Proof of irrationality of square root 2 and square root 3

Since \(3\) is prime, \(3\mid p^2\) implies \(3\mid p\). Put \(p=3k\) in \(p^2=3q^2\) to get \(3\mid q\), contradicting coprimality. Exam tip: explicitly cite the prime-divisibility rule.

Every prime exponent in a perfect square is even. In \(12=2^2\times3\), 3 has an odd exponent, so \(\sqrt{12}\) is irrational. Being even or prime alone is insufficient; first check whether the number is a perfect square.

From \(p^2=3q^2\), 3 divides \(p^2\), so 3 divides \(p\). Put \(p=3k\); then \(q^2=3k^2\), so 3 also divides \(q\). Thus they cannot be coprime. Exam tip: if a prime divides a square, it divides the number itself.

A contradiction arises only when \(\frac{m}{n}\) is assumed to be in lowest terms, so \(m,n\) are coprime. Otherwise, both may be even, as in \(\frac{6}{8}\). Exam tip: always state the lowest-terms condition first.

From \(a^2=2b^2\), \(a^2\) is even because it is a multiple of 2. If the square of an integer is even, then the integer itself is even; therefore, write \(a=2k\), where \(k\) is an integer. Writing \(a=2b\) directly is incorrect because \(\sqrt{2b^2}=b\sqrt2\), not \(2b\). Exam tip: In irrationality proofs, first state that an even square implies an even integer.

Since \(p^2=3q^2\), \(p^2\) is divisible by 3. As 3 is prime, if \(p^2\) is divisible by 3, then \(p\) must also be divisible by 3. Hence, write \(p=3r\), where \(r\) is an integer. Writing \(p=3q\) is unjustified because it incorrectly makes \(p\) directly three times \(q\). Exam tip: when a square is divisible by a prime, first conclude that its base is divisible by that prime.

The proof begins by assuming that √2 can be written as a fraction a/b in lowest terms, where a and b have no common factor. Squaring gives a² = 2b², so a² is even and therefore a is even; write a = 2k. Substitution gives b² = 2k², so b is also even. Thus both numerator and denominator have a common factor 2, contradicting the lowest-terms assumption. This is the essence of infinite descent: the assumed fraction produces another equivalent representation with a smaller reducible pair, and the process cannot continue indefinitely. Option B captures this contradiction; the other options do not describe the proof.

The proof that \(\sqrt{3}\) is irrational can be described using infinite descent. Assume, for contradiction, that \(\sqrt{3}=p/q\) is a fraction in lowest terms, with integers p and q and \(q\ne0\). Squaring gives \(p^2=3q^2\). This shows that 3 divides \(p^2\), so 3 divides p. Substituting \(p=3k\) then shows that 3 also divides q. Thus both numerator and denominator have a common factor 3.

Dividing both by 3 produces a smaller positive fraction representing the same number, which contradicts the assumption that the original fraction was already in lowest terms. If one repeatedly applies the same reasoning, it would create an endless chain of smaller positive integer pairs, which is impossible. Therefore option A captures the descent idea. Options B, C, and D do not follow from the proof and do not establish irrationality.

The relevant prime-factor principle is that every prime occurs to an even exponent in the prime factorisation of a perfect square. For example, if x = 2ᵏ times other prime factors, then x² contains 2²ᵏ, whose exponent is even. In the irrationality proof, assuming √2 = a/b in lowest terms leads to a² = 2b². Since the right side has one extra factor 2 beyond the square b², the parity of the exponent of 2 becomes impossible: the left side is a square and must have an even exponent, whereas the right side has an odd one. Hence option A states the governing fact. The other statements are false or irrelevant.

A perfect square has an even exponent for every prime in its prime factorisation. This applies to the prime 3 as well. In the usual contradiction proof, suppose √3 = p/q in lowest terms. Squaring gives p² = 3q². Because the right side is divisible by 3, p is divisible by 3; write p = 3k. Substitution gives q² = 3k², so q is also divisible by 3. The resulting common factor contradicts the assumption that p/q was in lowest terms. Equivalently, the square p² must have an even exponent of 3, while the factor 3q² introduces an odd exponent. Therefore option C is the precise useful statement; the remaining options are false.

