Proof of irrationality of √2, √3, √5

Step 1: In (a^2=2b^2), the right side has factor (2). Step 2: So first (a^2) is called even, and then (a) is proved even. Step 3: Do not change the order of conclusions in exams.

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: Use the prime rule to move from square to original number.

Step 1: (m=5k) shows factor (5) in (m). Step 2: Substituting it in (m^2=5n^2) gives (n^2=5k^2). Step 3: Then (n) is also proved divisible by (5), giving contradiction.

Step 1: If (a=2r), then (a^2=4r^2). Step 2: From (4r^2=2b^2), dividing by (2) gives (b^2=2r^2). Step 3: This becomes the basis for proving (b) even.

Step 1: Squaring (p=3k) gives (p^2=9k^2). Step 2: Substitution in (p^2=3q^2) gives (9k^2=3q^2). Step 3: Do not write ((3k)^2) as (3k^2).

Step 1: The square of (p=5k) is ((5k)^2). Step 2: ((5k)^2=25k^2), so writing (5k^2) is wrong. Step 3: Never forget to square the coefficient.

Step 1: Coprime numbers have no common factor except (1). Step 2: If both are divisible by (3), then (3) is a common factor. Step 3: So it contradicts their being coprime.

Step 1: (a) being even does not automatically make (b) even. Step 2: After substituting (a=2k), we get (b^2=2k^2). Step 3: Only then can (b) be proved even.

Step 1: (p=3r) and (q=3s) show factor (3) in both. Step 2: So their greatest common divisor cannot remain (1) and is at least (3). Step 3: This contradicts lowest form.

Step 1: From (p^2=5q^2), first (p^2), then (p), is proved divisible by (5). Step 2: Only after putting (p=5k) do we get (q^2=5k^2). Step 3: So divisibility of (q) is not immediate.

Step 1: After assuming rationality, the number is written in lowest-form fraction. Step 2: The proof finds the same factor in both numerator and denominator. Step 3: This cannot happen in lowest form, so contradiction occurs.

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: Therefore (q=3r) is written.

Step 1: If (p=5k) and (q=5r), both numerator and denominator share (5). Step 2: (\frac{5k}{5r}) can be reduced to (\frac{k}{r}). Step 3: This shows the fraction was not in lowest form.

Step 1: In lowest form, (a) and (b) were assumed coprime. Step 2: The proof shows both are even, so both have common factor (2). Step 3: This is the correct final contradiction.

Step 1: To get (3) from (\sqrt{3}), both sides must be squared. Step 2: The square of a fraction is (\frac{p^2}{q^2}). Step 3: So the correct form is (3=\frac{p^2}{q^2}).

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: Therefore (p=5k) is valid.

Step 1: (a=2k) and (b=2r) show both are divisible by (2). Step 2: So their greatest common divisor cannot remain (1). Step 3: This breaks the initial coprime condition.

Step 1: If a prime divides a square, it also divides the original number. Step 2: In (\sqrt{3}), this is used for (3); in (\sqrt{5}), it is used for (5). Step 3: This gives a common factor in numerator and denominator.

Step 1: (p^2=5q^2) is only a middle step. Step 2: After this, both (p) and (q) must be shown divisible by (5). Step 3: Without contradiction and final conclusion, the proof is incomplete.

Step 1: From (a^2=2b^2), (a) is proved even. Step 2: But to complete the proof, (b) must also be proved even. Step 3: Only when both are even does contradiction arise with the coprime condition.

Step 1: From (p^2=3q^2), first (p) is concluded divisible by (3). Step 2: After substituting (p=3k), (q^2=3k^2) is obtained. Step 3: Therefore jumping directly to (q) is an order mistake.

Step 1: (p=5k) and (q=5r) show factor (5) in both (p) and (q). Step 2: So they cannot be coprime. Step 3: This breaks the initial lowest-form condition.

Step 1: (a=2m) and (b=2n) show common factor (2) in numerator and denominator. Step 2: So (\frac{2m}{2n}=\frac{m}{n}). Step 3: This contradicts lowest form.

Step 1: Squaring (p=3k) gives ((3k)^2). Step 2: The correct value is (9k^2), not (3k^2). Step 3: Square the whole expression.

Step 1: Assume (\sqrt{5}) rational and write it in lowest-form fraction. Step 2: The proof shows both numerator and denominator divisible by (5). Step 3: This contradicts the coprime condition, so (\sqrt{5}) is irrational.