Proof of irrationality of √2, √3, √5

Step 1: In proof by contradiction, the opposite statement is assumed first. Step 2: Then that assumption leads to a result against the given condition. Step 3: The proofs of (\sqrt{2}), (\sqrt{3}), and (\sqrt{5}) are written by this method.

Step 1: In (3q^2), (3) is clearly a factor. Step 2: Since (p^2) equals it, (p^2) is also a multiple of (3). Step 3: Then use the prime rule to write (3\mid p).

Step 1: The proof obtains a contradiction from the rational assumption. Step 2: The contradiction shows that the starting assumption was false. Step 3: Therefore the final sentence should clearly state that (\sqrt{2}) is irrational.

Step 1: The step from (5\mid b^2) to (5\mid b) uses the prime-factor rule. Step 2: This rule applies because (5) is prime. Step 3: Mentioning this reason makes the proof complete in exams.

Step 1: Both have (3) as a common factor. Step 2: So the fraction can be reduced by (3). Step 3: Such a situation is impossible in lowest form.

Step 1: First (p) is proved even, so (p=2k). Step 2: Substituting in (p^2=2q^2) gives (q^2=2k^2). Step 3: Thus (q^2) is even and hence (q) is even.

Step 1: From (a^2=5b^2), we get (5\mid a^2). Step 2: Since (5) is prime, (5\mid a). Step 3: Then (a=5k) can be written.

Step 1: In the proof for (\sqrt{2}), (p^2) is found even. Step 2: If (p) were odd, (p^2) would be odd; so (p) is even. Step 3: The same parity idea is then used for (q).

Step 1: In (9k^2=3q^2), the common factor is (3). Step 2: Dividing by (3) gives (3k^2=q^2), that is (q^2=3k^2). Step 3: Remove only valid common factors while simplifying.

Step 1: For contradiction, first assume (\sqrt{5}) is rational. Step 2: Write the rational form as a lowest-form fraction with (b\neq0). Step 3: This start makes the later contradiction strong.

Step 1: The square of an odd integer is always odd. Step 2: Here (p^2) is even, so (p) cannot be odd. Step 3: Thus (p) is proved even.

Step 1: In the rational assumption, (\sqrt{3}=\frac{p}{q}) is taken in lowest form. Step 2: The proof gives both (3\mid p) and (3\mid q). Step 3: This is impossible in lowest form, so the assumption is false.

Step 1: By the prime rule, (5\mid a^2) gives (5\mid a). Step 2: So (a=5k) is correct, but (a=25k) is not necessary. Step 3: Avoid making extra claims in proofs.

Step 1: A rational number is written as (\frac{p}{q}). Step 2: In lowest form, (p) and (q) are coprime. Step 3: This property is used to create the contradiction.

Step 1: (p=2m) means (2\mid p). Step 2: (q=2n) means (2\mid q). Step 3: Together, they give a common factor against coprimality.

Step 1: A long decimal value is not a necessary part of the proof. Step 2: The proof is based on rational assumption, squaring, and prime divisibility. Step 3: Avoid unnecessary decimals in exams.

Step 1: At the beginning, (a) and (b) were assumed coprime. Step 2: Both being divisible by (5) gives a common factor. Step 3: Therefore this is a contradictory result, and the rational assumption is false.

Step 1: (p,q) were assumed coprime in lowest form. Step 2: Both being even makes (2) a common factor. Step 3: Therefore the rational assumption is proved false.

Step 1: (3\mid p) and (3\mid q) make (3) a common factor. Step 2: With a common factor, the two numbers cannot be coprime. Step 3: Therefore the statement that they are coprime becomes false.

Step 1: First assume (\sqrt{r}=\frac{p}{q}) in lowest form. Step 2: Square and use the related prime (r) to show (r\mid p) and (r\mid q). Step 3: Finally write the contradiction with coprimality.