Proof of irrationality of √2, √3, √5

Step 1: In contradiction, first assume (\sqrt{2}) rational. Step 2: Squaring gives evenness conclusions. Step 3: Finding both even contradicts the coprime condition.

Step 1: A rational number is written as a ratio of two integers. Step 2: In lowest form, numerator and denominator are coprime and denominator is non-zero. Step 3: This form starts the contradiction proof.

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

Step 1: From (q^2=2k^2), (q^2) is even. Step 2: If a square is even, the original integer is even. Step 3: (p) was already even and (q) is also even, creating contradiction.

Step 1: From (p^2=3q^2), first (p^2) and then (p) are found divisible by (3). Step 2: After substituting (p=3k), we get (q^2=3k^2). Step 3: Then (q) is concluded divisible by (3).

Step 1: In all three, the number is assumed rational and written as a lowest-form fraction. Step 2: At the end, a common factor is found in numerator and denominator. Step 3: This contradicts lowest form.

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: Hence (p=5k) is written.

Step 1: (p=2k) and (q=2r) mean numerator and denominator are divisible by (2). Step 2: So the fraction can be reduced by (2). Step 3: This contradicts the lowest-form assumption.

Step 1: (p=3k) means (p) is divisible by (3). Step 2: (q=3r) means (q) is also divisible by (3). Step 3: Common factor (3) contradicts the coprime condition.

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

Step 1: In the proof of (\sqrt{2}), (p^2=2q^2) is obtained. Step 2: This proves both (p) and (q) even. Step 3: Both even contradict the coprime condition.

Step 1: The main factor in the equation is (3). Step 2: (p^2=3q^2) usually comes from the proof of (\sqrt{3}). Step 3: To identify the proof, look at the factor in the equation.

Step 1: The square of an odd number is always odd. Step 2: Here (a^2) is even, so (a) cannot be odd. Step 3: Therefore (a) must be even.

Step 1: (p^2=5q^2) tells us divisibility of (p^2). Step 2: By the prime rule, (p) is divisible by (5), but (p=5q) does not follow directly. Step 3: In exams, writing (p=5k) is correct.

Step 1: In (\sqrt{n}=\frac{p}{q}), we square to remove the square root. Step 2: This gives an equation like (p^2=nq^2). Step 3: This equation gives divisibility and contradiction later.

Step 1: From (q^2=3k^2), (q^2) is divisible by (3). Step 2: Since (3) is prime, (q) is divisible by (3). Step 3: Therefore (q=3r) can be written.

Step 1: If both are even, both have common factor (2). Step 2: Coprime numbers should not have any common factor other than (1). Step 3: So the statement that they are coprime becomes false.

Step 1: In contradiction method, the opposite assumption is taken. Step 2: If the rational assumption becomes impossible, it is false. Step 3: Therefore (\sqrt{5}) is proved irrational.

Step 1: In (\sqrt{2})'s proof, factor (2) comes from (p^2=2q^2). Step 2: In (\sqrt{5})'s proof, factor (5) comes from (p^2=5q^2). Step 3: The number under the root becomes the key factor.

Step 1: In (\sqrt{3}), (3) is prime, and in (\sqrt{5}), (5) is prime. Step 2: In both, if the square is divisible by the prime, the original number is also divisible by it. Step 3: This common logic moves the proof forward.