Proof of irrationality of √2, √3, √5

Step 1: In (\sqrt{3}), (3) is prime; in (\sqrt{5}), (5) is prime. Step 2: Both proofs use divisibility from square to original number. Step 3: This is their main common logic.

Step 1: If (p) were odd, then (p^2) would also be odd. Step 2: But (p^2) is even, so (p) cannot be odd. Step 3: Thus (p) must be even.

Step 1: In contradiction, 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: After substituting (a=3k), we get (b^2=3k^2). Step 2: Hence (b^2) is divisible by (3). Step 3: By the prime rule, (b) is also divisible by (3).

Step 1: Both even means numerator and denominator have common factor (2). Step 2: Such a fraction can be reduced by (2). Step 3: So it cannot be in lowest form.

Step 1: A rational number is written as (\frac{p}{q}). Step 2: The denominator of a fraction cannot be zero. Step 3: Therefore (q\neq 0) must be written.

Step 1: Assuming (\sqrt{n}=\frac{p}{q}) and squaring gives (p^2=nq^2). Step 2: Here (n=3), so it relates to (\sqrt{3}). Step 3: Identify the square root from the factor in the equation.

Step 1: (p=2r) and (q=2s) mean both are divisible by (2). Step 2: So they cannot be coprime. Step 3: But they were assumed coprime at the start, which is the contradiction.

Step 1: First (p) is found to have factor (5). Step 2: Substituting (p=5k) in the equation gives (q^2=5k^2). Step 3: This proves (q) is also divisible by (5).

Step 1: (p^2=3q^2) is a middle step, not the end. Step 2: After this, both (p) and (q) must be shown divisible by (3). Step 3: The proof is incomplete without contradiction and conclusion.

Step 1: A rational number is written as a ratio of two integers. Step 2: The denominator cannot be zero, so (q\neq 0) is necessary. Step 3: In lowest form, (p) and (q) are also taken coprime.

Step 1: In lowest form, (p) and (q) are coprime. Step 2: When the proof shows both divisible by (3), this becomes impossible. Step 3: Thus lowest form helps show contradiction.

Step 1: From (p^2=5q^2), (p) is divisible by (5), so (p=5k). Step 2: Substitution gives (25k^2=5q^2), then (q^2=5k^2). Step 3: This proves (q) is also divisible by (5).

Step 1: If (p=2k), then (p^2=(2k)^2). Step 2: Its correct value is (4k^2), not (2k^2). Step 3: Square the coefficient while squaring.

Step 1: From (p^2=3q^2), (p^2) is divisible by (3). Step 2: (3) is prime, so the prime divisibility rule applies. Step 3: Therefore (p) is also divisible by (3).

Step 1: In lowest form, (p) and (q) should be coprime. Step 2: If both are proved even, both have common factor (2). Step 3: This is impossible, so the rational assumption is false.

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, and (\sqrt{5}) is irrational. Step 3: Check a number and its square root separately.

Step 1: Assuming (\sqrt{3}) rational gives (p^2=3q^2). Step 2: This proves both (p) and (q) divisible by (3). Step 3: This contradicts coprime condition, so (\sqrt{3}) is irrational.

Step 1: Treating (\sqrt{2}), (\sqrt{3}), and (\sqrt{5}) as (2), (3), and (5) is wrong. Step 2: The correct method assumes rationality, writes a fraction, and squares. Step 3: Do not write a square root equal to the number under it.

Step 1: The proof starts with the rational assumption. Step 2: At the end, a contradiction appears with the coprime condition. Step 3: The final line should clearly state the contradiction and irrationality conclusion.