Proof of irrationality of square root 2 and square root 3

Since \(p^2=3q^2\), 3 divides \(p^2\). As 3 is prime, it must divide \(p\). Putting \(p=3k\) then shows that 3 divides \(q\) too, contradicting coprimality. Exam tip: use the prime-divisor property for a square.

The tempting mistake is to treat a square equation as though taking square roots preserved the same linear relationship. From \(p^2=3q^2\), it is not valid to conclude directly that \(p=3q\). Taking square roots formally would suggest a factor of \(\sqrt{3}\), not 3, so that proposed step has no logical basis.

The correct argument uses divisibility. Since \(3\mid p^2\) and 3 is prime, it follows that \(3\mid p\). Write \(p=3r\) and substitute into the equation. Then \(9r^2=3q^2\), so \(q^2=3r^2\), which gives \(3\mid q\) as well. Thus both numerator and denominator share 3. Option A is the attractive but wrong statement.

In the standard proof that \(\sqrt{2}\) is irrational, we first suppose that \(\sqrt{2}=a/b\), where \(a\) and \(b\) have no common factor. Squaring gives \(a^2=2b^2\), so \(a^2\) is even and therefore \(a\) is even. Writing \(a=2k\) then shows that \(b\) is also even. This is an intermediate step in the chain of reasoning, not the final statement.

Option B is correct because “\(a\) is even” is reached during the proof. The final contradiction is that both \(a\) and \(b\) are even, which conflicts with the assumption that \(a/b\) was in lowest terms. Consequently, the original rational assumption is false and \(\sqrt{2}\) is irrational. Options A and C express final conclusions, while D describes the contradiction itself.

The proof assumes, for contradiction, that √3 = p/q in lowest terms. From p² = 3q², one first concludes that 3 divides p, so p = 3k. This is only an intermediate inference, represented by option C. Substituting p = 3k then gives 9k² = 3q² and hence q² = 3k², which shows that 3 also divides q. Since both p and q are divisible by 3, the fraction was not in lowest terms. That contradiction rejects the original rational assumption, and the final conclusion is that √3 is irrational. Options A, B, and D describe the concluding part rather than the requested middle step.

Without lowest form, the final contradiction becomes weak. This is the most important line in exams.

Since 3 is prime, \(3\mid n^2=n\times n\) implies \(3\mid n\) by Euclid’s lemma. It does not necessarily imply \(9\mid n\). Exam tip: state the prime-divisor lemma before applying it in the contradiction proof.

In a contradiction proof, we assume that \(\sqrt{2}=p/q\) is rational, where \(p\) and \(q\) are coprime integers. The argument shows that both \(p\) and \(q\) must be even, contradicting their being coprime. Hence the original assumption is false, so \(\sqrt{2}\) is irrational. Exam tip: A contradiction rejects the initial assumption, not the statement being proved.

The fraction must be in lowest terms. From \(p^2=2q^2\), \(p\) is even; putting \(p=2k\) shows that \(q\) is also even, contradicting coprimality. In exams, always state that the fraction is in lowest terms.

A rational number has a terminating or repeating decimal expansion. Hence \(\sqrt{2}\) must be non-terminating and non-repeating. Exam tip: many displayed digits alone are not proof.

Let \(S=\sqrt{2}+\sqrt{3}\) be rational. Then \(S^2=5+2\sqrt{6}\), giving \(\sqrt{6}=(S^2-5)/2\) as rational, a contradiction. Exam tip: square a sum of surds to isolate the mixed radical.

From \(m^2=3n^2\), \(m^2\) is divisible by 3. Since 3 is prime, \(m\) must be divisible by 3; then \(n\) also becomes divisible by 3, contradicting coprimality. Exam tip: use the prime-divides-a-square rule in such proofs.

Starting from m² = 3n² and using the earlier conclusion that 3 divides m, write m = 3k. Substitution gives (3k)² = 3n², so 9k² = 3n². Dividing both sides by 3 yields 3k² = n², or n² = 3k². Thus n² is divisible by 3. Since 3 is prime, Euclid’s lemma implies that n itself is divisible by 3; write n = 3l if continuing the proof. This creates a common factor 3 in m and n, contradicting their assumed coprimality. Option D contains the necessary algebraic step. The other choices either assert an unsupported equality or give an irrelevant value.

The contradiction with lowest fraction forms only when numerator and denominator both have common factor (2). Only (r^2) being even is a middle step.

Contradiction with the coprime condition forms only when both (m) and (n) are divisible by (3). (m) alone is not enough.

A fraction in lowest form is defined by the condition gcd(r,s) = 1, with s nonzero. If the proof establishes that 2 divides both r and s, then 2 is a common divisor of the pair. Consequently gcd(r,s) is at least 2, not 1. These two conclusions cannot simultaneously be true, so the assumption that √2 equals a fraction r/s in lowest terms is impossible. This is the exact logical contradiction, stated in option C. It is more precise than merely saying the numerator and denominator are both even, because it explicitly compares the defining lowest-form condition with the new divisibility result. Options A, B, and D do not follow from the proof.

Since \(p^2=2q^2\), \(p^2\) is even. By A, \(p=2k\); substitution gives \(q^2=2k^2\), so \(q\) is also even, contradicting coprimality. Exam tip: use the parity property of a square.

Assume \(2+\sqrt{3}\) is rational. Subtracting 2 then makes \(\sqrt{3}\) rational, contradicting its irrationality. Tip: adding a rational number cannot make an irrational number rational.

The equation is obtained while proving that
sqrt{3}
cannot be rational. From the equality of squares, it is not valid to remove the squares and claim that the two expressions are equal in the same form. In particular, the statement
m=3n
is much stronger than what the equation gives and is generally false.

The correct number-theory conclusion is that 3 divides
m^2
, because
m^2=3n^2
. Since 3 is prime, this implies that 3 divides
m
. Writing
m=3k
and substituting then produces the required contradiction with coprimality. Thus option B identifies the error: it takes an incorrect linear conclusion from a squared equation. The denominator is not being assumed zero, and the contradiction method itself is not the mistake.

From \(a^2=3b^2\), prime divisibility gives \(3\mid a\). Put \(a=3k\): then \(b^2=3k^2\), so \(3\mid b\) too. This contradicts coprimality. Exam tip: explicitly state the prime-divisor rule.

From m² = 3n², the right side is divisible by 3, so m² and therefore m are divisible by 3. Write m = 3k. Substitution gives 9k² = 3n²; after division by 3, n² = 3k². Hence n² is divisible by 3, and because 3 is prime, n is divisible by 3. This complete chain is given only in option A. It is important that the conclusion about n comes after introducing m = 3k and simplifying the substituted equation; it does not follow merely from the original equation without those steps. Once both m and n are divisible by 3, their assumed coprimality is contradicted.