Since 3 is prime, \(3\mid a^2\) implies \(3\mid a\). Put \(a=3k\); then \(b^2=3k^2\), so \(3\mid b\), contradicting coprimality. Exam tip: use prime divisibility carefully.

If (p) has no factor (3), then (p^2) also has no factor (3). But the equation shows (p^2) divisible by (3).

An infinite decimal is not sufficient evidence: \(0.333\ldots=\frac13\) is rational. To prove \(\sqrt{3}\) irrational, assume \(\frac pq\) is in lowest terms and derive a divisibility contradiction. Exam tip: distinguish recurring from non-recurring decimals.

In lowest terms, \(p\) and \(q\) are coprime. If the proof shows both are even, 2 becomes a common factor, giving a contradiction; they need not both be odd. Exam tip: lowest terms means coprime.

The two conditions serve different purposes. The statement \(b\neq0\) is required because \(a/b\) must be a defined fraction; division by zero has no meaning. It does not say that the fraction is reduced or that its numerator and denominator have no common factor. The condition \(\gcd(a,b)=1\), on the other hand, says that the fraction is in lowest form.

In the proof, assume \(\sqrt{2}=a/b\) with these conditions. The algebra eventually shows that both \(a\) and \(b\) are even, so they share the factor 2. That contradicts \(\gcd(a,b)=1\), producing the desired contradiction. Thus option B is correct: one condition keeps the fraction defined, while the other supports the contradiction.

(q\neq0) is necessary for a rational fraction. (\gcd(p,q)=1) breaks when both are divisible by (3).

The standard proof starts with rational assumption and derives contradiction that both are even. This is the correct order.

Since \(p^2=3q^2\), 3 divides \(p\). On writing \(p=3k\), we get \(q^2=3k^2\), so 3 divides \(q\) too. This contradicts coprimality. Tip: prove divisibility of \(p\) before substituting.

If \(r\sqrt{3}=q\) were rational, then since \(r\ne0\), \(\sqrt{3}=q/r\) would also be rational, a contradiction. Thus A is correct. Tip: a non-zero rational multiplier preserves irrationality.

Lowest form makes (p) and (q) coprime. This condition makes both divisible by (3) a contradiction.

From \(a^2=3b^2\), 3 divides \(a^2\), so the prime-divisibility property gives \(3\mid a\). Put \(a=3k\) to obtain \(3\mid b\) too, contradicting coprimality. In exams, state the prime-divisibility step clearly.

Both divisible by (3) is the contradiction found at the end. Initially (\gcd(p,q)=1) is assumed.

Assume \(a\sqrt{3}\) is rational, where \(a\ne0\) is rational. Then \(\frac{a\sqrt{3}}{a}=\sqrt{3}\) would be rational, contradicting its irrationality. Exam tip: always check that division is by a non-zero number.

A proof by contradiction temporarily assumes the opposite of the statement that is to be proved. Here the target is to show that \(\sqrt{3}\) is irrational, so the proof begins by assuming that \(\sqrt{3}\) is rational. It is then written as a fraction \(p/q\) in lowest terms, with \(q\neq0\). The algebraic argument eventually forces both \(p\) and \(q\) to be divisible by 3.

That conclusion conflicts with the choice of a lowest-terms fraction, so the temporary assumption cannot be true. The contradiction does not reject the fact that \(\sqrt{3}>0\), that it is real, or that the denominator is non-zero; those facts remain valid. It rejects only the assumption of rationality. Therefore option A is the correct choice.

Let \(5+\sqrt{3}=r\) be rational. Then \(\sqrt{3}=r-5\) is rational, a contradiction. Adding rational 5 cannot make it rational. Exam tip: state the subtraction step.