Proof of irrationality of √2, √3, √5

Step 1: In the proof, a rational number is written as a fraction in lowest form. Step 2: In lowest form, numerator and denominator are coprime. Step 3: Later, finding a common factor creates the contradiction.

Step 1: Squaring both sides gives (3=\frac{p^2}{q^2}). Step 2: Clearing the denominator gives (p^2=3q^2). Step 3: After squaring, do not forget to multiply by (q^2).

Step 1: In (m^2=5n^2), the right side has factor (5). Step 2: Therefore (m^2) is divisible by (5). Step 3: First write divisibility of the square, then conclude divisibility of (m).

Step 1: From (a^2=2b^2), (a^2) is even. Step 2: If a square is even, the original integer is even. Step 3: Therefore we write (a=2k), where (k) is an integer.

Step 1: Since (p=3k), (p^2=(3k)^2). Step 2: ((3k)^2=9k^2), so we get (9k^2=3q^2). Step 3: Always remember to square the coefficient.

Step 1: If (p=5k), then (p^2=25k^2). Step 2: From (25k^2=5q^2), we get (q^2=5k^2). Step 3: This leads to the conclusion that (q) is divisible by (5).

Step 1: From (a^2=2b^2), (a^2) is even. Step 2: This gives (a) even, but not directly (a=2b). Step 3: The correct step is to write (a=2k).

Step 1: Prime factors in a square occur in pairs. Step 2: If a prime divides (x^2), it also divides (x). Step 3: This rule is essential in the proofs of (\sqrt{3}) and (\sqrt{5}).

Step 1: Coprime numbers have no common factor except (1). Step 2: If both are divisible by (3), common factor (3) exists. Step 3: Therefore it contradicts the coprime condition.

Step 1: From (q^2=5k^2), (q^2) is divisible by (5). Step 2: Since (5) is prime, (q) is also divisible by (5). Step 3: This shows a common factor in (p) and (q).

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.