Proof of irrationality of square root 2 and square root 3

If (u) has no factor (3), then (u^2) also has none. But (u^2=3v^2) shows it divisible by (3).

From \(p^2=3q^2\), \(p^2\) is divisible by 3. Since 3 is prime, \(p\) must also be divisible by 3. Let \(p=3k\). Then \(9k^2=3q^2\), so \(q^2=3k^2\); hence \(q\) is also divisible by 3. This contradicts the fact that \(p\) and \(q\) are coprime, and this contradiction is used to prove that \(\sqrt{3}\) is irrational. Option B is incomplete because once \(p\) is divisible by 3, \(q\) must be divisible by 3 as well. Exam tip: If a prime divides a square, it also divides the original number.

Since \(3\) is prime, if it divides the square \(p^2\), it must divide \(p\); hence \(3\mid p\). On putting \(p=3k\), we also obtain \(3\mid q\), contradicting coprimality. Exam tip: state this prime-divides-a-square property explicitly.

Since 3 is prime, the prime-divisor property gives \(3\mid p\) whenever \(3\mid p^2\). Substituting \(p=3k\) later shows that \(q\) is also divisible by 3, contradicting lowest terms. Exam tip: state the prime-divisor step explicitly.

The lowest coprime form creates contradiction with the final common factor (3). Write it in exams.

From \(3q^2=p^2\), \(p^2\), and hence \(p\), is divisible by 3. Put \(p=3k\) to show that \(q\) is also divisible by 3, contradicting coprimality. Exam tip: begin such proofs with a fraction in lowest terms.

Since u=3t, u is divisible by 3. Substituting this in u²=3v² gives 9t²=3v², so v²=3t². Hence 3 divides v². As 3 is prime, if it divides v², it must also divide v. Therefore, both u and v are divisible by 3. Exam tip: If a prime p divides x², then p divides x.

From
\(p^2=3q^2\), 3 divides
\(p^2\), so 3 divides
\(p\). Put
\(p=3k\); then 3 also divides
\(q\). This contradicts coprimality. Exam tip: identify the common prime factor causing the contradiction.

The governing concept is the prime-factor property of a square. If a prime divides the square of an integer, that prime must divide the integer itself. In prime factorisation, every factor in u² occurs with twice its exponent. Therefore, if 3 ∣ u², the exponent of 3 in u² is positive and even, so the exponent of 3 in u is also positive; hence 3 ∣ u. This is exactly the step used in the contradiction proof for √3. Option A states this valid implication. Option B changes the prime from 3 to 2, while C incorrectly says u must be zero and D asserts an unrelated equality.

If 3 divides
\(p^2\), then 3 must divide
\(p\). Put
\(p=3k\); this gives
\(q^2=3k^2\), so 3 also divides
\(q\). This contradicts coprimality. Exam tip: apply this prime-divisibility rule in irrationality proofs.

The governing reasoning must proceed in two linked divisibility steps. From u² = 3v², the right-hand side is divisible by 3, so 3 ∣ u². Since 3 is prime, 3 ∣ u; therefore write u = 3t for some integer t. Substitution gives 9t² = 3v², and division by 3 gives v² = 3t². This now shows 3 ∣ v², and the prime-square property gives 3 ∣ v. Thus v is eventually proved divisible by 3, but it does not follow directly from the original equation without showing the intermediate steps. Hence A is correct; B skips reasoning, and C and D are unrelated.

The claim is false. If \(3\sqrt{2}\) were rational, division by the non-zero rational number 3 would make \(\sqrt{2}\) rational, a contradiction. Exam tip: divide by a non-zero rational factor to test such claims.

The rational assumption should be written in lowest fraction. (\gcd(u,v)=1) creates the final contradiction.

From \(p^2=3q^2\), 3 divides \(p^2\). A prime dividing a square must divide its base, so \(3\mid p\). On putting \(p=3k\), one also gets \(3\mid q\), contradicting lowest terms. Exam tip: explicitly state the prime-divisibility rule.

From \(a^2=3b^2\), \(a^2\) is divisible by 3. Since 3 is prime, \(a\) must be divisible by 3. Then put \(a=3k\) to show that \(b\) is also divisible by 3. Exam tip: if a prime divides a square, it divides the number itself.

Since \(p^2=3q^2\), \(p^2\) 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 3 is prime, \(3\mid p^2\) implies \(3\mid p\). Substituting \(p=3k\) in \(p^2=3q^2\) gives \(3\mid q\), contradicting coprimality. Exam tip: remember this prime-divisibility rule.

A rational number must be expressible as u/v, where u and v are integers, v ≠ 0, and the fraction is in lowest terms, so gcd(u,v) = 1. In the proof, assuming √3 = u/v leads to u² = 3v². The divisibility argument first gives 3 ∣ u and then, after writing u = 3t and substituting, gives 3 ∣ v. Thus u and v have the common factor 3, contradicting their coprime, lowest-form condition. This contradiction rejects the rational assumption. Option B contains the complete sufficient statement; A and C give only partial information, and D does not contradict rationality.

The governing concept is unique prime factorisation. In the square of an integer, every prime exponent is doubled, so every exponent is even. Suppose √2 were rational and write it as a fraction in lowest terms, a/b. Squaring gives a² = 2b². The prime 2 occurs to an odd extra power on the right-hand side, while the left-hand side is a square and must have even prime exponents. Equivalently, divisibility forces 2 ∣ a and then 2 ∣ b, contradicting lowest terms. Therefore option A expresses the key idea. The other options are false or unrelated to the proof.

In a perfect square, all prime exponents are even. This idea strengthens the contradiction for (\sqrt{3}).