If both are even, (2) is a common factor. Step 3: Therefore it directly contradicts their being coprime.

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

Step 1: In all three, the number is assumed rational and written in lowest form. Step 2: At the end, numerator and denominator share (2), (3), or (5). Step 3: This is the common contradiction.

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: Directly writing (p=5q) is not correct.

Step 1: (p=3r) means (p) is divisible by (3). Step 2: (q=3s) means (q) is also divisible by (3). Step 3: Both have common factor (3), so the coprime condition breaks.

Step 1: First (p) is proved even in the proof. Step 2: If (q=2s), then (q) is also even. Step 3: Both even gives common factor (2) and creates a contradiction.

Step 1: In (p^2=5q^2), (p^2) equals (5q^2). Step 2: So (p^2) definitely has factor (5). Step 3: Divisibility of (q^2) comes in a later step.

Step 1: Coprime means two numbers have no common factor except (1). Step 2: So finding both even breaks this meaning. Step 3: Understanding the definition makes the proof easier.

Step 1: (\sqrt{3}) contains a square root. Step 2: To remove it, we square both sides and get (3=\frac{p^2}{q^2}). Step 3: Choose the correct algebraic operation to remove the radical.

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

Step 1: In the proof of (\sqrt{2}), (2) is the key factor. Step 2: In the proof of (\sqrt{5}), (5) is the key factor. Step 3: Pay attention to the number under the root in each proof.