Proof of irrationality of square root 2 and square root 3

Since 3 is prime, \(3\mid a^2\) gives \(3\mid a\). Set \(a=3k\); then \(b^2=3k^2\), so \(3\mid b\), contradicting lowest terms. Exam tip: state the prime-divisor rule clearly.

Assuming \(\sqrt{3}=a/b\) gives \(a^2=3b^2\). Thus \(a^2\), hence \(a\), is divisible by 3; putting \(a=3k\) shows that \(b\) is also divisible by 3. This contradicts coprimality. Exam tip: if a square is divisible by 3, its root number is divisible by 3.

The proof uses contradiction. It begins by assuming that √2 is rational and can be written as a/b in lowest terms. The equation a² = 2b² then forces a to be even. Substituting a = 2r gives b² = 2r², so b is also even. This means that a and b share the factor 2, contradicting the claim that a/b was in lowest terms. Consequently, the assumption that √2 is rational is false, and the correct conclusion is that √2 is irrational. Positivity and reality are not disproved: √2 is a positive real number. Thus only option A identifies the statement rejected by the proof.

The rational assumption makes both (p) and (q) divisible by (3). Thus the assumption is rejected.

From \(a^2=3b^2\), 3 divides \(a^2\), so since 3 is prime, it divides \(a\) first. Put \(a=3k\); then 3 also divides \(b\), contradicting coprimality. Exam tip: track the numerator first.

Assume √3 = p/q in lowest terms, with q ≠ 0. Squaring gives p² = 3q². Since 3 divides p², the prime-divisibility property implies that 3 divides p. Let p = 3k; substitution gives q² = 3k², so 3 divides q as well. This creates a common factor 3 in p and q, contradicting the lowest-terms assumption. Therefore the initial assumption that √3 is rational is false, and √3 is irrational. Option C merely repeats an intermediate equation and does not establish rationality. Options A and D confuse the contradiction with an unrelated conclusion. Hence option B gives both the correct conclusion and its reason.

Coprime numbers have 1 as their only common factor, so their HCF is 1. If both a and b are divisible by 2, then 2 is a common factor and they cannot be coprime. In contrast, one number may be even and the other odd, such as 2 and 3, which are coprime. Exam tip: For coprime numbers, check that no common factor other than 1 divides both numbers.

From \(2q^2=p^2\), \(p\) is even. Putting \(p=2k\) then shows \(q\) is also even, contradicting lowest terms. Exam tip: this contradiction is the key proof step.

Since 3 is prime, \(3\mid p^2\) implies \(3\mid p\). Put \(p=3k\); then \(q^2=3k^2\), so \(3\mid q\) too. This contradicts coprime terms. Exam tip: apply prime divisibility carefully.

Assume \(\sqrt{2}=p/q\) with coprime integers \(p,q\). Then \(p^2=2q^2\), so \(p^2\), and hence \(p\), is even. Putting \(p=2k\) makes \(q\) even too, contradicting coprimality. Exam tip: always begin the contradiction proof with the fraction in lowest terms.

A rational number is written as a ratio of two integers in lowest form. The denominator is not zero.

When assumed rational, (\sqrt{3}) is written in lowest fractional form. Therefore (p) and (q) are coprime.

From \(p^2=3q^2\), 3 divides \(p^2\), so prime-factor reasoning gives \(3\mid p\). Put \(p=3k\) to obtain \(3\mid q\) too. This contradicts lowest terms. In exams, show both divisibility steps.

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. Thus both share factor 2, contradicting coprimality. Exam tip: an even square implies the number itself is even.

The standard proof of irrationality assumes, for contradiction, that the square root can be written as a fraction in lowest form. For \(\sqrt{2}\), this assumption leads to both numerator and denominator being divisible by 2. For \(\sqrt{3}\), it leads to both being divisible by 3. In each case, this contradicts the requirement that the fraction is already in lowest form.

Therefore neither \(\sqrt{2}\) nor \(\sqrt{3}\) can be rational. Both are irrational numbers, so option B is correct. They are also not integers, because 2 and 3 are not perfect squares. The supplied explanation correctly refers to the contradiction involving coprime numerator and denominator and reaches the correct common conclusion.

From b² = 2r², the square b² is divisible by 2, so b² is even. A fundamental parity property states that an integer whose square is even must itself be even. Therefore b can be written as b = 2s for some integer s. Since a was already written as a = 2r, both a and b are even. That is precisely the result needed to contradict the assumption that a/b was in lowest terms. The equation does not imply that b is odd, zero, or equal to a. It only establishes the parity of b, so option A is the correct next conclusion in the proof.

(q^2) is divisible by (3) so (q) is also divisible by (3). This contradicts the coprime assumption.

If \(1+\sqrt{3}\) were rational, subtracting 1 would make \(\sqrt{3}\) rational, contradicting its irrationality. Option D is false since sums are not always irrational. Exam tip: isolate the square root in contradiction proofs.

Having common factor (3) breaks the condition of lowest fraction. Therefore (\sqrt{3}) cannot be rational.

Since \(p^2\) is divisible by 3, \(p=3k\). Substitution gives \(9k^2=3q^2\), so \(q^2=3k^2\); hence \(q\) is also divisible by 3. This contradicts coprimality. Exam tip: establish divisibility of both integers.