Proof of irrationality of √2, √3, √5

Step 1: In lowest form, numerator and denominator are coprime. Step 2: If both are divisible by (5), they have a common factor. Step 3: This proves the original rational assumption false.

Step 1: We assume (\sqrt{2}=\frac{p}{q}), where (p,q) are coprime. Step 2: The proof shows both (p) and (q) are even. Step 3: Thus (2) becomes a common factor, contradicting coprimality.

Step 1: From (p^2=3q^2), we only get that (p^2) is divisible by (3). Step 2: The correct conclusion is (3\mid p), not (p=3q). Step 3: Be careful when removing squares in a proof.

Step 1: In proof by contradiction, we first assume the opposite of what we want to prove. Step 2: So (\sqrt{5}) is assumed rational and written as (\frac{a}{b}). Step 3: Writing the method clearly at the start strengthens the answer.

Step 1: An odd integer can be written as (2k+1). Step 2: Its square becomes (4k^2+4k+1), which is odd. Step 3: This fact helps prove that if (p^2) is even, then (p) is even in the (\sqrt{2}) proof.

Step 1: Coprime numbers have only (1) as a common factor. Step 2: If both are divisible by (5), then (5) becomes a common factor. Step 3: This impossible situation appears in the proof for (\sqrt{5}).

Step 1: The proof reaches a contradiction from the rational assumption. Step 2: Once a contradiction is reached, the original assumption is false. Step 3: End clearly by stating that (\sqrt{3}) is irrational.

Step 1: Putting (p=5k) gives (p^2=25k^2). Step 2: Then (25k^2=5q^2), so (q^2=5k^2). Step 3: Simplify algebra carefully, or the proof will break.

Step 1: (2) is a prime number. Step 2: If a prime factor divides (p^2), it must divide (p). Step 3: Writing this rule makes the proof logical.

Step 1: In proof by contradiction, we begin with the opposite assumption. Step 2: Then we reach a result that conflicts with the given condition. Step 3: The proofs of (\sqrt{2}), (\sqrt{3}), and (\sqrt{5}) follow this structure.

Step 1: Coprime numbers cannot both be even. Step 2: Both being even means (2) is a common factor. Step 3: Hence the rational assumption is proved false.

Step 1: Squaring (p=3k) gives (p^2=(3k)^2). Step 2: Therefore (p^2=9k^2). Step 3: Do not forget to square the coefficient; it leads to (q^2=3k^2) next.

Step 1: Assuming rationality, (\sqrt{5}=\frac{a}{b}) is taken in lowest form. Step 2: The proof shows that both (a) and (b) are divisible by (5). Step 3: This cannot happen in lowest form, so the assumption is false.

Step 1: (4,9,25) are perfect squares, so their square roots are integers. Step 2: (5) is not a perfect square, and (\sqrt{5}) is irrational. Step 3: In options, identify perfect squares first.

Step 1: (\frac{p}{q}) is valid only when (q\neq0). Step 2: This condition is necessary when writing the rational form. Step 3: Small conditions make the proof complete.

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

Step 1: In these proofs, the square root is first assumed rational. Step 2: Then an impossible situation appears because numerator and denominator get a common factor. Step 3: Hence this is called proof by contradiction.

Step 1: Both have (3) as a common factor. Step 2: So the fraction can be reduced by (3), meaning it is not in lowest form. Step 3: This becomes the contradiction in the proof for (\sqrt{3}).

Step 1: (\sqrt{2}) is assumed rational and written as a fraction in lowest form. Step 2: The proof shows that numerator and denominator are both even. Step 3: This is impossible in lowest form, so (\sqrt{2}) is irrational.

Step 1: If a prime number is a factor of a square, it is also a factor of the original number. Step 2: Therefore (5\mid b^2) gives (5\mid b). Step 3: Instead of writing it without reason, mention that (5) is prime.

Step 1: (\frac{p}{q}) was taken in lowest form. Step 2: The proof shows that both (p) and (q) are divisible by (2). Step 3: So the form is no longer lowest, and the assumption fails.

Step 1: In (\sqrt{3}), the common factor obtained is (3). Step 2: In (\sqrt{5}), the common factor obtained is (5). Step 3: The idea is the same; only the prime number changes.

Step 1: If (x) were odd, then (x^2) would be odd. Step 2: Since (x^2) is given even, (x) must be even. Step 3: This rule is used immediately in the proof of (\sqrt{2}).

Step 1: (5\mid a) only tells divisibility of (a). Step 2: Later (5\mid b) is also obtained, creating a common factor. Step 3: So coprimality is not proved; a contradiction is obtained.

Step 1: (2,3,5) are prime numbers and not perfect squares. Step 2: Assuming their square roots rational creates a common factor in the coprime numerator and denominator. Step 3: Therefore (\sqrt{2}), (\sqrt{3}), and (\sqrt{5}) are all irrational.