Proof of irrationality of √2, √3, √5

Step 1: For (\sqrt{3}), the equation is (p^2=3q^2). Step 2: For (\sqrt{5}), the equation is (p^2=5q^2). Step 3: The structure is the same; only the prime factor changes.

Step 1: The contradiction depends on (p) and (q) being coprime. Step 2: If lowest form is not taken, getting a common factor will not be a contradiction. Step 3: Therefore lowest form is essential at the start.

Step 1: (3\mid p) means (p) is a multiple of (3). Step 2: So we write (p=3k), where (k) is an integer. Step 3: Converting divisibility into a multiple form helps in the proof.

Step 1: From (5\mid p), it is proper to write (p=5k). Step 2: Substituting it into the original equation gives (q^2=5k^2). Step 3: Then prove (5\mid q) and complete the contradiction.

Step 1: Squares of integers are like (0,1,4,9). Step 2: No integer has square (2). Step 3: Still, to prove irrationality, the full rational-form proof is needed.

Step 1: Both have (3) as a common factor. Step 2: So the fraction could be reduced by (3). Step 3: This directly contradicts the lowest-form assumption.

Step 1: (2\mid q^2) means (q^2) is even. Step 2: If the square of an integer is even, the integer is also even. Step 3: Therefore (q) is even and the contradiction is completed.

Step 1: In lowest form, numerator and denominator are coprime. Step 2: Both being divisible by (5) gives a common factor. Step 3: So this situation goes against the starting condition.

Step 1: The proof gets a contradiction from the rational assumption. Step 2: When a contradiction occurs, that assumption is false. Step 3: In the final sentence, clearly write that (\sqrt{2}) is irrational.

Step 1: When squaring (p=3r), both (3) and (r) are squared. Step 2: Therefore (p^2=(3r)^2=9r^2). Step 3: Do not forget to square the coefficient, or the proof will go wrong.

Step 1: Putting (p=5r) gives (25r^2=5q^2). Step 2: Dividing both sides by (5) gives (q^2=5r^2). Step 3: Reduce factors correctly during simplification.

Step 1: (2,3,5) are not perfect squares and are prime. Step 2: Assuming their square roots rational creates a common-factor contradiction. Step 3: Identifying perfect squares is the first task in such questions.

Step 1: (p=5m) means (5\mid p). Step 2: (q=5n) means (5\mid q). Step 3: Together, these break coprimality.

Step 1: Proof by contradiction assumes the opposite statement. Step 2: Then that assumption gives an impossible result. Step 3: In (\sqrt{2}), the common factor (2) is that impossible result.

Step 1: From (3\mid p), write (p=3k). Step 2: Substituting in (p^2=3q^2) gives (q^2=3k^2). Step 3: Then (3\mid q) is proved.

Step 1: A long decimal value is not a necessary part of the proof. Step 2: The real proof is based on rational assumption and divisibility. Step 3: To save time, write only the logical steps.

Step 1: From (q^2=2r^2), (q^2) is a multiple of (2). Step 2: So (q^2) is even and the integer (q) is also even. Step 3: This is the second evenness conclusion in the proof.

Step 1: In the rational assumption, the fraction is in lowest form. Step 2: The proof shows that both numerator and denominator are divisible by (3). Step 3: This inconsistency shows that the assumption was false.

Step 1: From (5\mid p^2), we only get (5\mid p). Step 2: This allows (p=5k), not necessarily (p=5q). Step 3: Do not create an unsupported relation between variables.

Step 1: First assume the square root is rational. Step 2: Then square and use prime divisibility to show a common factor in numerator and denominator. Step 3: In exams, this order makes a clear full-mark answer.