Proof of irrationality of square root 2 and square root 3

From \(p^2=3q^2\), 3 divides \(p\). Put \(p=3k\); then \(q^2=3k^2\), so 3 also divides \(q\). This contradicts their being coprime. Exam tip: use the prime-factor property carefully.

Taking p/q in lowest terms makes p and q coprime. The proof reaches a contradiction when both are divisible by 3; merely being odd is irrelevant. Exam tip: “lowest terms” means coprime numerator and denominator.

If both \(p\) and \(q\) are divisible by 3, they have the common factor 3. This contradicts the initial condition that they are coprime, so \(\sqrt{3}\) cannot be rational. Exam tip: state clearly that the fraction was assumed to be in lowest terms.

Write an odd integer as \(2n+1\). Its square is \(4n^2+4n+1=2(2n^2+2n)+1\), which is odd. Hence, if a square is even, the integer must be even. Exam tip: use the contrapositive argument.

Lowest terms means that \(p\) and \(q\) have no common factor greater than 1, so \(\gcd(p,q)=1\). From \(p^2=3q^2\), the proof shows that both are divisible by 3, contradicting this condition. Exam tip: connect “lowest terms” with coprime integers.

The equation \(q^2=2m^2\) makes \(q^2\) even, so \(q\) must be even. Since \(p\) was already even, both share factor 2, contradicting coprimality. Exam tip: if a square is even, its integer root is even.

The governing concept is the parity argument used in the proof that √2 is irrational. Starting from c²=2d², the right side is even, so c² is even; an integer whose square is even must itself be even. Writing c=2u and substituting gives 4u²=2d², hence d²=2u², so d is also even. If c/d was assumed to be in lowest terms, both numerator and denominator being even contradicts that assumption. However, c²=2d² does not imply c=2d. Equality of squares with a factor of 2 does not permit taking such a direct linear conclusion. Therefore option C is the invalid shortcut; A, B, and D represent valid steps or conclusions in the proof.

From \(p^2=3q^2\), \(p^2\) is divisible by 3, so \(p\) is divisible by 3. Put \(p=3k\); then \(q\) is also divisible by 3, contradicting lowest terms. Exam tip: use the prime-divisibility property of squares.

Since
\(p^2=3q^2\),
\(p^2\) is divisible by 3, so
\(p\) is divisible by 3. Put
\(p=3k\); then
\(q^2=3k^2\), so
\(q\) is also divisible by 3. This contradicts lowest terms. Exam tip: if a prime divides a square, it divides the number itself.

From h² = 3k², h² is divisible by 3. Since 3 is prime, if it divides the square of an integer, it must also divide that integer. Hence 3 divides h, contradicting the given condition that h is not divisible by 3. Option A is not the required contradiction; the contradiction at this step concerns h directly. Exam tip: For a prime p, use p | n² ⇒ p | n.

\(\sqrt{3}\) is irrational because 3 is not a perfect square: its prime factor 3 has exponent 1, which is odd. In contrast, \(\sqrt{9}=3\) is rational. Exam tip: a perfect square has only even prime exponents.

From \(p^2=2q^2\), \(p^2\), and hence \(p\), is even. Put \(p=2r\); then \(q^2=2r^2\), so \(q\) is also even. This contradicts coprimality. Exam tip: an even square has an even root.

Since 3 is prime, \(p^2=3q^2\) implies that 3 divides \(p\). Put \(p=3k\); then \(q^2=3k^2\), so 3 divides \(q\) as well. This contradicts that \(p\) and \(q\) are coprime. Exam tip: if a prime divides a square, it divides the number itself.

Since 3 is prime, \(3\mid p^2=p\times p\) implies \(3\mid p\) by the prime-divisor property. It does not imply that \(p\) is divisible by 9. Exam tip: apply this rule only when the divisor is prime.

A fraction \(c/d\) represents a number only when its denominator is non-zero, so \(d\neq0\) is required to make the expression defined. This condition does not say that the fraction is reduced, and it does not imply any equality between \(c\) and \(d\). The condition \(\gcd(c,d)=1\) has a different purpose: it says that numerator and denominator have no common factor and that the fraction is in lowest terms.

Assuming \(\sqrt{2}=c/d\), squaring gives \(c^2=2d^2\). The parity argument shows that \(c\) is even and then that \(d\) is even. Thus both have the common factor 2, contradicting \(\gcd(c,d)=1\). The contradiction rests on the lowest-terms condition, while the non-zero condition only keeps the fraction meaningful. Hence option B is correct.

(k\neq0) is necessary for a rational fraction. (\gcd(h,k)=1) breaks when both are divisible by (3).

Assume \(\sqrt{3}=\frac{p}{q}\), where \(p\) and \(q\) are coprime integers. Then \(p^2=3q^2\), so \(3\mid p^2\). Since 3 is prime, \(3\mid p^2\) implies \(3\mid p\). On writing \(p=3k\), we also obtain \(3\mid q\), which contradicts the fact that \(p\) and \(q\) are coprime. Option B is incorrect because divisibility of \(n^2\) by 3 does not mean that \(n=3\); it means that \(n\) is divisible by 3. Exam tip: In irrationality proofs, look for the prime-divisibility property that forces both numerator and denominator to have a common factor.

From \(p^2=3q^2\), 3 divides \(p^2\), so the prime-factor property gives 3 divides \(p\). Put \(p=3k\); then 3 also divides \(q\), contradicting coprimality. Exam tip: transfer prime divisibility from a square back to its base.

The governing idea is the role of lowest terms in a contradiction proof. If √2 = c/d, squaring still gives c² = 2d² whether or not the fraction is reduced. The positivity statement √2 > 0 is also unaffected. However, the conclusion that both c and d are even is decisive only when gcd(c,d) = 1 was assumed at the start. Without lowest terms, a fraction may legitimately have an even numerator and denominator, such as 2/4, so their common divisibility is not itself a contradiction. Therefore option A is correct. The other choices describe steps that remain valid independently of reduction.

From \(3p^2=q^2\), \(q^2\) is divisible by 3, so \(q=3r\). Substituting gives \(3p^2=9r^2\), hence \(p^2=3r^2\); therefore, \(p\) is divisible by 3. Exam tip: if a prime divides a square, it divides the number itself.