Proof of irrationality of √2, √3, √5

Step 1: In the contradiction method, we begin by assuming the opposite statement. Step 2: So we assume (\sqrt{2}) is rational and write it as (\frac{p}{q}), where (p) and (q) are coprime. Step 3: In exams, write the starting assumption clearly.

Step 1: Square both sides of (\sqrt{2}=\frac{p}{q}). Step 2: The left side becomes (2) and the right side becomes (\frac{p^2}{q^2}), so (p^2=2q^2). Step 3: After squaring, multiply by (q^2) to clear the denominator.

Step 1: In (p^2=2q^2), the right side has a factor (2). Step 2: Therefore (p^2) is divisible by (2) and is even. Step 3: A number with factor (2) is even.

Step 1: If the square of an integer is even, then the integer itself is even. Step 2: So if (p^2) is even, (p) is also even. Step 3: This small fact is very important in the proof of (\sqrt{2}).

Step 1: An even number is completely divisible by (2). Step 2: Therefore if (p) is even, we can write (p=2k). Step 3: Writing an even number as (2k) makes the proof easier.

Step 1: If (p=2k), then (p^2=4k^2). Step 2: From (4k^2=2q^2), we get (q^2=2k^2), so (q^2) is even. Step 3: Simplify step by step to avoid mistakes.

Step 1: At the start, (p) and (q) were assumed coprime. Step 2: The proof shows both (p) and (q) are even, so they have common factor (2). Step 3: This contradiction shows that (\sqrt{2}) is not rational.

Step 1: Assuming rationality gives a common factor in (p) and (q). Step 2: This contradicts the condition that they are coprime. Step 3: Therefore the original assumption is false and (\sqrt{2}) is irrational.

Step 1: A rational number can be written as a ratio of two integers. Step 2: In the simplest form, (p) and (q) are coprime. Step 3: This form is used later to create a contradiction.

Step 1: Squaring both sides gives (3=\frac{p^2}{q^2}). Step 2: Clearing the denominator gives (p^2=3q^2). Step 3: In the proof of (\sqrt{3}), the factor (3) plays the main role.

Step 1: In (p^2=3q^2), the right side has factor (3). Step 2: So (p^2) is divisible by (3). Step 3: If a square is divisible by a prime, the original number is also divisible by that prime.

Step 1: (3) is a prime number. Step 2: If the square of an integer is divisible by (3), then the integer is also divisible by (3). Step 3: This rule is the main step in the proof of (\sqrt{3}).

Step 1: A number divisible by (3) has (3) as a factor. Step 2: So it can be written as (p=3k). Step 3: This form helps prove the same thing for (q) in the next step.

Step 1: Putting (p=3k) gives (p^2=9k^2). Step 2: From (9k^2=3q^2), we get (q^2=3k^2), so (q) is also divisible by (3). Step 3: A common factor breaks the coprime condition.

Step 1: At the start, (p) and (q) are taken as coprime. Step 2: The proof shows both are divisible by (3), so they have common factor (3). Step 3: This contradiction proves (\sqrt{3}) is irrational.

Step 1: Assuming rationality makes both (p) and (q) divisible by (3). Step 2: This goes against their being coprime. Step 3: Therefore (\sqrt{3}) is not rational, but irrational.

Step 1: In the contradiction method, we assume the opposite. Step 2: So (\sqrt{5}) is assumed rational and written as (\frac{p}{q}). Step 3: Do not forget to mention that (p) and (q) are coprime.

Step 1: Squaring both sides gives (5=\frac{p^2}{q^2}). Step 2: Multiplying by (q^2) gives (p^2=5q^2). Step 3: Write the denominator-clearing step clearly.

Step 1: In (p^2=5q^2), the right side has factor (5). Step 2: Therefore (p^2) is divisible by (5). Step 3: This gives the next conclusion about (p) in the proof.

Step 1: (5) is a prime number. Step 2: If the square of an integer is divisible by (5), then the integer is also divisible by (5). Step 3: This rule moves the proof forward.