Proof of irrationality of √2, √3, √5

Step 1: A number divisible by (5) has (5) as a factor. Step 2: So (p) can be written as (p=5k). Step 3: Substituting this form in the original equation gives the same conclusion for (q).

Step 1: If (p=5k), then (p^2=25k^2). Step 2: From (25k^2=5q^2), we get (q^2=5k^2), so (q) is also divisible by (5). Step 3: Getting a common factor creates the contradiction.

Step 1: We started by taking (p) and (q) as coprime. Step 2: The proof shows both are divisible by (5). Step 3: Two coprime numbers cannot have (5) as a common factor, so this is a contradiction.

Step 1: Assuming rationality makes both (p) and (q) divisible by (5). Step 2: This contradicts the coprime condition. Step 3: Hence (\sqrt{5}) is irrational.

Step 1: In all three proofs, the number is first assumed rational. Step 2: Then a contradiction appears through a common factor. Step 3: This type of proof is called the method of contradiction.

Step 1: A rational number is written as (\frac{p}{q}) in its simplest form. Step 2: In simplest form, (p) and (q) have only (1) as a common factor. Step 3: Getting another common factor later contradicts this condition.

Step 1: In contradiction, we assume the opposite statement. Step 2: If the opposite becomes impossible, the original statement is true. Step 3: That is why breaking the rational assumption proves irrationality.

Step 1: (3) and (5) are prime numbers. Step 2: If a prime factor divides a square, it also divides the original number. Step 3: This helps prove a common factor in (p) and (q).

Step 1: A rational number is written in the form (\frac{p}{q}). Step 2: The denominator of a fraction cannot be zero. Step 3: Therefore (q\neq 0) must be written in the proof.

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 do not have a common factor other than (1).

Step 1: Being divisible by (3) means both have (3) as a factor. Step 2: Coprime numbers should not have a common factor other than (1). Step 3: Therefore it gives a contradiction in the proof of (\sqrt{3}).

Step 1: If both are divisible by (5), then (5) is a common factor. Step 2: Coprime numbers cannot have such a common factor. Step 3: This creates the contradiction in the proof of (\sqrt{5}).

Step 1: (4), (9), and (25) are perfect squares. Step 2: (2) is not a perfect square, so (\sqrt{2}) is proved irrational. Step 3: First identify perfect and non-perfect squares.

Step 1: (4) is a perfect square. Step 2: (\sqrt{4}=2), which is rational. Step 3: The square root of a perfect square is not proved irrational.

Step 1: From (p^2=2q^2), we get that (p^2) is even. Step 2: Then (p) is even and can be written as (p=2k). Step 3: Saying (p=q) from this equation is a wrong step.

Step 1: (p=3k). Step 2: Squaring gives (p^2=(3k)^2=9k^2). Step 3: Do not forget to square the coefficient also.

Step 1: We have (p=5k). Step 2: Squaring gives (p^2=(5k)^2=25k^2). Step 3: Do this simplification carefully in the proof of (\sqrt{5}).

Step 1: (p=2k). Step 2: Squaring gives (p^2=(2k)^2=4k^2). Step 3: Writing ((2k)^2) as (2k^2) is a common mistake.

Step 1: In the proof of (\sqrt{2}), we get (p^2=2q^2). Step 2: This makes both (p) and (q) even. Step 3: The common factor (2) creates the contradiction.

Step 1: In the proof of (\sqrt{3}), we get (p^2=3q^2). Step 2: This proves both (p) and (q) are divisible by (3). Step 3: The prime under the root becomes the common factor.