Proof of irrationality of √2, √3, √5

Step 1: In lowest form, the numerator and denominator of a fraction are coprime. Step 2: This means their greatest common divisor is (1). Step 3: This condition breaks when a common factor is found.

Step 1: At the beginning, (a) and (b) were assumed coprime. Step 2: Finding both divisible by (3) contradicts this. Step 3: Therefore assuming (\sqrt{3}) rational is false.

Step 1: The rational assumption makes both (a) and (b) divisible by (5). Step 2: This contradicts the coprime condition. Step 3: Therefore the conclusion is that (\sqrt{5}) is irrational.

Step 1: From (a^2=2b^2), (a^2) is even. Step 2: By the rule, (a) is even, and later (b) is also even. Step 3: Understanding this rule clearly is important.

Step 1: (3) and (5) are both prime. Step 2: If they divide (a^2), they also divide (a). Step 3: This rule leads to the common factor.

Step 1: Divide both sides of (4k^2=2b^2) by (2). Step 2: This gives (2k^2=b^2), that is (b^2=2k^2). Step 3: In such steps, divide both sides by the same number.

Step 1: Divide both sides of (9k^2=3b^2) by (3). Step 2: We get (3k^2=b^2), that is (b^2=3k^2). Step 3: Then conclude that (b) is divisible by (3).

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 helps show that (b) is divisible by (5).

Step 1: Lowest form means numerator and denominator are coprime. Step 2: If there is a common factor other than (1), the fraction can be reduced further. Step 3: This becomes the contradiction in irrationality proofs.

Step 1: At the beginning, the fraction is taken in lowest form. Step 2: At the end, a common factor is found in numerator and denominator. Step 3: This is impossible for a lowest-form fraction, so a contradiction occurs.

Step 1: To prove irrationality, we take the opposite assumption. Step 2: So first we assume (\sqrt{3}) is rational. Step 3: Then we write it as (\frac{a}{b}) in lowest form.

Step 1: From (a^2=5b^2), (a^2) is divisible by (5). Step 2: Since (5) is prime, (a) is also divisible by (5). Step 3: This is the correct middle step, not directly (a=5b).

Step 1: An even number is always divisible by (2). Step 2: Both are even, so (2) is their common factor. Step 3: Finding a common factor contradicts the lowest-form condition.

Step 1: (a=3k) means (a) is divisible by (3). Step 2: (b=3l) means (b) is also divisible by (3). Step 3: So both have (3) as a common factor.

Step 1: (a=5k) and (b=5l) mean both are divisible by (5). Step 2: But (a) and (b) were assumed coprime at the beginning. Step 3: Hence this result gives a contradiction.

Step 1: In the proof of (\sqrt{2}), we get (a^2=2b^2). Step 2: So the factor (2) plays the main role. Step 3: The number under the square root appears as the key factor in the proof.

Step 1: Assuming (\sqrt{3}) rational gives (a^2=3b^2). Step 2: So factor (3) becomes the main base of the proof. Step 3: It shows both (a) and (b) divisible by (3).

Step 1: Assuming (\sqrt{5}) rational gives (a^2=5b^2). Step 2: Here prime factor (5) is the key. Step 3: It leads to common factor (5) in both numbers.

Step 1: (\sqrt{2}=2) is false because (2^2=4). Step 2: The correct proof uses (a^2=2b^2) to get evenness and contradiction. Step 3: Avoid writing false equalities.

Step 1: (3) is not a perfect square. Step 2: In the proof of (\sqrt{3}), the fact that (3) is prime is useful. Step 3: Understand the difference between perfect square and prime.

Step 1: (\sqrt{5}=5) is wrong because (5^2=25). Step 2: In the correct proof, (\sqrt{5}) is assumed rational and a contradiction is obtained. Step 3: Do not treat a square root as equal to the number under it.

Step 1: Assuming rational, we write (\sqrt{2}=\frac{a}{b}) in lowest form. Step 2: The proof shows both (a) and (b) are even. Step 3: This contradicts lowest form, so (\sqrt{2}) cannot be rational.

Step 1: Assuming (\sqrt{3}=\frac{a}{b}) and squaring gives (a^2=3b^2). Step 2: This makes both (a) and (b) divisible by (3). Step 3: This is impossible in a lowest-form fraction.

Step 1: We assume (\sqrt{5}) rational and write it in lowest form. Step 2: The proof shows numerator and denominator both divisible by (5). Step 3: This contradicts lowest form, so (\sqrt{5}) is irrational.

Step 1: The proof starts with the rational assumption. Step 2: At the end, this assumption contradicts the coprime condition. Step 3: In the final line, clearly write the contradiction and the irrational conclusion.