Proof of irrationality of √2, √3, √5

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.

Step 1: In the proof, we assume (\sqrt{2}=\frac{a}{b}). Step 2: To remove the square root, we square both sides. Step 3: In exams, do not skip the squaring step.

Step 1: The denominator of a fraction cannot be zero. Step 2: So while writing (\frac{a}{b}), the condition (b\neq 0) is necessary. Step 3: Write this condition when expressing a rational number.

Step 1: A rational number is written as a ratio of two integers. Step 2: In the proof, it is taken in lowest form, so (m) and (n) are coprime. Step 3: Later, finding a common factor gives the contradiction.

Step 1: The right side of the equation is (2b^2). Step 2: So (a^2) has factor (2) and is divisible by (2). Step 3: Use the factor to decide divisibility.

Step 1: From (a^2=3b^2), (a^2) is divisible by (3). Step 2: Since (3) is prime, (a) is also divisible by (3). Step 3: Remember this rule from square to original number.

Step 1: From (a^2=5b^2), (a^2) is divisible by (5). Step 2: Therefore (a) is also divisible by (5), so (a=5k). Step 3: After divisibility, write the number using that factor.

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

Step 1: (a=3k). Step 2: Squaring gives ((3k)^2=9k^2). Step 3: Write this algebraic step carefully in the proof of (\sqrt{3}).

Step 1: We write (a=5k). Step 2: Squaring gives (a^2=(5k)^2=25k^2). Step 3: A small algebra mistake can spoil the whole proof.

Step 1: An even number is divisible by (2). Step 2: If both (a) and (b) are even, both have (2) as a common factor. Step 3: This contradicts the coprime condition.