Proof of irrationality of √2, √3, √5

Step 1: Assuming (\sqrt{2}=\frac{p}{q}) gives (p^2=2q^2). Step 2: So (p^2) is even, hence (p) is even, and then (q) also becomes even. Step 3: In exams, remember that coprime numbers cannot both be even.

Step 1: Assuming (\sqrt{3}=\frac{p}{q}) gives (p^2=3q^2). Step 2: The key step is using (3\mid p^2\Rightarrow 3\mid p). Step 3: Writing this prime-number property clearly helps in scoring well.

Step 1: The equation (a^2=5b^2) shows that (a^2) is divisible by (5). Step 2: Since (5) is prime, (a) must also be divisible by (5). Step 3: In the proof, write the conclusion about (a) first and then move to (b).

Step 1: From (p^2=2q^2) and (p=2k), we get (4k^2=2q^2). Step 2: Simplifying gives (q^2=2k^2), so (q^2) and (q) are even. Step 3: This second evenness completes the contradiction.

Step 1: From (\sqrt{3}=\frac{m}{n}), we get (m^2=3n^2). Step 2: This leads to (3\mid m) and then (3\mid n). Step 3: Coprime numbers cannot have such a common factor.

Step 1: (5) is a whole number, but its square root need not be rational unless it is a perfect square. Step 2: Since (5) is not the square of an integer, (\sqrt{5}) is not rational. Step 3: In exams, distinguish a number from its square root.

Step 1: A rational number is written in lowest form as (\frac{p}{q}). Step 2: The proof shows that both (p) and (q) become even, which contradicts lowest form. Step 3: Always mention coprime at the start of the proof.

Step 1: Taking (\sqrt{3}=\frac{p}{q}) gives (p^2=3q^2). Step 2: This forces both (p) and (q) to have (3) as a common factor. Step 3: That contradicts the condition of being coprime.

Step 1: (5) is a prime number. Step 2: If (5\mid a^2), then (5\mid a), because a prime factor in a square must occur in the base. Step 3: This rule is the backbone of the proof for (\sqrt{5}).

Step 1: A rational number is written as (\frac{p}{q}), where (q\neq0). Step 2: The fraction is taken in lowest form, so (p,q) are coprime. Step 3: This condition is what creates the contradiction later.

Step 1: (2,3,5) are prime numbers and not perfect squares. Step 2: Assuming their square roots rational creates the same prime as a common factor of numerator and denominator. Step 3: Understand the difference between a perfect square and a prime number.

Step 1: From (\sqrt{3}=\frac{p}{q}), we get (p^2=3q^2). Step 2: Substituting (p=3r) gives (9r^2=3q^2), so (q^2=3r^2). Step 3: This shows the path to proving (q) is divisible by (3).

Step 1: In (a^2=5b^2), putting (a=5k) gives (25k^2=5b^2). Step 2: Simplifying gives (b^2=5k^2), so (5\mid b^2) and (5\mid b). Step 3: This is the final step against coprimality.

Step 1: From (p^2=2q^2), it is correct that (p) is even. Step 2: After putting (p=2k), (q) also becomes even, not odd. Step 3: In error-identification questions, match every step with the equation.

Step 1: A prime factor appears in a square only if it appears in the base. Step 2: Therefore (r\mid x^2) implies (r\mid x). Step 3: This general rule works for the proofs of (2,3,5).

Step 1: The correct proof starts by assuming the number is rational. Step 2: Squaring gives (a^2=5b^2), and divisibility by (5) is then forced on both variables. Step 3: Keeping the order correct makes the proof clear.

Step 1: If (p) were odd, then (p^2) would also be odd. Step 2: Since (p^2) is even, (p) cannot be odd, so (p) is even. Step 3: This parity fact is very important in the proof.

Step 1: A rational number is written as the ratio of two integers. Step 2: The denominator cannot be zero, and the fraction is taken in lowest form. Step 3: Write this complete form at the beginning of the proof.

Step 1: In all three proofs, the number is first assumed rational. Step 2: Then the related prime number is forced to divide both numerator and denominator. Step 3: Understanding this common structure makes all three proofs easier to remember.

Step 1: In lowest form, (p) and (q) should be coprime. Step 2: The proof shows that both are divisible by (3). Step 3: The contradiction is the clash between coprimality and a common factor.