Proof of irrationality of √2, √3, √5

Step 1: (p=3m) and (q=3n) show that (3) is a common factor of both. Step 2: This contradicts the lowest-form condition. Step 3: Therefore assuming (\sqrt{3}) rational is proved false.

Step 1: From (a^2=5b^2), (5\mid a^2), so (5\mid a). Step 2: This does not necessarily mean (a) is divisible by (25). Step 3: Write only the conclusion that is actually proved.

Step 1: Proving (p) even is only half of the proof. Step 2: We must next put (p=2k) and show (q) is also even. Step 3: Without reaching the final contradiction, the answer is incomplete.

Step 1: From (p^2=rq^2), we get (r\mid p^2). Step 2: Since (r) is prime, (r\mid p). Step 3: After this, put (p=rk) to show (r\mid q).

Step 1: From (p^2=3q^2), we get (3\mid p^2), not directly (p=3q). Step 2: The correct conclusion is (3\mid p), then (p=3k). Step 3: Do not create an unsupported equality while removing squares.

Step 1: From (a^2=5b^2), (5\mid a). Step 2: Putting (a=5k) gives (b^2=5k^2), so (5\mid b). Step 3: Now (5) becomes a common factor and gives the contradiction.

Step 1: In all three proofs, the rational assumption is made first. Step 2: Then the related prime number becomes common to numerator and denominator. Step 3: The structure is the same; only the prime factor changes.

Step 1: Even means divisible by (2). Step 2: If both numerator and denominator are divisible by (2), the fraction can be reduced by (2). Step 3: This contradicts the lowest-form assumption.

Step 1: The rational form (\frac{p}{q}) is valid only when (q\neq0). Step 2: If the denominator is zero, the fraction is not defined. Step 3: This condition must be written at the beginning of the proof.

Step 1: Substituting (a=5k) in (a^2=5b^2) 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 shows the final common factor.

Step 1: The square of an odd integer is odd. Step 2: So if (p^2) is even, (p) cannot be odd and must be even. Step 3: This parity rule is a key step in the proof.

Step 1: (3\mid p) means (p) is a multiple of (3). Step 2: Therefore (p=3k), where (k) is some integer. Step 3: Do not assume (k=q) without reason.

Step 1: At the beginning, (\frac{a}{b}) was taken in lowest form. Step 2: This means (a) and (b) are coprime. Step 3: (5) being common to both breaks this condition.

Step 1: The contradiction depends on (p) and (q) being coprime. Step 2: Without stating lowest form, both being even is not a decisive contradiction. Step 3: Therefore mention lowest form at the start.

Step 1: The proof starts by assuming (\sqrt{3}) rational. Step 2: That assumption gives a common factor against coprimality. Step 3: Therefore the final conclusion is that (\sqrt{3}) is irrational.

Step 1: In (a^2=5b^2), the right side is a multiple of (5). Step 2: Since both sides are equal, (a^2) is also divisible by (5). Step 3: Then the prime rule gives (5\mid a).

Step 1: From (q^2=2k^2), (q^2) is even. Step 2: If the square of an integer is even, the integer is also even. Step 3: Thus both (p) and (q) are found even.

Step 1: In (\sqrt{3}), the common factor is (3). Step 2: In (\sqrt{5}), the common factor is (5). Step 3: The prime factor changes, but the contradiction structure is the same.

Step 1: Since (2) is not a perfect square, (\sqrt{2}) is not an integer. Step 2: But irrationality needs proving it is not any rational fraction. Step 3: Therefore write the contradiction proof using a coprime fraction.

Step 1: Divide both sides of (25k^2=5b^2) by (5). Step 2: We get (5k^2=b^2), that is (b^2=5k^2). Step 3: This gives (5\mid b^2) and then (5\mid b).