Proof of irrationality of square root 2 and square root 3

In the proof of \(\sqrt{2}\), the number 2 is important because it is prime. Assuming \(\sqrt{2}=a/b\) in lowest form gives \(a^2=2b^2\). Since the square \(a^2\) is divisible by the prime 2, \(a\) itself must be divisible by 2. Writing \(a=2k\) and substituting back shows that \(b^2\), and hence \(b\), is also divisible by 2.

Therefore 2 becomes a common factor of both numerator and denominator. This contradicts the statement that \(a/b\) was in lowest form, or that \(\gcd(a,b)=1\). It does not make the denominator zero, prove \(a=b\), or prove rationality. Hence option A correctly describes the role of 2.

To prove that \(\sqrt{3}\) is irrational, assume the opposite and write \(\sqrt{3}=\frac{p}{q}\), where \(pq\) are coprime integers and \(q\neq0\). Squaring gives \(p^2=3q^2\). This shows that 3 divides \(p^2\), and because 3 is prime, it must divide \(p\). Write \(p=3k\).

Substitution gives \(9k^2=3q^2\), so \(q^2=3k^2\). Hence 3 also divides \(q\). Thus 3 is a common factor of both \(p\) and \(q\), contradicting their being coprime. Therefore option A correctly describes the role of 3: it produces the common factor and the final contradiction.

In the contradiction proof, we assume \(\sqrt{2}=a/b\), where \(a\) and \(b\) are coprime. The equation first shows that \(a\) is even; substituting this result back then shows that \(b\) is also even. Saying only that \(a\) is even does not contradict the coprime condition. The contradiction arises because both numbers would have 2 as a common factor. Exam tip: always state the final contradiction explicitly in a proof.

The proof begins by assuming √3 = p/q in lowest terms, so gcd(p,q) = 1. After squaring, p² = 3q², which implies that 3 divides p. However, that fact alone is not a contradiction: a numerator may be divisible by 3 while the denominator is not. The argument must continue by writing p = 3k and substituting into the equation. This leads to q² being divisible by 3, and hence q is divisible by 3 as well. Then p and q share the factor 3, contradicting gcd(p,q) = 1. Thus option A identifies the missing step; the other options either deny a valid result or introduce irrelevant conditions.

Since \(p^2=3q^2\), \(3\mid p^2\), and as 3 is prime, \(3\mid p\). Writing \(p=3k\) gives \(3\mid q\) too. Thus \(p\) and \(q\) are not coprime. Exam tip: start with a fraction in lowest terms.

Common factor (3) in both contradicts the coprime condition. Therefore (\sqrt{3}) is irrational.

Since 3 is prime, Euclid’s lemma gives \(3\mid p\) from \(3\mid p^2\). A conclusion about \(q\) follows only later after substituting \(p=3k\). In exams, state this prime-divisibility step clearly.

Since 3 is prime and \(3\mid a^2\), Euclid’s lemma gives \(3\mid a\). Writing \(a=3k\) then gives \(3\mid b\), contradicting coprimality. Exam tip: apply this converse directly only when the divisor is prime.

The contradiction comes from the way the rational number is initially represented. In a standard irrationality proof, assume √3 = p/q, where p and q are integers, q ≠ 0, and the fraction is in lowest terms. The last condition is expressed as gcd(p,q) = 1. If the algebra later proves that both p and q are divisible by 3, then 3 is a common divisor and gcd(p,q) is at least 3, not 1. This directly contradicts the initial lowest-form condition. Positivity of √3 is true but irrelevant, p = q was never assumed, and q = 0 is forbidden rather than an initial condition. Hence option C is correct.

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.