Proof of irrationality of √2, √3, √5

Step 1: In contradiction, we assume the opposite of what we want to prove. Step 2: Here we want to prove (\sqrt{2}) irrational, so we first assume it rational. Step 3: In exams, write the opposite assumption clearly.

Step 1: The denominator of any fraction cannot be zero. Step 2: So in (\frac{p}{q}), we must write (q\neq 0). Step 3: Do not forget this small condition while writing the rational form.

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

Step 1: If the square of an integer is even, the integer itself is even. Step 2: So if (p^2) is divisible by (2), then (p) is also divisible by (2). Step 3: This is the key rule in the proof of (\sqrt{2}).

Step 1: The right side of the equation is (3q^2). Step 2: So (p^2) has factor (3) and is divisible by (3). Step 3: In such questions, identify the factor on the right side.

Step 1: In (p^2=5q^2), the right side has factor (5). Step 2: Therefore (p^2) is divisible by (5). Step 3: First write divisibility of the square, then conclude about (p).

Step 1: A number divisible by (3) has (3) as a factor. Step 2: So it is written as (p=3k), where (k) is an integer. Step 3: In proofs, write this type of form after getting divisibility.

Step 1: We have (p=2k). Step 2: Squaring gives (p^2=(2k)^2=4k^2). Step 3: Do not forget to square the coefficient.

Step 1: Square (p=5k). Step 2: Since ((5k)^2=25k^2), (p^2=25k^2). Step 3: Do this algebraic step carefully in the proof of (\sqrt{5}).

Step 1: Divide both sides of (9k^2=3q^2) by (3). Step 2: We get (3k^2=q^2), that is (q^2=3k^2). Step 3: This leads to (q) being divisible by (3).

Step 1: An even number is divisible by (2). Step 2: If both (p) and (q) are even, both have (2) as a common factor. Step 3: Coprime numbers have no common factor except (1).

Step 1: In the proof of (\sqrt{5}), (p^2=5q^2) makes (p) divisible by (5). Step 2: Then (q) is also found divisible by (5). Step 3: Having common factor (5) contradicts the coprime condition.

Step 1: Assuming (\sqrt{3}) rational gives (p^2=3q^2). Step 2: The factor (3) appears commonly in (p) and (q). Step 3: The number under the square root becomes the key factor.

Step 1: In all three proofs, the number is first assumed rational. Step 2: Then a contradiction is shown using the coprime condition. Step 3: Therefore it is called the method of contradiction.

Step 1: Lowest form means the fraction cannot be reduced further. Step 2: So the numerator and denominator have only (1) as a common factor. Step 3: This fact is important in irrationality proofs.

Step 1: From (p^2=2q^2), we get only that (p^2) is even. Step 2: Then by rule, (p) is even and can be written as (p=2k). Step 3: Writing (p=2q) directly from it is wrong.

Step 1: (\sqrt{5}=5) is false 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: To prove irrationality, we assume the opposite statement. Step 2: So at the beginning, (\sqrt{3}) is assumed rational. Step 3: Then it is written as a fraction in lowest form.

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 is used in the proof of (\sqrt{5}).

Step 1: Divide both sides of (4k^2=2q^2) by (2). Step 2: We get (2k^2=q^2), that is (q^2=2k^2). Step 3: This indicates that (q) is even later.