Proof of irrationality of √2, √3, √5

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.

Step 1: In the proof of (\sqrt{5}), we get (p^2=5q^2). Step 2: This proves both (p) and (q) are divisible by (5). Step 3: The common factor (5) breaks the coprime condition.

Step 1: The proof begins by assuming rationality. Step 2: So first we write (\sqrt{2}=\frac{p}{q}). Step 3: Conclusions about (p) and (q) being even come later.

Step 1: The rational assumption and squaring steps come first. Step 2: Near the end, both (p) and (q) are found divisible by (3). Step 3: This creates a contradiction against the coprime condition.

Step 1: From (p^2=5q^2), we get that (p^2) is divisible by (5). Step 2: We cannot directly write (p=5q). Step 3: The correct step is to say (p) is divisible by (5), then write (p=5k).

Step 1: A rational number is written as the ratio of two integers. Step 2: In lowest form, those integers are coprime. Step 3: Therefore (p) and (q) are taken as integers and coprime.

Step 1: Putting (n=5) gives the number (\sqrt{5}). Step 2: Squaring gives (p^2=5q^2). Step 3: This starts the irrationality proof of (\sqrt{5}).

Step 1: In the proof of (\sqrt{2}), (p^2) is found even. Step 2: By the correct rule, (p) is also even. Step 3: Then writing (p=2k) gives the same result for (q).

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 property creates the common-factor contradiction.

Step 1: The rational assumption leads to a contradiction in the proof. Step 2: When the assumption is false, the given number is proved irrational. Step 3: In the final sentence, clearly write both the contradiction and the conclusion.

Step 1: A rational number is written as (\frac{p}{q}) in its lowest form. Step 2: In lowest form, (p) and (q) have no common factor except (1). Step 3: Later, finding both even creates the contradiction.