Proof of irrationality of square root 2 and square root 3

The rational assumption makes both (p) and (q) divisible by (3). This contradiction proves (\sqrt{3}) irrational.

In both proofs, the opposite is assumed and contradiction with coprime condition is derived. Hence contradiction method is correct.

\(1.732\) is a terminating decimal and hence rational, but it is only an approximation to \(\sqrt{3}\). Check: \(1.732^2=2.999824\), not \(3\). In exams, carefully distinguish an exact equality from an approximation.

Although \((\sqrt{2})^2=2\) is rational, the converse statement is invalid. Assuming \(\sqrt{2}=p/q\) in lowest terms gives \(p^2=2q^2\), making both \(p\) and \(q\) even—a contradiction. Exam tip: check whether a converse is justified.

Putting (m=2r) gives (n^2=2r^2). This makes (n) even too and creates the contradiction of both even.

From \(a^2=2b^2\), \(a\) is even; write \(a=2k\). Substitution gives \(b^2=2k^2\), so \(b\) is also even. Thus both share factor 2, contradicting coprimality. Exam tip: state the lowest-terms contradiction clearly.

Assume \(\sqrt{3}=p/q\) with coprime integers \(p,q\). From \(p^2=3q^2\), first \(p\), and then \(q\), is divisible by 3. This contradicts coprimality. Exam tip: identify the common factor obtained in both numerator and denominator.

From \(3q^2=p^2\), 3 divides \(p^2\), so it divides \(p\). Putting \(p=3k\) then shows that 3 also divides \(q\). This contradicts coprimality. Exam tip: always assume the fraction is in lowest terms first.

In both proofs, the rational assumption clashes with the coprime condition. Therefore both (\sqrt{2}) and (\sqrt{3}) are irrational.

From \(3q^2=p^2\), \(p^2\) is divisible by 3, so \(p\) is divisible by 3. Substitution then shows that \(q\) is also divisible by 3, contradicting lowest terms. Exam tip: check for a common factor in both numerator and denominator.

Put \(p=3k\). Then \(3q^2=p^2=9k^2\), so \(q^2=3k^2\) and \(q\) is also divisible by 3. Thus \(p\) and \(q\) have a common factor, contradicting lowest terms. Exam tip: state the common-factor contradiction clearly.

In option C, the number of zeros between 1s keeps increasing, so no fixed repeating block occurs. An irrational number has a non-terminating, non-repeating decimal expansion. A and D terminate, while B repeats. Exam tip: terminating or recurring decimals are rational.

Assume \(\sqrt{2}=p/q\) in lowest terms. From \(p^2=2q^2\), p is even; writing \(p=2k\) then shows q is even too. This contradicts coprimality. Exam tip: begin with lowest terms.

For contradiction, assume \(\sqrt{3}\) is rational in lowest form \(p/q\). Squaring gives \(p^2=3q^2\), which ultimately makes both \(p\) and \(q\) divisible by 3. Exam tip: the coprime condition creates the contradiction.

From p² = 3q², p is divisible by 3. Substituting p = 3k gives q² = 3k², so q is also divisible by 3. This contradicts p and q being coprime. Exam tip: always state that p/q is in lowest terms.

Since \(\sqrt{12}=2\sqrt{3}\), we get \(\sqrt{3}=\frac{\sqrt{12}}{2}\). If \(\sqrt{12}\) were rational, division by 2 would make \(\sqrt{3}\) rational, a contradiction. Exam tip: state the contradiction explicitly.

(q^2) is divisible by (3), so (q) is also divisible by (3). This leads to the final contradiction.

In a contradiction proof, assume \(\sqrt{2}=p/q\) in lowest terms, with coprime integers \(p,q\). From \(p^2=2q^2\), both \(p\) and then \(q\) are even, contradicting coprimality. Exam tip: always state that the fraction is in lowest terms.

In lowest form, there should be no common factor except (1). If both are even, (2) is common.

To prove √3 irrational, assume that √3 = p/q, where p and q are integers, q ≠ 0, and the fraction is in lowest form, so gcd(p,q) = 1. Squaring gives p² = 3q². Because 3 is prime and divides p², it must divide p; write p = 3k. Substitution then gives 9k² = 3q², so q² = 3k², which implies that 3 divides q as well. Thus both p and q have the common factor 3, contradicting the lowest-form condition. Therefore option D identifies the contradiction. Options A, B, and C are assumptions used in the proof, not contradictions.