Proof of irrationality of √2, √3, √5

Step 1: From (a^2=2b^2), first (a^2), then (a), is proved even. Step 2: To prove (b) even, (a=2k) must be substituted. Step 3: Jumping directly to (b) is an order error.

Step 1: If both are divisible by (5), the fraction has common factor (5). Step 2: Such a fraction can be reduced. Step 3: Therefore it cannot be in lowest form, which is the contradiction.

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

Step 1: At the beginning, (\frac{a}{b}) was assumed in lowest form. Step 2: If the proof shows it can be reduced, the initial assumption is impossible. Step 3: Therefore (\sqrt{2}) is irrational.

Step 1: Assuming (\sqrt{3}=\frac{p}{q}) and squaring gives (p^2=3q^2). Step 2: Later, common factor (3) is found in both (p) and (q). Step 3: This identifies the proof of (\sqrt{3}).

Step 1: (a) is an integer and has been proved even. Step 2: An even integer is written as (2k), where (k) is an integer. Step 3: Clearly mentioning the type of the new variable strengthens the proof.

Step 1: (p) being divisible by (5) can be a true conclusion. Step 2: But its reason is not the positivity of (\sqrt{5}). Step 3: The correct reason is (p^2=5q^2) and (5) being prime.

Step 1: First (p) is proved divisible by (3). Step 2: Substituting (p=3k) in the equation gives (q^2=3k^2). Step 3: This proves (q) is also divisible by (3).

Step 1: Writing (\sqrt{2}=2), (\sqrt{3}=3), or (\sqrt{5}=5) is wrong. Step 2: In the correct proof, rationality is assumed and fraction form is taken. Step 3: Do not treat the square root and the number inside as the same.

Step 1: From (p^2=5q^2), (p=5k) is obtained. Step 2: Substitution gives (25k^2=5q^2), then (q^2=5k^2). Step 3: Then (q) is proved divisible by (5).

Step 1: (a=2m) and (b=2n) show common factor (2) in both. Step 2: So (\gcd(a,b)) cannot be (1). Step 3: This goes against the lowest-form condition.

Step 1: From (p^2=3q^2), (p) is proved divisible by (3). Step 2: But to complete the proof, (q) must also be shown divisible by (3). Step 3: Only then does contradiction arise with the coprime condition.

Step 1: (5) is rational, but it is not a perfect square. Step 2: The square root of a non-perfect square need not be rational. Step 3: The proof of (\sqrt{5}) shows it is irrational.

Step 1: (a=2b) does not directly follow from (a^2=2b^2). Step 2: The correct conclusion is that (a^2) is even and (a) is even. Step 3: Do not hastily make a root-level equation from a squared equation.

Step 1: First (p) is proved divisible by (5). Step 2: After substitution, (q) is also proved divisible by (5). Step 3: After this, contradiction is written using common factor (5).

Step 1: All three proofs start with a rational assumption. Step 2: At the end, a contradiction is obtained from the coprime condition. Step 3: Therefore the final line should clearly state contradiction and irrationality.

Step 1: First assume the number rational and write it as (\frac{p}{q}) in lowest form. Step 2: Squaring gives a divisibility equation. Step 3: Finally, a common factor gives contradiction and proves irrationality.

Step 1: Getting only (a) even does not create contradiction with the coprime condition. Step 2: Substitute (a=2k) in (a^2=2b^2) to get (b^2=2k^2), then prove (b) even. Step 3: Contradiction occurs only when both have common factor (2).

Step 1: In a perfect square, the exponent of every prime factor is even. Step 2: In (p^2=3q^2), the right side adds one extra factor (3) to (q^2), disturbing the exponent balance. Step 3: This idea explains why (3) finally appears in both numerator and denominator.

Step 1: If (p=5m) and (q=5n), both numerator and denominator have common factor (5). Step 2: So (\frac{p}{q}=\frac{5m}{5n}=\frac{m}{n}), meaning the fraction can be reduced. Step 3: This contradicts the lowest-form assumption, so (\sqrt{5}) is proved irrational.