Proof of irrationality of √2, √3, √5

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.

Step 1: First, by the prime rule, (5\mid x). Step 2: Divisibility is written in multiple form, so (x=5m). Step 3: In the proof, write these two small steps clearly.

Step 1: In (\sqrt{3}), the prime nature of (3) gives the common factor. Step 2: The same method can be applied to any prime (r). Step 3: While generalizing, do not forget the condition that (r) is prime.

Step 1: Putting (p=2k) gives (4k^2=2q^2). Step 2: Dividing both sides by (2) gives (2k^2=q^2), that is (q^2=2k^2). Step 3: A simplification error can spoil the proof.

Step 1: Being even means being divisible by (2). Step 2: If both (p) and (q) are even, (\frac{p}{q}) can be reduced by (2). Step 3: This contradicts lowest form.

Step 1: The rational assumption begins with a lowest-form fraction. Step 2: Squaring gives (p^2=3q^2), and then (3) divides first (p), then (q). Step 3: This order makes the answer organized.

Step 1: (a=3m) and (b=3n) show that both are divisible by (3). Step 2: Thus (3) becomes a common factor. Step 3: This conflicts with the starting condition of coprimality.

Step 1: Since (5) is not a perfect square, (\sqrt{5}) cannot be an integer. Step 2: But to prove irrationality, we must also show it is not any rational fraction. Step 3: That is why the contradiction proof is written.

Step 1: We initially assumed that (\sqrt{2}) is rational. Step 2: That assumption led to a common factor in a lowest-form fraction. Step 3: Therefore the initial rational assumption is proved false.

Step 1: Substitute (p=3k) in (p^2=3q^2). Step 2: Simplifying gives (q^2=3k^2), so (3\mid q^2) and (3\mid q). Step 3: This is the second divisibility step.

Step 1: The fact that (2) is even is not enough by itself. Step 2: The real proof assumes (\sqrt{2}) rational and shows numerator and denominator both even. Step 3: Write the full reason, not a short guess.

Step 1: In a rational number (\frac{x}{y}), the denominator cannot be zero. Step 2: So along with (x,y) being coprime integers, (y\neq0) must also be written. Step 3: Complete conditions make the proof stronger.

Step 1: (3\mid a) and (3\mid b) mean both are multiples of (3). Step 2: So the fraction can be reduced by (3). Step 3: This is not possible in lowest form.

Step 1: The square of an odd number is always odd. Step 2: When the square is even, the original number cannot be odd. Step 3: This idea proves both (p) and (q) even.

Step 1: The step from (5\mid y^2) to (5\mid y) uses the prime-divisibility rule. Step 2: Since (5) is prime, the conclusion is valid. Step 3: Without mentioning primality, this step looks incomplete.

Step 1: In (3b^2), (3) is clearly a factor. Step 2: Since (a^2) equals this, (a^2) is also a multiple of (3). Step 3: Then the prime rule gives divisibility of (a).