Proof of irrationality of √2, √3, √5

Step 1: Divide both sides of (25k^2=5q^2) by (5). Step 2: We get (5k^2=q^2), that is (q^2=5k^2). Step 3: This later shows that (q) is divisible by (5).

Step 1: (4) is a perfect square. Step 2: (\sqrt{4}=2), which is rational. Step 3: Square roots of perfect squares are rational.

Step 1: The square root of a perfect square is an integer. Step 2: (2), (3), and (5) are not perfect squares. Step 3: Therefore their square roots are proved irrational.

Step 1: A rational number can be written as a ratio of two integers. Step 2: So assuming (\sqrt{2}) rational, we write it as (\frac{p}{q}). Step 3: Start the proof from the definition.

Step 1: While squaring a fraction, both numerator and denominator are squared. Step 2: Hence \(\left(\frac{p}{q}\right)^2=\frac{p^2}{q^2}\). Step 3: Squaring only the numerator is a common mistake.

Step 1: The square of (\sqrt{5}) is (5). Step 2: So the left side becomes ((\sqrt{5})^2=5). Step 3: A square and square root cancel each other.

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

Step 1: Assuming rationality makes both (p) and (q) even. Step 2: This contradicts their being coprime. Step 3: Therefore the initial assumption is false and (\sqrt{2}) is irrational.

Step 1: In the proof of (\sqrt{3}), both (p) and (q) are found divisible by (3). Step 2: But they were assumed coprime at the beginning. Step 3: This is the final contradiction.

Step 1: First, (p) is found divisible by (5). Step 2: Substituting (p=5k) gives divisibility by (5) for (q) too. Step 3: A common factor in both creates the contradiction.

Step 1: A rational number is taken as a fraction in lowest form. Step 2: In lowest form, numerator and denominator are coprime. Step 3: Finding a common factor later contradicts this condition.

Step 1: If both are divisible by (5), then (5) is a common factor. Step 2: Coprime numbers should not have a common factor other than (1). Step 3: So this situation goes against being coprime.

Step 1: Assume (\sqrt{2}) rational and write it in lowest form. Step 2: The proof gives both numerator and denominator even. Step 3: This contradicts lowest form, so (\sqrt{2}) is irrational.

Step 1: From (p^2=3q^2), (p^2) is divisible by (3). Step 2: This gives (p) divisible by (3), but we cannot directly write (p=3q). Step 3: The correct way is to write (p=3k).

Step 1: An even integer is written as (2) times an integer. Step 2: Therefore in (p=2k), (k) is an integer. Step 3: Mentioning the type of (k) makes the proof clear.

Step 1: First assume (\sqrt{2}) is rational. Step 2: After squaring, both (p) and (q) are found even. Step 3: Both being even contradicts the coprime condition.

Step 1: From (p^2=5q^2), (p^2) is divisible by (5). Step 2: So (p) is also divisible by (5) and is written as (p=5k). Step 3: Choose the correct factor according to the number.

Step 1: In contradiction, we work with the opposite assumption. Step 2: If that assumption becomes impossible, the original statement is true. Step 3: So when rationality fails, irrationality is proved.

Step 1: In the proof of (\sqrt{2}), we get (p^2=2q^2). Step 2: This makes both (p) and (q) divisible by (2), that is even. Step 3: The common factor (2) creates the contradiction.

Step 1: The square of an even number is even and the square of an odd number is odd. Step 2: So if (p^2) is even, (p) is also even. Step 3: This rule is used in the proof of (\sqrt{2}).