Proof of irrationality of square root 2 and square root 3

Option B is correct. If \(n\) is not a perfect square, its prime factorisation has at least one odd exponent, so \(\sqrt{n}\) cannot be rational. For example, \(12=2^2\times3\). Exam tip: first check whether the number is a perfect square.

If prime (3) divides a square, it divides the original number too. Therefore (u) is a multiple of (3).

Let \(p=3k\). Substituting in \(p^2=3q^2\) gives \(9k^2=3q^2\), so \(q^2=3k^2\). Hence \(q\) is also divisible by 3, contradicting that \(p\) and \(q\) are coprime. Exam tip: if a prime divides a square, it divides the number itself.

Assume \(\sqrt{3}=p/q\) in lowest terms. From \(p^2=3q^2\), 3 divides \(p\); putting \(p=3k\) then shows that 3 also divides \(q\). This contradicts coprimality. Exam tip: if a prime divides a square, it divides the number itself.

The contradiction completes when numerator and denominator both have common factor (2). This opposes the coprime condition.

In lowest terms, \(p\) and \(q\) have no common factor. From \(p^2=2q^2\), \(p\) is even; substituting \(p=2k\) shows that \(q\) is also even, contradicting coprimality. Exam tip: always begin with a fraction in lowest form.

From \(a^2=3b^2\), \(a^2\) is divisible by 3. If the square of an integer is divisible by 3, the integer itself is divisible by 3, so write \(a=3k\). Substitution then makes \(b\) divisible by 3 too, contradicting coprimality. Exam tip: check squares of remainders 0, 1, and 2 modulo 3.

\(2+\sqrt{3}\) is irrational. If it were rational, subtracting the rational number 2 would make \(\sqrt{3}\) rational, which is impossible. Also, \(\sqrt{9}=3\), and terminating or recurring decimals are rational. Exam tip: adding a rational number to an irrational number gives an irrational result.

From \(3q^2=p^2\), \(p^2\), hence \(p\), is divisible by 3. Put \(p=3k\); then \(q^2=3k^2\), so \(q\) is also divisible by 3. This contradicts coprimality. Exam tip: check common divisibility of both terms.

From \(p^2=2q^2\), \(p^2\) is even, so \(p\) is even. Put \(p=2k\); then \(q^2=2k^2\), making \(q\) even too. This contradicts coprimality. Exam tip: an even square always has an even base.

From \(p^2=3q^2\), we get \(3\mid p^2\). Since 3 is prime, \(3\mid p\); putting \(p=3k\) then gives \(3\mid q\) too. Thus \(p,q\) are not coprime. Exam tip: use the prime-divisor property in such proofs.

The direct answer is A. If u is not divisible by 3, then its square u squared is also not divisible by 3, because squaring cannot create a missing prime factor. But the equation u squared equals 3v squared has an explicit factor 3 on the right-hand side, so the right side is divisible by 3. Equality would then force u squared to be divisible by 3, producing a contradiction. Option A correctly states this conflict. Option B, v=0, does not follow; in a fraction denominator is nonzero anyway. Option C, u=v, is unrelated to the equation and is not required. Option D, square root of 3 equals zero, is false and is not the contradiction used. The proof works by comparing divisibility facts, not by claiming arbitrary values are zero or equal. Memory cue: no factor 3 in u means no factor 3 in u squared, but the equation supplies one.

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-divisor property of a square. If a prime p divides the square of an integer u, then p must divide u itself. This follows from prime factorisation: every prime factor in u² appears with twice its exponent, so the presence of 3 in u² means that 3 already occurs in the factorisation of u. Consequently, 3 ∣ u² implies 3 ∣ u. In the proof of √3, an equation such as u²=3v² first shows that 3 divides u², and this rule then gives 3 ∣ u. Option A states exactly this valid implication. Option B uses the wrong prime, option C wrongly requires u to be zero, and option D introduces 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 idea is that the divisibility argument must be carried out in two justified stages. From u²=3v², the right-hand side is divisible by 3, so 3 divides u². Because 3 is prime, 3 divides u; therefore write u=3t for an integer t. Substituting gives 9t²=3v², and dividing by 3 gives v²=3t². Only now is v² visibly divisible by 3, allowing the prime-square rule to prove 3 divides v. Thus v is not obtained directly from the original equation without explanation. Option A correctly identifies the missing intermediate step. The other choices either accept an unsupported conclusion or discuss unrelated denominator and decimal issues.

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 direct answer is C: square root of 3 equals u/v, where u and v are coprime integers and v is not zero. To prove irrationality by contradiction, assume the number is rational. By the definition of a rational number, it can be written as a quotient of integers with a nonzero denominator. We may reduce that fraction to lowest form, so gcd(u,v)=1. Squaring gives u squared equals 3v squared. Then 3 divides u squared, so the prime property gives 3 divides u; substitution then gives 3 divides v. This contradicts coprimality. Option A is impossible because division by zero is undefined. Option B is not the required general form; a rational number need not be an integer sum. Option C is correct and contains every needed condition. Option D is too restrictive and does not represent every rational assumption. Memory cue: rational number = coprime integer fraction with nonzero denominator.

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.