Proof of irrationality of √2, √3, √5

Step 1: Coprimality means there is no common factor. Step 2: When both (p) and (q) are proved even, (2) becomes a common factor. Step 3: At this point, coprimality gives the decisive contradiction.

Step 1: The proof gets a common-factor contradiction from the rational assumption. Step 2: The contradiction proves that assumption false. Step 3: End clearly by writing that (\sqrt{5}) is irrational.

Step 1: In (b^2=3k^2), the right side is a multiple of (3). Step 2: Therefore (b^2) is divisible by (3). Step 3: Then use (3\mid b) to complete the contradiction.

Step 1: (4) is a perfect square. Step 2: (\sqrt{4}=2), which is rational and an integer. Step 3: The irrationality contradiction proof is not applied to perfect squares.

Step 1: Both being divisible by (5) shows that (5) is a common factor. Step 2: Coprime numbers cannot have such a common factor. Step 3: Therefore the statement that they are coprime is proved false.

Step 1: From (p^2=3q^2), (3) first appears in (p). Step 2: Then putting (p=3k) makes (3) appear in (q) too. Step 3: This gives a common factor in numerator and denominator.

Step 1: (p=2m) shows (2\mid p). Step 2: (q=2n) shows (2\mid q). Step 3: Together, they make (2) a common factor.

Step 1: From (x^2=5y^2), we get (5\mid x^2) and then (5\mid x). Step 2: This does not necessarily mean that (x) is divisible by (25). Step 3: Write only what is proved.

Step 1: First write (\frac{p}{q}) in lowest form. Step 2: Then square and use the related prime factor to show divisibility of both numerator and denominator. Step 3: Finally state the contradiction with coprimality clearly.

Step 1: Assuming (\sqrt{3}) rational, (\frac{p}{q}) is taken in lowest form. Step 2: The proof gives (3\mid p) and (3\mid q), so (3) is a common factor of both. Step 3: A lowest-form fraction cannot have a common factor, so (\sqrt{3}) is proved irrational.

Step 1: From (p^2=2q^2), (p^2) is even, so (p) is even. Step 2: Putting (p=2k) gives (q^2=2k^2), so (q) is even. Step 3: Both being even contradicts coprimality of the lowest-form fraction.

Step 1: From (p^2=3q^2), we get (3\mid p^2). Step 2: Since (3) is prime, (3\mid p) is a valid conclusion. Step 3: Do not say only odd; mention primality for a complete proof.

Step 1: A rational number is written as a ratio of two integers. Step 2: In the proof, the fraction is taken in lowest form, so (a,b) are coprime and (b\neq0). Step 3: This condition later creates the contradiction with a common factor.

Step 1: Substituting (p=2k) in (p^2=2q^2) gives (4k^2=2q^2). Step 2: Dividing both sides by (2) gives (q^2=2k^2). Step 3: Reduce factors carefully during algebraic simplification.

Step 1: Substitute (p=3r) in the original equation. Step 2: From (9r^2=3q^2), we get (q^2=3r^2), so (3\mid q). Step 3: Do not conclude about (q) without substitution.

Step 1: (5\mid a) means (a) is divisible by (5). Step 2: Divisibility is written in multiple form, so (a=5t). Step 3: This form helps prove divisibility of (b) next.

Step 1: Coprime numbers have no common factor except (1). Step 2: (2\mid p) and (2\mid q) make (2) a common factor. Step 3: This is the final contradiction.

Step 1: (1.732) is only an approximate value, not the full value. Step 2: To prove irrationality, we must assume rationality and show a contradiction with coprimality. Step 3: In exams, write a logical proof, not an approximation.

Step 1: Both numbers are divisible by (5). Step 2: Therefore their greatest common divisor cannot remain (1); it will be at least (5). Step 3: This breaks the coprimality condition.

Step 1: (\sqrt{8}=\sqrt{4\cdot2}=2\sqrt{2}). Step 2: (\sqrt{2}) is irrational and (2) is a nonzero rational number, so (2\sqrt{2}) remains irrational. Step 3: Separate perfect-square factors while simplifying roots.