Proof of irrationality of √2, √3, √5

Step 1: From (p^2=2q^2), (p^2) is even. Step 2: If the square of an integer is even, the integer itself is even, so (p=2k) can be written. Step 3: In exams, give the reason for evenness before writing (p=2k).

Step 1: The contradiction depends on (a) and (b) being coprime in lowest form. Step 2: Without this condition, finding (3) common to both will not be a real contradiction. Step 3: Therefore lowest form must be stated at the beginning.

Step 1: Squaring (\sqrt{5}=\frac{x}{y}) gives (5=\frac{x^2}{y^2}). Step 2: Multiplying both sides by (y^2) gives (x^2=5y^2). Step 3: Remember the condition (y\neq0) while removing the denominator.

Step 1: (q^2=2k^2) shows that (q^2) is even. Step 2: If a square is even, the integer itself is even, so (q) is even. Step 3: Now both (p) and (q) are even, completing the contradiction.

Step 1: (3) is a prime number. Step 2: If a prime number divides a square, it also divides the original number. Step 3: This principle plays the main role in the proof for (\sqrt{3}).

Step 1: Coprime numbers have no common factor except (1). Step 2: (5\mid p) and (5\mid q) make (5) a common factor. Step 3: Therefore the rational assumption is proved false.

Step 1: If (p^2) is even, then (p) must be even. Step 2: Calling (p) odd violates the parity rule. Step 3: In proofs, a small logical error can change the whole argument.

Step 1: Squaring (a=3m) gives (a^2=9m^2). Step 2: Substituting in (a^2=3b^2) gives (9m^2=3b^2). Step 3: Squaring the coefficient correctly is necessary for the next conclusion.

Step 1: (2.236) is only an approximate value, not the full value. Step 2: To prove irrationality, we must assume rationality and show a contradiction. Step 3: In exams, do not write a decimal approximation in place of proof.

Step 1: In all three, the square root is first assumed rational. Step 2: Then squaring and prime divisibility give a common factor. Step 3: This common factor contradicts coprimality.

Step 1: Squaring (\sqrt{r}=\frac{p}{q}) gives (r=\frac{p^2}{q^2}). Step 2: Multiplying both sides by (q^2) gives (p^2=rq^2). Step 3: This general equation applies to (2,3,5).

Step 1: From (b^2=3m^2), we get (3\mid b^2). Step 2: Since (3) is prime, (3\mid b). Step 3: This shows (3) common to both (a) and (b).

Step 1: In (25n^2=5y^2), both sides can be divided by (5). Step 2: This gives (5n^2=y^2), that is (y^2=5n^2). Step 3: While simplifying, remove only the common factor, not the whole (25).

Step 1: Both being even means both have (2) as a common factor. Step 2: A fraction in lowest form cannot have such a common factor. Step 3: This breaks the rational assumption.

Step 1: Coprime means there is no common factor except (1). Step 2: Both being divisible by (3) gives a common factor. Step 3: This contradiction proves (\sqrt{3}) irrational.

Step 1: Putting (x=5n) gives (y^2=5n^2). Step 2: So (5\mid y^2), and by the prime rule (5\mid y). Step 3: Then (5) becomes common to both (x) and (y).

Step 1: A decimal approximation of (\sqrt{2}) does not prove irrationality. Step 2: The real proof assumes (\sqrt{2}=\frac{p}{q}) and derives a contradiction. Step 3: In exams, give priority to logical proof.

Step 1: From (3\mid a^2), we get (3\mid a). Step 2: So (a=3k) is correct, where (k) is an integer; it is not necessary that (k=b). Step 3: Using a new helper variable is safer.

Step 1: (5) is a prime number. Step 2: If a prime number divides a square, it also divides the original number. Step 3: This rule gives the divisibility of (x) and later (y).

Step 1: First (p^2) even proves (p) even. Step 2: Then (q^2) even proves (q) even. Step 3: Therefore this fact is used twice in an important way.

Step 1: For (\sqrt{3}), the equation is (p^2=3q^2), so (3) is used. Step 2: For (\sqrt{5}), the equation is (p^2=5q^2), so (5) is used. Step 3: Identify the related prime in each proof.

Step 1: A rational number is written as a ratio of two integers. Step 2: The denominator cannot be zero, and the ratio should be in lowest form. Step 3: This complete opening sentence sets the proof correctly.

Step 1: Both (x) and (y) are divisible by (5). Step 2: Therefore their greatest common divisor is at least (5). Step 3: This goes against the condition of being coprime.

Step 1: The contradiction works only when numerator and denominator are first assumed coprime. Step 2: If lowest form is missing, a common factor will not be decisive. Step 3: So write the fraction in lowest form at the start.

Step 1: (2\mid p) and (2\mid q) mean both have (2) as a common factor. Step 2: This cannot happen for coprime numbers. Step 3: This conflict is the decisive point of the proof.