Proof of irrationality of square root 2 and square root 3

Since \(3\mid p^2\) and 3 is prime, \(3\mid p\). Put \(p=3k\); then \(q^2=3k^2\), so \(3\mid q\) as well. This contradicts that \(p\) and \(q\) are coprime. Exam tip: state the prime-factor property explicitly.

The assumption gives \(p^2=3q^2\). Thus 3 divides \(p^2\), so it divides \(p\); substituting back shows that 3 also divides \(q\). This contradicts coprimality. Exam tip: state the prime-divisor property clearly.

An approximate decimal is useful for estimating the size of \(\sqrt{3}\), but it cannot prove whether the number is rational or irrational. Any finite decimal is rational, and a displayed decimal approximation leaves infinitely many possible later digits. Even a long nonterminating-looking calculation cannot establish an exact mathematical property by itself. A proof needs a conclusion that follows with certainty.

The standard proof assumes \(\sqrt{3}=p/q\) in lowest form and obtains an equation such as \(p^2=3q^2\). Divisibility by 3 then forces a corresponding divisibility conclusion for the integers, eventually making both numerator and denominator divisible by 3. That contradicts lowest form. Therefore option A is correct: approximation alone does not provide a complete proof.

Assume coprime \(p,q\) in \(\sqrt{3}=p/q\). Then \(p^2=3q^2\) makes both divisible by 3, a contradiction. Taking both as multiples of 3 initially is invalid. Tip: use lowest terms.

From \(p^2=3q^2\), 3 divides p. Put \(p=3k\); then 3 also divides q. This contradicts p and q being coprime. The denominator need not be prime. Exam tip: state that the fraction is in lowest terms before deriving the contradiction.

Since 3 is prime, \(3\mid p^2\) implies \(3\mid p\). Putting \(p=3k\) then shows that \(q\) is also divisible by 3, contradicting coprimality. Exam tip: use the prime-factor rule for squares.

From \(p^2=2q^2\), \(p^2\), and hence \(p\), is even. Substituting \(p=2k\) shows that \(q\) is also even, contradicting coprimality. Both being odd gives no contradiction. Tip: always begin with lowest terms.

Because \(1.732^2=2.999824\ne3\), 1.732 is an approximation, not the exact value. A decimal point does not make a number irrational. Exam tip: square a claimed exact value.

Since \(2q^2\) is even, \(p^2\) is even. The square of an integer is even only if the integer itself is even; hence \(p\) is even. An even value of \(2q^2\) does not directly imply that \(q\) is even. Exam tip: apply parity rules to the correct factor.

From \(p^2=2q^2\), \(p\) is even. Put \(p=2k\); then \(q^2=2k^2\), so \(q\) is also even. This contradicts only a coprime pair. Exam tip: always begin with a fraction in lowest terms.

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.

The correct answer is option A: write a/b in lowest form with gcd(a,b)=1. To prove √2 irrational, first assume √2 is rational. Then write √2=a/b, where a and b are integers, b≠0, and the fraction is already reduced. Squaring gives a²=2b². Hence a² is even, so a is even; let a=2k. Substitution gives 4k²=2b², so b²=2k², and b is also even. Thus both a and b are divisible by 2, contradicting gcd(a,b)=1. The lowest-form condition is essential because without it, both numerator and denominator might already contain a common factor, and the final contradiction would not be valid. Option A is correct. Option B is wrong because a denominator of zero does not define a fraction. Option C is wrong because decimal expansions do not provide this required contradiction and can be misleading. Option D is wrong because there is no reason that a=b; it is not part of the proof. Exam cue: always write “in lowest form” and “gcd(a,b)=1” before beginning.

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.