Proof of irrationality of √2, √3, √5

Step 1: Coprime numbers have no common factor except (1). Step 2: If both (a) and (b) are divisible by (3), they have common factor (3). Step 3: This is the contradiction in the proof.

Step 1: If both are divisible by (5), both have (5) as a common factor. Step 2: This cannot happen for coprime numbers. Step 3: Thus the initial rational assumption becomes false.

Step 1: First assume (\sqrt{2}) is rational. Step 2: Then square and use (a^2=2b^2) to get evenness results. Step 3: Finally, the coprime condition gives a contradiction.

Step 1: From (a^2=3b^2), (a^2) is divisible by (3). Step 2: Then (a) is divisible by (3), so (a=3k). Step 3: Directly writing (a=3b) from the equation is wrong.

Step 1: The right side of the equation is (5b^2). Step 2: Therefore the left side (a^2) is also divisible by (5). Step 3: First write divisibility of the square, then of the original number.

Step 1: In contradiction, we take the opposite assumption. Step 2: If it leads to an impossible result, the original statement is proved true. Step 3: This method is very useful in irrationality proofs.

Step 1: An even integer is written as (2) times an integer. Step 2: So in (a=2k), (k) is an integer. Step 3: It is good to mention the type of (k) in such forms.

Step 1: From (a^2=3b^2), (a) is found divisible by (3). Step 2: A number divisible by (3) is written as (3k). Step 3: This form helps show divisibility of (b) later.

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

Step 1: At the beginning, (\frac{a}{b}) is taken in lowest form. Step 2: The proof shows both (a) and (b) are even. Step 3: Being coprime and both even is impossible.

Step 1: Square roots of perfect squares are integers. Step 2: (2), (3), and (5) are not perfect squares. Step 3: That is why irrationality proofs are studied for their square roots.

Step 1: (9) is a perfect square. Step 2: (\sqrt{9}=3), which is rational. Step 3: A square root of a perfect square does not need an irrationality proof.

Step 1: A rational number can be written as a ratio of two integers. Step 2: So after assuming rationality, we write (\sqrt{2}=\frac{a}{b}). Step 3: The definition of rationality starts the proof.

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

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

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.